2017-18 100 &

op kers
Class 12 T
By E ran culty
-JE Fa r
IIT enior emie .
S fP r es
o titut
Ins

CHEMISTRY
FOR JEE MAIN & ADVANCED
SECOND
EDITION

Exhaustive Theory
(Now Revised)

Formula Sheet
9000+ Problems
based on latest JEE pattern

2500 + 1000 (New) Problems

of previous 35 years of
AIEEE (JEE Main) and IIT-JEE (JEE Adv)

5000+Illustrations and Solved Examples

Detailed Solutions
of all problems available

Plancess Concepts
Topic Covered Tips & Tricks, Facts, Notes, Misconceptions,
Key Take Aways, Problem Solving Tactics
Biomolecules and Polymers
PlancEssential
Questions recommended for revision
 26. BIOMOLECULES AND
POLYMERS

1. INTRODUCTION
The study of chemical make-up and structure of living matter and of the chemical changes that takes place within
them is called biochemistry.
The various activities of living organisms are regulated by complex organic molecules, such as carbohydrates,
lipids, proteins and nucleic acids, called biomolecules.

2. CARBOHYDRATES
Carbohydrates are principally plants products and are a part of an extremely large group of naturally occuring
organic compounds. Cane sugar, glucose, starch and so on are a few examples of carbohydrates. The general
formula for carbohydrates is Cz(H2O)y. Carbohydrates are generally hydrates of carbon, which is where the name
was derived. So, carbohydrates on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

2.1 Classification of Carbohydrates

(a) On the basis of Physical Characteristics

(i) Sugar: Characteristics of sugars are crystalline substances, taste sweet and readily water soluble. Because
of their fixed molecular weight, sugars have sharp melting points. A few examples of sugars are glucose,
fructose, sucrose, lactose, etc.

(ii) Non-Sugars: Amorphous, Tasteless, waster insoluble substances with variable melting points e.g., Starch.

(b) On the basis of Hydrolysis

On the basis of hydrolysis

Monosaccharides Oligosaccharides Polysaccharides

(Mono-one) (Oligo-few) (Poly-many)

Flowchart 26.1: Classification based on Hydrolysis

Monosaccharaides’: A carbohydrate that can be hydrolyzed only once to break down into simpler units of
polyhydroxy aldehyde or ketone is called monosaccharide. These include glucose, mannose, etc.
2 6 . 2 | Biomolecules and Polymers

Monosaccharide classification
 O 
 || 
(i) Based on location of  − C − 
 

Monosaccharide classification 
 
Based on location of C=O
H CH2OH

C=O C=O

H-C-OH HO-C-H

H-C-OH H-C-OH

H-C-OH H-C-OH

CH2OH CH2OH
Aldose Ketone
Aldehyde (-CHO), ketone (C=O)
 O 
 || 
Aldehyde (-CHO), ketone  − C − 
 
 
 
(ii) Based on number of carbon atoms in the chain
H

H C=O

H C=O H-C-OH
H C=O H-C-OH H-C-OH

C=O H-C-OH H-C-OH H-C-OH

H-C-OH H-C-OH H-C-OH H-C-OH

CH2OH CH2OH CH2OH CH2OH

Triose Tetrose Pentose Hexose

Can be either aldose or ketose sugar.

Oligosaccharides: Sugars that on hydrolysis produce two or more molecules of monosaccharides are called
oligosaccharides. These are further classified as di-, tri- or tetrasaccharides, etc.

•• Disaccharides: These are sugars that produce two molecules of the same or different monosaccharides on
hydrolysis. Examples are sucrose, maltose and lactose. An example for disaccharides is sucrose: C12H22O11.
Invertas e
C 12 H 22 O 11 + H 2 O  → C 6 H12 O 6 + C 6H12 O 6
S ucros e or H + G lucos e F ructos e

Maltas e
C 12 H 22 O 11 + H 2 O  →2 C 6 H12 O 6
Maltos e or H + G lucos e
Lactas e
C 12 H 22 O 11 + 2H 2 O  → C 6 H11O 6 + C6H12O6
Lactos e or H + G lucos e Galactose

•• Trisaccharides: Sugars that yield three molecules of the same or different monosaccharides on hydrolysis are
called trisacchardies. An example of trisaccharides is Raffinose C18H32O16
C 18 H 32 O 16 + 2H 2 O → C 6 H12 O 6 + C6H12O6 + C6H12O6
R affinos e G lucos e Galactose Fructose
 Chem i str y | 26.3

•• Polysachharides: On hydrolysis polysaccharides yield large number of monosaccharides units. An example of

polysaccharide is starch cellulose.

( C6H10O5 )n + nH2O → nC6H12O6

Strach cellulose Glucose

(c) On basis of test with reagents (like Benedict’s solution, Tollen’s reagent and Fehling’s solution):
(i) Reducing Sugars:
 O 
 || 
•• These have a free aldehyde (-CHO) or ketone  − C −  group.
 
 
 
•• These have the ability to reduce the cupric ions (Cu2+; blue) in Fehling’s or Benedict’s Solution to
cuprous ions (Cu+; reddish) that separates out as cuprous oxide (Cu2O) from the solution.
•• Examples include maltose, lactose, melibiose, gentiobiose, cellobiose, mannotriose, rhamnotriose.

(ii) Non-reducing sugars:

•• A free aldehyde or ketonic group is absent.
•• No cuprous oxide (Cu2O) producing chemical reaction takes place.
•• Examples are sucrose, trehalose, raffinose, gentiarose, melezitose.

Table 26.1: The common oxidizing agents used to test for the presence of a reducing sugar

Oxidizing Reagent Benedict’s Solution Fehling’s Solution Tollen’s Reagent

Composition Copper sulfate in alkaline Copper sulfate in alkaline Silver nitrate in aqueous
citrate tartrate ammonia
Non-reducing sugar Deep blue Deep blue Colorless
Color after reaction with Brick red precipitate Cu2O(s) Brick red precipitate Cu2O(s) Silver mirror forms Ag(s)
reducing sugar
Species being reduced Cu2+ Cu2+ Ag⊕
(the oxidant)
Cu2++e- → Cu+ Cu2+ + e- → Cu+ Ag+ + e‒ → Ag(s)
Species being oxidized Reducing sugar oxidized to Reducing sugar oxidized to Reducing sugar oxidized to
(the reductant) carboxylate carboxylate carboxylate

2.2 Classification of Aldols

In the given chemical formulae, symbols D and L refer to the relative configuration of the (–OH) group at the
penultimate carbon, where glyceraldehyde is taken as standard. D and L refer to the (–OH) groups that lies on right
hand side and left hand, respectively.

1 1 1
CHO 1 CHO
2 2
H OH or 2
HO OH or 2

3
3CH2OH
3 3CH2OH

D(+) Glyceraldehyde L(-) Glyceraldehyde

Pair of C-2, Epimer as well as enantiomers

2 6 . 4 | Biomolecules and Polymers

2.3 Epimers
Epimers are essentially diastereomers that contain multiple chiral centers that are absolutely separate from each
other in configuration at only one chiral center.

Example: CHO CHO

H OH HO H

H OH H OH

H OH H OH

CH2OH CH2OH

1 2

2.4 Anomers
Cyclic monosaccharides or glycosides that are epimers, differing from each other in the configuration of C-1 if they
are aldoses or of C-2 if they are ketoses are called anomers. The epimeric carbon in such compounds is known as
anomeric carbon or anomeric center.
Example 1: α-D-Glucopyranose and β-D-glucopyranose are anomers.
6 6
CH2OH CH2OH
Anomeric 5 O OH
5 O carbon
4 1
4 OH 1 OH

OH OH Anomeric
OH 3 2
3 2 carbon
OH OH
-D-glucopyranose -D-glucopyranose

Example 2: α-D-Fructofuranose and β-D-fructofuranose are anomers.

1 1
6 6 OH
O CH2OH O
HOCH2 HOCH2

5 2 Anomeric 5 2 Anomeric
HO carbon HO carbon
4 3
OH 4 3 CH2OH
OH OH
-D-fructofuranose -D-fructofuranose

Example 3: Methyl α-D-glucopyranoside and methyl β-D-glucopyranoside are anomers.

6
6 CH2OH
CH2OH
Anomeric 5 O OCH3
5 O carbon
4 1
4 OH
OH 1
OH Anomeric
OH OCH3 3
3 2
2 carbon
OH
OH
Merthl -D-glucopyranoside Methyl -D-glucopyranoside
 Chem i str y | 26.5

2.5 Mutarotation
Mutarotation is commonly used in carbohydrate chemistry to describe the change in specific rotation of a chiral
compound due to epimerization.
For example, the monosaccharide D-glucose can be found in two cyclic forms, α-D-glucose ([α]D25 = +112) and
β-D-glucose ([α]D25 = +18.7), which are epimers and are available as pure compounds.
CH2OH CH2OH

O O OH

OH OH
OH OH OH

OH OH
-D-glucose -D-glucose

On adding water to one of the cyclic forms of D-glucose, through reversible epimerization it changes to the other
via open-chain form, while the specific rotation of the solution gradually changes, until it reaches the equilibrium
value +52.7°.
CH2OH CHO CH2OH

O H OH O OH
H2O
HO H
OH H2O OH
H OH
OH OH OH
H OH
OH OH
CH2OH

2.6 Representation of Structures of Glucose and Fructose

(a) Open chain structures (Fischer projection):

1
1 CH2OH
CHO
2 2 C=O
H OH
3 3
HO H HO H
4 4 OH
H OH H
5 5 OH
H OH H
6 6
CH2OH CH2OH
D (+) Glucose D (-) Fructose

D symbolizes the comparative configuration of C-5 -OH group with relation to D-glyceraldehyde, which these
are prepared from. Here, (+) and (–) refer to optical rotation. Naturally, D-glucose is (+) or dextrorotatory but
D-fructose is (–) or laevorotatory.
(b) Cyclic structures of glucose: The CHO group of glucose reacts with either C-5 -OH group or with C-4 -OH
group to form stable six- and five-membered cyclic rings, respectively, having hemiacetalic linkage. When
-CHO group reacts with C-6 -OH group, seven-membered ring is formed, or four-membered ring is formed
on reaction with C-3 -OH; both are unstable.
(i) -CHO reaction with C-5 -OH group produces two anomeric glucose: In such a case, a six-membered ring,
pyranose, is formed.
2 6 . 6 | Biomolecules and Polymers

1 1
H C OH HO C H

 2  2
H OH H OH
3 3
 HO H O + Enantiomer  HO H O + Enantiomer
4  H 4 OH
 H OH
 H 5  H 5

6 6
CH2OH CH2OH
(C-1) (OH) on R.H.S. is (C-1) (OH) on L.H.S. is
called -glucose called -glucose
-D (+) Glucopyranose -D (+) Glucopyranose

O
Pyranose refers to six-membered ring, like in pyran , or to δ-linkages, since there is hemiacetalic
linkage is between C-1 and C-5 δ-C atom. δ-C atom is formed next to the functional group -CHO and β next
to the α-C atom and so on.
(ii) Haworth representation: English chemist W. N. Haworth (awarded the Nobel Prize in chemistry for his study
of carbohydrate in 1937) established cyclic structure of glucose. Below, R.H.S. shows the -OH group above
the plane of the ring in Haworth structures while as per Fischer projection, on the L.H.S., it is shown below the
plane of the ring .
6 6
CH2OH CH2OH

H 5 O H H 5 O OH
H H
4 OH H 1 4 OH H 1

OH OH OH 3 2 H
3 2
H H OH
OH
-D (+) Glucopyranose -D (+) Glucopyranose

(iii) Chair-form conformation structures: Haworth projection structures are transformed to the chair conformation.
The -OH groups present below the plane of the ring in Haworth structures remain below the plane in the chair
conformation as well.
H H
6 6
CH2OH O 4 CH2OH O
4
HO 5 HO 5 H
H
H 2 HO H 2 OH
HO H
OH 3 OH
3 1 1

H OH H H

-D (+) Glucopyranose -D (+) Glucopyranose

(Less stable one OH (more stable since all OH
Group is in axial position) Groups is in equatorial Position)

(iv) -CHO reaction with C-4 -OH group produces two anomeric glucose: In this case, five-membered ring (furanose)

is formed. Furanose refers to five-membered ring, like in furan, or O to γ-linkage, as the hemiacetal linkage

is between C-1 and C-4 γ-C atom. Not all carbohydrates exist in equilibrium as six-membered hemiacetal

rings; in several compounds, the ring is five-membered, as in fructose. However, glucose naturally occurs
 Chem i str y | 26.7

only in pyranose form (six-membered), and in small extent in furanose form, which is in equilibrium with five-
membered hemiacetal ring,.

1 1
H C OH HO C H

 2  2
H OH H OH
3 3
 HO H O + Enantiomer  HO H O + Enantiomer
4  H 4
 H
5 5
H OH H OH
6 6
CH2OH CH2OH
-D(+) Glucopyranose
(C-1) (OH) on L.H.S.
Haworth representation

6 CH OH 6 CH OH
2 2

6CHOH 6CHOH
O O
H OH
54 1 54 1

HOH H OH HOH HH
2 2
3 3
H OH H OH
-D (+) Glucofuranose -D (+) Glucofuranose

Structures of Glucosides
D-Glucose on reaction with MeOH + HCl gives α- and β- D-glucopyranoside.
CH2OH CH2OH
O O
H H H H OMe
MeOH/HCl
H
D-Glucose +
OH H OH H
HO OMe HO H

H OH H OH
Methyl--D (+) Methyl--D (+)
glucopyranoside glucopyranoside

Methyl glucoside reacts with an excess of Me2SO4/NaOH to produce pentamethyl derivatives. Presence of many
electronegative O atoms in the -OH groups of monosaccharides makes them more acidic than alcohols, and all
of them exert –I effect on the nearby -OH groups. In aqueous NaOH, the -OH groups get converted to alkoxide
ion (RO–) that reacts with Me2SO4 by SN2 mechanism and forms methyl ether. This process is called exhaustive
methylation.
The OMe groups at C-2, C-3, C-4 and C-6 of the pentamethyl derivative are ordinary ether groups. These groups
are stable in diluted aqueous acid, since ether groups are cleaved by heating with concentrated HBr or HI.
The OMe groups at C-1 are formed partly of acetal linkage (it is glucosidic) and hence are different from others.
CH2OMe CH2OMe 1
CHO
H O H O 2
H H3 O
+
H H OMe
OMe H OMe OMe H OH 3
MeO H
MeO MeO 4
H OMe
H OMe H OMe 5
H OH
Pentamethyl 6
derivative CH2OMe
(I)
2 6 . 8 | Biomolecules and Polymers

Hydrolysis of the glycosidic -OMe group occurs when pentamethyl derivatives are treated with dilute aqueous acid
to give 2, 3, 4, 6-tetra-O-methyl-D-glucose. (O in the name refers that the Me groups that are attached to O atoms.)
In the open-chain structure, absence of Me group at C-5 because it was originally a part of the cyclic hemiacetal
linkage of D-glucose.

Representation of Structure of Fructose

1. Haworth representation
1
1 CH2OH
CH2OH2
2 CO
C  O3
HO H
HO 3 H4
4 H OH
H OH
5
H 5H OH OH
6
6 CH OH
CH2OH 2
Open chain
Open chain
Structure
Structure
of D(-) fructose
of D(-) fructose




1 2 1 2
1 2  2C C OH
HOH 1 2 HO C CH OH
HOH2 C C OH3 HO C2CH23OH2 2
HO H O + Enantiorner HO H O + Enantiorner
 HO

3 H4 O + Enantiorner HO 3 H4 O + Enantiorner
H OH H OH
 H 4 OH5 H 4 OH
5
 H 5H OH 5 H OH
 OH 6 H OH6
 6 6
-D(-) Fructopyranose
-D(-) Fructopyranose -D(-) Fructopyranose
(C=O) reacts with C-6 OH -D(-) Fructopyranose
(C=O) reacts with C-6 OH OH group on L.H.S. is 
Six-membered pyranose OH group on L.H.S. is 
Six-membered pyranose
Ring of -linkage
Ring of -linkage
OH group on R.H.S. is 
OH group on R.H.S. is 





H H H H
1 O
1 O H 6 HO 6
H 6 H O 6CH CH2OH OH
2OH
H H HOH
5 H 5 2 OH 2
2OH 2 5 5
OH H OH H
OH HOH OH 3 OH OH HOH CH2OH CH OH
3 1 2
4 34 4 3 41
OH HOH H OH HOH H
-D (-) Fructopyranose
-D (-) Fructopyranose -D (-) Fructopyranose
-D (-) Fructopyranose

2. Chair-form conformational structures

H H
H O H O
HO OH 1 OH
CH2OH HO OH
H 3 H 3
H H 2 H H 2
4 4 1
OH CH2OH
OH OH
-D(-) Fructopyranose -D(-) Fructopyranose
More stable, since Less stable, since
bulky (CH2OH) group bulky (CH2OH) group
on equatorial position on axial position
 Chem i str y | 26.9

3. Fructose occurs in nature in furanose form (five-membered cyclic ring) and in small extent in pyranose form
which is in equilibrium with six-membered hemiacetal ring.
 O 
 || 
It also exists in two cyclic forms that are obtained by the reaction of  − C −  group with C-5 -OH group.
 
 
 

1 2 2
HOH2C C OH HO C 1CH2OH
3
 HO H O + Enantiomer HO 3 H O + Enantiomer
4
 H OH H 4 OH
 H 5 5
H
 6 6
CH2OH CH2OH
-D (-) Fructopyranose -D (-) Fructopyranose

Haworth representation

O O
6 1 6
HOH2C O CH2OH HOH2C O OH
6 5 12 65 2
HOH2C CH
2OH HOH2C OH
H H HO OH H H HO CH 2OH
5 2 5 2 1
H H4 HO 3OH H H
4 HO3 CH2OH
OH H OH H 1
3 3
4 (-) Fructofuranose
-D -D4(-) Fructofuranose
H
OH H OH
O O
-D (-) Fructofuranose -D (-) Fructofuranose
HOH2C CH2OH HOH2C OMe
O O
HOH2CH CH2OH HOH2HC OMe
H HO OMe H HO CH2OH

H H H OMe
HO H OH
H H CH2OH
HO
OH
-D (-) Fructofuranose -D (-) Fructofuranose
OH H OH H
-D (-) Fructofuranose -D (-) Fructofuranose

2.7 Glucose

Preparation:

(a) From sucrose (cane sugar): Boiling sucrose with diluted HCl or H2SO4 in alcoholic solution produces glucose
and fructose in equal proportions.
H ⊕
C12H22O11 + H2O → C6H12O6 + C6H12O6
(Sucrose) Glucose Fructose

(b) From starch: Industrially, glucose is manufactured by the hydrolysis of starch by boiling it with dil. H2SO4 at
393 K under pressure.
H ⊕
(C6H12O5 )n + nH2O → nC6H12O6
Starchor Gellolose 393K;2−3bar Glucose
2 6 . 1 0 | Biomolecules and Polymers

PLANCESS CONCEPTS

(a) Carbohydrates: Polyhydroxy aldehydes or ketones with general formula Cx(H2O)y.

(b) Classification based on hydrolysis:
(i) Monosaccharides: Simplest carbohydrates that cannot be hydrolyzed any further.
(ii) Oligosaccharides: On hydrolysis gives two to nine units of monosaccharides.
(iii) Polysaccharides: On hydrolysis gives many units of monosaccharides.

(c) Reducing sugar: Sugars that reduce Fehling’s and Tollen’s solutions.
(d) D-L configuration: Refers to relative configuration of the (–OH) group with respect to glyceraldehydes.
(e) Epimers: Have more than one stereocenter but differ in configuration about only stereocenter.
(f) Anomers: Differ in configuration about the acetal or hemiacetal carbon.
(g) All anomers are epimers but the reverse is not true.
(h) Mutarotation: Change in specific rotation of an optically active compound when dissolved in solution is
called mutarotation.
(i) β-D glucose is more stable compared to α-D glucose, since all –OH groups are in equatorial position.
( j) In solution, β-D glucose and α-glucose are both in equilibrium.
(k) Fructose being a polyhydroxy ketone still gives positive test for Tollen’s, Benedict’s and Fehling’s reagent
and thus is a reducing sugar.
Vaibhav Krishnan (JEE 2009, AIR 22)

Chemical Reactions: Glucose has one aldehyde group, one primary (—CH2OH) group and four secondary
(—CHOH) hydroxyl groups, and gives the following reactions:
(a) Acetylation of glucose with acetic anhydride forms a penta-acetate, proving the presence of five hydroxyl
groups in glucose.
(CH COO) O
3 2 → OHC — (CHOCOCH ) — CH OOCCH
OHC — (CHOH)4 — CH2OH  3 4 2 3

(b) Glucose reacts with hydroxylamine to form monoxime and adds up a molecule of hydrogen cyanide to form
a cyanohydrin.
HOH2C — (CHOH)4 — CHO + HONH2 → HOCH2 — (CHOH)4 — CH =
NOH

Glucose monoxime

HOH2C — (CHOH)4 — CHO + HCN→ HOCH2 — (CHOH)4 — CH(OH)CN

Glucosecyanohydrin

The above reactions validate the occurrence of a carbonyl group in glucose.

(c) Ammonical silver nitrate solution (Tollen’s reagent) is a reducing sugar reducing to metallic silver.
Similarly, Fehling’s solution is reduced to reddish brown cuprous oxide. On the other hand, it itself gets
oxidized to gluconic acid. This proves the presence of an aldehydic group in glucose.

(d) HOCH — (CHOH) — CHO + Ag O → HOCH — (CHOH) — COOH + 2Ag

2 4 2 2 4
Gluconic acid
 Chem i str y | 26.11

H O
COOH C COOH

H C OH H C OH H C OH

OH C H Bromine water HO C H Nitric acid HO C H

H C OH H C OH H C OH

H C OH H C OH H C OH

CH2OH CH2OH COOH

Gluconic acid Glucose Glucaric acid

(e) On prolonged heating with HI, glucose forms n-hexane, indicating that all the six carbon atoms in glucose
are linked linearly.
HI
HOCH2 — (CHOH)4 — CHO → H3C — CH2 — CH2 — CH3
n−hexane

(f) D-Glucose reacts with phenyl hydrazine and gives glucose phenylhydrazone, which is soluble.
Excess use of phenylhydrazine produces, a dihydrazone, called osazone.
1 1
CHO CH = NHPh
2 PhNHNH2 2 CH = NHPh
H C OH H C OH
5 -H2O 5 2H2N.NPh
(CHOH)3 (CHOH) 3 PhNH2 + NH3 + C = NHPh
6 6 (CH2OH)3
CH2OH CH2OH
D-Glucose D-Glucose CH2OH
phenylhydrazone D-Glucosazone

H
H H
C=O
C=O H NH2 C=O
H OH HO H H OH
HO H H OH HO H
H OSO3 H OH H OH
H OH CH2OH H OH
CH2OH D - Glucosamine CH3
D - Glucose -4-Sulfate Re 6 - Deoxy -D- Glucose
p lac
wit eO Replace OH with NH3
hs H
ulp
ate H H
OH C=O C=O
C=O H OH H OH
H OH Oxidation of the C - 1 Oxidation of the C - 6
HO H HO H
HO H aldehyde to a H OH position H OH
H OH carboxilic acid H OH H OH
H OH CH2OH COOH
CH2OH D - Glucose D - Glucuronic Acid
D - Gluconic Acid H te (An Aldonic Acid)
(An Aldonic Acid) O ha Reduction of the anomeric carbon
c e sp
a o
H pl Ph
Re ith CH2OH CO2H
C=O w O OH
OH H OH H
H H
HO H HO H OH H
H OH H OH OH H

H OH H OH H OH
H2C-PO4 CH2OH Look! this too
Sorbital can be drawn in a
D- Glucose -6- Phosphate Haworth projection
(An alditol, Polyol)

Lobry de bruyn van ekenstein rearrangement reaction

2 6 . 1 2 | Biomolecules and Polymers

Mechanism
 to a common intermediate enediol, establishing the following
(a) The ketose and aldose tautomerise in OH
equilibrium: H
1 1 1
H C=O H C = ÖH H C = OH
2 2 2
H C OH C ÖH C OH
3 3 3
H C OH H C OH H C OH

Aldose Ketose
(Glucose) (Fructose)

(b) When the aldose reforms from the enediol, H+ can add to C-2 from either face of the (C=C) to give C2
aldohexose epimers.
H C OH H C = OH H C = OH
C OH HO C H + HO C H

Enediol Glucose Mannose

(C2 epimeric aldoses)

2.8 Some Disaccharides

Maltose: (a) Preparation: Fractional hydrolysis of starch by enzyme diastase.
Diastase 6 6
2(C6H10 O5 )n + nH2O  → nC12H22O11 CH2OH CH2OH
Starch Maltose
5 O
H O H H H
5
H H
4 1
OH H 1 4 OH H 
O OH
HO 3 2 -Linkage 3 2
H OH H OH

-D-glucose (Maltose) -D-glucose

(b) Units: Two units of α-D glucose.

(c) Reducing sugar
(d) Linkage: α glycosidic linkage between C1 and C4 carbon atoms.

Sucrose: (a) Preparation: Prepared from sugarcane and beetroot 6

CH2OH
(b) Units: α-D glucose and α-D fructose.
5 O
H H
(c) Non-reducing sugar
4
H 1 -link (Glucose unit)
(d) Linkage: α gylcosidic linkage with reference to glucose and β OH H
gylcosidic linkage to fructose, both linked at C1 carbon. HO 3 2
O
H OH
6
HOH2C O
5 -link
H H HO (Fructose unit)
CH2OH
4 31
HO H
 Chem i str y | 26.13

Lactose: 6 6
CH2OH CH2OH 
(a) Preparation: Found in milk.
5 5 HO
O O
(b) Units: β-D glucose and β-D galactose. HO H 1
H H
(c) Reducing sugar. 4
OH
1 O OH
H H
(d) Linkage: β glycosidic linkage. H
-Linkage 4 H
H 3 2 3 2
The laevo rotation of fructose (–92.4°) is more
H OH H OH
than rotation of glucose (+52.5º), so the mixture
is laevorotatory. Sucrose hydrolysis brings about a
change in the sign of rotation, from dextro (+) to
laevo (–) and the product is called invert sugar.

2.9 Polysaccharides
(a) Starch: It is the main storage poly saccharide of plants having general formula (C8H16O5)n. The main source is
maize, wheat, barley, rice and potatoes. It is a polymer of α-glucose and consists of two components – Amylose
and Amylopectin. Amylose is made up of long, unbranched chain of α-D-(+)-glucose linkage. Amylopectin is
a branched chain polymer of α-D-glucose units, in which chain is formed by C1–C6 glycosidic linkage whereas
branching occurs by C1–C6 glycosidic linkage.
H +
(C6H10O5 )n + H2O → nC6H12O6
Starch Glucose

CH2OH CH2OH CH2OH

6 O 6 O 6 O
H H H H H H H H H
 4 1 4 1
H H H
HO OH O OH
H
O OH O

H OH H OH H OH

-link -link
Amylopectin
CH2OH CH2OH
O O
H H H H H5 H
4 1
OH H 1 4 H
O O OH
-link
H OH H OH
O
Amylose
CH2OH CH2 CH2OH
O O 6 O
H H H H H5 H H H H
4 1 4 1
OH H 1 4 H 1 4 H
O O OH O OH O
H
H OH H OH H OH
-link -link
2 6 . 1 4 | Biomolecules and Polymers

Table 26.2: Difference between Amylose and Amylopectin

S.No. Amylose Amylopectin

1. It is water soluble fraction for starch It is water insoluble fraction of starch.
2. It is 20% of starch It is 80% of starch.
3. It is straight chain polymer of D-glucose units. It is branched chain polymer of D glucose units.
4. Glucose units are joined by α-1, 4 glycosidic linkage In amylopectin, the glucose units are joined by α-1,
6 glycosidic linkage.
5. Its molecular mass lies in the range of 10,000–50,000 Its molecular mass is in the range of 50,000–1,00,000.

(b) It occurs exclusively in plants and is a main constituent of cell wall of plant cell. It is a linear polymer of β-D-
glucose in which glucose units are linked together by C1–C4 glycosidic linkage. It is a non-reducing sugar.

CH2OH CH2OH CH2OH

6 O 6 O 6 O
H H H H H H H H H
1 4 1 4 1
O OH H O OH H O OH H O
H

H OH H OH H OH

Table 26.3: Relative sweetness of sugar

Sugar Relative Sweetness Sugar Relative Sweetness

Sucrose 100 (standard) Glucose 74
Lactose 16 Invert sugar 130
Galactose 32 Fructose 173
Maltose 33

Chemical treatment of cellulose, the most important natural polymer, gives various useful derivatives
•• Rayon: Cellulose acetate and cellulose xanthate are used as a fibre.
•• Celluloid: Cellulose dinitrate or pyroxylin, combined with plasticizer and alcohol, is used for the manufacturing
of photographic film, spectacle frames, piano keys, etc. It is known as artificial ivory.
•• Explosive: Cellulose trinitrate is used extensively as a blasting and propellant explosive.
•• Lacquer: Collodion is used for manufacturing washable cellulose paints.
•• Water proofing: Solution of cellulose acetate is used to provide anti-shrink property to textile fabric.
•• Methyl cellulose: is used in fabric sizing, paste and cosmetics. Ethyl cellulose is used for manufacturing of rain
coats and plastic films.

PLANCESS CONCEPTS

•• Both glucose and gluconic acid on oxidation with nitric acid yield a dicarboxylic acid.
•• Glucose on heating with HI gives n-hexane.
•• D-glucose reacts with phenylhydrazone to give glucose phenylhydrazine and excess use gives
osazone. But three molecules per molecule of glucose are used for oxidation while the other two are
attached to the molecule.
 Chem i str y | 26.15

PLANCESS CONCEPTS

•• Glucose when heated with concentrated NaOH, establishes an equilibrium of D-glucose, D mannose
and D-fructose.
•• Sucrose on hydrolysis changes from Dextrorotatory to Levorotatory and hence is called invert sugar.
•• Sucrose is a non-reducing sugar since the hemiacetal hydroxyl groups are linked by glycosidic
linkage.
•• Sucrose on hydrolysis gives glucose and fructose.
•• Maltose is a reducing sugar that on hydrolysis gives two molecules of D-glucose.
•• Starch is non-reducing polysaccharides that on hydrolysis give D-glucose.
•• Starch is mixture of two polysaccharides: amylose and amylopectin.
•• Amylase is straight-chain polysaccharide that is soluble in water and gives blue colour with iodine.
•• Amylopectin is a branched chain polysaccharide insoluble in water and doesn’t give blue colour with
iodine.
•• Cellulose is a straight chain polysaccharide comprising of only D-glucose units.
•• The difference between starch and cellulose is that starch has α-glycosidic linkage while cellulose
has β glycosidic linkage.
Nikhil Khandelwal (JEE 2009, AIR 94)

Illustration 1: Write the names of monomers of the following polymers: (JEE MAIN)

H H O H O H
(i) (ii) (iii) CF2 CF2
N (CH2)6 N C (CH2)4 C C (CH2)5 N n
n n

Sol: (i) Hexamethylenediamine, H2N—(CH2)6—NH2 and adipic acid, HOOC—(CH2)4—COOH

(ii) Caprolactam (iii) Tetrafluoroethene, F2C = CF2.

Illustration 2: Describe the term D-and L-configuration used for amino acids with examples. (JEE ADVANCED)

Sol: Consider the following configurations (I and II) of a-amino acids

COOH COOH
 
H2N C H H C NH2
R R
I II
L-Amino acid D-Amino acid

NH2 group on the α-carbon oriented toward left (as in structure I) is referred to as L-amino acid and NH2 group
oriented toward right (as in structure II) is referred to as D-amino acid.

Illustration 3: The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicates the correlation of
configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde.
Predict whether the following compound has ‘D’ or ‘L’ configuration.
2 6 . 1 6 | Biomolecules and Polymers

1
CHO
2
HO H
3
H OH
4
HO H
5
HO H
6
CH2OH

Sol: The orientation of the OH group at the penultimate chiral carbon (i.e. last but one or C5) toward left gives the
compound it’s L-configuration.

Illustration 4: How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
 (JEE ADVANCED)

Sol: The –OH group on the terminal carbon atom (i.e. C6) is called the 1° hydroxyl while the rest of the four
remaining OH group present on C2, C3, C4 and C5 are called 2° hydroxyl groups. While 1° Hydroxyl groups are
easily oxidized to give carboxylic acids but 2° hydroxyl groups undergo oxidation only under drastic conditions.
For example, glucose on oxidation with HNO3 gives a dicarboxylic acid, saccharic acid having the same number of
carbon atoms as glucose. This indicates that glucose contains one 1° hydroxyl group while the remaining four are
2° hydroxyl groups.
CHO COOH
2 (CHOH)4 HNO3
(CHOH)4
[O]
1 CH2OH COOH
Glucose Saccharic acid

3. AMINO ACIDS
α-Amino Acids: Carboxylic acids in which one α-hydrogen atoms of alkyl group is substituted by amino (–NH2)
group are called α-Amino acids. The general formula is

R CH COOH
NH2

Where R = H or alkyl group

Structure of α-amino acids: The amino acids containing one carboxylic group and one amino group behave
like a neutral molecule. This is because in aqueous solutions the acidic carboxylic group and basic amino group
neutralize each other intramolecularly to produce an internal salt structure known as zwitter ion or dipolar ions.
+
NH2 NH3
R C COOH R C COO-
Zwitter ions

However, the neutral zwitter ion (dipolar ions) changes to cation in acidic solution and exist as anion in alkaline
medium. Thus amino acids exhibit amphoteric character.

H2N CH COO Alkali H3N CH COO Acid H3N CH COOH

R R R
Anion Zwitter ion Cation

Therefore, amino acid exists as zwitter ion when the solution is neutral or pH-7. The pH at which the structure of
an amino acid has no net charge is called its isoelectric point.
 Chem i str y | 26.17

Classification of Amino Acids: Based on the relative number of –NH2 and –COOH group, α-amino acids are
classified in three main groups
(a) Neutral Amino acids: Amino acids containing one –NH2 group and one –COOH group. For example, glycine,
valine, alanine etc.
(b) Basic amino acids: These contain one –COOH group and two –NH2 groups, such as lysine and arginine.
(c) Acidic amino acids: Amino acids that containing two –COOH groups and one –NH2 group are called acidic
amino acids; for example, aspartic acid and glutamic acid, etc.

Isoelectronic Point: In acidic solution, an amino acid being a positive ion moves toward the cathode in an electric
field. On the other hand in alkaline solution, it is available as a negative ion that migrates toward anode. At a
specific hydrogen ion concentration (pH), the dipolar ion exists as neutral ion and does not show migration toward
any electrode. This pH is termed as the isoelectric point of the amino acid.
The isoelectric point is dependent on other functional groups in the amino acids. Neutral amino acids have the
range between pH 5.5 and 6.3. At isoelectric points, the amino acids are least soluble in water. This property is
utilized in the separation of various amino acids formed by the hydrolysis of proteins.

Table 26.4: List of 20 naturally occurring α- amino acids

Nature α-Amino acid Abbreviation or code Structure

NEUTRAL
1. Glycine Gly H2N–CH2–COOH
2. Alanine Ala H2N CH COOH
CH3

3. Valine Val H2N CH COOH

CH(CH3)2

4. Proline Pro
COOH
N
H
ACIDIC
5. Serine Ser H2N CH COOH
CH2 OH

6. Aspartic acid Asp H2N CH COOH

CH2 COOH

7. Glutamic acid Glu H2 N CH COOH

CH2 CH2 COOH

BASIC

8. Lysine* Lys H2N CH COOH

CH2 CH2CH2 CH2NH2

9. Arginine Arg H2N CH COOH

CH2 (CH2)2NH C NH2
NH
2 6 . 1 8 | Biomolecules and Polymers

4. PROTEINS
Proteins are complex nitrogeneous organic substance that occurs in all animals and plants. Proteins are called the
most vital chemical substance and are necessary for the normal growth and maintenance of life. Protein serves
following functions in our body:
•• Proteins promote growth
•• Proteins supply essential amino acids to blood.
•• Proteins help maintain body tissues.
•• Proteins synthesize various enzymes.
•• Proteins protect body from infections.

Protein content of food stuffs

Food stuff % Protein Food stuff % Protein
Milk 5 Meat 24-26
Wheat 14 Egg yolk 16
Peas 21 Egg (white) 12.5
Maize 10 Cheese 33

Classification of proteins: Proteins are classified on the basis of two different methods. The first mode of
classification, proteins are of two types is based on their shape and functions:
(a) Fibrous proteins (b) Globular proteins

(a) Fibrous proteins: They are thread like molecules that lie side by side to form fibers. They are held together by
hydrogen bonds. These are insoluble in water but soluble in concentrated acids and alkalis. A few examples
are keratin (present in hair, nails, wood, feathers and horns). Muscles have myosin, silk is composed of fibroin,
bones and cartilages have collagen.

(b) Globular proteins: These proteins have molecules folded into compact units that often acquire spheroidal
shape. Such proteins are soluble in water, diluted acids and alkalis, such as insulin, haemoglobin, albumin, etc.

Structure of Proteins: The structure of proteins is quite complex. Study of its structure is carried out under the
following headings.
(a) Primary structure of protein: The primary structure of protein refers to its covalent structure, that is, the
sequence in which various α-amino acid are arranged in protein or in the polypeptide structure of protein.
The linkage (–CO–NH–) is known as peptide linkage.

H2N CH COOH+H2N CH COOH

R R’
H2N CH CO NH CH COOH+H2O
R Peptide bond R’
The dipeptide still has free amino and carboxyl groups that can react with other molecules of amino acid
resulting in polypeptide formation.
O H R’ O H R’’’
C N CH C N CH
CH C N CH C N
R O H R’’ O H
C-terminal N-terminal

In the polypeptide chain, the free amino end is termed as N–terminal and the free carboxyl end is called C–
terminal end.
 Chem i str y | 26.19

(b) Secondary structure of protein: This refers to the arrangement of polypeptide chains into a definite three-
dimensional structure that protein assumes as a result of hydrogen bonding. Depending upon the size of the
R-group of the amino acids in polypeptides, two different types of secondary structure are possible:
(i) α-helix structure (ii) β-Pleated structure
(i) α-Helix structure: This type of secondary structure is possible when the alkyl groups present in amino
acids are large and involved in coiling of the polypeptide chains. The intramolecular hydrogen bond
between the >C = O group of one amino acid and –NH group of the fourth amino acid stabilizes the
helical pattern in right-handed coil and the shape.
(ii) β-Pleated structure: Such secondary structure is acquired when the alkyl groups of amino acids are
small. In this kind of structure, the linear polypeptide chains are arranged side by side and are held
 O 
 || 
together by intermolecular hydrogen bond between the  − C −  and –NH group.
 
 
 
R
C N O
H H C C
C C O H R C
N H N
R C R N H
O C
H H N O
C
C
H O C C R
N N H
C R
C H N
O O H R O C O
N H C
H C C N N R C
C H N H
R C R
O C
H H C N O C O R
H C
R C H N
C N H N O
C R R C
O C
H H H C O
C R C
C N R H N N H
R C C R
O C
O O O
O C C R
H H C N N H N
R C H C H C O
C N R
C R C O
O N R C
H
C R

-helix
-pleated sheet

(c) Tertiary structure of protein: The tertiary structure of protein is the most stable shape that a protein assumes
under the normal temperature and pH conditions. Attractive forces between the amino acid chains are
involved in acquiring tertiary structure,. These attractive forces, like hydrogen bond, disulphide bonds, ionic,
chemical and hydrophobic bonds, result in a complex and compact structure of protein. The two important
tertiary structures of proteins are fibrous structures and globular structure. Fibrous proteins have largely
helical structure and are rigid molecules of rod-like shape. Globular proteins show a polypeptide chain that
consists partly of helical sections and partly β-pleated structure and remaining in random coil form. These
different segments of secondary structure then fold up to give protein a spherical shape.

(d) Quaternary structure of proteins: The quaternary structure of proteins develops when the polypeptide
chains, which may or may not be identical, are held together by hydrogen bonds. It results in the increase of
molecular mass of protein greater than 50,000 amu. For example, hemoglobin contains four subunits, two
identical α-chains containing 141 amino acids each and the other two identical β-chains containing 161 amino
acids each.
2 6 . 2 0 | Biomolecules and Polymers

(e) Denaturation of proteins: Proteins when subjected to the action of heat, mineral acids or alkali, the water
soluble form of globular protein changes to water insoluble fibrous protein resulting in the precipitation or
coagulation of protein. This is called denaturation of proteins.

5. NUCLEIC ACIDS
Nucleic acids are vital biomolecules that present in the nuclei of all living cells as nucleoproteins. These are long-
chain polymers with a high molecular mass. Also called biopolymer, they have nucleotide as a repeating structural
unit (monomer). These play an important role in transmission of the heredity characteristics from one generation to
the next and also in the biosynthesis of proteins. Therefore, the genetic information coded in nucleic acid governs
the structure of protein during its biosynthesis and hence controls the metabolism in the living system.

Structure of nucleic acids: The nucleic acid is the prosthetic component of nucleoproteins. Nucleic acid on
stepwise hydrolysis gives following products as shown in the chart.
Nucleic acid
Hydrolysis

Nucleotides

Hydrolysis

Nucelosides Phosphoric acid

Hydrolysis

Pentose sugar Purine bases Pyrimidine bases

Ribose or (Adenine and guanine) (Thymine and cytosine) from DNA
Deoxyribose (Uracil and cytosine) from RNA
Flowchart 26.1: Components of nucleic acid

Difference between DNA and RNA

The main points of difference between the two types of nucleic acids are given in the table.

Table 26.5: Difference between DNA and RNA

DNA RNA

1 The pentose sugar present in it is 2- deoxy D(–) ribose. It has D(–) ribose sugar.

2. It contains cytosine and thymine as pyrimidine bases. It contains cytosine and uracil as pyrimidine bases.

3. DNA is double strand and pairing of bases is present It is a single strand molecule looped back on itself. The
throughout the molecule. pairing of bases in present only in helical part.

4. It occurs in the molecules of the cell. It mainly occurs in the cytoplasm of the cell.

5. It is a very large molecule and the molecular weight It is a much smaller molecule and its molecular weight
varies from 6 million to 16 million amu. ranges from 20 thousand to 40 thousand amu.

6. It has a characteristic property of replication. It does not replicate.

7. DNA controls the heredity character. RNA only governs the biosynthesis of proteins.
 Chem i str y | 26.21

Ribonucleic acids
(a) The pentose sugar in RNA is ribose.
(b) Adenine and guanine represent the purine bases of RNA; the pyrimidine bases are uracil and cytosine.
(c) The thymine in DNA is replaced by uracil in RNA.
(d) RNA is single stranded, but double stranded RNA is present in Rheovirus and wound tumor virus.
(e) There are three major classes of RNA, each with specific functions in protein synthesis

(i) mRNA
•• Messenger RNA is produced by DNA; the process is called transcription.
•• Messenger RNA encodes the amino acid sequence of a protein in their nucleotide base sequence.
•• A triplet of nitrogenous bases specifying an amino acid in mRNA is called codon.

(ii) tRNA
•• tRNA is also known as soluble RNA (sRNA) as it is soluble in 1 molar solution of sodium chloride.
•• tRNA identifies amino acids in the cytoplasm and transports them to the ribosome.
•• Molecules of tRNA are single-stranded and relatively very small.
•• Anitcodon is a three-base sequence in a tRNA molecule that forms complementary base pairs with
a codon of mRNA.
•• All transfer RNA possess the sequence CCA at their three ends; the amino acid is attached to the
terminal as residue.

(iii) rRNA
•• Ribosomal RNA is found in ribosomes of cell and is also called insoluble RNA.
•• The main function of rRNA is to attract provide large surface for spreading of mRNA over ribosome’s
during translocation process of protein synthesis.

Table 26.6: Types of RNA

Sedimentation Number of nucleotide Percentage of total cell

Type Mol. Wt.
Coefficient residues RNA
mRNA 6 to 25 25,000-1000,000 75-3000 2
tRNA 4 23,000-23,000 75-90 16
rRNA 5 35,000 100

16 550,000 1500 82

23 1100,000 3100

•• The relationship between the sequence of amino acids in polypeptide with base sequence of DNA
or mRNA is genetic code.
•• Genetic code determines the sequence of amino acids in a protein.
•• A triplet would code for a given amino acid as long as three bases are present in a particular sequence.
•• Later in a cell-free system. Marshall Nirenberg and Philip (1964) were able to show that GUU codes
for the amino acid valine.
•• The spellings of further codons were discovered by R. Holley. H. Khorana and M. Nirenberg.
•• They have been awarded the Nobel Prize in 1968 for researches in genetic code.
2 6 . 2 2 | Biomolecules and Polymers

Lipids: The term lipids represent a group of biomolecules that are insoluble in water but soluble in organic solvents
of low polarity, such as chloroform, toluene, ether, carbon tetrachloride.
(a) They serve as the energy reserve for living cell.
(b) Lipids are classified in three groups:
(i) Triglyceride ester of higher fatty acids or oils and fats
(ii) Phospholipids
(iii) Waxes

PLANCESS CONCEPTS

•• Proteins: Complex biopolymers of amino acids. Proteins on hydrolysis give peptides that on hydrolysis
give amino acids. Almost all amino acids found in our body is α-amino carboxylic acid.
•• Amino acids are generally colorless, water soluble, have high melting point and are crystalline solids.
•• In aqueous solutions, carboxylic group loses a proton and the amino group accepts it and thus it exists in
form of dipolar ion called zwitter ion.
•• At a particular pH, the dipolar ion exists as a neutral ion and does not show migration toward any electrode,
and this pH is known as isoelectric point.
•• At isoelectric point, the amino acids are least soluble in water and this property is used for separation of
various amino acids formed by hydrolysis of proteins.
•• –CO–NH– bond in peptide are called the peptide linkage. There is restriction in rotation about the peptide
bonds and thus free rotation is not possible.
•• Test for protein:
(i) Biurets’ test: the blue color reagent turns violet in presence of protein.
(ii) Xanthoprotic test: protein + concentrated HNO3 gives orange color in alkaline solution.
(iii) Millons test: protein + Million’s reagent gives white precipitate that on heating gives red precipitate.
(iv) Ninhydrin test: proteins or amino acids in presence of Ninhydrin reagent give a colored product.

Saurabh Gupta (JEE 2010, AIR 443)

Illustration 5: Two samples of DNA, A and B have melting temperature, Tm 340 K and 350 K, respectively. Can you
draw any conclusion from these data regarding their base content?  (JEE MAIN)

Sol: We know that CG base pair has three H-bonds and AT base pair has two H-bonds; therefore, CG base pair is
more stable than AT base pair. Since sample B has higher melting temperature than sample A; therefore, sample B
has higher CG content as compared to sample A.

Illustration 6: Which of the following will reduce Tollens’ reagent? Explain.  (JEE MAIN)
6 6 6 6
CH2OH CH2OH CH2OH CH2OH
5 O 5 5 O
HO HO O HO HO O 5 OH
H H H H
4 H O 4 H 1 4 H O 1 OH 4
HO 1 HO HO 1 H
H 2 H 2 OH H 2 H H
3 3 3 2 3
HO H H OH H H OH
(P) (Q)
 Chem i str y | 26.23

Sol: In disaccharide (Q), both the monsaccharides are linked through their reducing centres (C1); therefore, it is
not a reducing disaccharide. In disaccharide (P), the reducing end (C1) of one monosaccharide is linked to non-
reducing end (C4) of the other monosaccharide. In other words, reducing end of one monosaccharide is free.
Therefore, it is a reducing disaccharide.

Illustration 7: An optically active amino acid (A) can exist in three form depending upon the pH of the medium. If
the molecular formula of (A) is C3H7NO2 write.  (JEE ADVANCED)
(i) Structure of compound (A) in aqueous medium. What are such ions called?
(ii) In which medium will the cationic form of compound (A) exist?
(iii) In alkaline medium, towards which electrode will the compound (A) migrate in electric field?

Sol: An optically active amino acid having the M.F. C3H7NO2 is alanine, i.e. CH3 — CH — COOH
|
NH2

Depending upon the pH of the medium, alanine can exist in the following three forms:
CH3 CH3 CH3
+ OH- + OH-
H3N CH COOH H3N CH COO- H2N CH COO-
H+ H+
Cationic form(I) Zwitterion(II) Anionic(III)
(exists in acid medium) (exists in aqueous medium) (exists in basic medium)

(i) In aqueous medium, the compound (A) exists as a zwitterion (II).

(ii) In acidic medium, it will exist in the cationic form (I).
(iii) In alkaline medium, it will exist in the anionic form (III) and will migrate towards anode under the influence of
an electric field.

Illustration 8: Following two amino acids, lysine and glutamine form dipeptide linkage. What are the two possible
dipeptides?
NH2 NH2

H2N COOH HOOC CONH2

Sol:
NH2 H
NH2 H NH2
CONH2 N
N
(i) H2N (ii) H2NOC C
C COOH
COOH O
O Glnyllysine
Lyslglutamine
(Gln-Lys or Q-K)
(Lys-Gln or K-Q)

6. POLYMERS
A polymer is a compound of high molecular mass formed by the combining large number of small molecules and
the process is called polymerization. The small molecules that constitute the repeating units in a polymer are called
monomer units. These large molecules have relative molecular masses in the range 104 – 106 amu.

E.g. nCH
=2 CH2 → CH2 – CH2 –  .
n
ethene polyethene

Where n is as high as 105. The number of monomer units in a polymer is called the degree of polymerization.
2 6 . 2 4 | Biomolecules and Polymers

7. CLASSIFICATION OF POLYMERS
(a) On the Basis of Source
(i) Natural polymers: These polymers are found in plants and animals. Examples are proteins, cellulose,
starch, some resin and rubber.
(ii) Semi-synthetic polymers: Cellulose derivatives as cellulose acetate (rayon), cellulose nitrate, etc. are the
usual examples of this sub category.
(iii) Synthetic polymers: A variety of synthetic polymers such as plastic (polythene), synthetic fibers (nylon
6,6) and synthetic rubbers (Buna –S) are examples of manmade polymers extensively used in daily life
as well as in industry.

(b) On the Basis of Structure: There are three different types based on the structure of the polymers.
(i) Linear polymers: These polymers consist of long and straight chains. The examples are high density
polythene, polyvinyl chloride, etc. These are represented as:

(ii) Branched chain polymers: These polymers contain linear chains having some branches, For example,
low density polythene. These are depicted as follows:

(iii) Cross-linked or Network polymers: These are usually formed from bifunctional and trifunctional
monomers and contain strong covalent bonds between various linear polymer chains. For example,
bakelite, melamine, etc. These polymers are depicted as follows:

(c) On the Basis of Synthesis: These are of two types based on synthesis –
(i) Condensation polymerization: In this, the monomer (same or different) units link with each other by
the elimination of a small molecule (for example, water, methyl alcohol) as a byproduct. The polymer
formed is known as condensation polymer. Nylon and terylene are the most common examples.
Since the condensation polymerization proceeds by a stepwise intermolecular condensation, it is also
known as step polymerization and the polymer formed is known as step growth polymer.

n H2N(CH2)6NH2 + n HOOC (CH2)4 COOH → – NH(CH2 )6 NHCO(CH2 )6 CO  – + nH2O

n
Nylon− 6,6

(ii) Addition polymerization: This involves self-addition of several unsaturated molecules of one or two
monomers without loss of any small molecule to form a single giant molecule. The polymer formed is
known as addition polymer. Polythene is the most common example. However, the addition polymers
formed by the polymerization of a single monomeric species is known as homopolymers, for example,
polyethene.

nCH
=2 CH2 → –(CH2 – CH2 ) –Homopolymer
Ethene n
Polyethene

The polymers made by addition polymerization from two different monomers are termed as copolymers,
for example, Buna-S, Buna-N, etc.
 Chem i str y | 26.25

(d) On the Basis of Molecular Forces: A large number of polymer applications in different fields depend on their
unique mechanical properties like tensile strength, elasticity, toughness, etc. These mechanical properties are
governed by intermolecular forces, for example, van der Waals forces and hydrogen bonds.
(i) Elastomer: In these polymers, polymer chains are held together by the weakest intermolecular force.
Weak binding forces permit the polymer to be stretched. A few “crosslinks” are introduced in between
the chains, which help the polymer to retract to its original position after the force is released as in
vulcanized rubber. The examples are Buna-N, neoprene, etc.
(ii) Fibers: In these polymers, polymer chain held together by strong intermolecular forces, like hydrogen
bonding. These strong forces also lead to close packing of chains and thus impart crystalline nature.
Fibers are the thread-forming solids that possess high tensile strength and high modulus. The examples
are polyamides (nylon -6, 6), polyesters (terylene), etc.
(iii) Thermoplastic polymers: These polymers possess intermolecular forces of attraction intermediate
between elastomers and fibers. These are the linear or slightly branched long-chain molecules capable
of repeatedly softening on heating and hardening on cooling. Some common thermoplastics are
polythene, polystyrene, polyvinyls, etc.
(iv) Thermosetting polymers: These polymers are cross linked or heavily branched molecules, which on
heating undergo extensive cross linking in molds and again become infusible. These cannot be reused.
Some common example are bakelite, urea-formaldelyde resins, etc.

(e) Classification on the Basis of Type of Polymerization: Synthetic polymers have been classified into definite
classes in various manners, for example,
(i) All the synthetic polymers may be classified into two groups based on of the type of the process involved
during their preparation, viz. condensation or addition polymers involving condensation or addition
polymerization processes, respectively, during their synthesis.
(ii) A more rational method of classification is based upon the sequence of synthesis of the polymer (i.e.,
mode of addition of the monomer units to the growing chain).
According to this method, polymers may be of two types, viz. chain growth polymers and step growth polymers.
•• Chain growth polymers (earlier called as addition polymers). These polymers are formed by the
successive addition of monomer units to the growing chain having a reactive intermediate (free
radical, carbocation or carbanion)
The most important free radical initiator is benzoyl peroxide.
• •
heat
E.g. RCO — O — O — OCR  → 2RCOO → 2R + 2CO2
Benzoylperoxide

• • •
heat CH =CH
R + H=
2C CH2  → R – CH2 CH2 
2 2 → R — CH — CH — CH →→ R  –CH – CH  —
2 2 2  2 2
n

Examples of chain growth polymers

Monomer(s) Polymer
Ethylene Polyethylene
Propylene Polypropylene
Tetrafluoroethylene Polytetrafluoroethylene (PTFE) or Teflon
Vinyl chloride Polyvinyl chloride (PVC)
Isoprene Polyisoprene (Natural rubber)
Butadiene Polybutadiene
Butadiene and styrene Buna-S
2 6 . 2 6 | Biomolecules and Polymers

•• Step growth polymers (earlier called as condensation polymers). These polymers are formed through
a series of independent steps (reactions). Each step involves the condensation (bond formation)
between two difunctional units (monomers) leading to the formation of small polymer (say dimer,
trimer, tetramer, etc.)
Step 1. A + B → A–B
Monomer Monomer Dimer
Step 2. A — B + A → A — B —A
Trimer
Step 3. A—B+A—B → A—B—A—B
Dimer Dimer Tetramer
Step 4. A — B — A — B + A — B → A — B — A — B — A — B or (A —B)3
Step 5. (A — B)3 → ( — A — B — )n
Polymer

Examples of step growth polymers

Monomers Polymers
Adipic acid hexamethylene diamine Nylon-66
Terepthalic acid and ethylene glycol Terylene
Phenol and formaldehyde Bakelite

(f) Based on the magnitude of molecular forces the polymers may be classified as:
(i) Elastomers
•• In which the intermolecular forces of attraction are the weakest. They are amorphous and
have high degree of elasticity, for example, Buna-S.
(ii) Fibers
•• In which the intermolecular forces of attraction are the strongest like H bonding or di-
pole-dipole attraction.
•• They have high tensile strength, high melting point and low solubility, for example, polyes-
ters, polyacrylonitriles
(iii) Thermoplastics
•• In which the intermolecular forces of attraction are in between those of elastomers and
fibers
•• They are hard at room temperature, soften on heating without any change in mechanical
properties of plastic, have little or no cross links, for example, polythene, polyacrylonitrile,
Teflon, etc.
•• Semi-fluid substances with low molecular weight that once set into particular mold cannot
be used again and again.
•• This is because of extensive crosslinking between two polymer chains to give a three-dimen-
sional network, for example, bakelite, urea-formaldehyde; melamine formaldehyde etc.
 Chem i str y | 26.27

8. MOLECULAR MASS OF POLYMERS

(a) There are two types of average molecular mass of polymers.

( )
(b) Number average molecular mass Mn : It is determined by methods depending upon the number of molecules
present in the polymer sample, for example, colligative properties like ∆Tf; ∆Tb, Osmotic pressure

=(c) Mn
N1M1 + N2M2 + N3M3
=
∑ N1M1
N1 + N2 + N3 ∑ N1
Where N1,N2, N3 are the number of molecules. M1, M2, M3 are the molecular masses.

Weight Average Molecular Mass: It is determined by using the methods depending upon the masses for the
individual molecules like light scattering, ultracentrifugation, sedimentation, etc.

Mw
=
∑
=
m1M1 ∑ N1M12
; M– is mass of macromolecule
∑ m1 ∑ N1m1
Example in a polymer sample, 30% molecules have molecular mass 20,000; 40% have molecular mass 30,000 and
the rest 30% have 60,000. Calculate the number average and mass average molecular masses.

30 × 20000 + 40 × 30,000 + 30 × 60,000 600000 + 1200000 + 1800000 3600000

Mn = = Mn = 36000
30 + 40 + 30 100 100

30(20000)2 + 40(30,000)2 + 30(60,000)2

Mw = 43,333
30 × 20000 + 40 × 30,000 + 30 × 60,000

9. VARIOUS TYPES OF POLYMERS

Table 26.7: Types of polymers

Name of the polymer Structure of Natural of polymer Properties Uses

along with abbreviation monomer
A. Addition Polymers
I. Polyolefins
1. Polymers or polythome CH2 = CH2 (a) Low-density Transparent, Packing material
polyethylene (LDPE) in an moderate tensile (plastic films, bags etc.)
CH2 CH2 addition or chain-growth strength, high insulation for electrical
homopolymer. It is a toughness. wires and cables.
highly branched polymer
Translucent, Manufacture buckets,
and is obtained by free-
chemical tubs, pipes, bottles and
radical polymerization
inert, growth toys.
(b) High-density toughness and
polyethylene (HDPE) is tensile strength.
obtained by coordination
polymerization. It is a
linear addition or chain-
growth homopolymer.
2. Polypropylene or CH3CH=CH2 Addition homopolymer Harder and Packing of textile and
Polypropene linear can be obtained by foods, for making liners
stronger than
free-radical or Ziegler- of bags, heat shrinkable
polythene
—CH–CH2— Natta polymerization. wraps, carpet fibers,
CH3 ropes, automotive
— molding, stronger pipes
and bottles.
2 6 . 2 8 | Biomolecules and Polymers

Name of the polymer Structure of Natural of polymer Properties Uses

along with abbreviation monomer
3. Polystyrene or Styron C6H5CH=CH2 Addition homopolymer, Transparent Plastic toys, household
linear chain wares, radio and
CH CH2 television bodies,
n refrigerator linings etc.
II Polydienes
1. Neoprene Cl Addition homopolymer Rubber like, Hoses, shoe heels,
CH2 CH C CH2 inferior to natural stoppers etc.
CH2 CH C CH2 rubber but
Cl Chloroprene or 2- superior in its
n
Chloro-1, 3-butadiene. stability to serial
oxidation and its
resistance to oils,
gasoline, etc.
III. Polyacrylates
1. Poly (methyl Cl Addition homopolymer Hard, transparent, Lenses, light covers,
methacrylate) Plexigas, CH2 C COOCH3 excellent light light shades, signboards,
Lucise, Acrylite Perspex transmission transparent domes,
(PMMA) Methyl methacrylate properties, skylights, aircraft
optical clarity windows, dentures and
better than glass. plastic jewelry.
2. Poly (ethyl acrylate) CH2=CHCOOC2H5 Addition homopolymer. Tough, rubber like —
product.
CH2 CH
COOH n

3. Polyacrylonitrile (PAN) CH2 = CH–CN Addition homopolymer. Hard and has Manufacture of fibers,
Acrylonitrile high melting orlon, acrilon used for
CH2 CH
material. making clothes, carpets,
CN n blankets.

IV. Polyhalolefines
1. Polyvinyl chloride (PVC) CH2=CH–Cl Homopolymer, chain Pliable (easily (i) When plasticized
Vinyl chloride growth molded) with high-boiling esters
used in rain coats, hand
CH2 CH
bags, shower curtains,
Cl n upholstery fabrics, shoe
soles, vinyl flooring

(ii) Good electrical

insulators for wires and
cables

(iii) Making hose pipes.

2.Polytetrafluoro ethylene F2C = CF2 Chain growth, Flexible and (i) For making nonstick
of Teflon (PTFE) Tetrafluoroethylene homopolymer inert to solvents, utensils coating
boiling acids,
F F (ii) Making baskets,
even aquaregia,
C CH pump packings, valves,
stable up to
seals, nonlubricated
F F 598 K.
n bearings.
Condensation Polymers
I. Polyesters
 Chem i str y | 26.29

Name of the polymer Structure of Natural of polymer Properties Uses

along with abbreviation monomer
1. Terylene or Dacron HO–CH2–CH2–OH Copolymer, linear, step- For making paints
growth. and lacquers.
Ethylene glycol or
O Ethane-1, 2-diol and Thermoplastic, dissolves
O O O in suitable solvents
HO C C OH and its solutions, on
evaporation leaves a
Terephthalic acid tough but non-flexible
or Benzene-1, a film.
O 4-dicarboxylic acid
O

2. Glyptal or Alkydresin HO–CH2–CH2–OH Copolymer, linear, step- For making paints

growth. and lacquers.
Ethylene glycol and
O
Thermoplastic, dissolves
HOOC HOOC
CH2 in suitable solvents
O O and its solutions, on
Phthalic acid or evaporation leaves a
Benzene-1, tough but non-flexible
a film.
2-dicarboxylic acid
O O
CH2
O

II. Polyamides
Nylon-6,6 O O Copolymer, step, growth, High tensile (i) Textile fabrics, carpets,
NH2 HO C (CH2)4 C OH linear strength, abrasion bristles for brushes
resistant,
Adipic acid and (ii) Substitute for metals
somewhat elastic.
in bearings and gears.
H2N—(CH2)4—NH2
(iii) Crinked nylon is
Hexamethylene used for making elastic
O
diamine hosiery.
OH
2. Nylon-610 H2N(CH2)6NH2 Copolymer, step- growth, For making (i) Textile fabrics, carpets,
linear points and bristles for brushes
H H O O Hexamethylene
lacquers.
diamine (ii) Substitute for metals
[ N (CH2)6 N N (CH2)8 C [
in bearings and gears.
and
(iii) Crimped nylon is
HOOC(CH2)8
used for making elastic
COOH hosiery.
Sebacic acid
3. Nylon-6 or Parlon H Homopolymer, step Mountaineering ropes,
O N growth, linear tire cords and fabrics
(CH2)5 C O
N
H E-Caprolactam
III Formaldehyde resins
2 6 . 3 0 | Biomolecules and Polymers

Name of the polymer Structure of Natural of polymer Properties Uses

along with abbreviation monomer
1. Phenol-formaldehyde Phenol and Copolymer, step (i) Bakelite with low
resin or Bakelite formaldehyde growth, highly branched degree of polymerization
thermosetting polymer is used as binding glue
for wooden planks and in
varnishes and lacquers

(ii) Bakelite with high

degree of polymerization
is used for making combs
and micatable tops,
fountain pen barrels,
electrical goods (switches
and plugs), gramophone
records etc.

PLANCESS CONCEPTS

•• Polymers are product of large number of small molecules, called monomers, chemically bonded to
each other.
•• The individual large polymer molecules are known as macromolecules.
•• Polymers are characterized by the average molecular mass of the chains and the number of repeating
units in such polymers is known as the degree of polymerization.
•• The physical properties of a polymer are determined by such factors as the flexibility of macromolecules,
the sizes and types of group attached to the polymer chains and the magnitude of intermolecular
forces.
•• Polymers may be linear, branched or cross linked.
•• Copolymers are produced from two monomers combined randomly or in a specific manner
•• To participate in polymerization, a molecule must be able to react at both ends.
•• The principal types of polymerization reaction are chain reaction polymerization (initiation
propagation and termination), which is undergone by monomers such as vinyl bromide and step
reaction polymerization, which involves reactions between functional group on different monomer
molecules like adipic acid and hexamethylene diamine to give nylon 66.

Neeraj Toshniwal (JEE 2009, AIR 21)

Illustration 9: Write the structures of the monomers of the following polymers:  (JEE MAIN)

(a) ( CH2CH2—O—CH2CH2 )n (b) [ = CH(CH2)4CH = N(CH2)6N = ]n

Sol: (a) CH2—CH2 (b) HO — CH —(CH2)4—CH2 — OH and H2N—(CH2)6—NH2

Hexane-1, 6 diol Hexamethylenediamine
O
(Monomer) (Monomer)
Ethylene oxide
(Monomer)
 Chem i str y | 26.31

Illustration 10: (a) Can a copolymer be formed in both addition and condensation polymerization? Explain.
(b) Can a homopolymer be formed in both addition and condensation polymerization? Explain. (JEE MAIN)

Sol: (a) Yes, copolymers can be formed both in addition and condensation, polymerization. For example, Buna-S
is an addition copolymer of styrene and 1, 3-butadiene while nylon-6,6, bakelite and polyester are condensation
copolymers.
(b) Yes, homopolymers can be formed both in addition and condensation polymerization. For example, polythene,
PVC, PMMA, PAN, neoprene, etc. are example of addition homopolymers while nylon-6 is an example of
condensation homopolymer.

Illustration 11: How does the presence of benzoquinone inhibit the free radical polymerization of a vinyl derivative?
 (JEE ADVANCED)
Sol: Benzoquinone react with radical of the growing O O OR
polymer chain (R᛫) to form a new radical (I) which is
extremely uncreative, since it is highly stabilized by R + etc.
resonance. Because of the lack of reactivity of this
new radical further growth of the polymer chain is O O O
interrupted and hence the reaction stops. p-Benzoquinone I

Illustration 12: Differentiate the following pairs of polymers based on the property mentioned against each. 
 (JEE ADVANCED)
(i) Novolac and bakelite (structure).
(ii) Buna-S and terylene (intermolecular forces of attraction).
- -
O O
Sol: (i) Novolac is a linear but
bakelite is a cross-linked polymer O CH2 CH2 O C
+
C
+
of phenol and formaldehyde. n
(ii) Terylene contains ester functional Dipole-dipole interaction Dipole-dipole interaction
groups that are polar in nature. - -
O O
Therefore, the intermolecular forces
of attraction involved in terylene are O CH2 CH2 O C
+
C
+
strong dipole–dipole interaction as n
shown below:
Terylene polymer chains
Buna-S, on the other hand, does
not have polar functional groups. It has only nonpolar hydrocarbon C6H5
chains and hence has only weak Van Der Waals forces of attraction as shown below:
Weak van der Waals
Forces of attraction

C6H5
Buna-S polymer Chains
2 6 . 3 2 | Biomolecules and Polymers

POINTS TO REMEMBER

Sugar
On the basis of physical
Non-sugar
characteristics
Monosaccharides
Carbohydrates On the basis of hydrolysis Oligosaccharides
Polysaccharides
Reducing
On the basis of test with reagents
Non-neducing
Maltose
Dissachharides Sucrose
Lactose

Open chain (Fischer Projection)

Representation
Structure Haworth
Cyclic structure
Chain form
Fibrous Proteins
Proteins
Globular Proteins

Primary structure
-helix
Structure Secondary structure
of -pleated
Protein Tertiary structure
Quaternary structure

Basic amino acids

Amino acids Neutral amino acids
Acidic amino acids

On the basis of source.

On the basis of structure.
Polymers On the basis of synthesis.

On the basis of molecular force.

On the basis of type of polymerization.
 Chem i str y | 26.33

Solved Examples

JEE Main/Boards Example 5: Define the terms:

(A) Gene (B) Genetic code
Example 1: What type of bonding occurs in globular (C) Transcription (D) Translation
protein?
(E) Codons
Sol: Globular protein is spherical in nature and are
Sol: (A) Gene: A gene is a sequence of base triplets in a
water soluble. Globular protein may have the following
strand of DNA helix that factions to code a polypeptide
types of bonding: hydrogen bonding, disulphide
chain. The polypeptide chain ultimately becomes the
bridges, ionic or salt bridges, and hydrophobic
part of the protein synthesized. Every protein in a cell
interactions.
has the corresponding gene.
(B) Genetic code: The relation between the amino acids
Example 2: What will be the sequence of bases on
and the nucleotide triple is called is genetic code.
mRNA molecules synthesized on the following strand
of DNA? (C) Transcription: The way the code on DNA is copied
to give the complementary code on RNA is called
TATCTACCTGGA transcription.
Sol: The opposite bases bind with the strand i.e. A-T (D) Translation: The way the four-base code in nucleic
and G-C. acid is turned into a 20 unit code needed to specify
Sequence of bases on mRNA molecule: The amino acid sequence in proteins during synthesis
DNA strand TAT CTA CCT GGA is called translation.
m-RNA AUA GAU GGA CCU (E) Codons: The nucleotide bases in RNA function in
groups of three (triplets) in coding amino acids. These
base triplets are called codons.
Example 3: Name one reducing and one nonreducing
disaccharide.
Example 6: a. Write the structure of histidine when pH
Sol: Reducing sugar has a free aldehydic group which < 1.82 and pH > 1.82.
can be easily oxidized and can thus reduce other
substances. Sol: a. HN

-H
Maltose or lactose (reducing) COO H
(COOHCOO-)
Sucrose (non-reducing)
:


N NH3
HN
H
COO-
pH>1.82
Example 4: Explain the functions of nucleic acids.
:


N NH3
Sol: Nucleic acids are large biomolecules consisting of H
RNA and DNA. pH<1.82

Nucleic acids have two important functions:

i) Replication: Due to its unique property of self- Example 7: ‘The two strands of DNA are not identical,
replication (process by which a single DNA molecule but are complementary’. Explain this statement.
produces two identical species of itself), it is responsible
for maintaining the heredity traits from one generation Sol: DNA consists of two strands of polynucleotides
to another. coiled around each other in the form of a double helix. The
nucleotides making up each strand of DNA are connected
ii) Protein synthesis: RNA helps in the biosynthesis of
by phosphate ester bonds. This forms the backbone of
proteins and is, in one way, responsible for the process
each DNA strand, from which the bases extend. The
of learning and memory storage. Furthermore, it
bases of one strand of DNA are paired with the bases
sends information and instructions to the cell for the
on the other strand by means of hydrogen bonding. This
manufacture of specific proteins.
2 6 . 3 4 | Biomolecules and Polymers

hydrogen bonding is very specific as the structure of (i) HO CH2 6

bases permits only one mode of pairing. Adenine pairs
only thymine via two hydrogen bonds and guanine pairs HO CH 5 O
Me
with cytosine through three hydrogen bonds. The two 4
OH
1 +2 O
strands of DNA are said to be complementary to each OH
Me
2
other in the sense that the sequence of bases in one 3

strand automatically determines that of the other. These OH (i) PhCH2Cl/OH
(i) -D-Glocofuranose (B)
strands are not identical. (A)
Acetone/H2SO4
(A) (i) H
-D-Glucopyranose (C)
Example 8: Which purine and pyrimidine bases are (Two pair of OH are cis)
present in DNA and RNA? (One pair at C-1 and C-2)

(Another pair at C-5 and C-6) (i) PhCH2Cl/OH
Sol: Adenine and guanine (purine base) in DNA and (ii) -D-Glocofuranose
Acetone/H
(E)
(D)  (i) H
RNA. Cytosine, thymine, and uracil (pyrimidine base) O CH2
6
(i) PhCH2Cl/OH Me (F )
(ii) H
Example 9: What are the monomers constituting
Me CH 5 O
O 
proteins? (i) PhCH2Cl/OH
(iii)HOCH
-D-Glucopyranose
6 (H)
2
Acetone/H 4 OH 1
(i) H O
Sol: The amino acids such as glycine and alanine are 5
(G)
Me
the monomers that constitute proteins. HOCH 2
O 3 (F )
O
4 1 OH Me
OCH Ph
Example 10: What is the effect of pH on the action of (iv) -D-Glucopyranose
2
(K) (B)
2 OH 
enzyme? 3 (J) Acetone/H
(Cyclic diketal) (Diacetonide)
OH
Sol: The low or high pH values can cause denaturation
of the protein and hence make enzyme’s protein inactive.
(ii) HO CH2 6

JEE Advanced/Boards HO CH 5 O OH Me H
 4 1 + O
OH
Example 1: Complete the following reactions.
(i) -D-Glocofuranose (B)
(i) PhCH2Cl/OH Me
Acetone/H2SO4 (i) PhCH(i)Cl/OH 
 3 2
(A) H
(i) -D-Glocofuranose (B) 2 OH
Acetone/H2SO4 (C)
(A) (i) H
(C) (D)
 -D-Glucofuranose
(i) PhCH2Cl/OH (One pair of cis OH group)
(ii) -D-Glocofuranose (E) 
(D) Acetone/H (i) PhCH(i) H
2Cl/OH (at C-5 and C-6)
(ii) -D-Glocofuranose (E)
(D) Acetone/H (i) H (F )
(F ) Me
 (i) 3PhCH2/Cl2 OH- O-CH26
(i) PhCH2Cl/OH
(iii) -D-Glucopyranose (H) 
Acetone/H (i) PhCH(i)2Cl/OH
(ii) H
(iii) -D-Glucopyranose
(G) (H) H Me O-CH 5
Acetone/H O
(G) (i) H (F ) 6
HOCH2 4 1
(F ) OH
(iv) -D-Glucopyranose 
(K) HOCH 5 O OCH2Ph 3 2
(iv) -D-Glucopyranose
(J) Acetone/H (K)
 4 OH
(J) Acetone/H OCH2Ph 1 (E)
(Cyclic ketal) (acetonide)
Sol: Acetone forms a cyclic ketal called an acetonide 2
3
with two cis (OH) groups. D-Glucopyranose is in OCH2Ph
6
equilibrium
(ii) HOwith
CHsome
2 D-glucofuranose (having two
(ii) HO CH2 6 (F)
pais of cis OH groups), the formation of diketal shift the
HO CH55 O
equilibrium toward
HO CH O the reaction
OH
OH D-glucofuranose.
Me 
H
4
OH 1 + O Me H
4 1 + O Me
OH
Me
3 2
3 2
OH
OH

(D)
(D)
-D-Glucofuranose
-D-Glucofuranose
(One pair of cis OH group)
 Chem i str y | 26.35

(iii) 1
CHO
1
COOH
6
CH2OH HO 2 [O] HO 2

3 OH HNO3 3 OH
5
O Me
H 4
CH2OH
4
COOH
4 1 + O
OH Me D-Threose D(-)Tartaric acid
HO 2 OH (Optically active enantiomer)
3
H2/N2
OH [H] or
(G)
NaBH4
-D-Glucopyranose
(One pair of cis OH group) 1
(at C-1 and C-2) CH2OH D-Buan-1, 2, 3, 4-tetraol
CH2O
OH HO 2 (Optically active enantiomer)
 5 3 OH
(i) PhCH2Cl/OH 1 4
4 OH CH2OH
(ii) H HO 2 Me
3
Me Example 3: a. Name the smallest aldose that forms
CH2OCH2Ph cyclic hemiacetal and the functional groups are involved
(H)
5
O in its formation.
4 OCH2Ph b. What is invert sugar?
PhHO2CO 3
OH c. Calculate the specific rotation of invert sugar.
OH Given, α|D of D-glucose = 52.7º
|α|D of D-fructose = –92.4º
(iv) 6
CH2OH
d. Give the mechanism of mutarotation of β-D-gluco-
5
O OH
Me pyranose in (i) aq. H⊕ and (ii) OHΘ.
4 OH 1 + O No reaction e. Why is the mutarotation faster in the presence of
Me
HO 2-pyridinol?
3 2

OH
Sol: a.
(J) 1
-D-Glucopyranose CH=O 1
O
H C OH
2 OH O 
(No pair of cis OH group) 2 OH
3 OH 3 OH OH
4
Example 2: Differentiate between D-erythrose and CH2OH 4
OH OH
D-threose by D-Erthrofuranose
1
(a) Mild oxidation and (b) Reduction. HO C OH O OH
2 OH O  4 1
Sol: 1
CHO
1
COOH
3 OH
4 3 2
2 OH [O] 2 OH OH OH
-D-Erthrofuranose
3 OH HNO3 3 OH 1
4 4 CH=O O
CH2OH COOH 2 H C OH
HO O 
D-Eryhrose Meso-tartaric acid HO 2
OH
Plane of symmetry 3 OH OH
H2/N2 3 OH
(Optically inactive) 4
CH2OH
[H] or 4
OH
NaBH4 -D-threofuranose
1 O
CH2OH HO C OH OH
2 O 
2 OH Meso-butan-1, 2, 3, 4-tetraol HO OH
(Optically inactive) 3 OH
3 OH
4
4 OH
CH2OH
-D-threofuranose
1 1
CHO COOH
HO 2 [O] HO 2
HNO
2 6 . 3 6 | Biomolecules and Polymers

b. Equimolar mixture of D-glucose and D-fructose Sol: (i)

obtained but the hydrolysis of sucrose is called invert 1 1 1
sugar. Since the specific rotation of sucrose is positive CHO CHO CHO
2 HO 2 HO HO 2
and after hydrolysis it changes to negative value and
Killiani
synthesis
this process is called inversion of cane sugar. HO 3
Refer to Section
3 HO 3 HO

c. The specific rotation is half the sum of specific 4 HO HO 4 HO 4

rotation of glucose and fructose. 5 HO 5 HO 5 HO

1
Specific rotation of invert sugar = [ +52.7º +(–92.4º )]
6
CH2OH 6 HO
= –19.99º 2 D-glucose 6 HO
7 7
(A) CH2OH CH2OH
d. i. In acidic medium: (C-2) Epimer of
(B) (B)
The smallest aldose is a tetraldose that has four C atoms HNO3 HNO3
and an O atom to form a five-membered ring. The CHO 1 1
group and 1º OH group of tetrose are involved in the COOH COOH
formation of ring. Both erythrose and threose form 2 HO HO 2

cyclic hemiacetals. The rate depends on the conversion Plane of symmetry 3 HO 3

of cyclic hemi- (ii)acetal to open-chain aldehyde (i). 4 4
HO H HO
e.2-Pyridinol is an acid-base catalyst [containing both
 5 HO 5 HO
basic (N) and acidic (OH) group] that gives H ion to
hemiacetalic O atom and simultaneously removes H⊕ 6 HO 6 HO
ions from the HO − group of the hemiacetal by the 7
COOH
7
COOH
formation a cyclic intermediate transition state. Meso-acid (C) (It is optically active)

(ii)
1 1 1
CHO CHO CHO
:

H O N
2 OH 2 OH HO 2
O H
Killiani
:

O O H 3 OH 3 OH 3 OH
synthesis
H OO
:

N
H (I) 4 OH 4 OH 4 OH
H
-Pyranonse
5 OH 5 OH 5 OH
O H 6
CH2OH 6 OH 6 OH
OH D-Allose
7 7
(A) CH2OH CH2OH
-Pyranonse
(C-2) Epimer of
(B) (B)
Example 4: Complete the following reactions: HNO3 HNO3

HNO 1 1
(i) Killiani HNO33
(i) Glucose Killiani Pair
(i) DD -- Glucose Pair COOH COOH
synthesis
synthesis
(A)
(A) (B)
(B) 2 OH HO 2
Meso-Heptaldaric
Meso-Heptaldaric acid
acid ((CC)) Plane of symmetry 3 OH 3 OH
4 4
(ii)
(ii)
(ii) DD -- Allose
Allose Killiani Pair HNO
Killiani HNO33 HO OH HO
Pair
synthesis
synthesis
(D)
(D) (B)
(B) 5 OH 5 OH
Meso-Heptaldaric
Meso-Heptaldaric acid
acid (F)
(F)
6 OH 6 OH
Explain whether the acids (C) and (F) are same or 7 7
COOH COOH
different. Which pair out of (B) and (E) gives meso-acids
Meso-acid (It is optically active)
(C) and (F)?
Although both acids (C) and (F) are heptaldric acid and
both are meso (optically inactive) but they are different.
 Chem i str y | 26.37

Example 5: a. How is the mixture of aspartic acid Ph3C - C1 only

CH2OCPh3
(A) histidine (B) and threonine (C) separated by CH2OH is etherified O 
Me2SO4OH
electrophoresis method? protection of OH OMe Methylation
HO of C-2, C-3, C-4.
pI (pH at isoelectric point) are given. (CH2OH)
(OH) gp.
OH
(C)
pI of (A), (B), and (C) are 2.77, 7.59 and 5.60, respectively.
Tritylated methyl-D-glucopyranoside
b. How are they separated by solubility method? 6
CH2OCPh3 CH2OCPh3
O O
Sol: a. Choose the intermediate pH of 5.60. This is the OMe
5
OMe H
OMe OMe
4
pI of threonin, which has a zero net charge and does eMe
1
MeO
2
not migrate in the electric field or in the electrophoresis
3
OMe OMe
experiment.
(E) (D)
Aspartic acid (pI = 2.77) [Refer to solved example No. 8 (d) 2,3,4-Trimethyl glucopyranoside
above] donates an H⊕ and is converted to anion (III), and
migrates to the anode. Histidine (pI = 7.59) accepts an H⊕
and is converted to a cation, and migrates to the cathode. Example 7: Write the names and structures of the
monomers of the following polymers:
b. At isoelectrical point (pI) the amino acids have least
solubility in water and this property is exploited in the (i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene.
separation of different amino acids obtained from the
hydrolysis of protein. Sol: The names and structures of the monomers are:
To the mixture of three amino acids (A, B and C) set the (i) Buna-S:
pH of the solution by adding acid upto 2.77, at which
CH2 CH — CH CH2 and C6H5 — CH CH2
(A) will be least soluble in H2O and will precipitate out.= =
1,3−Butadiene
=
Styrene
It is followed by the separation of amino acid (A).
(ii) Buna-N:
Increase the pH of the remaining solution by adding
base upto 5.60, at which amino acid (c) will precipitate= CH2 CH — CH CH2 and CH
= =2 — CH CN
out. Similarly, (B) will precipitate out at the pH of 7.59. 1,3−Butadiene Acrylonitrile

(iii) Neoprene:
Example 6: Convert D-glucopyranose to 2, 3,
Cl
4-trimethyl glucopyranoside. CH2
CH2
Sol: First convert D-glucose to methyl glucopyranoside
Chloroprene or 2 − chloro − 1,3 − butadiene
with MeOH/HCl. Then CH2OH is protected by reacting
methyl glucopyranoside with Ph3C—Cl (trityl chloride), (iv) Dacron:

CH2OH CH2OH HO — CH2 — CH2 — OH and

Ethyleneglycol
O O
MeOH/HCl
OH OH OH OMe
HO HO HOOC HOOC
OH OH
(A) (B)
Terephthalic acid
(-or -D-glucopyrnose) Methyl-D-glucopyranoside

Only CH2OH is etherified giving CH2OCPh3 group. The

Example 8: Identify the monomer in the following
2º OH groups are sterically hindered and do not react
polymeric structures:
with the bulkyl (Ph3C—Cl) group.
(i) H H
The free OH groups of tritylated methyl glucosides are
now methylated with Me2SO4/NaOH followed by acidic
hydrolysis removes both trityl group and the glycosidic
Me group leaving OMe group at C2, C3, and C4 intact.
N
 6
N
 8
The hydrolysis of –OCPh3, groups proceeds by the O O
formation of stable Ph3 C⊕ (triphenyl methyl carbocation). n
2 6 . 3 8 | Biomolecules and Polymers

Example 9: How does the presence of double bonds

(ii) N
HN NH CH2 in rubber molecules influence their structure and
reactivity?
N N
Sol: Natural rubber is cis-polyisoprene and is obtained
NH n by 1, 4-polymerization of isoprene units. In this polymer,
double bonds are located between C2 and C3 of each
isoprene unit. These cis-double bonds do not allow the
Sol: Monomers are:
polymer chains to come closer for effective interactions
and hence intermolecular forces are quite weak. As
(i) HOOC — (CH2 )8 — COOH
Decanoic acid or Sebacic acid
a result, natural rubber, i.e., cis-polyisoprene has a
randomly coiled structure and hence shows elasticity.
and H2N — (CH2 )6 — NH2
Hexamthylenediamine 2 4
1 n
6 N 2 1 3
(ii) H2N NH2
Isoprene
3
N N
5 4
NH2
Melamine or 2 3 2 3
1
2.4.6.-Triamino-1,3,5-triazine CH2
4 1 CH2
and 4
H
C=O cis-Polyisoprene
(Natural rubber)
H
Formaldehyde

JEE Main/Boards

Exercise 1 Q.5 How do you explain the amphoteric behavior of

amino acids?
Q.1 What happens when D-glucose in treated with
the following reagents? Q.6 What is the effect of denaturation on the structure
(i) HI (ii) Bromine water (iii) HNO3 of proteins?

Q.2 Define the following terms in relation to proteins Q.7 Describe the following: (i) Glycosidic linkage

(i) Peptide linkage (ii) Denaturation

Q.8 Enumerate the reactions of D-glucose which cannot
be explained by its open chain structure.
Q.3 Define the following as related to proteins
(i) Peptide linkage (ii) Primary structure Q.9 What are essential and non-essential amino Acids?
(iii) Denaturation Give two examples of each type.

Q.4 What are the common types of secondary structure Q.10 List any four vitamins. Mention the chief sources
of proteins? and functions of two of them.
 Chem i str y | 26.39

Q.11 (i) What are essential and non-essential amino Q.24 Distinguish between α-glucose and β-glucose.
acids? Give two example of each.
(ii) What is a denatured protein? Q.25 What happens when L-glucose is treated with the
following reagents?
Q.12 How are vitamins classified? Name the vitamin (i) HI (ii) Bromine water (iii) HNO3
responsible for the coagulation of blood.
Q.26 Write the important structural and functional
Q.13 Why are vitamins A and vitamin C essential to us? differences between DNA and RNA.
Give their important sources.
Q.27 What are the different types of RNA found in the cell?
Q.14 Draw open chain structure of aldopentose and
aldohexose. How many asymmetric carbons are present Q.28 Define the following and give one example of
in each? each
(a) Isoelectric point (b) Mutarotation
Q.15 (a) Describe the following giving one example: (c) Enzymes
Nucleotide.
(b) List four functions of carbohydrates in living Q.29 Answer the following queries about proteins?
organisms.
(i) How are proteins related to amino acid?
Q.16 What type of bonding helps in stabilizing the (ii) How are oligopeptides different from polypeptides?
α-helix structure of proteins? (iii) When is a protein said to be denatured?

Q.17 Differentiate between globular and fibrous

Q.30 (a) Name the three major classes of carbohydrates
proteins.
and give the distinctive characteristic of each class.
Q.18 (a) Give reasons for the following statements: (b) What are nucleotides? Name two classes of nitrogen
containing bases found amongst nucleotides.
(i) Amino acids have comparatively higher melting
points than the corresponding haloacids.
(b) What deficiency diseases are caused due to lack to Exercise 2
lack of vitamins A, B1, B6 and K in human diet?
Biomolecules
Q.19 The two strands in DNA are not identical but are
complementary. Explain. Single Correct Choice Type

Q.1 The chromophore in the dye

Q.20 State difference between the following pair
(i) α-helix and β-pleated structures. HO3S N=N N(CH3)2 is

(ii) Primary and secondary structures of a protein.

(A) –N(CH3)2 (B) –SO3H
Q.21 What are nucleic acids? Mention their two (C) –C6H5 (D) –N = N–
important functions.
Q.2 At the isoelectric point for an amino acid the
Q.22 What is the difference between a nucleoside and species present are
a nucleotide?
(A) R – CH – COOH (B) R – CH – COOH
| |
+
NH2 NH3
Q.23 What are reducing and non-reducing sugar?
What is the structural feature characterizing reducing –
(C) R – CH – COO – (D) R – CH – COO
sugars? | |
+
NH2 NH3
2 6 . 4 0 | Biomolecules and Polymers

Q.3 Secondary structure of a protein refers to Q.12 The reagent used in Ruff’s degradation is
(A) Mainly denatured proteins and structures of (A) Baeyer’s reagent (B) Tollen’s reagent
prosthetic groups
(C) Fentons’ reagent (D) Benedict’s reagent
(B) Regular folding patterns of contiguous portions of
the polypeptide chain Q.13 If Ka1 and Ka2 are the ionization constants of
(C) Linear sequence of amino acid residues in the H3N+CHICOOH and H3N+CHICOO–, respectively, the pH
polypeptide chain of the solution at the isoelectric point is

(D) None of these (A) pH = pKa1 + pKa2 (B) pH = (pKa1 + pKa2)1/2

(pK a1 + pK a2 )
(C) pH = (pKa1 + pKa2)1/2 (D) pH =
Q.4 The general formula of carbohydrates is: 2

(A) CnH2n+1O (B) CnH2nO Q.14 Coordination polymerization was developed by

(C) Cn(H2O)n or Cx(H2O)y (D) Cn(H2O)2n (A) Zeigler and Natta (B) Linus Pauling
(C) Beckamann (D) None of these
Q.5 Which of the following is a disaccharide?
(A) Sucrose (B) Glucose Q.15 Teflon, polystyrene and neoprene are all
(C) Fructose (D) Starch (A) Copolymers (B) Condensation polymers
(C) Homopolymers (D) Monomers
Q.6 The iron in hemoglobin is bound by
(A) Hydrogen bonds (B) Chelation Q.16 Carbohydrates which differ in configuration at the
glycoside carbon (i.e. C1 in aldose and C2 in ketoses) are
(C) Ionic bonds (D) Covalent bonds called
(A) Anomers (B) Epimers
Q.7 Anomers have different
(C) Diastereomers (D) Enantiomers
(A) Properties (B) Melting points
(C) Specific rotation (D) All of these Q.17 Choose the correct relationship for α-D-glucose
(1) and β-D-glucose (2)
Q.8 Peptide bond is (A) A and B are epimers
(A) – CO – NH – (B) NH2 – CO – NH – R (B) A and B are crystal modification
(C) A is a pyranose sugar and B is furanose sugar
(C) R – CO – NH – R (D) – CONH2
(D) A is an aldose and B is a ketose.

Q.9 Glucose and Fructose are

Q.18 Natural rubber is a polymer of
(A) Tautomers (B) Chain isomers
(A) Chloroprene (B) Isoprene
(C) Functional isomers (D) Geometrical isomers
(C) 1, 3-Butadiene (d) None

Q.10 Glucose is
Q.19 Hydrolysis of sucrose is called
(A) Aldopentose (B) Aldohexose
(A) Saponification (B) Inversion
(C) Ketopentose (D) Ketohexose
(C) Esterification (D) Hydration

Q.11 A substance which can act both as an antiseptic

and disinfectant is Q.20 In vulcanization of rubber

(A) Aspirin (B) Chloroxylenol (A) Sulphur reacts to form a new compound
(C) Bithinal (D) Phenol (B) Sulphur cross-links are introduced
(C) Sulphur forms a very thin protective layer over rubber
(D) All statements are correct
 Chem i str y | 26.41

Q.21 The simplest amino acid is Q.30 Main structural unit of protein is
(A) Glycine (B) Alanine (A) Ester linkage (B) Ether linkage
(C) Guanine (D) All of the above (C) Peptide linkage (D) All the above

Q.22 Which of the following belong to the class of Q.31 Which of the following statements is true of
natural polymers? proteins?
(A) Proteins (B) Cellulose (A) They catalyse the biochemical reactions
(C) Rubber (D) All of the above (B) They act as antibodies
(C) They perform all these functions
Q.23 D-Glucose and β-D glucose differ from each other
due to difference in one of carbon with respect to its (D) They perform all these functions
(A) Size of hemiacetal ring
(B) Number of —OH groups Q.32Which of the following is a polysaccharide?

(C) Configuration (A) Glucose (B) Galactose

(D) Conformation (C) Sucrose (D) Pectines

Q.24 Glucose gives the silver mirror test with ammoniacal Q.33 Starch can be used as an indicator for the detection
solution of silver nitrate because it contains of traces of
(A) Aldehydes group (B) Ester group (A) Glucose in aqueous solution
(c) Ketone group (D) Amide group (B) Proteins in blood
(C) Iodine in aqueous solution
Q.25 Oligosaccharides contain ……………. Simple sugar
units (D) Urea in blood
(A) 2 to 10 (B) 4 to 8 (C) 6 to 12 (D) 6 to 10
Q.34 Which of the following statements about ribose
Q.26 A pair of diastereomers that differ only in the in incorrect?
configuration about a single carbon atom are called (A) It is polyhydroxy compound
(A) Anomers (B) Epimers (B) It is an aldehyde sugar
(C) Conformes (D) Enantiomers (C) It has six carbon atoms
(D) It exhibits optical activity
Q.27 Pick out the incorrect statement about ATP.
(A) It is a nucleotide
Q.35 Which of the following is the structure of D-xylose?
(B) It contains the purine, adenine
CHO CHO
(C) The enzyme-catalysed hydrolysis of ATP to ADP and
AMP is accompanied by absorption of energy HO H HO H
(D) Energy is stored in the cell in the form of ATP. (A) H OH (B) HO H
H OH H OH
Q.28 Cellulose is a linear polymer of
CH2OH CH2OH
(A) Glucose (B) Glucose
(C) Fructose (D) None of these
CHO CHO

Q.29 Glucose molecule reacts with X number of H H H OH

molecules of phenylhydrazine to yield osazone. The (C) OH H (D) H OH
value of X is H OH H OH
(A) Three (B) Two (C) One (D) Four
CH2OH CH2OH
2 6 . 4 2 | Biomolecules and Polymers

Q.36 Glucose gives the silver mirror test with Q.44 P.V.C. is formed by polymerization of-
ammoniacal solution of silver nitrate because it contains
(A) 1-Chloroethene (B) Ethene
the group
(C) Propene (D) 1-Chloropropene
(A) Aldehyde (B) Ester
(C) Ketone (D) Amide Q.45 Polyacrylonitrile, characterized by the repeating
unit, is made from which of the following monomer?
Q.37 A condensation polymer among the following is-
(A) CH3CH2CN (B) HOCH2CH2CH3
(A) Dacron (B) PVC
(C) CH3CH = CHCN (D) CH2 = CHCN.
(C) Polystyrene (D) Teflon
Q.46 On the basis of intermolecular forces, polymers
Q.38 Melamine polymer is copolymer of are classified as
(A) Melamine and acetaldehyde (A) Elastomers, Fibres, Thermoplastics and Thermosetting
(B) Melamine and formaldehyde (B) Elastomers, Fibres, Chain growth and Step growth
(C) Phenol and formaldehyde (C) Addition polymers and Condensation polymers

(D) None of the above (D) None of these

Q.47 Which of the following polymers do not involve

Q.39 Which one of the following pairs is not correctly
cross linkages-
matched?
(A) Melmac (B) Bakelite
(A) Terylene: Condensation polymer of terephthalic
acid and ethylene glycol (C) Polythene (D) Vulcanised rubber
(B) Perspex: A homopolymer of methymethacrylate
Q.48 Ziegler-Natta catalyst is-
(C) Teflon: Thermally stable cross-linked polymer of
phenol and formaldehyde (A) K[KPtCl3(C2H4)] (B) (Ph3P)3RhCl
(D) Synthetic rubber: A copolymer of butadiene and (C) Al(C2H5)3+TiCl4 (D) Fe(C5H5)2
styrene
Q.49 Which one of the following monomers give the
Q.40 Orlon is a polymer of- polymer neoprene on polymerization?
(A) Styrene (B) Tetrafluoroethylene (A) CH2 = CHCl (B) CH2 = CCl – CH = CH2
(C) Vinyl chloride (D) Acrylonitrile (C) CF2 = CF2 (D) CCl2 = CCl2

Q.41 Which one of the following is not an example of Q.50 Which of the following pairs is not correctly
chain growth polymer- matched?
(A) Neoprene (B) Buna-S (A) Terylene-condensation polymer of terephthalic acid
and ethylene glycol
(C) PMMA (D) Glyptal
(B) Teflon-thermally stable cross linked polymer of
phenol and formaldehyde
Q.42 Ebonite is-
(A) Natural rubber (C) Perspex-a homopolymer of methyl methacrylate

(B) Synthetic rubber (D) Synthetic rubber –a copolymer of butadiene and

styrene
(C) Highly vulcanized rubber
(D) Polypropene
Q.51 Which one of the following is used to make
‘non-stick’ cookware?
Q.43 F2C = CF2 is a monomer of-
(A) PVC
(A) Teflon (B) Glyptal (C) Nylon-6 (D) Buna-S
(B) Polystyrene
 Chem i str y | 26.43

(C) Polyethylene terephthalate Q.5 The two functional groups present in a typical
carbohydrate are: (2009)
(D) Polytetrafluoroethylene
(A) –OH and –COOH (B) –CHO and –COOH
Q.52 Which compound/set of compounds is used in (C) > C = O and –OH (D) –OH and –CHO
the manufacture of nylon-66?
(A) HOOC(CH2)4COOH+H2N(CH2)6NH2 Q.6 Buna-N synthetic rubber is a copolymer of:(2009)
Cl
(B) CH2 = CH–CH(CH) = CH2 |
(A) H2C = CH − C = CH2 and H2C = CH − CH = CH2
(C) CH2 = CH2 HOOC COOH
(B) H2C = CH − CH = CH2 and H5C6 − CH = CH2
(D) HOOC COOH + HOCH2 – CH2OH
(C) H2C = CH − CN and H2C = CH − CH = CH2

+ HOCH2 – CH2OH (D) H2C =CH − CN and H2C =CH − C =CH2

Q.53 Teflon, styron and neoprene are all- |
CH3
(A) Copolymers (B) Condensation polymers
(C) Homopolymers (D) Monomers Q.7 Biuret test is not given by  (2010)
(A) Carbohydrates (B) Polypeptides
(C) Urea (D) Proteins
Previous Years’ Questions
Q.8 The polymer containing strong intermolecular
Q.1 Which of the following pairs give positive Tollen’s
forces e.g. hydrogen bonding, is (2010)
test?  (2004)
(A) Teflon (B) Nylon 6, 6
(A) Glucose, sucrose (B) Glucose, fructose
(C) Polystyrene (D) Natural rubber
(C) Hexanal, acetophenone (D) Fructose, sucrose

Q.9 The presence or absence of hydroxyl group on

Q.2 Two forms of D-glucopyranose, are called  (2005) which carbon atom of sugar differentiates RNA and
(A) Enantiomers (B) Anomers DNA? (2011)
(C) Epimers (D) Diastereomers (A) 2nd (B) 3rd (C) 4th (D) 1st

Read the following questions and answer as per the Q.10 Which one of the following statements is correct?
direction given below:  (2012)
(a) Statement-I is true; statement-II is true; statement-II (A) All amino acids except lysine are optically active
is not the correct explanation of statement-I.
(B) All amino acids are optically active
(b) Statement-I is true; statement-II is true; statement-II
is not the correct explanation of statement-I. (C) All amino acids except glycine are optically active
(c) Statement-I is true; statement-II is false. (D) All amino acids except glutamic acid are optically
active
(d) Statement-I is false; statement-II is true.

Q.11 Synthesis of each molecule of glucose in

Q.3 Statement-I: Glucose gives a reddish-brown
photosynthesis involves: (2013)
precipitate with Fehling’s solution.
(A) 18 molecules of ATP (B) 10 molecules of ATP
Statement-II: Reaction of glucose with Fehling’s
solution gives CuO and gluconic acid. (2007) (C) 8 molecules of ATP (D) 6 molecules of ATP

Q.4 α − D − ( + ) - glucose and β − D − ( + ) - glucose are  Q.12 Which one is classified as a condensation
 (2008) polymer?  (2014)
(A) Conformers (B) Epimers (A) Dacron (B) Neoprene
(C) Anomers (D) Enantiomers (C) Teflon (D) Acrylonitrile
2 6 . 4 4 | Biomolecules and Polymers

Q.13 Which one of the following bases is not present in (A) It is a poor conductor of electricity.
DNA? (2014)
(B) Its synthesis required dioxygen or aperoxide initiator
(A) Quinoline (B) Adenine as a catalyst.
(C) Cytosine (D) Thymine (C) It is used in the manufacture of buckets, dust-bins
etc.
Q.14 Which polymer is used in the manufacture of (D) Its synthesis requires high pressure.
paints and lacquers?  (2015)
(A) Bakelite (B) Glyptal Q.17 Which of the following is an anionic detergent?
 (2016)
(C) Polypropene (D) Poly vinyl chloride
(A) Sodium lauryl sulphate
Q.15 Which of the vitamins given below is water (B) Cetyltrimethyl ammonium bromide
soluble? (2015)
(C) Glyceryl oleate
(A) Vitamin C (B) Vitamin D
(D) Sodium stearate
(C) Vitamin E (D) Vitamin K
Q.18 Thiol group is present in:  (2016)
Q.16 Which of the following statements about low
(A) Cystine (B) Cysteine
density polythene is FALSE?  (2016)
(C) Methionine (D) Cytosine

JEE Advanced/Boards

Exercise 1 Q.7 What are essential and non-essential amino acids?

Give two examples of each.
Q.1 State a use for the enzyme streptokinase in
medicine. Q.8 (a) Define the following term:
(i) Co-enzymes
Q.2 Why is cellulose in our diet not nourishing?
Q.9 (a) Answer the following questions briefly
Q.3 Explain muta-rotation taking D-glucose as an
example. (i) What are any two good sources of vitamin A?
(ii) What are nucleotides?
Q.4 Enumerate the structural difference between DNA
and RNA. Write down the structure of sugar present in (b) How are carbohydrates classified?
DNA.
Q.10 Write two main functions of carbohydrates in
Q.5 Answer the following queries about proteins- plants.

(i) How are proteins related to amino acids? Q.11 What happens when D-glucose is treated with the
(ii) How are oligopeptides different from polypeptides? following reagents?

(iii) When is a protein said to be denatured? (i) alk.KMnO4 (ii) Br2 + CS2 (iii) H2SO4

Q.6 (a) Define and classify vitamins. Give at least two Q.12 Name the four bases present in DNA. Which one
example of each type. of these is not present in RNA?

(b) Define an enzyme and comment on the specificity Q.13 Name two fat soluble vitamins that sources and
in action of an enzyme. Illustrate with an example. diseases caused due to their deficiency in diet.
 Chem i str y | 26.45

Q.14 State a use for the enzyme streptokinase in Q.2 During hydrogenation of oils, higher melting
medicine. ‘vegetable ghee’ is formed because
(A) Hydrogen is dissolved in the oil
Q.15 Write the major classes in which the carbohydrates
are divided? (B) Hydrogen combines with oxygen of the oil
(C) Ester of unsaturated fatty acids are reduced to those
Q.16 Aspartame, an artificial sweetener, is a peptide
of saturated acids
and has the following structures:
NH2 CH2C6H5 (D) Hydrogen drives off the impurities from the oil
| |
HOOC – CH2CH – CH – COOCH3
Q.3 Structurally a biodegradable detergent should
(a) Identify the four functional groups. contain a
(b) Write the zwitter ionic structure (A) Normal alkyl chain (B) Branched alkyl chain
(c) Write the structure of the amino acids obtained from (C) Phenyl side chain (D) Cyclohexyl side chain
the hydrolysis of aspartame.
(d) Which of the two amino acids is more hydrophobic? Q.4 Thrust imparted to the rocket is governed by the
(A) Third law of thermodynamics
Q.17 Give the chemical name of vitamin B12.
(B) Gravitational law
Q.18 What are the following substances?
(C) Newton’s third law
(i) Invert sugar (ii) Polypeptides
(D) None of these
Q.19 Which forces are responsible for the stability of
α-helix? Why is it named as 3.613 helix? Q.5 Which of the following represent a bi-liquid
propellant?
Q.20 What are complementary bases? Draw structure (A) N2O4 + unsymmetrical dimethylhydrazine
to show hydrogen bonding between adenine and
thymine and between guanine and cytosine. (B) N2O4 + acrylic rubber
(C) Nitroglycerine + nitrocellulose
Q.21 Give reasons for the following:
(D) Polybutadiene + ammonium perchlorate
(i) On electrolysis in acidic solution amino acids migrate
towards cathode, while in alkaline solution these Q.6 ‘Placedo’ is often given to patients. It is
migrate towards anode.
(A) An antidepressant
(ii) The monoamino monocarboxylic acids have two pK
values. (B) A broad spectrum antibiotic
(C) A sugar pill
Q.22 Glycine exists as a Zwitter ion but anthranilic acid
does not comment. (D) A tonic

Q.23 Write the difference between DNA and RNA? Q.7 An aldohexose (e.g., glucose) and 2-oxohexose
(e.g., fructose) can be distinguished with the help of
Q.24 Explain structure of protein.
(A) Tollen’s reagent (B) Fehling’s solution
(C) Benedict solution (D) Br2 / H2O
Exercise 2
Single Correct Choice Type Q.8 The open-chain glucose on oxidation with HIO4
gives
Q.1 If the sequence of bases in one strand of DNA (A) 5 HCOOH + H2C = O
is ATGACTGTC, then the sequence of bases in its
complementary strand is (B) 4 HCOOH + 2 H2C = O

(A) TACTGACAG (B) TUCTGUCUG (C) 3 HCOOH + 3 H2C = O

(C) GUACTUAUG (D) None of these (D) 2 HCOOH + 4 H2C = O

2 6 . 4 6 | Biomolecules and Polymers

Q.9 Glucose and fructose give the same osazone. One Q.14 The best way to prepare polyisobutylene is
may, therefore, conclude that
(A) Coordination polymerization
(A) Glucose and fructose have identical structures
(B) Free radial polymerization
(B) Glucose and fructose are anomers
(C) Cationic polymerization
(C) The structure of glucose and fructose have mirror-
(D) Anionic polymerization
image relationship
(D) The structure of glucose and fructose differ only Q.15 Which of the following is not correctly matched?
in those carbon atoms which take part in osazone
formation.
 
 —CH — C = CH — CH — 
(A) Neoprene;  2
| 2 
Q.10 For α-amino acid having the structure
 Cl 
R — CH — CO2H  n

|
NH2 O O
(B) Nylon-6; NH (CH2)6 NH CO (CH2)4 C O
Which of the following statements are true? n
(a) Water solubility is maximum at a pH when O O
concentration of anions and cations are equal.
(C) Terylene; OCH2 CH2 O C C
(b) They give ninhydrin test
(c) On reacting with nitrous acid give of N2 n
CH3
(A) All (B) b and c
(C) a and b (D) None of these (D) PMMA; CH2 C

COOCH3
n
Q.11 Bakelite is obtained from phenol and
formaldehyde. The initial reaction between the two
compounds is an example of Q.16 Acrilan is a hard, horny and a high melting
material. Which of the following represent its structure?
(A) Aromatic electrophilic substitution
(B) Aromatic nucleophilic Substitution  
 CH3 
   —CH2 — CH — 
(C) Free radical reaction |  | 
(A)  —CH2 —C—  (B)  COOC2H5 
(D) Aldol reaction  | 
  
COOCH3  n
 
 n
Q.12 If N1, N2, N3,………. are number of molecules with
molecular masses M1, M2, M3, …………. respectively, then
(C)  —CH2 — CH —  (D)  —CH2 — CH
average molecular mass is expressed as  |   | 
 Cl   CN 
2
ΣNM
i i ΣNM
i i



n



n
(A) (B)
NM
i i ΣNi

(C) Both (A) and (B) (D) None of these Q.17 Which of the following statement/s is (are) correct?
(A) Vinyon is a copolymer of vinyl chloride and vinyl
Q.13 The ratio of weight average molecular mass to acetate
number average molecular mass is called as (B) Saran is copolymer of vinyl chloride and vinylidene
(A) Planck’s disposal index chloride

(B) Polydiagonal index (C) Butyl rubber is a copolymer of isobutylene and

isoprene
(C) Polydispersity index
(D) All are correct
(D) None of these
 Chem i str y | 26.47

Q.18 Plexiglass (perspex) is Q.22 Assertion: The enzyme amylase hydrolyses starch
to maltose.
(A) Polyacrylonitrile (B) Polyethylacrylate
Reason: Starch is polymer containing glycosidic
(C) Polystyrene (D) Polymethylmethacrylate
linkages.

Multiple Correct Choice Type Q.23 Assertion: A solution of sucrose in water is

dextrorotatory but on hydrolysis in the presence of
Q.19 Which of the following statement (s) is (are) true?
small amount of dil. HCl, it becomes laevorotatory.
(A) All amino acids contain one chiral centre
Reason: Sucrose on hydrolysis gives unequal amounts
(B) Some amino acids contain one, while some contain of glucose and fructose as a result of which change in
more chiral center or even no chiral center sign of rotation is observed.
(C) All amino acids found in proteins have L configuration
Q.24 Assertion: Each turn of the α-helix structure of
(D) All amino acids found in proteins have 1º amino protein forms a 13 membered ring the containing 3.6
group amino acids.
Reason: α-helix is a secondary of protein which gets
Q.20 Pick out correct statements. stabilized via hydrogen bonding and disulphide linkages.
(A) In an electrolysis experiment, amino acids migrate
at the isoelectric point towards electrodes Q.25 Assertion: Styrene is more reactive than propylene
towards cationic polymerization.
(B) p-aminobenzensulphonic acid is a dipolar ion: while
p-aminobenzoic acid is not Reason: The carbocation resulting from styrene is more
stable than that resulting from propylene.
(C) Sulphanilic acid is soluble in base, but not in acid
⊕
(D) H3 N CH2COOH(pka = 2.4) is more acidic than Q.26 Assertion: Natural rubber is all cis-polyisoprene.

RCH2COOH (pKa = 4–5) Reason: trans-Polyisoprene cannot be formed.

Q.27 Assertion: PMMA is used for making lenses and

Q.21 Which of the following statements is/ are correct? light covers.
(A) Polyethylene contains double bonds Reason: It has excellent light transmission properties.
(B) The monomer used to make Teflon is C2F4
(C) Condensation polymers are known as copolymers Q.28 Assertion: Polyvinyl alcohol is obtained by
polymerization of vinyl alcohol
(D) A denatured protein could have the same primary
structure as the active protein. Reason: Polyvinyl alcohol is prepared by hydrolysis of
polyvinyl acetate.

Assertion Reasoning Type

Q.29 Assertion: Nylon-6 is obtained by polymerization
Questions 22-29 of caprolactum
Each of the questions given below consists of two Reason: It is a polyamide.
statements, an assertion (A) and reason (R). Select the
number corresponding to the appropriate alternative Comprehension Type
as follows
Paragraph 1: The utility of the polymers in various
(A) If both assertion and reason are true and reason is
fields is due to their mechanical properties like tensile
the correct explanation of assertion.
strength, elasticity, toughness etc.
(B) If both assertion and reason are true but reason is
These properties mainly depend upon intermolecular
not the correct explanation of assertion.
forces like van der Waal’s forces and hydrogen bonding
(C) If assertion is true but reason is false. operating in polymer molecules. Polymers have been
(D) If both assertion and reason are false. classified on this basis, e.g.,
2 6 . 4 8 | Biomolecules and Polymers

(A) Elastomers (B) Fibers

List I List II
(C) Thermoplastics (D) Thermosetting II. Nucleic acid (q) Sex hormone
III. Ascorbic acid (r) Vitamin C
Q.30 The molecular forces of attraction are weakest in
IV. Testosterone (s) Digestive enzyme
(A) Elastomers (B) Fibers
(A) I → s, II → p, III → r, IV → q
(C) Thermoplastics (D) Thermosetting polymers
(B) I → s, II → p, III → q, IV → r
Q.31 Which of the following have usually a linear structure? (C) I → s, II → p, III → r, IV → q
(A) Thermoplastics (B) Thermosetting polymers (D) None of these

(C) Polyethylene (D) Nylon-66

Q.37 Match list I with list II and select the correct answer
using the codes given below the lists.
Q.32 Which of the following is hard?
(A) Elastomer (B) Fibre List I List II

(C) Thermoplastic (D) Thermosetting polymers (Polymer) (Polymerizing units)

I. Bakelite (p) Butadiene and styrene
Paragraph 2: A natural elastomer polymer when with
II. Dacron (q) Phenol and methanal
sulphur got stiff and resistant to action of common
solvents and wear & tear. III. Nylon-66 (r) 1, 2-dihydroxyethane and
dimethylterephthalate
Q.33 The natural polymer is:
IV. Buna-S (s) Urea and methanol
(A) Rubber (B) Cellulose (C) Silk (D) Starch
(t) 1, 6-hexanedioic acid and
1, 6-dimino hexane
Q.34 Heating of polymer with sulphur is called:
(A) I → s, II → r, III → t, IV → p
(A) Galvanisation (B) Saponification
(B) I → q, II → r, III → t, IV → p
(C) Vulcanization (D) None of these
(C) I → t, II → q, III → r, IV → p
Match the Columns (D) None of these

Q.35 Match list I with list II and select the correct answer Q.38 Match list I with list II and select the correct answer
using the codes given below the lists. using the codes given below the lists.
List I List II
List I List II
I. Nucleic acids (p) D.N.A.
I. Phenol + formaldehyde. (p) Synthec rubber
II. Uracil (q) Hormones
II. Terephthalic acid (q) Bakelite + ethylene
III. Thymine (r) Polynucleotides glycol

IV. Double-helix structure (s) R.N.A. III. Caprolactam (r) Nylon-6

(A) I → s, II → r, III → p, IV → p IV. Butadiene+styrene (s) terylene

(B) I → r, II → s, III → p, IV → p (A) I → q, II → r, III → s, IV → p

(C) I → r, II → p, III → s, IV → p (B) I → r, II → p, III → q, IV → s

(D) None of these (C) I → q, II → s, III → r, IV → p

(D) None of these
Q.36
Q.39 Match list-I (Monomer) with list-II (Polymer) and select
List I List II
the correct answer using the codes given below the lists:
I. Pepsin (p) Genetic material
 Chem i str y | 26.49

(A) Nylon (B) Poly (vinyl chloride)

List I List II
I. Hexamethylenediamine (p) Bakelite
(C) Cellulose (D) Natural rubber

II. Phenol (q) Dacron

Q.3 The correct statement about the following
III. Phthalic acid (r) Glyptal disaccharide is  (2012)
IV. Terephtalic acid (s) Melamine CH2OH
H
(t) Nylon 6, 6 H O H2COH O H
H
(A) I → t, II → p, III → q, IV → r OH H HO
(B) I → t, II → p, III → r, IV → q HO CH2OH
OCH2CH2O
(C) I → s, II → r, III → p, IV → q H OH OH H
(D) I → s, II → r, III → p, IV → q (a) (b)

(A) Ring (a) is pyranose with α-glycosidic link

Previous Years’ Questions (B) Ring (a) is furanose with α-glycosidic link
Q.1 Cellulose upon acetylation with excess acetic (C) Ring (b) is furanose with α-glycosidic link
anhydride/H2SO4 (catalytic) givesAcO
cellulose triacetate
O (D) Ring (b) is pyranose with β-glycosidic link
whose structure is  H O (2008)
AcO H
AcO O AcH
O
H O O
H Q.4 The following carbohydrate is  (2011)
AcO H H O
AcO O AcH HH OAc
O O H O AcH
(A) H O H OH
H H O
H
H OAc H2C
AcO O AcH H OAc
O H O AcH O
O H
(A) H H OAc O HO
H H OAc
O AcH HO
O H AcO OH
H OAc O OH
H O
AcO
AcO
H
OH H (A) A ketohexose (B) An aldohexose
O O O H
AcO
H
H H O (C) An α-furanose (D) An α-pyranose
AcO OH H HH OH
O H OH H
H O O O H
(B) H
H
AcO OH H HH OH H OH Q.5 The correct statement (s) about the following
OH H
O
H
O H
O H sugars X and Y is (are):  (2011)
(B) H OH
H H OH
OH H
O H CH2OH
H OH H O
H HOH2C H
H
AcO AcO AcO
H H H OH H H HO
O O OO CH2OH
H OH
(C) H H H O
AcO O AcH AcO O AcH AcO O AcH H
O O
H O H
O O H OO H OH OH OH
H
(C) H
H OAc HH OAc HH OAc
O AcH O AcH O AcH H X
O O O
(A) X is a reducing sugar and Y is a non-reducing sugar
H OAc H OAc H OAc

AcO
(B) X is a non-reducing sugar and Y is a reducing sugar
AcO AcO
H H H
H
O O OO (C) The glucosidic linkage in X and Y are α and β
H
(D) AcO OAc H
H H AcO OAc H
H
AcO OAc
HH respectively
O O O H O O H OO
H
(D) H
H HH OAc HH OAc (D) The glucosidic linkage in X and Y are β and α,
OAc H
O O
OAc H
O
OAc H H
respectively
H H OAc H OAc
Q.2 Among cellulose, poly (vinyl chloride), nylon and
Q.6 For ‘invert sugar’, the correct statement(s) is (are)
natural rubber, the polymer in which the intermolecular
(Given: specific rotations of (+)-sucrose, (+)-maltose,
force of attraction is weakest is  (2012)
2 6 . 5 0 | Biomolecules and Polymers

L-(-)-glucose and L-(+)-fructose in aqueous solution Q.13 Aspartame, an artificial sweetener, is a peptide
are +66°, + 140°, −52° and +92° , respectively. )(2016) and has the following structure  (2001)
(A) ‘invert sugar’ is prepared by acid catalyzed hydrolysis
CH2C6H5
of maltose |
H2N — CH — CONH — CH — COOCH3
(B) ‘invert sugar’ is an equimolar mixture of D-(+)- |
glucose and D-(-)-fructose CH2 — COOH
(C) Specific rotation of ‘invert sugar’ is −20° (i) Identify the four functional groups.
(D) On reaction with Br2 water, ‘invert sugar’ forms (ii) Write the Zwitter ionic structure.
saccharic acid as one of the products
(iii) Write the structures of the amino acids obtained
from the hydrolysis of aspartame.
Q.7 Match the chemical substances in column I with
type of polymers/type of bonds in column II. Indicate (iv) Which of the two amino acids in more hydrophobic?
your answer by darkening the appropriate bubbles of
the 4 × 4 matrix given in the ORS. (2007) Q.14 Name the heterogenous catalyst used in the
polymerization of ethylene.  (2003)
Column I Column II
(A) Cellulose (p) Natural polymer
Q.15 Statement-I: Glucose gives a reddish-brown
(B) Nylon-6, 6 (q) Synthetic polymer precipitate with Fehling’s solution.  (2007)
(C) Protein (r) Amide linkage Statement-II: Reaction of glucose with Fehling’s
solution gives CuO and gluconic acid.
(D) Sucrose (s) Glycoside linkage

(A) Statement-I is True, statement-II is True;

Q.8 The total number of basic group in the following statement-II is a correct explanation for statement-I
form of lysine is  (2010)
(B) Statement-I is True, statement-II is True;
H3N

CH2 CH2 CH2 H2C O statement-II is NOT a correct explanation for statement-I
(C) Statement-I is True, statement-II is False
CH C

(D) Statement-I is False, statement-II is True
H2N O

Q.9 A decapeptide (Mol. Wt. 796) on complete Q.16 Cellulose upon acetylation with excess acetic
hydrolysis gives glycine (Mol. Wt. 75), alanine and anhydride / H2SO4 (catalytic) gives cellulose triacetate
phenylalanine. Glycine contributes 47.0 % to the total whose structure is  (2008)
weight of the hydrolysed products. The number of AcO
glycine units present in the decapeptide is  (2011) O
H O
AcO H
O AcH
Q.10 Write the structure of alanine at pH = 2 and H
O O H
H
pH = 10.  (2000) (A) AcO
O AcH
H OAc
O O H
H
H H OAc
Q.11 Give the structure of the products in the following O AcH
O H
reaction (2000)
+ H OAc
H
Sucrose →A +B
AcO
O
Q.12 Give the structure of the products in the following H
H
O
AcO
reaction (2000) OH H
O O H
NOH H
(B) AcO H H OH
OH H
O O H
H
H+ Polymerisation H H
C * D * OH H OH
n O H
H OH
 Chem i str y | 26.51

O
AcO AcO AcO
H H
O H O H OO H H N
H O O H O
(C) H H H N N N N
O AcH O AcH O AcH H N N N
O O O
N CH2 O H H CH2 O
O
H OAc H OAc H OAc H
O
AcO AcO AcO
H H H
O O OO
H
H H H Q.22 The structure of D-(+)-glucose is  (2015)
(D) H H H H H H H
O O O CHOH
OAc OAc OAc OAc OAc OAc H OH
HO H
Q.17 The substituents R1 and R2 for nine peptides are
H OH
listed in the table given below. How many of these
peptides are positively charged at pH = 7.0?  (2012) H OH
⊕ Θ CH2OH
H3 N− CH − CO − NH − C H − CO − NH − C H − CO − NH − CH − CO O
| | | |
H R1 R2 H The structure of L-(–)-glucose is

Peptide R1 R2 CHO CHO

HO H H OH
I H H
H OH HO H
II H CH3 (A) (B)
HO H H OH
III CH2COOH H
HO H HO H
IV CH2CONH2 (CH2)4NH2 CH2OH CH2OH
V CH2CONH2 CH2CONH2
VI (CH2)4NH2 (CH2)4NH2 CHO CHO
VII CH2COOH CH2CONH2 HO H HO H

VIII CH2OH (CH2)4NH2 HO H HO H

(C) (D)
H OH HO H
IX (CH2)4NH2 CH3
HO H H OH
CH2OH CH2OH
Q.18 When the following aldohexose exists in its
D-configuration, the total number of stereoisomers in
Q.23 Under hydrolytic conditions, the compounds
its pyranose form is  (2012)
used for preparation of linear polymer and for chain
CHO − CH2 − CHOH − CHOH − CH2OH termination, respectively, are (2015)

Q.19 A tetrapeptide has –COOH group on alanine. (A) CH3SiCl3 and Si ( CH3 )
4
This produces glycine (Gly), valine (Val), phenyl alanine (B) ( CH3 ) SiCl2 and ( CH3 ) SiCl
(Phe) and alanine (Ala), on complete hydrolysis. For 2 3

this tetrapeptide, the number of possible sequences (C) ( CH3 ) SiCl2 and CH3SiCl3
2
(primary structures) with −NH2 group attached to a
chiral center is (2013) (D) SiCl4 and ( CH3 ) SiCl
3
Q.24 On complete hydrogenation, natural rubber
Q.20 The total number of lone-pairs of electrons in produces (2016)
melamine is  (2013) (A) Ethylene-propylene copolymer
(B) Vulcanised rubber
Q.21 The total number of distinct naturally occurring
amino acids obtained by complete acidic hydrolysis of (C) Polypropylene
the peptide shown below is  (2014) (D) Polybutylene
2 6 . 5 2 | Biomolecules and Polymers

Q.25 The correct functional group X and the reagent/ (A) X = COOCH3, Y = H2/Ni/heat
reaction conditions Y in the following schemes are
(B) X = CONH2, Y = H2/Ni/heat
 (2011)
(i) Y (C) X = CONH2, Y = Br2/NaOH
X (CH2)4 X Condensation polymer
O O (D) –X = CN, Y = H2/Ni/heat

(ii) C (CH2)4 C
heat
HO OH

PlancEssential Questions
JEE Main/Boards JEE Advanced/Boards
Exercise 1 Exercise 1
Q.5 Q.14 Q.18 (i) Q.1 Q.11 (a) Q.14
Q.23 Q.24 Q.22 Q.25

Exercise 2 Exercise 2
Q.1 Q.11 Q.20 Q.2 Q.8 Q.19
Q.27 Q.33 Q.39 Q.13 Q.15 Q.27
Q.43 Q.38 Q.40
Q.48 Q.51 Previous Years’ Questions
Q.1 Q.5 Q.12
Previous Years’ Questions
Q.13
Q.3

Answer Key

JEE Main/Boards
Exercise 2
Single Correct Choice Type

Q.1 D Q.2 D Q.3 C Q.4 C Q.5 A Q.6 B

Q.7 C Q.8 A Q.9 C Q.10 B Q.11 D Q.12 C

Q.13 D Q.14 A Q.15 C Q.16 A Q.17 A Q.18 B

Q.19B Q.20 B Q.21 A Q.22 D Q.23 C Q.24 A

 Chem i str y | 26.53

Q.25 A Q.26 B Q.27 C Q.28 B Q.29 A Q.30 C

Q.31 D Q.32 D Q.33 C Q.34 C Q.35 C Q.36 A

Q.37 A Q.38 B Q.39 C Q.40 D Q.41 D Q.42 C

Q.43 A Q.44 A Q.45 D Q.46 A Q.47 C Q.48 C

Q.49 B Q.50 B Q.51 D Q.52 A Q.53 C

Previous Years’ Questions

Q.1 B Q.2 B Q.3 C Q.4 C Q.5 C Q.6 C

Q.7 A Q.8 B Q.9 A Q.10 C Q.11 A Q.12 A

Q.13 A Q.14 B Q.15 A Q.16 C Q.17 A Q.18 B

JEE Advanced/Boards
Exercise 2
Single Correct Choice Type

Q.1 A Q.2 C Q.3 D Q.4 C Q.5 A Q.6 C

Q.7 D Q.8 A Q.9 D Q.10 B Q.11 A Q.12 A

Q.13 C Q.14 C Q.15 B Q.16 D Q.17 D Q.18 D

Multiple Correct Choice Type

Q.19 B, C Q.20 B, C, D Q.21 B, C, D

Assertion Reasoning Type

Q.22 A Q.23 C Q.24 C Q.25 C Q.26 C Q.27 A

Q.28 D Q.29 B

Comprehension Type

Paragraph 1: Q.30 A Q.31 A Q.32 D

Paragraph 2: Q.33 A Q.34 C

Match the Columns

Q.35 B Q.36 C Q.37 B Q.38 C Q.39 B

Previous Years' Questions

Q.1 A Q.2 B Q.3 A Q.4 B Q.5 B, C Q.6 B,C

Q.7 A → p, s; B → q, r; C → p, r; D → s Q.8 2 Q.9 6 Q.15 C Q.16 A

Q.17 D Q.18 8 Q.19 4 Q.20 6 Q.21 1 Q.22 A

Q.23 B Q.24 A Q.25 C, D

2 6 . 5 4 | Biomolecules and Polymers

Solutions

JEE Main/Boards of protein.

(iii) Denaturation – when the proteins are subjected
Exercise 1 to the action of heat, mineral acids or alkali, the water
soluble form of globular protein changes to water
insoluble fibrous protein resulting in the precipitation or
Sol 1: (i) coagulation of protein, called denaturation of proteins.
CHO

H C OH Sol 4: (i) α ‒ Helix structure: This structure is acquired

when the alkyl groups in amino acids are large and are
OH C H
HI H 3C CH2 CH2 CH2 CH2 CH3 involved in the coiling of the polypeptide chain. This
H C OH is stabilized by the intermolecular hydrogen bond
H C OH n-hexane between the C = O group of one amino acid and
CH2OH
−NH group of the fourth amino acid.
(ii) (ii) β ‒ pleatedstructure: This structure is acquired
CHO CHO when the alkyl group are small in this structure. Linear
Br2H2O polypeptide chains are arranged side by side and
(CH OH)4 (CH OH)4 stabilised by intermolecular hydrogen bond between

CH2OH CH2OH C = O and −NH group.

D-Gluconic Acid Sol 5: The amino acids containing one carboxylic group
and one amino group behave like a neutral molecule.
(iii) This is due to the formation of a zurtter ion structure.
CHO COOH
+
HNO3 NH2 NH3
(CH OH)4 (CH OH)4
R-CH-COOH R-CH-COOH
CH2OH COOH
This zwitter ion changes to cation in acidic solution
Glucaric acid and anion in alkaline medium. making it amphoteric in
nature.

Sol 2: (i) Peptide Linkage: The linkage ( −CO − NH−) is H2N -CH-COO
Alkali
H3N
+
CH-COO-
known as peptide linkage. This linkage is found in the
primary structure of proteins. R R
(ii) Denaturation – when the proteins are sulgected to
the action of heat, mineral acids or alkali, the water Acid +
soluble form of globular protein changes to water H3N CH COOH
insoluble fibrous protein resulting in the precipitation or
R
coagulation of protein, called denaturation of proteins.
Sol 6: Due to denaturation, water soluble form of
Sol 3: (i) Peptide Linkage: The linkage ( −CO − NH−) is globular protein changes to water insoluble fibrous
known as peptide linkage. This linkage is found in the protein, resulting in the precipitation or coagulation of
primary structure of proteins. protein.

(ii) The primary structure of proteins refer to its covalent

Sol 7: (i) A glycosidic linkage is a type of covalent bond
structure, i.e sequence in which various α − amino acids
that joins a carbohydrate molecule to another group
are arranged in protein or in the polypeptide structure
which may or may not be another carbohydrate.
 Chem i str y | 26.55

Sol 8: (i)Glucose does not undergo certain reactions (v) Antioxidant activity
of aldehydes, for ex reaction with NaHSO3 schiff’s test
Deficiency causes night blindness.
etc.
Source: Liver, orange, carrots, pumpkin
(ii) Reaction of glucose with NH2OH
Vitamin C
(iii) Mutarotation of α and β glucose
Function: Highly effective antioxidant
(iv) Formation of two isomeric methyl glycosides when
treated with methanol. Lessen oxidative stress
Natural antihistamine
Sol 9: Essential amino acids: Amino acids which cannot
Functioning of immune system
by synthesized by the body and therefore needs to
taken through external diet. For ex. Phenylalanine, Deficiency causes scurvy, bleeding gums
valine Sources: Virus fruits, amla, green leafy vegetable
Non-Essential amino acids: - These amino acids can by
synthesized by the body and therefore. need not be Sol 14: Aldopentose
supplied by an external diet. For ex. Alanine, aspartic
acid. CHO

H * OH
Sol 10: (i) Vitamin A: Chief source orange, ripe yellow
H * OH Asymmetric carbons are 3
fruits, leafy vegetables, carrots, pumpkin Function.
helps in vision, gene transcription, bone metabolism, H * OH
antioxidant, activity.
CH2OH
(ii) Vitamin B: Chief source pork, oatmeal, brown rise,
potatoes, eggs Aldohexose

Function. growth, regulation of apetite Functioning of CHO

heart, muscles and nervous system.
H * OH
(iii) Vitamin D
H * OH
(iv) Vitamin K Asymmetric carbons are 4
H * OH

Sol 11: (i) Refer solution 9. H * OH

(ii) Refer solution 2. CH2OH

Sol 12: Vitamin are classified into two groups. Sol 15: Nucleotides are organic molecules that serves
Depending upon their solubility in water or fat. as the monomers. Or subunits of nucleic acids like DNA
or RNA
(i) Fat soluble Vitamin: Vitamin which are soluble in fats
and oils, but insoluble in water. for ex. Vitamin A,P,E O
For E.g.
and K O
O- P O CH2 Base
(ii) Water soluble vitamins: Vitamins which are soluble
in water. For ex B group vitamins and vitamin C. O-
H H
H H
Vitamin K is responsible for coagulation of blood.

OH OH
Sol 13: Function of vitamin A
(i) Vision (b) (i) As a source of energy (more than 50–80% of
energy in the diet is supplied by carbohydrates)
(ii) Gene transcription
(ii) Protein sparing action: As carbohydrates are mainly
(iii) Immune function
used for energy need of body, proteins are spared for
(iv) Bone metabolism tissue building and repairing.
2 6 . 5 6 | Biomolecules and Polymers

(iii) Essential for fat oxidation

C = O and −NH group.
(iv) Gastro intestinal function.
(ii) (a) Primary Structure: This structure refer to its
Sol 16: α Helix structure is stabilized by the intra covalent structure which various sequence in which
molecular hydrogen bond between the C ==0O group various α − amino acids are arranged in protein or in
the polypeptide structure of protein.
of one amino acid and −NH group of the fourth amino
(b) Secondary Structure: This refers to the arrangement
acid.
of polypeptide chains into a defined three dimension
structure which protein assumes as a result of hydrogen
Sol 17: Fibrous proteins have largely helical structure bonding.
and are rigid molecules of rod like shape. Globular
proteins, on the other hand have a polypeptide chain
Sol 21: The particles in nucleus of the cell, responsible
which consist partly of helical section and partly β
for heredity, are called chromosomes, which are made
pleated structure and Remaining random coil form.
up of proteins, and another type of biomolecules called
nucleic acids.
Sol 18: (i) Amino acids have very strong intermolecular
forces due to highly effective hydrogen bonding Function:
between (i) Nucleic acid serves as chemical basis of heredity
C ==0O and −NH groups. and may be regards regarded as reserve of genetic
information.
O (ii) Protein synthesis in cells.
H O
R C N H
H C N Sol 22: Nucleoside:
H
R (a) Consist of a nitrogenous base covalently attached
to a sugar (ribose or deoxyribose) but without the
(ii) Vitamin A Night blindness Vitamin B1 Beri Beri phosphate group.
Vitamin B6 Convulsions
Nucleoside = sugar + base
Vitamin K increasing blood clotting time
(b) Used as antiviral or anticancer agents

Sol 19: In DNA, two nucleic acid chains are wound about (c) E.g. Cytidine, uridine, adenosine
each other held together by hydrogen bonds between Nucleotide:
pair of bases. The two strands are complementary to
(a) Consists of a nitrogenous base, a sugar (ribose or
each other because the hydrogen bonds are formed
deoxyribose ) and one to three phosphate groups.
between specific pair of bases. Adenine forms hydrogen
bonds with thymine whereas cytosine forms hydrogen (b) Nucleotide sugar + base + phosphate
bonds with guanine.
(c) Malfunctioning nucleotides are one of the main
causes of cancer.
Sol 20: (i) α − Helix
E.g. S-uridine monophosphate
(a) Alkyl groups in amino Acids are large.
(b)Polypeptide chains are coiled leading to right Sol 23: Reducing sugar: sugar that contain aldehyde
handed helical coil group that are oxidized to carboxylic acids are classified
as reducing sugar.
(c)Stablised by intermolecular H – bonding between
groups on one amino acid and –NH group of 4 amino For e.g. Sucrose act as reducing agents.
acid.
Non reducing sugar: they cannot act as a reducing
β − Helix group due to absence of an aldehydic group.
(a) Alkyl groups in amino acids are small Reducing sugars must either contain aldehyde group
or in is capable of forming one in solution through
(b) Polypeptide chains are arranged side by side
isomerism.
(c) Stabilized by intermolecular H-Bonding between the
 Chem i str y | 26.57

Sol 24: α glucose Sol 29: Refer theory.

(a) CH2OH
O Sol 30: (a) Refer text Pg. 2
H H H
(b) A nucleotide is an organic molecule made up of
nucleotide base, a five carbon sugar and at least one
HO OH H OH
phosphate group.

H OH The two classes are

(b) −OH group lies below the ring on carbon (i) Purine: contains adenine and guanine

(c) α glucose folds up into a helix. (ii) Pyrimidine: contains cytosine, thymine and uracil.

β glucose
Exercise 2
(a) CH2OH
Single Correct Choice Type
O
H H OH Sol 1: (D) Chromophore is the colouring agent, which
is diazo group (–N = N -)
OH OH H H
Sol 2: (D) Zwitter ion is present
H OH
R CH COO-
(b) −OH group lies above the ring on carbon
(c) β glucose folds up into a pleated sheet. *NH3

Sol 25: Refer solution 1. Sol 3: (C) This is secondary structure of a protein
Same reaction as D-Glucose, only structural difference is
in the configuration of atoms around different carbons. Sol 4: (C) General formula of carbohydrates:

Cn (H2O)n or Cx (H2O)y
Sol 26: Refer text pg. 15
Sol 5: (A) Rest are all monosaccharide except sucrose.
Sol 27: Different types:
(i) mRVA Sol 6: (B) Iron is bonded by co-ordination ring
formation (chelation)
(ii) fRNA
(iii) rRNA
Sol 7: (C) Anomers have different specific relation.

Sol 28: (i) This pH at which the structure of amino acid

Sol 8: (A) Amide linkage ( −CO − NH−) is called peptide
has no net charge is called it’s isoelectronic point for ex.
bond
Aspartic acid hs isoelectric point of 2.77
(ii) Mutarotation is the change in the optical rotation Sol 9: (C) Glucose is a hydroxyl aldehyde whereas
that occurs by epimerization (change in equilibrium Fructose is a ketone.
between two epimers, when the corresponding
stereocentres interconvert) for ex. β − D glucose .
Sol 10: (B) Glucose is a hydroxy aldehyde with 6 carbon
With specific rotation of +18.7°, when dissolved in or aldehexose.
water undergo mutarotation and attains a final specific
rotation of 52.50
Sol 11: (D) Phenol acts both as an antiseptic and
(iii) An enzyme is a substance produced by a living disinfectant.
organism which acts as a catalyst to bring about a
specific biochemical reaction for ex. Streptokinase
2 6 . 5 8 | Biomolecules and Polymers

Sol 12: (C) Fenton’s reagent (H2O2 + Fe) converts

Sol 28: (B) Cellulose is a linear polymer of β glucose.
D -glucose (6 carbon) to D arabinose (5 carbon)
Sol 29: (A)
Sol 13: (D) At isoelectronic point.
H H
pk a1 + pk a2 C=O
pH = C = N NH C6H5
2 H C OH H C OH

Sol 14: (A) Zeigler and Natta developed zeigler Natta HO C H C6H5NH NH2 HO C H

Catalysts (Al(C2H5 )3 + TiCl3 ) which are used for H C OH H C OH

coordination polymerization. H C OH H C OH
CH2OH C
Sol 15: (C) Teflon is homopolymer of tetraflourethylene. CH2OH
D-Glucose
Polystyrene is a homopolymer of styrene. Neoprene is a
homopolymer of chloroprene. C6H5NH NH2
H
C = N NH C6H5
Sol 16: (A) Anomer have different configurations at H
Glycosidic carbon and hence different specific rotation. C = N NH C6H5 C = N NH C6H5
HO C H C=O
Sol 17: (A) They are called epimers. C6H5NH NH2
H C OH HO C OH

Sol 18: (B) Natural rubber is polymer of isoprene H C OH H C OH

CH2OH H C OH


H3C
C-C
CH2  3 molecules of phenylhydrazine is used.
CH2OH

H2C H

Sol 30: (C) Main structural unit is peptide linkage

Sol 19: (B) It is called inversion as sucrose, which is
( −CO − NH−)
dextrorotatory gives a laevorotatory mixture on dilution.
Sol 31: (D) All these are function of the proteins
Sol 20: (B) Sulphur cross links are present in vulcanised
rubber.
Sol 32: (D) Glucose and galactose are monosaccharide
while sucrose is disaccharide.
Sol 21: (A) Glycine (H 2 N − CH 2 − COOH) is the
simplest amino acid.
Sol 33: (C) Starch on reaction with iodine gives blue
colour which serves as a test for presence of I2
Sol 22: (D) All of these are natural polymers.
Sol 34: (C) Ribose has five carbon atoms, rest all
Sol 23: (C) They differ in configuration of Cl options are correct.
Ribose
Sol 24: (A) Aldehydic group gives silver mirror test with
tollen’s reagent. H C O
H C OH
Sol 25: (A) This is definition of oligosaccharides.
H C OH

Sol 26: (B) This is definition of epimer. H C OH

CH2OH
Sol 27: (C) Energy is evolved with ATP is hydrolysed to
ADP and AMP. Rest are correct.
 Chem i str y | 26.59

Sol 35: (C) D – xylose is a diastereomer of ribose with Sol 43: (A) Teflon is a polymer of F2C = CF2
formula.
CHO F F
nCF2 = CF2 → ( C C (n
H OH
F F
OH H
H OH

CH2OH Cl Cl
Sol 44: (A) H3C CH2 ( H2C CH (n
Sol 36: (A) Glucose gives the silver mirror test with PVC
ammoniacal solution of silver nitrate because it contains
the group Aldehyde. Sol 45: (D) Polyacrylonitrile is a polymer of acrylonitrile
(CH2 = CHCN)
Sol 37: (A) Dacron is a copolymer of ethylene glycol
and terephthalic acid. Sol 46: (A) This is the classification based on inter
molecular forces.
PVC is a polymer of : Vinyl chloride
Polystyrene is a polymer of styrene Sol 47: (C) Polyethene involves linear linkage between
Teflon is a polymer of tetraflouroethylene chains (CH2 – CH2)n

Sol 38: (B) Melamine resin: Sol 48: (C) Al (C2H5)3 + TiCl4 is Zieglar – Nata catalyst

HN Sol 49: (B) Neoprene is a polymer of chloroprene

N N CH2 =C CH=CH2 CH2 CH=C CH2

Cl Cl
NH N NH n
Sol 50: (B)Teflon is a polymer of CH2 = CF2.
F F
NH2 Sol 51: (D) Polytetraflouroethylene or Teflon ( C C (n
F F
N N

+ 3HCHO is used because of its being a thermoplastic polymer.

N NH2
H2N Sol 52: (A)
nHOOC ( CH2 )4 COOH + nH2N ( CH2 ) 6 NH2 →
F F
Sol 39: (C) Teflon is a polymer of C C O O
(CH2)6
F F
(CH2)4 N N
Sol 40: (D) Acrylonitrile n H2C = C – C ≡ N →
H H n
(HC = C – C = N )n

Sol 41: (D) Glyptal is a condensation polymer. Sol 53: (C) They are made from a single compound i.e.
they contain a single repeating unit.
Sol 42: (C) Highly vulcanised rubber is called ebonite.
2 6 . 6 0 | Biomolecules and Polymers

Previous Years’ Questions Acrylonitrile, Neoprene and Teflon are addition

polymers of acrylonitrile, isoprene and tetrafluoro
Sol 1: (B) Both glucose and fructose are reducing ethylene respectively.
sugars, reduces Tollen’s reagent to metallic silver.
Sol 13: (A) DNA contains ATGC bases
Sol 2: (B) “ α ” and “ β ” cyclic hemiacetals of D-glucose A – Adenine
having difference in configuration at C-1 only are called
anomers. T – Thymine
G – Guanine
Sol 3: (C) Statement I is Correct: Presence of -CHO
C – Cytocine
group in glucose is tested by Fehling solution test
where a reddish-brown precipitate of Cu2O is formed. So quinoline is not present.
Hence, statement II is incorrect.
Sol 14: (B) Glyptal is used in the manufacture of paints
Sol 4: (C) α − D ( + ) glucose and β − D ( + ) glucose are and lacquers.
anomers.
Sol 15: (A) Vitamin B and C are water soluble and
Sol 5: (C) Carbohydrates are polyhydroxy carbonyl Vitamin A, D, E and K are water insoluble.
compounds.
Sol 16: (C) Low density polythene is not used in the
Sol 6: (C) Buna-N synthetic rubber is a copolymer of manufacturing of buckets, dust-bins etc. because buckets,
acrylonitrile (ACN) and butadiene. dustbins are manufactured by high density polythene.

Sol 7: (A) It is a test characteristic of amide linkage. Sol 17: (A)

Urea also has amide linkage like proteins.
Sodium lauryl sulphate = detergent, anionic
Cetyltrimethyl ammonium bromide = detergent,
Sol 8: (B) Nylon 6,6 is a polymer of adipic acid and
cationic
hexamethylene diamine
Glyceryl oleate = detergent, non-ionic
O O
Sodium stearate = soap, anionic
 C (CH2)4 C NH (CH2)6 NH 
n
Sol 18: (B) NH2
Sol 9: (A) In RNA, the sugar is β − D -Ribose, where as
in DNA the Sugar is β − D − 2 - deoxy Ribose. Cystine HO S COOH
C S
O NH2
NH2
Sol 10: (C) Glycine CH2 Cysteine COOH
COOH HS
NH2
Sol 11: (A) Fact
Thiol group (SH) is present in cysteine.
6CO2+12NADPH+18ATP → C6H12O6+12NADP+18ADP O
||
Methionine CH3 − S − CH2 − CH2 − C H − C − OH
Sol 12: (A) Dacron is polyester formed by condensation |
polymerisation of terephthalic acid and ethylene glycol NH
2
Cytosine NH2
HOOC COOH +HO CH2 CH2 OH
N

CO CH2 CH2 O O N
n H
Dacron
 Chem i str y | 26.61

JEE Advanced/Boards Sol 5: (i) Proteins are polymer of α‒amino acids and
they are connected to each other by peptide bond or
peptide linkage.
Exercise 1
(ii) A polypeptide is a single linear chain of amino acids
Sol 1: Streptokinase is used as a medicine for blood whereas an oligopeptide is a polypeptide less then
clots. 30-50 amino acids long.
(iii) Refer Exercise I, Q.6
Sol 2: Cellulose in our diet is not nourishing as because,
it is a complex from of carbohydrate, and no mammal
makes the necessary enzyme to break down cellulose. Sol 6: A vitamin is an organic compound required by an
organism as a vital nutrients in limited amounts.
Sol 3: Mutarotation is the change in the optical (a) Two types Fat Soluble Vitamins: Vitamins which
rotation that occurs by epimerization (that is the are soluble in fat and oil, but insoluble in water for
change in he equilibrium between two epimers, when E.g. Vitamin A,D,E And K
the corresponding stereo centers interconvert epimers (b) Water Soluble Vitamin: Vitamins which are soluble
of D-glucose in water, for Ex. B group vitamins and vitamins C
CH2OH
(b) Refer theory part.
O
H H H
Sol 7: (a) (i) A coenzyme is a substance that works with
HO OH H OH an enzyme to initiate or aid the function of enzyme.
They can not function on their own and require the
H OH presence of an enzyme.
α ‒ D ‒ (+) ‒ Glycopyranose
Sol 8: (a) (i) Oranges, carrots, pumpkins
θ = +112°C
CH2OH Sol 9: (i) Cellular Respiration: Like other organisms,
O plants store corbohydrates and burn them for energy.
H H OH
(ii) Mechanical Strength: Certain carbohydrats, like
OH OH H
cellulose helps plants in enhancing mechanical strength.
H

H OH Sol 10: (i)

CHO COOH
β ‒ D ‒ (+) ‒ Glycopyranose alk.MnO4-
(i) (CHOH)4 (CHOH)4
θ = 19°C
CH2OH COOH
When either of these forms of D-glucose is dissolved Saccharic acid
in water and allowed to stand, a gradual change in
specific rotation occurs the specific rotation of the α
form falls and that of the β form rises until a constant CHO COOH
values of 53° is obtained. (CHOH)4
Br2+CS2
(CHOH)4
(ii)
CH2OH COOH
Sol 4: For differences, refer text. Gluconic acid
Sugar: 2 ‒ deoxy D(‒)ribose

O CHO
HO CH2 OH H2+SO4
(iii) (CHOH)4 6C+6H2O
H CH2OH
H H

OH H
2 6 . 6 2 | Biomolecules and Polymers

Sol 11: (i) Adenine (ii) Thymine (iii) Cystosine C=O group of one amino acid and ‒HN group of fourth
amino acid stabilises α‒helix structure.
(iv) Guanine
In 3.613‒Helix, 3.6 is the number of residues per turn
Thymine is not present in RNA.
and 13 is the number of atoms in the hydrogen bonded
loop.
Sol 12: Two fat soluble vitamin are vitamin A and D.
Vitamin A: Deficiency disease night blindness.
Sol 19: Complementary bases are specific pairs that
Sources: (i) Carrots (ii) Liver (iii) Pumpkin (iv) Orange join up the two strands of double stranded DNA Via
hydrogen bonds.
Vitamin D: Deficiency disease: Rickets sources. H
(i) Sunlight ( chief source ) N O H N

(ii) Some mushrooms

N N-H N
R
Sol 13: Refer theory. N N
N-H O
R
Sol 14: Refer Exercise I, two (ii) H
Guanine Cytosine

Sol 15: (a) Functional group H

N N-H
(i) Carboxylic acid (ii) Amine (GC)
O (A =T)
Amide Ester N N
NH3+ CH2C6C5 R
(b) N H-N N
-OOC-CH -CH-CONH-CH-COO CH R
2 3
O
Adenine Thymine
(c) (i) NH2

HOOC - CH 2CH - COOH Sol 20: (i) In acidic medium,

(ii) CH2 C6 H5 R-CH-NH3+ +

H R-CH-NH+
3

H2 N - CH - COOH COO - COOH

Since this is a positively charged ion (cation), they

(d) (ii) is more hydrophobic as (i) is more polar due to migrate towards the cathode.
presence of 2 COOH groups,
In basic medium,
Helping, (i) to make hydrogen bonds with water,
increasing its hydrophilicity R-CH-NH3+ OH
+
R-CH-NH2
COO COO -
Sol 16: α ‒ (5, 6 ‒dimenthylbenzimidazolyl)
Since this is an anion (negatively charged), they migrate
Cobamidcyanide more commonly called cobalmin. towards the anode.
(ii)Monoamino carboxylic acids exits as zwitter ions
Sol 17: (a) A mixture of equal parts of glucose and and exhibit two different k a values and hence two pk a
fructose resulting from the hydrolysis of sucrose is values.
called invert sugar.
(b) Polypeptide is a linear organic polymer consisting R-CH-COOH PKa1 R-CH-COO- Pka2 R-COO-
of a large number of amino acid residues bonded NH+
3 NH+
3 NH2
together in a chain, forming part of (or the whole of) a
protein molecule.
Sol 21: Glycine (H2N ‒ CH2 ‒ COOH) exists as zwitter
ion (H3N ‒ CH2 ‒ COO‒) as it contains both acid and
Sol 18: Intramolecular Hydrogen bonding is between
 Chem i str y | 26.63

amino group on the same α ‒ Carbon (unsymmetrical dimethyl hydrazine )

COOH Is used as bi liquid propellant in rocket fuels.
NH2
Anthranilic acid
 
Cannot exist as zwitter ion, because
Sol 6: (C) In ‘placebo’, a group of patients is given an
actual medicine, while the other group is given an
ordinary sugar pill. The ordinary pill is called placebo.
COO-
+
NH3 Sol 7: (D)
The resulting zwitter ion
  CHO

(CHOH)2
Br2, H2O
COOH

(CHOH)2
is highly destabilised by the strong +I And no m-effect
of NH3+ group. CH2OH CH2OH
Glucose
Sol 22: Refer theory Part.
Fructose does not undergo any reaction.

Sol 23: Refer theory Part.

Sol 8: (A)
CHO
Sol 24: Refer theory Part.
HIO4
(CHOH)4 5HCOOH + H2C=O

Exercise 2 CH2OH

Biomolecules Sol 9: (D) This is correct.

Single Correct Choice Type Sol 10: (B) Amino acids give minhydrin test as they are
primary amines.
Sol 1: (A) In complementary strand, complementary NH2
With HNO2, |
bases will be present, that is according to pairing ( A =
HNO2
T ) and ( C = G ). Now R CHCOOH  → R CH2COOH+N2
DNA strand.
Sol 11: (A)
OH OH
A T G A C T G T C CH2OH
T A C T G A C A G (A) + HCHO

Complementary strand
This is an electrophilic substitution.
Sol 2: (C) Saturated acids have high melting point due
to low bond polarity ΣN i M :2
Sol 12: (A) Average molecular mass =
ΣN i M :
Sol 3: (D) Biodegradable detergent should contain a
cyclohexyl side chain. Sol 13: (C) Polydispersity index
ΣNi Mi2
Sol 4: (C) The gases ejected by the rocket exert an Weight Average Molecular Mass ΣNi Mi
= =
equal and opposite thrust on the rocket. Number Average Molecular Mass ΣNi Mi
ΣNi
ΣNi × ΣNi Mi2
Sol 5: (A) =
( ΣNi Mi )
2
H CH3
N2O4+ N=N
CH3 H
2 6 . 6 4 | Biomolecules and Polymers

Sol 14: (C) Isobutane CH3 − CH2 − CH3 Due to very high acidity of ‒SO3H group and stabilisation
| of resulting dipolar ion due to charge distribution. In
CH3 p – aminobenzoic acid
COOH
+
We see then its cation CH3 − CH − CH3 is very stable,
|
CH3

as it is tertiary, therefore the best way to polymer it is

  NH2
by cationic polymerization.
the resulting dipolar ion
O COO-
Sol 15: (B) Nylon-6 : N – (CH2 )6 – C
H n   NH2
Sol 16: (D) Acrilan CH2 – CH
Does not exist as relatively low acidity of ‒COOH group
CN n and less stablisation of dipolar ion, as negative charge
is distributed between only 20 atoms.
Sol 17: (D) They are all correct.
SO3H
Sol 18: (D) Commercial name for

polymethylmethacrylate, is
1
O O
(c) Sulphanilic acid
  NH2

plexiglass (orperpox)
  n Exists as dipolar ion, SO-3

NH+3
Multiple Correct Choice Type
Which is an insoluble salt.
Sol 19: (B, C) Self explanatory.
Now in base,
Sol 20: (B, C, D) SO-3 SO-3
(a) At isoelectric point, amino acid solution is neutral.
OH-
(b) φ-amino benzesulphonic acid
SO3H
NH+3 NH2

  NH2
Reacts with base as OH − is more basic than
group.
−
NH 2

exists as dipolar ion In acid,

SO-3
SO-3

  H-
No Reaction

NH+3
NH+3
 Chem i str y | 26.65

No reaction, as H+ is less acidic than ‒SO3H group. Trans-Polyisoprene can be formed but is generally not
found naturally.
(d) Since pKa of H3N+CH2COOH is less than pKa of
RCH2COOH, it is more acidic
Sol 27: (A) PMMA (Poly (Methyl methacrylate)
Sol 21: (B, C, D) Polyethylene contains single bonds has excellent light transmission properties as it is
only. transparent and thus used for making lanses end light
covers.
Polyethylene : (CH2 – CH2)n
F F Sol 28: (D) Vinyl alcohol, H2C = CH – OH is unstable
Teflon : ( C C )n O
||
F F and converts to H3C − CH . Therefore, poly vinyl alcohol
(C) and (D) are correct is prepared by hydrolysis on polyvinyl acetate.

CH2OH CH2OH
O O
H H
Assertion Reasoning Type MeOH/HCl
H H H
D-Glucose +
OH H OH H
Sol 22: (A) Strach Maltose HO OMe HO
Amylase

H OH H O
Starch contains glycosidic linkage which are broken
down by amylase. Methyl--D (+) Methyl--D
glucopyranoside glucopyran

Sol 23: (C) On hydrolysis, sucrose (θ = 66.47°) gives

CH2OH CH2OH
equal amount of glucose (θ = 52.5°) and fructose O O
(‒92.4°). Thus total specific rotation of mixture is H H H H H OMe
D-Glucose MeOH/HCl
‒39.9° which makes it laevorotatory, whereas the initial +
OH H OH H
solution was dextrorotatory. HO OMe HO H
Sucrose → Glucose + Fructose OH OH
H H
θ = 66.470 θ = 52.50 θ = −92.40
Methyl--D (+) Methyl--D (+)
glucopyranoside glucopyranoside
Sol 24: (C) A is correct
Helix is stabilised via intramolecular
R is wrong as α − Helix Sol29: (B) It is a polyamide
hydrogen bonding between -CO of one amino acid
H
and -NH group of fourth amino acid O
N O

+ N (CH2)6 C
Sol 25: (C)
H
n
Carbonation form styrene Caprolactum
Carbocation form styrene Nylon-6

+
propylene CH3 -HC-CH3 (ii)
Carbocation from prokylene Comprehension Type

(i) is more than (ii) as it is resonance stabilised through Paragraph 1:

conjugation with benzene ring.
Sol 30: (A) Elastomers have weakest forces of attraction
e.g. rubber.
Sol 26: (C) Natural rubber is all cis-polyisoprene.

CH3 H2C Sol 31: (A) Thermoplastics usually have a linear

CH2= C HC = C structure which allows them to be repeatedly softened
H2C - H2O H2C (or hardened) by an increase (decrease) in temperature.
n
2 6 . 6 6 | Biomolecules and Polymers

Sol 32: (D) Thermosetting polymers are hard due to Previous Years’ Questions
three dimensional network of bonds.
Sol 1: (A) Cellulose is biopolymer of
Paragraph 2:
β -D –glucopyranose as:
Sol 33: (A) Rubber on reaction with sulphur forms
CH2OH CH2OH CH2OH
vulcanized rubber which is shift and resistant to action
O O O
of common solvents and wear and tear. H
H O H
H
O H
H
O

OH H H OH H OH H H
H
Sol 34: (C) This process is called vulcanisation. O
H OH H OH H OH
Match the Columns cellusose
(CH3CO)2O
+
Sol 35: (B) Nucleic acids are polynucheotides uracil is H

found only in RNA. CH2OAc CH2OAc CH2OAc

Thymine is found only in DNA, DNA has double helix H
O O
H O
O H O O
structure. H H H
OAc H H OAc H H OAc H H
O
Sol 36: (C) Pepsin is a digestive enzyme Nucleic acid
H OH H OAc H OAc
contains genetic material Ascorbic acid is chemical
name for vitamin C. Testosterone is a sex hormone. Tri-acetylated cellulose

Sol 37: (B) Bakelite is copolymer of phenol and Sol 2: (B) Cellulose and nylons have H-bonding type
methanol dacron is copolymer of 1, 2-dihydroxyethane of intermolecular attraction while poly (vinyl chloride)
and dimethylterephthalate. is polar. Natural rubber is hydrocarbon and has the
weakest intermolecular force of attraction, ie, van der
Nylon-66 is a copolymer of 1, 6-hexanedioic acid and 1, Waals’ force of attraction.
6-diaminol hexane.
(o) Buna-S is copolymer of butadiene and styrene. Sol 3: (A) The six-membered cyclic ether is known
as pyranose while the five membered cyclic ether is
Sol 38: (C) known as furanose. Hence, ring (a) is a pyranose and
it has ether linkage at α-Position that is known as
Polymer Monomer α- glycosidic linkage in carbohydrate chemistry.
Bakelite Phenol and formaldehyde
Terylene Terephthalic acid and ethylene glycol Sol 4: (B) Here, the –OH of hemiacetal group is equatorial
Nylon-6 Caprolactam therefore, it is a β -pyranose of an aldohexose.
Synthetic rubber Butadiene and styrene
Sol 5: (B, C) X is acetal, has no free hemiacetal, hence
Sol 39: (B) a non-reducing sugar while Y has a free hemiacetal
Polymer Monomer group, it is reducing sugar. Also, glucosidic linkage of X
is ‘ α ’ while that of Y is β -linkage.
Nylon-66 Hexamethylene diamine + adipic acid
Bakelite Phenol and formaldehyde Sol 6: (B, C)
Glyptal Phthalic acid and ethylene glycol (A) False (B) Factual
Dacron Terephthalic acid and ethylene glycol. H
(C) C12H22O11 + H2O →
+
C6H12O6 + C6H12O6
Invertase
Sucrose D-Glucose D-Fructose

Net specific Rotation of an equimolar mixture of

52 − 92 −40
Invert = = = −20
2 2
 Chem i str y | 26.67

Sol 7: A → p, s; B → q, r; C → p, r; D → s = 796 + 18 × 9= 958

(A) Cellulose 958 × 47
⇒ Mass of glycine in hydrolysis product
= = 450
100
O O O ⇒ Number of glycine molecule in one molecule of
O O 450
decapeptide
= = 6
75

(Glycoside linkage) Sol 10:

pH = 10
–
CHO CH2OH H2N CH COO H2N CH COOH
H OH C O
CH3 CH3
HO H HO H alanine

H OH H OH
pH=2 +
H2N CH COOH H3N CH COOH
H OH H OH
CH2OH CH3 CH3
CH2OH
Alanine
(B) Nylon 6, 6
H+
Sol 11: Sucrose  → D-glucose + D –fructose
H O
HO 2
N C
C N
Sol 12:
O
(Amide linkage)

(C) Protein
H O
R H

C N C
N C N
Sol 13:
R’ H
H O H O CH2C6H5
(Amide linkage) H2N CH C NH CH COOCH3
(D) Sucrose CH2 COOH
CH2OH
aspartame
O
H H H
O CH2OH
(i) Aspartame has amine, acid, amide and ester groups.
HO OH H O HO H O CH2C6H5
H +
H OH H
(ii) H3N CH C NH CH COOCH3
OH

(Glycoside linkage)
CH2 COO-

Sol 8: –COO- and –NH2 are basic groups in lysine. (iii) CH2C6H5
+
H
Aspartame H2N CH COOH + H2N CH COOH + CH3OH
H2O
Sol 9: A decapeptide has nine peptide (amide) linkage as - II
CH2COOH
Therefore, on hydrolysis, it will absorb nine water I
molecules.
(iv) II is more hydrophobic due to presence of phenyl
Hence, total mass of hydrolysis product group.
2 6 . 6 8 | Biomolecules and Polymers

Sol 14: Zeigler-Natta catalyst, which is a mixture of Phenyl alanine (P)

triethylaluminium “(C2 H5) 3Al” and TiCl4, is used as
Alanine (A)
heterogeneous catalyst in polymerization of ethylene.
So the number of possible sequence are 4.
Sol 15: (C)
Sol 20: lone pairs
C6H12O6 + Fehling solution → ( C6H11O7 ) + Cu2O ↓
−

(Red ppt. )

:
:
NH2 N NH2

Sol 16: (A) As in cellulose β 1 − 4 glycosidic linkage is

N N
present.

:
Sol 17: (D) Peptides with isoelectric point (pI) > 7, NH2

:
would exist as cation in neutral solution (pH = 7). Melamine
IV, VI, VIII and IX Sol 21: This peptide on complete hydrolysis produced
4 distinct amino acids which are given below:
Sol 18: O O
(1) H2N CH2 C OH (2) HO C NH2
6
Fixed in D-configuration Glycine
1 CHO CH2 OH
(natural)
H
2
H
5* O CH2
3 H H H
H OH 4 * * 1
4
H OH HO H H OH
5 *
H OH 3 2 O
6 OH H O
OH O
H2C C OH
(3) HO C (4) HO C NH
Hence total number of stereoisomers in pyranose form NH2
3
= 2=
of D-configuration 8
CH2

Sol 19: Because –COOH group of tetrapeptide is

intact on alanine, its NH2 must be participating in
Only glycine is naturally occurring amino acid.
condensation.
∴ Alanine is at one terminus, – – –A.
Sol 22: (A)
To fill the 3 blanks, possible options are:
CHO CHO
(i) When NH2 group attached to non chiral carbon
H OH HO H
G V P
OH H H OH
G P V
H OH HO H
(ii) When NH2 group attached to chiral carbon
H OH HO H
V G P P V G
CH2OH CH2OH
V P G P G V
D(+)glucose L(-)glucose
where, Glycine (G)
Valine (V)
 Chem i str y | 26.69

Sol 23: (B)

 
CH3 CH3 CH3 CH3
H2O
Cl Si Cl HO Si OH H O Si O Si O H
CH3 CH3 CH3 CH3 n

Me3SiCl, H2O

 
Me CH3 CH3
Me
Me Si O Si O Si O Si Me
Me Me
CH3 CH3 n

Sol 24: (A)

n CH2 = C CH = CH2 ( CH2 CH = CH2)n

CH3 CH3
Natural Rubber

Sol 25: (C, D)

(a) When X = COOCH3

H2 /Ni HOOC—(CH2)4 —COOH
CH3OOC—(CH2)4 —COOCH3 HOCH2 —(CH2)4—CH2OH+2CH3OH
heat heat

O O

— O—(CH2)6—O—C—(CH2)4—C —
n
Ester, condensation polymer

(b) When X = CONH2

H2 /Ni HOOC—(CH2)4 —COOH

H2NOC — (CH2)4 — CONH2 H2N — (CH2)6 — NH2
heat heat

O O

—HN—(CH2)6—NH—C—(CH2)4—C —
Nylon, condensation polymer n

Br2 HOOC—(CH2)4 —COOH

(c) H2NOC — (CH2)4 — CONH2 H2N —(CH2)4 — NH2
NaOH heat
Hofmann’s bromamide reaction

O O

—HN—(CH2)4—NH—C—(CH2)4—C —
Nylon, condensation polymer n

(d) When X = CONH2

H2 /Ni HOOC—(CH2)4 —COOH

NC—(CH2)4 —CN H2N —(CH2)6 — NH2
heat heat

O O

—HN—(CH2)6—NH—C—(CH2)4—C —
Nylon, condensation polymer n