JEE - 2014

XII

BIOMOLECULES
 CONTENT

S.No Pages

1. Carbohydrates 1 – 34
2. Exercise-1 (Objective Questions) 35 – 47
3. Exercise-2 (Subjective Questions) 48 – 48
4. Exercise-3 (Section-A) 49 – 52
5. Exercise-3 (Section-B) 53 – 54
6. Exercise-4 (Section-A) 55 – 57
7. Exercise-4 (Section-B) 57
8. Answer Key 58 – 67

JEE - 2014
XII

BIOMOLECULES
 BIOMOLECULES

CARBOHYDRATES
Carbohydrates (hydrates of carbon) are naturally occuring compounds having general formula Cx(H2O)y,
which are constantly produced in nature & participate in many important bio-chemical reactions.
Ex. Glucose C6H12O6 C6 (H2O)6
Fructose C6H12O6 C6 (H2O)6
Cellulose and Starch (C6H10O5)n
Sucrose (Cane sugar) - C12H22O11, and Maltose (Malt Suger) C12(H2O)11
But some compounds which have formula according to Cx(H2O)y are not known as carbohydrate
Ex. CH2O Formaldehyde
C2(H2O)2 Acetic acid
C3(H2O)3 lactic acid

There are many compounds, which shows chemical behaviour of carbohydrate but do not confirm the
general formula Cx(H2O)y such as - C5H10O4 (2–deoxyribose), C6H12O5 (Rahmnose)
C7H14O6 (Rahmnohexose)
Modern definition of carbohydrate: Carbohydrates are polyhydroxy aldehyde or ketone
or
Substances which yield these (polyhydroxy aldehyde or ketone)
on hydrolysis
Carbohydrates H / H 2O Poly hydroxy aldehyde or ketone
Carbohydrates are also known as Saccharides.

CLASSIFICATION OF CARBOHYDRATES

Carbohydrates

Sugar Non Sugar (Polysacch.)

Sweet in taste, crystalline & soluble in water taste less, amorphous solid
(non crystalline) insoluble in
water
Monosaccharides Oligosaccharides These yields many sweet
monosacch. on acidic hydrolysis
Tetra (C6H12O5)n+nH2O nC6 H12 O6
Aldose Ketose Disacch. Trisacch.
Sucrose, maltose Refinose
lactose

Monosaccharides : (simple sugars)

These are the sugars which cannot be hydrolysed into smaller molecules. General formula is CnH2nOn .
Ex. - Glucose, Fructose, Ribose
Oligosaccharides :
These are the sugars which yields 2–10 monosaccharides units on hydrolysis. Such as.
(a) Disaccharides : Two monosaccharide unit on hydrolysis (may or may not be same).
Ex. - Sucrose, Maltose
(b) Trisaccharides : Three monosaccharide unit on hydrolysis.
Polysaccharides : These are the non sugars which yield a large no of monosaccharide units on
hydrolysis General formula - (C6H10O5)n. Ex.- Starch, Cellulose.

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Note :- A group of polysaccharides which are not so widely used in nature is pentosans (C5H8O4)n
Monosaccharides, General formula Cx(H2O)y x = 3 – 8. Nomenclature of monosaccharides are given
according to the no. of carbons present in them.

If –CHO group is present in monosaccharide, then it is known as aldose.

If C group is present in monosaccharide , then it is known as ketose.
||
O

Aldoses Ketoses
3C Tropose or Triose Aldotriose Ketotriose
4C Tetrose Aldotetrose Ketotetrose
5C Pentose Aldopentose Ketopentose
H C O

5C including –CHO Aldopentose (Ribose) H—C—OH

H—C—OH
H—C—OH
CH2 OH

5C including C Ketopentose (Ributose)

||
O

6C Hexose Aldohexose Ketohexose

CHO
|
H C OH
|
HO C H
|
6C including –CHO Aldohexose (Glucose) H C OH
|
H C OH
|
CH 2OH

D glucose

CH 2 OH
|
C O
|
HO C H
|
H C OH
6C including C Ketohexose (Fructose) |
|| H C OH
O |
CH 2 OH

D fructose

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STEREOCHEMISTRY OF CARBOHYDRATES :

D & L-Sugars : The series of aldoses or ketoses in which the configuration of the penultimate C-atom
(C-next to CH2–OH group) is described as D-sugars if –OH is towards RHS & L-sugars if it is towards
LHS.

Smallest carbohydrate Aldotriose CHO Glyceraldehyde

|
CH OH
|
CH 2OH

Fischer projection

Fisher formula of compounds containing many asymmetric carbon.

a5 b5 a3 b3 CHO
a2 b2
CHO b3 a3
a4 b4
CH2OH b5 a5
a4 b4 a2 b2
CH2OH

Classification of Aldotetros : (i) Erythrose

(ii) Threoese
C-4

C-5

No. of C* = 3 (inAldopentose)
No. of optical isomers 23 = 8
No. of D Sugars 4
No. of L Sugars 4
D-Aldopentose :

All Isomeric D-sugars are diastereomers.

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Aldohexose :

No. of C* = 4
No. of stereoisomers = 24 = 16
No. of D-sugars = 8
No. of L-sugars = 8

The D–family aldoses

CHO
|
H C OH
|
CH 2OH

D-glyceraldehyde
CHO CHO
| |
H C OH HO C H
| |
H C OH H C OH
| |
CH 2OH CH2OH

D-erythrose D-threose

CHO CHO CHO CHO

| | | |
H C OH HO C H H C OH HO C H
| | | |
H C OH H C OH HO C H HO C H
| | | |
H C OH H C OH H C OH H C OH
| | | |
CH 2OH CH2OH CH2OH CH 2OH

D-ribose D-arabinose D-xylose D-lyxose

CHO CHO CHO CHO CHO CHO

| | | | | | CHO CHO
H C OH HO C H H C OH HO C H H C OH HO C H | |
| | | | | | H C OH HO C H
| |
H C OH H C OH HO C H HO C H H C OH H C OH HO C H HO C H
| | | | | | | |
H C OH H C OH H C OH H C OH HO C H HO C H HO C H HO C H
| | | | | | | |
H C OH H C OH H C OH H C OH H C OH H C OH H C OH H C OH
| | | | | | | |
CH 2 OH CH 2 OH CH 2 OH CH 2 OH CH 2OH CH 2OH CH 2 OH CH 2OH

D( ) allose D( ) altrose D( ) glucose D( )mannose D( )gulose D( )idose D( ) galactose D( ) talose

Epimers: Apair of diastereomers that differ only in the configuration about of a single carbon atom are
said to be epimers. D(+)-glucose is epimeric with D(+)- mannose and D(+)-galactose as shown below.

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Epimers Epimers

CHO CHO CHO

| | |
H – C – OH H – C – OH HO – C – H
| | |
HO – C – H HO – C – H HO – C – H
| | |
HO – C – H H – C – OH H – C – OH
| | |
H – C – OH H – C – OH H – C – OH
| | |
CH2OH CH2OH CH2OH

D(+) galactose D(+) glucose D(+) mannose

C4 is epimeric carbon C2 is epimeric carbon

Another example with C2 epimeric carbon is

&

Anomers:Anomers are the stereoisomers which differs at a single chiral centre out of many & are ring
chain tautomer of the same open chain compound.
The two sugars that differs in configuration only on the carbon that was the carbonyl carbon in the open
chain form is called as anomers glucose and glucose are known as anomers their equilibrium mixture
contains 36 % –D–glucose , 63.8 % -D-glucose and 0.2 % open chain form.
C1 Carbon is known as anomeric carbon.

Haworth suggested to write glucose and glucose in pyran structure

O
CHO
HO H OH HO H
H
HO HO H H OH
H O
H O H OH HO H
HO
OH H OH H OH
H
H H
CH2OH CH2OH CH2OH

–D–glucose D–glucose (open structure) –D–glucose

(sp rotation 112°) (sp rotation 19°)

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CH2OH CH2OH
O O
H H H H H OH
HO OH H OH HO OH H H
H OH H OH
Haworth formula Haworth formula
–D–glucopyranose –D–glucopyranose

Anomers are epimers but epimers may not be anomers.

CYCLIC STRUCTURES OF MONOSACCHARIDES

Many five membered and six membered monosaccharides occur in cyclic form. Cyclic structures of
monosaccharides are established by many experiments. The cyclic structure is due to intramolecular
hemiacetal formation between aldo / keto group and OH of any one carbon. The ring formed are
generally six membered (pyranose) or five membered (furanose). Each cyclization results in creation of
a new asymmetric centre apart from the existing ones. The isomers resulting from cyclizations are called
anomers. Example, when D-glucose (open structure) cyclise, it gives -D-glucose and -D-glucose.

Haworth projection :
Many of monosaccharides form cyclic structures. The actual structure is almost planer and be represented
by Haworth projection, which is a way of depicting three - dimensional cyclic structure.
Rule -1 : In a Haworth projection draw a fisher projection in which ring oxygen is in a down position.
Rule -2 : Imagine that carbon chain of fisher projection is folded around a barrel or drum, which
provide a ring lies in a plane to the page.
Rule -3 : Now plane of ring is turned 90° so that anomeric carbon is on the right and the ring oxygen is
in the rear. Obtained projection is a Haworth projection.
Example : (D-glucose)
OH H
OH H H OH H
CHO OH CH2OH

CH2OH OH
CHO
H OH H OH
H
OH H

Projection :
H CH2OH H CH2OH
HO HO

H OH H O

H H
CHO
OH
OH OH
H
OH H OH H

Hawarth projection
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Chair conformation of D–glucose

CH2OH
O H
H H
CH2OH H HO OH
HO H
H
O
H HO H
HO HO H
H OH
OH
Chair forms of (conformation) and D-Glucose :
CH2OH
CH2OH O
OH OH OH
HO anomeric O
OH carbon
HO OH
OH
-D-Glucose (most stable glucose form) all groups are equatorial.
CH2OH
CH2OH O OH OH
HO anomeric O OH
HO carbon
HO OH
OH
-D-Glucose –OH group at anomeric carbon is axial.

Mutarotation

Specific rotation of glucose + 112° Specific rotation of glucose + 19°

Equilibrium mixture[ ]D = 52.5 degree mL g–1 dm–1

Fresh -glucose 52.5 Fresh -glucose

112° 36 % glucose 19°
63.8 % glucose
When pure -D glucose is dissolved in water its specific rotation is found to be +112° with time,
however the specific rotation of the solution decreases ultimately reaches stable value of +52.5°. When
D-glucose is dissolved in water, it has a specific rotation of 19°. The specific rotation of this solution
increases with time also to + 52.5 °.
This change of optical rotation with time is called mutarotation. It is caused by the conversion of and
glucopyranose anomers into an equilibrium mixture of both. Mutarotation is catalyzed by both acid
and base, but also occurs is even in pure water. Mutarotation is characteristic of the cyclic hemiacetal
form of glucose.

Mutarotation occurs first by opening of the pyranose ring to the free aldehyde form.

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120°

Structure of fructose

CH2–OH HO–C–CH2OH
HOH2C–C–OH
C=O HO–C–H
HO–C–H O
O HO–C–H H–C–OH
H–C–OH
H–C–OH H–C–OH
H–C–OH
H–C–OH H–C
H–C
CH2–OH H
H
fructose fructose (more stable)

Ring structure of fructose C, Pyranose structures 6 membered ring, C2 – C6 linkage

H
H C O CH2 OH
H
C H OH
OH C C OH
OH H
Chair conformations

-D-fructopyranose -D-fructopyranose

Furanose structure (5 membered ring)

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Mutarotation: Fructose undergo complex mutarotation. The structure of the cyclic hemiacetal form of
d-fructose can be derived from it’s carbonyl (Ketone) form using the methods described as follows.

CH2OH
C=O
HO H CH2OH
H OH C=O
H OH HO H
CH2OH H OH
H OH This oxygen is
This structure is involved in
CH2OH
involved in pyranose furanose formation
formation

It happens that the crystalline form of D-fructose is -D-Fructopyranose. When crystals of this form are
dissolved in water, it equilibrates to both pyranose & furanose forms.

HO
H2C–C–OH HOH 2C O OH O
anomeric HOH2C CH2OH
carbon
HOH2C–C–OH anomeric
HO–C–H O H HO
carbon
H CH2OH HO–C–H O H HO
H–C–OH H OH
H–C–OH
H–C OH H
H–C OH H
CH2OH -D-fructofuranose
CH2OH -D-fructofuranose
-D-fructose

* All monosaccharides are reducing sugars, hence they show mutarotation.

* Starch, cellulose are Polymers of Glucose
* Lactose and sucrose are disaccharides
* Sucrose is a non reducing sugar, gives negative test for Benedict and tollen's reagent, they do not form
osazone and do not show mutarotation.
* Acetals of carbohydrates are called as GLYCOSIDE
FORMATION OF GLYCOSIDES

Glucose reacts with methyl alcohol in presence of dry HCl to form and -methyl glycoside of glucose.
The reaction takes place only on OH of hemi-acetylic carbon. Other hydroxyl groups are unreactive.
H OCH3 CH3O H
C C
(CHOH)3 O (CHOH)3 O
CH CH
CH2OH CH2OH
–Methyl glucose –Methyl glucose

To methylate all the OH groups, methylating agent used is dimethyl sulphate.

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+ CH3OH HCl

Such compounds are called glucoside (cyclic acetals). They are special type of acetals in which one of
the oxygen of the acetal linkage is the ring oxygen of the pyranose or furanose.
H–C

Ring structure of glucose : H–C–OH

(i) Glucose does not give pink colour with schief reagent. HO–C–H O
(ii) Does not form adduct with NaHSO3, NH3 H–C–OH
(iii) Glucose exist in two isomeric form H–C
(iv) It show mutarotation CH2OH
Since there is no free aldehyde group, so it does not react with weak reagent (NH3, NaHSO3) but
strong reagent (HCN, NH2OH, C6H5NH – NH2) break up ring

Hydrolysis
H 2O

REACTIONS OF GLUCOSE

COOH
(CHOH)4 Br2 / H2O Red P / HI n Hexane
CH 2OH or Alkaline
Gluconic acid solution of I2 CH3COCl
Pentaacetate
+
COONH 4 (5-OH group)
(CHOH)4 Tollen’s NH2–OH (1 eq)
C6H12 O6 Oxime
Salts of CH 2OH
gluconic acids HCN
COONa Glucose cyanohydrin
Fehling’s
(CHOH)4
5HIO 4
CH 2OH HCHO + 5HCOOH

COOH Schief's reagent, NaHSO 3

HNO3 No reaction
(CHOH)4 NH3
COOH
(Saccharic acid)

These reactions indicate that glucose has 6-C straight chain with one –CHO group & 5-OH group.
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GENERAL REACTIONS OF MONOSACCHARIDES

1. Acetylation :

+ 5CH3COCl / (CH3CO)2O + 5HCl

This reaction suggests presence of 5 ( OH ) group.

Q. The penta acetate of glucose give –ve test with Tollen’s reagent & Fehling solution, explain?
No hemiacetal linkage, So no
CHO group in aq.
alkane solution –ve test
Tollen's –ve
Sol. 5 AcCl
Fehling
–ve
Benedict
–ve
Shifts –ve
NaHSO 3
–ve

2. Red by HI / Red P:

Re d P / HI Re d P / HI
n-Hexane n-Hexane

3. Reaction with HCN:

+ HCN + HCN

4. Reaction with NH2–OH (hydroxyl amine):

+ NH2–OH + NH2OH

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5. Reaction with phenyl hydrazine: Both glucose and fructose give “osazone”.
Reaction with glucose :

+ 3 C6H5 NH – NH2 Osazone

Mechanism :

O O H
+ HC– N–NHPh
HC HC–NH2NHPh HC= N–NHPh
HC–NH–NH–Ph
NH2NHPh IMPE
CHOH CH–OH CH–OH C – OH + +
–H2O –H C O H
R R R R
R
+
–H

HC=NNHPh HC=NH
2PhNHNH2
C NNHPh –NH3 C O

R R

Both compounds give same product because structure of last four C is same in both (glucose & fructose)
Only C-1 and C-2 in glucose and fructose are involved in osazone formation addition reaction do not
run through out the chain. The failure to undergo further reaction has been explained by stabilization of
the osazone by chelation.

Osazone :
NHPh
HOH2C O Chelation
N
H
N
HO N Ph
OH

or

So we do not get hexaphenyl hydrazone.

6. Catalytic reduction:

CH2OH CH2–OH

(CHOH)4 HO–C–H
H 2 / Ni or H 2 / Ni or
D-sorbital + (CHOH)3
NaBH 4 CH2OH NaBH 4
D-sorbital CH2OH
D-Manitol

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7. Oxidation:
Tollen 's ,Fehling
Glucose Glucose dil.HNO3
Br2 water

Glucose HIO4 5HCOOH + HCHO + 5HIO3

Oxidation of fructose :
Fructose Tollen's,Fehling
Fructose reduces tollen’s & fehling reagent because in basic medium fructose isomerises to glucose.
Fructose Br2 water No reaction

Fructose HNO3 + +

8. Reaction with enzyme:

Yeast
Glucose or Fructose 2C2H5OH + 2CO2
Zymase enzyme

9. Reaction with dil NaOH / Ca(OH)2

Glucose or fructose dil NaOH Glucose + Fructose + Mannose
H–C–OH

C–OH

Base-catalyzed isomerisation of aldoses and Ketoses:

Although glucose in solution exists mostly in its cyclic hemiacetal forms it is also in equilibrium
with a small amount of it’s acyclic aldehyde form.

0.02 M
Ca ( OH ) 2
+ + + Traces of other compound

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Mechanism : Like other aldehyde with -hydrogen, glucose ionise to give small amount of its inolate
ion in base. Protonation of this enolate ion at one face of the double bond gives back glucose & protonation
at the other face gives mannose.
CH=O
CH–O–
CHO CH= O
HO–C–H
H OH OH C–OH
OH— H2O HO–C–H
HO H HO H HO H
H OH H OH H OH H–C–OH
H OH H OH H OH H–C–OH
CH2OH CH2 OH CH2OH CH2OH
D-mannose
The enolate ion can also be protonated on oxygen to give a new enol called enediol this enediol
converts to fructose as follows.

H–C=O H–C=O H–C–O

H OH OH – C–OH C–OH H–OH

CH2–OH
H–C–OH H–C–OH H–C–OH
C=O HO–H C=O C–O OH–
C–OH
Fructose
enediol
Method of ascending the sugar series : An aldose may be converted into it’s next higher aldose eg.
an aldopentose into an aldohexose.

SOME IMPORTANT CARBOHYDRATES

1. Sucr ose (C12H22O11) : It is white, crystalline & sweet substance soluble in water obtained from the
sugar cane. When heated above its melting point, it forms a brown substance known as caramel.
It's aqueous solution is dextrorotatory [ ]D = 66.5°
dextro laevo
H+
C12H22O11 + H 2O C6H12O6 + C6H12O6
D-Glucose D-Frucose
[D] = 52.7° [ ]D = –92.4°

mixture is laevorotatory
(–)
Thus hydrolysis of sucrose brings about a change in the sign of rotation, from dextrol (+) to leavo (–) &
such a change is known as inversion of cane sugar and the mixture is known as invert sugar.
The inversion of cane-sugar may also be effected by the enzyme invertase which is found in yeast.
Sucrose is non-reducing sugar because it has stable acetal linkage & in aq. solution it can not give free
carbonyl group and so it does not reduces Tollen's & Fehling's solution.
This indicates that neither the aldehyde group of glucose nor the ketonic group of fructose is free in
sucrose.

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anomeric C
6 1
CH 2OH CH2OH
O O
5 H
H H H 2 5
4 1 CH2OH
HO OH H O H HO
3 2 3 4
H OH HO H
–D–Glucopyranose –D–Fructofuranose

Structure of sucrose
( –D–glucopyranosyl– –D–fructofuranoside)

All non reducing sugars do not show mutarotation .

There is no free carbonyl group so it is non reducing sugar
-glycoside bond to anometric O
O CH2OH
position of D-glucose H
HOH2C
HO OH linkage
HO CH2OH O at glucose
OH HO
HO OH O
HO O linkage
O CH2OH H HO
b-Glycoside at fructose
bond to anomeric H CH2OH
positon of D-fructose CH2OH OH H

2. Maltose: It is obtained by partial hydrolysis of starch by the enzyme diastase present in malt i.e.,
sprouted barely seeds.
2 (C6H10O5)n + n H2O Diastase n C12H22O11
Starch Maltose
As stated above, hydrolysis of one mole of maltose yields two moles of D-glucose. Maltose is a reducing
sugar since it forms an osazone, undergoes mutarotation and also reduces Tollen’s and Fehling’s solutions,
Methylation studies have revealed that
(i) both glucose units are present in the pyranose form.
(ii) C1 of one glucose unit is linked to C4 of the other
Further since maltose is hydrolysed by the enzyme maltase which specifically hydrolyses -glycosidic
linkage, therefore, the non-reducing glucose unit in maltose must be present in the -form. In other
words, C1 – of non-reducing glucose unit is attached to C4 of the reducing glucose unit as shown in the
figure on next page.

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3. Lactose (Milk sugar) C12H22O11

Lactose occurs in milk and that is why it is called milk sugar.
Lactose on hydrolysis with dilute acid or by enzyme lactase, yields an equimolar mixture of D-glucose
and D-galactose. It is a reducing sugar since it forms an osazone, undergoes mutarotation and also
reduces Tollen’s or Fehling’s solution. Methylation studies have revealed that
(i) both glucose and galactose are present in the pyranose form.
(ii) glucose is the reducing half while -galactose is the non-reducing half.
(iii) C1 of galactose unit is connected to C4 of glucose unit.

Further since emulsin, as, enzyme which specifically hydrolyses -glycosidic linkages also hydrolyses
lactose, therefore, galactose must be present in the -form. In other words, in lactose, C1 – of galactose
is attached to C4 of glucose as shown in figure.

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

4 H H OH
OH H
H GLUCOSE
3 H (Reducing half)
2

H OH
GALACTOSE
(Non-reducing half)

4. Starch Amylum, (C6H10O5)n

Occurrence : The value of n (100 – 3000) varies from source to source. It is the chief food reserve
material or storage polysaccharide of plants and is found mainly in seeds, roots tubers etc. Wheat,
maize, rice, potatoes, barley, bananas and sorghum are the main sources of starch. Starch occurs in the
form of granules, which vary in shape and size depending upon their plant source.
Occurs in all green plants. Starch consists of two fractions, one being known as –amylose, which gives
blue colour with iodine. This blue colour is believed to be due to the formation of an inclusion complex.
An aqueous solution of –amylose slowly forms a precipitate, since -amylose has a strong tendency to
'revert' to the insoluble state in solution. Amylopectin is insoluble in water and is stable towards both
hydrolysis to maltose by the enzyme diastase and to D(+)-glucose by dilute acids (amylopectin gives
about 50 percent of maltose).
6 6
CH2OH 6
CH 2OH CH2OH
5 O O
H H H H 5 5 O
H H H H H
4 1 4 1
HO O O
4 1
OH H OH H
3 2 2
OH H OH
3 3 2
H OH H OH n–2 H OH
–D–glucopyranose –D–glucopyranose

Structure of Starch ( –D–glucoamylose)

–amylose consists of an unbranched chain, with a molecular weight varying between 10,000(n 60)
and 10,00,000(n 6,000), The value of n depends on the source and treatment of –amylose.

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Properties : (i) Starch is a white amorphous powder sparingly soluble in water. Its aqueous solution
gives a blue colour with iodine solution due to the formation of an inclusion complex. The blue pears on
cooling.
(ii) On hydrolysis with dilute mineral acids or enzymes, starch beaks down first to smaller molecules
(n > n’), then to maltose and finally to D-glucose.
H / H 2O
(C6H10O5)n H / H 2O (C6H10O5)n’ H / H 2O C12H22O11 C6H12O6
or Maltase
Starch Maltose D-Glucose
(iii) Starch is a non-reducing saccharide. It neither reduces Tollen’s reagent or Fehling’s solution nor
forms an osazone. This suggests that all hemiacetal OH groups of glucose units at C1 are not free but are
involved in glycosidic linkages.
Composition : Starch is not a single compound but is a mixture of two components–a water soluble
component called amylose (10-20%) and a water insoluble component called amylopectin (80-90%).
Both amylose and amylopectin are polymers of -D-glucose.

Structure of amylose :Amylose is water soluble and gives blue colour with iodine solution. It may have
100-3000 glucose units, i.e., its molecular mass can vary from 10,000 to 500,000. It is a linear polymer
of -D-glucose in which C1 of one glucose unit is attached to C4 of the other through -glycosidic
linkage as shown in figure.

Pectins
Pectins are found in plant and fruit juices. Their characteristic property is the ability of their solutions to
gelate, i.e. form jellies. They have a high molecular weight and are polygalacturonic acid (linked 1,4) with
the carboxyl groups partially esterified with methanol.
Glycogen (C6H10O5)n :
Glycogen is found in nearly all animals cells, occurring mainly in liver. It is the reserve carbohydrate of
animals and so is often known as 'animal starch'. It has also been isolated from plant sources.
Glycogen is a white powder, soluble in water, the solution giving a purplish-red colour with iodine. On
hydrolysis with dilute acid, glycogen gives D(+)-glucose. The molecular weight of glycogen has been
given as 10,00,000 to 50,00,000 and glycogen contains highly branched chains. Glycogen has a structure
similar to amylopectin, except that it has more cross-linking.
5. Cellulose:
Cellulose is colourless, solid which is insoluble in water & organic solvents. But it is soluble in ammonical
cupric hydroxide (Schweizer's reagent) or in conc. HCl cellulose is a regular polymer of d-glucopyranose
residues connected by -1,4 glycosidic linkages. It is straight chain polymer.
6 6
CH2OH CH2OH
5 O H 5 O
H H H
4 1 O 4 1 O
O
OH H OH H
3 2 H 3 2 H
H OH H OH
–D–glucopyranose –D–glucopyranose

Structure of Cellulose

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Some points about cellulose :

1. General empirical formula (C6H10O5)
H
2. Cellulose + H2O 96% of crystalline D-glucose
3. No. of monomer units in cellulose are 1000 – 1500 in one molecule.
4. Cellulose doesn't show mutarotation (like starch)
5. It is non reducing sugar because there is no hemiacetal linkage.
6. Acetylations nitration & methylation of cellulose give trisubstituted cellulose which suggest that only three
– OH groups are free.

Tests for carbohydrates:

(i) When heated in a dry test tube, it melts, turns brown and finally black, giving a characteristic smell of
burning sugar.
(ii) When warmed with a little concentrated H2SO4, it leaves a charred reside of carbon.
(iii) Molisch’s Test (named afterAustrian botanist Hans Molisch) is a sensitive chemical test for the presence
of carbohydrates, based on the dehydration of the carbohydrate by sulfuric acid to produce an aldehyde,

thymol) also give colored products) resulting in a red- or purple-colored compound.

in a test tube. After mixing, a small amount of concentrated sulfuric acid is slowly added down the sides
of the sloping test-tube, without mixing, to form a bottom layer. A positive reaction is indicated by
appearance of a purple ring at the interface between the acid and test layers.

All carbohydrates — monosaccharides, disaccharides, and polysaccharides — should give a positive

reaction, and nucleic acids and glycoproteins also give a positive reaction, as all these compounds are
eventually hydrolyzed to monosaccharides by strong mineral acids. Pentoses are then dehydrated to
furfural, while hexoses are dehydrated to 5-hydroxymethylfurfural. Either of these aldehydes, if present,
will condense with two molecules of naphthol to form a purple-colored product, as illustrated below by
the example of glucose:

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BARFOED'S TEST
Barfoed's Test is a chemical test used for detecting the presence of monosaccharides. It is based on
the reduction of copper (II) acetate to copper (I) oxide (Cu2O), which forms a brick-red precipitate.
Barfoed's reagent consists of a 0.33 molar solution of neutral copper acetate in 1% acetic acid solution.
The reagent does not keep well and it is therefore advisable to make it up when it is actually required.
Reducing monosaccharides are oxidized by the copper ion in solution to form a carboxylic acid and a
reddish precipitate of copper (I) oxide within three minutes. Reducing disaccharides undergo the same
reaction, but do so at a slower rate and ppt. will come after 10 min. For Non reducing saccharides ppt.
will not form.
H O HO O
C +2 C +
+ 2 Cu + 2 H2O + Cu2O + 4H
R R
The aldehyde group of the monosaccharide which normally forms a cyclic hemiacetal is oxidized to the
carboxylate. Anumber of other substances, including sodium chloride may interfere.

STARCH

Plants store glucose as the polysaccharide starch. The cereal grains (wheat, rice, corn, oats, barley) as
well as tubers such as potatoes are rich in starch.
Starch can be separated into two fractions--amylose and amylopectin. Natural starches are mixtures of
amylose (10-20%) and amylopectin (80-90%).

Iodine - KI Reagent: Iodine is not very soluble in water, therefore the iodine reagent is made by
dissolving iodine in water in the presence of potassium iodide. This makes a linear triodide ion complex
with is soluble. The triodide ion slips into the coil of the starch causing an intense blue-black color.

AMINO ACIDS AND PROTEINS

NH 2
|
Bifunctional compounds R CH COOH having an acidic corboxylic group & a basic amino group.
There are 20 amino acids commonly found in proteins and are standard amino acids. All are amino
acids. Most of them have 1° amino group.(– NH2). However proline is a 2° amino

CH2 —COOH Me—CH—COOH Ph—CH2—CH—COOH N COOH

NH2 NH2 NH2 H

Proline [P]
(Gly) Glycine [G] Alanine [A] Phenyl alanine [F] Proline [P] ( -imino acid)
All amino acids are chiral molecules with atleast one chiral carbon except glycine, H3N CH2COO¯.
All other amino acids are optically active & can be assigned D & L configuration.
\\\\\\\\\\\\\\\\\\\\

COOH COOH
H NH2 H2N H
CH3 CH3
D-Alanine L-Alanine

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 BIOMOLECULES

Classification of Amino acids

Occurance Requirement Chemical Nature

Natural Essential Neutral
Synthetic Non Essential Acidic

Semi Essential Basic

Based on requirement
1. Essential amino acids can not be synthesized in human body so dietary intake is required. For any human
being 1 gm a day is required.
2. Semi essential amino acids can be synthesized in human body but dietary intake is required during
growing stages (when more of cell division is required).
For example : Early childhood, pregnancy and lactating mother.
3. Non essential amino acid - Body can synthesize them.

Chemical classification
Neutral - Amino acid having equal number of NH2 and COOH.
Neutral amino acids are further classified as polar and nonpolar depending on whether their side chains
have polar substituents (for example, asparagine with an NH2CO group) or are completely hydrocarbon
in nature (for example alanine, valine etc.).
Acidic - Amino acid having more COOH than NH2
Aspartic acid and glutamic acids, each with a second CO2H in their side chain are acidic amino acids.

Basic - Amino acid having more NH2 than COOH

Lysine, arginine and histidine, each with a basic site in their side chain are basic amino acids.

Proteins: The name protein is taken from the Greek word "proteios", which means "first". Of all
chemical compounds, proteins must almost certainly be ranked first, for they are the substance of life.
Proteins make up a large part of the animal body, they hold it together and they run it. They are found in
all livingcells.

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Chemically, proteins are high polymers. They are polyamides and the monomers from which they are
derived are the - amino carboxylic acids.Asingle protein molecule contains hundreds or even thousands
of amino acid units. These units can be of twenty-odd different kinds. The number of different combinations,
i.e., the number of different protein molecules that are possible, is almost infinite.

1. Neutral amino acids (with nonpolar side chains)

ISOELECTRIC
NAME ABBREVIATIONS STRUCTURAL FORMULAE
POINT[pI]

@Glycine Gly(G) 6.0

Alanine Ala(A) 6.0

Valine* Val(V)
6.0

Leucine* Leu(L) 6.0

CH 3 N H 3
| |
Isoleucine* Ile(I) CH 3CH 2 CH CHCO 2 6.0

N H3
|
Methionine* Met(M) CH 3SCH 2CH CHCO 2 5.7

@@Proline Pro(P) 6.3

Phenylalanine* Phe(F) 5.5

Tryptophan* Trp(W) 5.9

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2. Neutral amino acids (with polar, but nonionized side chains)

NAME ABBREVIATIONS STRUCTURAL FORMULAE ISOELECTRIC

POINT[pI]

Asparagine Asn(N) 5.4

Glutamine Gln(Q) 5.7

Serine Ser(S) 5.7

Threonine* Thr OH N H 3
| | 5.6
CH 3CH CHCO 2

Tyrosine Tyr(Y) 5.7

N H3
| 5.1
Cysteine Cys HSCH 2 CHCO 2

N H3 N H3
| |
| Cystine Cys–Cys
OOCCHCH 2S SCH 2CHCOO

3. Acidic amino acids (side chain with carboxylic acid group)

ISOELECTRIC
STRUCTURAL FORMULAE POINT[pI]
NAME ABBREVIATIONS

Aspartic acid Asp(D) 2.8

GlutamicAcid Glu(E) 3.2

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4. Basic amino acids (side chain with nitrogenous basic group)

NAME ABBREVIATIONS STRUCTURAL FORMULAE ISOELECTRIC
POINT[pI]

Lystine* Lys(K)
9.7

Arginine* Arg(R)
10.8

Histidine* His(H)
7.6

Note :
* Amino acids with an asterisk are essential amino acids, that must be supplemented through diet.
| At pH = 7, Asp and Glu have a net negative charge and exist as anions. At pH = 7, Lys and Arg have a
net positive charge and exist as cations. Rest of the amino acids at this pH exist in the neutral form.
| Structurally, in cystine, the two cysteine molecules are joined through sulfur (disulfide linkage).
@@ Proline is an -imino acid, all amino acids are primary amines except proline and 4-hydroxyproline,
which are 2° amines.
@ Except Glycine all other amino acids are optically active.

Preparation of amino acids

(a) Gabriel Phthalimide synthesis
Better yields are generally obtained by the gabriel phthalimide synthesis ; the - halo esters are used
instead of - halo acids .

O O
C C
+ – KCl
N¯ K + Cl–CH2COOC2H5 N– CH 2COOC2H5
C C
O O
Potasium
phthalimide HCl , H2O

+
Cl¯ H3N – CH2COOH + phthalic acid
Glycine hydrochloride

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(b) Amination of - Halo acids

Sometimes an - chloro or - bromo acid is subjected to direct ammonolysis with a large excess of
concentrated ammonia. For example.
Br2, P
CH3CH2COOH CH3CHCOOH NH3(excess) CH 3CHCOO¯
+
Br NH3
-Bromopropionic acid Alanine

(c) From diethyl malonate

COOC2H5 COOC2H5 COOH

C6H5 CH2Cl H3 O
Na
+ CH HC– CH 2C 6H 5 H–C– CH2C6H5

COOC2H5 COOC2H5 COOH

Sodiomalonic ester Ethyl benzylmalonate Benzylmalonic acid

Br2/ OH

COOH
NH 3(excess) heat Br–C– CH C H
C6H5CH 2CHCOOH C6H5CH 2CHCOOH 2 6 5
–CO 2
NH3+ Br COOH

Phenylalanine
35% overall yield

(d) Strecker's synthesis

Strecker's synthesis is also used for preparing - amino acids

+
O N–H N–H NH2 NH2
+
–H2O CN¯ H2 O H3 O
R–C–H + NH3 R–C–H R–C–H R–C–H R–C–H
Imine
CN CN COO¯
- amino nitrile - amino acid
In this reaction general aldehyde is treated with mixture of ammonium chloride and KCN in aqueous
solution which forms NH 3 and HCN, NH4Cl + KCN aqueous NH4CN + KCl,
NH4CN aqueous NH3 + HCN

(e) Using KOOP synthesis

Br NH NH2
O
HVZ DMSO NH3 SBH
CH3COOH CH2—COOH CH—COONa CH—COONa CH2
NaHCO3 Xs

COONa

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Properties of Amino acids

Although the amino acids are commonly shown as containing an amino group and a carboxyl group,
H2NCHRCOOH, certain properties, (both physical and chemical) are not consistent with this structure
I. Physical properties
In contrast to amines and carboxylic acids, the amino acids are nonvolatile crystalline solids, which melt
with decomposition at fairly high temperatures.
They are insoluble in non-polar solvents like petroleum ether, benzene or ether and are appreciably
soluble in water.
Their aqueous solutions behave like solutions of substances of high dipole moment.
Amino acids as dipolar ions
Acidity and basicity constant are ridiculously low for –COOH and –NH2 groups. Glycine, for example,
has Ka = 1.6 × 10–10 and Kb = 2.5 × 10–12, whereas most carboxylic acids have Ka values of about
10–5 and most aliphatic amines have Kb values of about 10–4.
All these properties are quite consistent with a dipolar ion structure for the amino acids (I)
+
H3N–CH–COO¯
R
Amino acids : dipolar ions (Zwitter ion)

Physical properties - melting point, solubility, high dipole moment - are just what would be expected of
such a salt. The acid-base properties also become understandable when it is realized that the measured
Ka actually refers to the acidity of an ammonium ion, RNH3+,
+ +
H3NCHCOO¯ + H2O H3O + H2NCHCOO¯
R R
Acid
+
[H3O ] [H2NCHRCOO ]
Ka = +
[H3NCHRCOO¯]

When the solution of an amino acid is made alkaline, the dipolar ion (I) is converted into the anion (II).
The stronger base, hydroxide ion, removes a proton from the ammonium ion and displaces the weaker
base, the amine.
+
H3NCHCOO¯ + OH¯ H2NCHCOO¯ + H2O
R (I) R (II)
Stronger Weaker Weaker
Stronger base acid
acid base

+
H
+
+ H +
H2NCHCOO¯ H3NCHCOO¯ H3NCHCOOH
OH¯ OH¯
R R R
(II) (I) (III)
Anion Ampholyte Cation

Wherever feasible, we can speed up a desired reaction by adjusting the acidity or basicity of the solution
in such a way as to increase the concentration of the reactive species.
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Isoelectric point of amino acids

What happens when a solution of an amino acid is placed in an electric field depends upon the acidity or
basicity of the solution.
+ +
H H
H2NCHCOO¯ H3NCHCOO¯ H3NCHCOOH
OH¯ OH¯
R R R
Ka 2 Ka 1
(II) A (I) DI (III) C

In quite alkaline solution, anions (II) exceed cations (III), and there is a net migration of amino acid
toward the anode. In quite acidic solution, cations (III) are in excess, and there is a net migration of
amino acid toward the cathode. If (II) and (III) are exactly balanced, there is no net migration ; under
such conditions any one molecule exists as a positive ion and as a negative ion for exactly the same
amount of time, and any small movement in the direction of one electrode is subsequently cancelled by
an equal movement back towards the other electrode. The hydrogen ion concentration of the solution
in which a particular amino acid does not migrate under the influence of an electric field is
called the isoelectric point (pI) of that amino acid. The isoelectric point (pI) is the pH at which
the amino acid exists only as a dipolar ion with net charge zero.
For glycine, for example, the isoelectric point is at pH 6.1.

An amino acid usually shows its lower solubility in a solution at the isoelectric point, since here there is
the highest concentration of the dipolar ion. As the solution is made more alkaline or more acidic, the
concentration of one of the more soluble ions, (II) or (III) increases.
[DI] [H ] [A ] [H ]
Ka1 = Ka2 = at pI [ A ] = [C ]
[C ] [DI]

[ DI] [ H ] Ka 2 [DI]
Ka1
=
[H ] [H ]2 = Ka1 × Ka2

P Ka1 P Ka 2
on taking antilog pI =
2
The pI of an amino acid that does not have an ionizable side chain. (e.g. alanine), is midway between its
two pKa values.
O

Alanine CH3–CH–C–OH pKa = 2.34

NH3 pKa = 9.69

2.34 9.69 12.03

pI = = = 6.02
2 2
If an amino acid has an ionizable side chain, its pI is the average of the pKa values of the similarly ionizing
groups (positive ionizing to uncharged or uncharged ionizing to negative.)
For example :
(i) Lysine
O pKa = 2.18
H3N –CH2 – CH2 – CH2 – CH2 – CH – C – OH

pKa = 10.79 NH3 pKa = 8.95

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10.79 8.95 19.74

pI = = 9.87
2 2
(ii) Glutamic acid

O O pKa = 2.19
HO–C–CH2 – CH2 – CH – C – OH

pKa = 4.25 NH3 pKa = 9.67

21.19 4.25 6.44

pI = = 3.22
2 2
An amino acid having more COOH than NH2 or more acidic COOH will have pI less than 7.
An amino acid having more – NH2 than COOH or more basic – NH2 will have pI more than 7.

Q. Write the structure of alanine at pH 2.5, 10.5 and 6.

Electrophoresis
The movement of charged molecules (like amino acid) under the influence of an electric field is called
electrophoresis. Electrophoresis separates amino acids on the basis of their pI values.
Amino acid is positively charged (moves towards cathode) if pH of the solution < pI
Amino acid is negatively charged (moves towards anode) if pH of the solution > pI

Q. How will you separate a ternary mixture of arginine, alanine & aspartic acid?
Ans. A few drops of a solution of an amino acid mixture are applied to the middle of a piece of filter paper.
When the paper is placed in a buffer solution (pH = 5) between the two electrodes and an electric field
is applied then arginine & alanine with pI > pH move towards the cathode and aspartic acid with pI < pH
moves towards the anode. Out of arginine & alanine, alanine will move slowly towards the cathode due
to lesser positive charge.
Before electrophoresis
After electrophoresis
– +
– +

Buffer solution
(pH = 5) A B C
Buffer solution
(pH = 5)

Mixture of arginine
+ alanine + aspartic acid Figure (ii)
Figure (i)

A = arginine (pI = 10.76) B = alanine (pI = 6.02) C = aspartic acid (pI = 2.98)

NH2 O OOCCH2CHCOO
|| CH3CHCOO
| |
H2NCNHCH2CH2CH2 CHCO NH3
| NH3
NH3

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General reactions of amino acids

(1) Reactions due to – NH2 group

H H
HBr COOH
COOH
NH3 Br
NH2 H CH2 —COOH
MeCOCl PhCOCl
COOH CH2 —COOH
NHCOMe NH2
H NH—C—Ph
NaNO2
COOH
HCl O
OH
Hipuric acid
H
NOCl COOH
Cl
H
NOBr COOH
Br

H H
HCHO
COOH COOH Sorison's titration method.

NH2 N = CH2
Reactions is used to block – NH2 group during volumetric analysis in.

(2) Reactions due to – COOH group.

H
H
COOH ROH COOR
H
NH3
NH2 H
NaOH
CaO/
NH2
NaHCO3
Na salt

H
LAH CH2OH
NH2
H
NH3 CONH2
NH2

(3) Heating Effect

(i) Heating of amino acids leads to intermolecular dehydration to form cyclic diamides.
O

CH2 –C–OH O
H
NH –2H2 O
NH NH N–H
H
CH2
HO –C O

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(ii) When alanine is heated, then two diastereomers are obtained. One of them (trans) is not resolvable.

O O
Me NH Me H NH Me

2MeCH(NH 2)COOH –2H O +

2
NH H NH H
H Me
O O
Cis (Racemic) Trans (Meso)

(iii) When - amino acids are heated, , - unsaturated salt are formed.

RCHCH2COOH RCH = CH–COONH 4

NH2

(iv) - amino acids when heated alone gives - lactam and polymer respectively. The reason for the
formation of polymer is that when - amino cyclises intramolecularly, it leads to large angle strain within
the compound
O O
O
O
OH Heat C– OH Heat
NH2 N–H + H2O N–H H2 O N–H

- lactam H - lactam

NH3(CH2)5COO –NH(CH2)5–C –NH(CH2)5–C–NH(CH2)5C–

O O O
Nylon 6
(4) Peptide
Peptides are amides formed by interaction between amino groups and carbonyl groups of amino acids.
The amino bonds NH C that link amino acid residues are called peptide bonds. So peptide
||
O

bonds are the only covalent bonds that hold amino acid residues together in a peptide or protein.
Depending upon the number of amino acid residues per molecule, they are known as dipeptides, tripeptides
and so on and finally polypeptides. (By convention, peptides of molecular weight upto 1000 are known
as polypeptides and above that as protein)

O O O

H3N–CH2–C–NH–CH 2COO H3N–CH2–C–NH–CH–C–NH–CH–COO

(Gly-Gly) CH3 CH2–Ph
Glycine-Alaninedipeptide (Gly-Ala-Phen Tripeptide)
Glycine-Alanine-phenylalanine
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O O
H3N–CH–C–NH–CH–C—NH–CH–COO
R R R
n
Polypeptide
Dipeptides are made from two amino acids where as oligopeptides are made from 3 to 10 amino acids.
If 11 to 100 amino acids are present together they called polypeptide and 100 onwards they are called
as macropeptide.

O O O O

H2N–CH2–C–OH + H2N–CH2–C–OH – NH–CH 2–C–NH–CH2–C –

n
+ peptide linkage
PhCOCl ROH/H
O O O O O

Ph–C–NH–CH2–C–OH H2N–CH2–C–OR Ph–O–C–NH–CH2–C–NH–CH2–C–OR

For the synthesis of polypeptides, the amino groups that are not to be linked in peptide bonds must be
blocked so as to be unreactive. Then all other reactive functional groups must be protected to prevent
their participation in the coupling produces. The coupling must be effected by a method that does not
cause racemization or chemical alternation of the side chains.
These type polyamide can be made only using part blocking technique.
Abbreviated name of amion acid with free NH2 is written first.
By convention peptide are written with the free amino group (the –N–terminal amino acid) on the left
and the free carbonyl group (the –C– terminal amino acid) on the right.
O O
H3N–CH–C–NH–CH–C—O
R R’
For the nomenclature of peptides abbreviated name of amino acid with free NH2 group is written first.
For example:
O
H2N–CH2–C–NH–CH–COOH H2N–CH–CO–NH–CH2–COOH
CH3 CH3
(Gly-Ala) (Ala-Gly)
Glycine-Alanine-dipeptide Alanine-Glycine-dipeptide
Polypeptide on hydrolysis give two amino acid are known as dipeptide
O
HOOC –CH –NH–C–CH 2–NH 2 Dipeptide
Me
GlyAla [GlycineAlanine]
O O
H 2N–CH–C–NH–CH 2–C–OH Ala - Gly [Alanine-Glycine ]
Me

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Total number of polypeptide possible = Xn [X = type of amino acid interacting,

n = number of amino acid molecule are interacting.]
Q. Glycine can form how many Dipeptide ? [Ans. One]
Q. Glycine can form how many Tripeptide ? [Ans. One]
Q. Glycine andAla can form how many Dipeptide ? [Ans. Four]
Q. Gly, Ala, and PhenylAla can form how many Dipeptide ? [Ans. Nine]
Q. Gly, Ala, can form how many Tripeptide ? [Ans. Eight]

When Macropeptide takes different shape due to intramolecular H - bonding between different layers is
known as proteins

Proteins

You have already read that proteins are the polymers of -amino acids and they are connected to each
other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between —
COOH group and — NH2 group. The reaction between two molecules of similar or different amino
acids, proceeds through the combination of the amino group of one molecule with the carboxy1 group of
the other. This results in the elimination of a water molecule and formation of a peptide bond — CO —
NH —. The product of the reaction is called a dipeptide because it is made up of two amino acids. For
example, when carboxy1 group of glycine combines with the amino group of alanine we get a dipeptide,
glycylalanine.
H2N — CH2 — COOH + H2 N — CH — COOH

– H2O CH3

H2N — CH2 — CO — NH — CH — COOH

CH3
Peptide linkage

Glycylalanine (Gly-Ala)
If a third amino acid combines to a dipeptide, the product is called a tripeptide. A tripeptide contains
three amino acids linked by two peptide linkages. Similarly when four, five or six amino acids are linked,
the respective products are known as tetrapeptide, pentapeptide or hexapeptide, respectively. When
the number of such amino acids is more than ten, then the products are called polypeptides.Apolypeptide
with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a
protein. However, the distinction between a polypeptide and a protein is not very sharp. Polypeptides
with fewer amino acids are likely to be called proteins they ordinarily have a well defined conformation
of a protein such as insulin which contains 51 amino acids.
Proteins can be classified into two types on the basis of their molecular shape.

(a) Fibrous proteins : When the polypeptide chains run parallel and are held together by hydrogen and
disulphide bonds, then fibre-like structure is formed. Such proteins are generally insoluble in water.
Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc.

(b) Globular proteins : This structure results when the chains of polypeptides coil around to give a spherical
shape. These are usually soluble in water. Insulin and albumins are the common examples of globular
proteins.

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Structure of Proteins

Structure and shape of proteins can be studied at four different levels, .i.e., primary, secondary, tertiary
and quaternary, each level being more complex than the previous one.
(i) Primary structure of proteins : Proteins may have one or more polypeptide chains. Each polypeptide
in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino
acids that is said to be the primary structure of the protein. Any change in this primary structure i.e., the
sequence of amino acids creates a different protein.
(ii) Secondary structure of proteins : The secondary structure of protein refers to the shape in which a
long polypeptide chain can exist. They are found to exist in two different types of structures viz. -helix
and -pleated sheet structure. These structures arise due to the regular folding of the backbone of the
O
polypeptide chain due to hydrogen bonding between — C — and — NH — groups of the peptide
bond.

O
C
C
O
H O
H
O
HO
C
H C N
O
O
H
H
H N
N

-Helix structure of proteins

-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen
bonds by twisting into a right handed screw (helix) with the — NH group of each amino acid residue
hydrogen bonded to the C = O of an adjacent turn of the helix as shown in figure.
In -structure all peptide chains are stretched out to nearly maximum extension and then laid side by side
which are held together, by intermolecular hydrogen bonds. The structure resembles the pleated folds of
drapery and therefore is known as -pleated sheet.
N N N
RCH RCH RCH
C O C O
H C O H N HN
N HCR
HCR HCR
O CN H O C O C H
NH N
RCH RCH RCH
C C C O
H N O
H N O HN
HCR HCR HCR
C C C
-Pleated sheet structure of proteins
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(i) Ionic bonding : between COO¯ and NH3+ at different sites.

(ii) H-bonding : mainly between side-chain NH2 and COOH, also involving OH's (Of serine, for example)
and the N–H of tryptophan.
(iii) Weakly hydrophobic Van der Waal's attractive forces engendered by side-chain R groups and
(iv) Disulfide cross linkages between loops of the polypeptide chain.
The same kind of attractive and repulsive forces responsible for the tertiary structure operate to hold
together and stabilize the subunits of the quaternary structure.
(iii) Tertiary structure of proteins : The tertiary structure of proteins represents overall folding of the
polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular
shapes viz. fibrous and globular. The main forces which stabilise the 2° and 3° structures of proteins are
hydrogen bonds, disulphide linkages, van der walls and electrostatic forces of attraction.
(iv) Quaternary structure of proteins : Some of the proteins are composed of two or more polypeptide
chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other is
known as quaternary structure.
According the their biological action, they are classified as enzymes, hormones, antibodies, etc.
Protein found in living system with definite configuration and biological activity is termed as native protein.
If a native protein is subjected to physical or chemical treatment, which may disrupt its higher structures
(conformations) without affecting its primary structure, the protein is said to be denatured. During
denaturation, the protein molecule uncoils form an ordered and specific conformation into a more random
conformation leading to precipitation. Thus denaturation leads to increase in entropy and loss of biological
activity of the protein. The denaturation may be reversible or irreversible. Thus, the coagulation of egg
white on boiling of egg protein is an example of irreversible protein denaturation. However, in certain
cases it is found that if the disruptive agent is removed the protein recovers its original physical and
chemical properties and biological activity the reverse of denaturation is known as renaturation.

TESTS OF PROTEINS

Biuret test : Addition of a very dilute solution of CuSO4 to an alkaline solution of a protein is done. A
positive test is indicated by the formation of a pink violet to purple violet color.
The name of test is derived from a specific compound, biuret, which gives a positive test with this reagent
O O
H2N—C—NH—C—NH2
biuret
When a protein reacts with copper (II) sulphate (blue), the positive test is the formation of a violet
colored complex.
H
O H
N N
R O
CH—C—N— + Cu+2 Cu
H (Blue)
N N
O H
H
(Violet)

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The biuret test works for any compound containing two or more of the following groups.
O O NH S
—C—NH— —C—NH 2 —CH2—NH2 —C—NH 2 —C—NH2

Ninhydrin Test : The ninhydrin test is a test for amino acids and proteins with a free –NH2 group.
Amino acids are detected by ninhydrin test.All amino acids give violet - coloured product with ninhydrin
(triketo hydroindene hydrate) except proline and 4 - hydroxy proline, which gives yellow colour with it.

When such an –NH2 group reacts with ninhydrin, a purple-blue complex is formed.

O O
O
R O
OH
H2N—CH—C—OH + 2 N
OH
O
O
O
(Purple-Blue)

The same violet coloured dye forms from all -AA's with 1° amino groups because only their nitrogen
is incorporated into it. The 2° amines proline and 4 - hydroxyproline give different adducts that absorb
light at a different and thus have a different yellow colour.

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EXERCISE-1 (Objective Questions)

[SINGLE CORRECT CHOICE TYPE]

Q.1 The commonest disaccharide has the molecular formula
(A) C10H18O9 (B) C10H20O10 (C) C18H22O11 (D) C12H22O11

Q.2 On complete hydrolysis of starch, we finally get

(A) Glucose (B) Fructose
(C) Glucose and fructose (D) Sucrose

Q.3 The term anomers of glucose refers to

(A) Isomers of glucose that differ in configurations at carbons one and four (C–1 and C-4)
(B) Amixture of (D)-glucose and (L)-glucose
(C) Enantiomers of glucose
(D) Isomers of glucose that differ in configuration at carbon one (C–1)

Q.4 Which of the following is an example of aldopentose?

(A) Erythrose (B) Ribose (C) Fructose (D) Dihydroxyacetone

Q.5 Glucose and fructose form

(A) Same osazone (B) Same acid on oxidation
(C) Same alcohol when reduced (D) Different osazone

Q.6 The change of optical rotation of glucose solution with time is refered to as:
(A) Mutarotation (B) Inversion (C) Specific rotation (D)Autorotation

Q.7 To become a carbohydrate a compound must contain at least:

(A) 2 carbon atoms (B) 3 carbon atoms (C) 4 carbon atoms (D) 6 carbon atoms

Q.8 Methyl- -D-glucoside and methyl- -D-glucoside are

(A) Epimers (B)Anomers
(C) Enantiomers (D) Conformational diastereomers

Q.9 The correct name of 'sucrose' is

(A) -D-glucopyranosyl- -D-fructofuranoside
(B) -D-glucopyranosyl- -D-fructofuranoside
(C) -D-glucopyranosyl- -D-fructofuranoside
(D) -D-glucopyranosyl- -D-fructofuranoside

Q.10 Which one of the following is laevorotatory

(A) Glucose (B) Sucrose (C) Fructose (D) None of these

Q.11 Which of the following is a disaccharide

(A) Lactose (B) Starch (C) Cellulose (D) Glucose

Q.12 The two functional groups present in a typical carbohydrate are

(A) – OH and – COOH (B) – CHO and – COOH
(C) > C = O and – OH (D) – OH and – CHO

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Q.13 The monomer of cellulose is
(A) Fructose (B) Galactose (C) Glucose (D) None of these

Q.14 Glucose is a
(A) Monosaccharide (B) Disaccharide (C) Trisaccharide (D) Polysaccharide

Q.15 Which pair is different for reaction with Fehling solution:

(A) Glucose, Fructose (B) HCHO, CH3CHO
(C) CH2COCH3, C6H5CHO (D) Glucose, Sucrose

Q.16 Glucose contains in addition to aldehyde group

(A) One secondary OH and four primary OH groups
(B) One primary OH and four secondary OH groups
(C) Two primary OH and three secondary OH groups
(D) Three primary OH and two secondary OH groups

Q.17 Glucose reacts with excess of phenyl hydrazine and forms

(A) Glucosazone (B) Glucose phenyl hydrazine
(C) Glucose oxime (D) Sorbitol

Q.18 Which set of terms correctly identifies the carbohydrate shown

H O OH
OH H
HOH2C H
H OH
1. Pentose 2. Hexose
3. Aldose 4. Ketose
5. Pyranose 6. Furanose
(A) 1, 3 and 6 (B) 1, 3 and 5 (C) 1, 4 and 6 (D) 2, 3 and 6
(E) 1,4 and 6

Q.19 Hydrolysis of sucrose is called:

(A) Thermal decomposition (B) Saponification
(C) Inversion (D) Hydration

Q.20 Which of the following pentoses will be optically active

CHO CHO CHO

| | |
HCOH HCOH HCOH
| | |
HOCH HCOH HCOH
| | |
HCOH HOCH HCOH
| | |
H 2COH H 2COH H 2COH
I II III

(A)All (B) II and III (C) I (D) II

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Q.21 A tripeptide is written as Glycine-Alanine-Glycine. The correct structure of the tripeptide is

O CH3
NH
(A) NH2 NH COOH
O CH3

CH3 O CH3
NH
(B) NH2 NH COOH
O
O
NH
(C) NH2 NH COOH
O CH3

CH3 O
NH
(D) NH2 NH COOH
O CH3

Q.22 Which compound can exist in a dipolar (zwitter ion) state

(A) C6H5CH2CH(N = CH2) COOH (B) (CH3)2CH·CH(NH2)COOH
(C) C6H5CONHCH2COOH (D) HOOC·CH2CH2COCOOH

Q.23 Which of the following statement is incorrect for maltose.

(A) It is a disaccharide (B) It undergoes mutarotation
(C) It is a reducing sugar (D) It does not have hemiacetal group.

Q.24 Identify the correct statement about lactose.

(A) It consists of one galactose and one glucose unit
(B) Mutarotation is not possible
(C) Anomeric carbon of galactose is attached to 4' carbon of glucose which is –1, 4'-glycoside bond.
(D) Lactase is not used to cleave the –1, 4'-glycoside bond.

Q.25 Naturally occuring (+) - sucrose is :

(A) -D-glucopyronoside- -D-fructofuranoside
(B) -D-glucopyronoside- -D-fructofuranoside
(C) -D-glucopyronoside- -D-fructofuranoside
(D) -D-glucopyronoside- -D-fructofuranoside

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CHO
H OH
Q.26 H OH D-Ribose when treated with dilute HNO3 forms.
H OH
CH2 OH

HO O
HO O
H OH
OH
(A) H , Achiral (B) HO H , Chiral
H OH
H OH
H OH
O O OH
HO

HO O HO O
H OH H OH
(C) HO H , Achiral (D) H OH , Chiral
H OH H OH
HO O O OH

Q.27 Consider the given process.

CHO CHOH CHO
H–C–OH C–OH HO–C–H
HO H OH¯ HO H OH¯ HO H
H OH H 2O H 2O
H OH H OH
H OH H OH H OH
CH2OH CH2OH CH2OH
(I) (II) (III)
and identify the incorrect statement.
(A) Configuration at C–2 is lost on enolisation
(B) I and III are epimers
(C) Proton transfer from water to C–1 converts ene diol to an aldose.
(D) D-glucose can isomerise to D-fructose through enol intermediate.
( i ) HCN / H 3O
Q.28 Glucose P
( ii ) P / HI
P is :
(A) n- heptanoic acid (B) 2-methyl hexanoic acid
(C) n-heptane (D) 2-methyl hexane

Q.29 The configuration of the C-2 epimer of D-glucose is-

(A) 2R, 3S, 4R, 5S (B) 2S, 3S, 4R, 5R
(C) 2S, 3R, 4S, 5R (D) 2R, 3S, 4R, 5R

Q.30 Mutarotation involves

(A) Racemisation (B) Diastereomerisation
(C) Optical resolution (D) Conformational inversion

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Q.31 Same osazone derivative is obtained in case of D-glucose, D-Mannose and D-Fructose due to
(A) the same configuration at C-5
(B) the same constitution.
(C) the same constitution at C-1 and C-2
(D) The same constitution and same configuration at C-3, C-4, C-5 and C-6 but different constitution
and configuration at C-1 and C-2 which becomes identical by osazone formation.

Q.32 Which of the following compounds will not show mutarotation:

(A) -D (+) glucopyranose (B) -D(+) glucopyranose
(C) Methyl- -D-glucopyranoside (D) -D(+) galactopyranose

Q.33 Amylose and cellulose both are linear polymers of glucose. The difference between them is:
(A) Amylose has (1 4') linkage and cellulose has (1 4') linkage
(B) Amylose has (1 4') linkage and cellulose has (1 4') linkage
(C) Amylose has (1 4') linkage and cellulose has (1 6') linkage
(D) Amylose has (1 4') linkage and cellulose has (1 6') linkage

Q.34 Glycogen on hydrolysis gives:

(A) Lactose and Glucose (B) Only Glucose
(C) Glucose and Fructose (D) Glucose and Maltose

Q.35 An example of disaccharide made up of two unit of the same monosaccharide.

(A) Maltose (B) Sucrose (C) lactose (D) None

Q.36 The colour of the precipitate formed when a reducing sugar is heated with Fehling solution is:
(A) Brown (B) Red (C) Blue (D) Green

Q.37 Osazone formation involves only 2 carbon atoms of glucose because of

(A) Oxidation (B) Chelation (C) Reduction (D) Hydrolysis

Q.38 Methyl- -D-glucose and methyl - -D-glucose are

(A) Epimers (B)Anomers (C) Enantiomers (D) Constitutional isomers

Q.39 Hydrolysis of lactose with dilute acid yield

(A) Equimolar mixture of D-glucose and D-glucose.
(B) Equimolar mixture of D-glucose and D-galactose.
(C) Equimolar mixture of D-glucose and D-fructose.
(D) Equimolar mixture of D-galactose and D-galactose.

Q.40 Celluose is a straight chain polysaccharide composed of

(A) D-glucose units joined by -glycosidic linkage
(B) D-glucose units joined by -glycosidic linkage
(C) D-galactose units joined by -glycosidic linkage
(D) D-galactose units joined by -glycosidic linkage

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COO¯

Q.41 The pH of the solution containing following zwitter ion species is NH3 H

R
(A) 4 (B) 6 (C) 8 (D) 9

Q.42 Peptide linkage is -

O O O O
(A) –C – O – (B) –C – NH2 (C) –C – NH– (D) –C – NH–NH 2

Q.43 Test used to identify peptide linkage in protein is:

(A) Biuret (B) Ninhydrin test (C) Molisch test (D) 2,4-DNP test

Q.44 Which one of the following structures represents the peptide chain:
H O H O H O
(A) –N–C–C–C–N–C–C–N–C–C–C– (B) –N–C–N–C–NH–C–NH–
H O H
H H H H
(C) –N–C–C–C–C–NH–C–C–C (D) –N–C–C–N–C–C–N–C–C–
O O O

Q.45 Among the following L-serine is:

COOH COOH

(A) H2N CH2OH (B) HOCH2 H

H NH2

NH2 CH2OH
(C) H COOH (D) H2 N H

CH2OH COOH

Q.46 Glucose does not react with

(A) 2,4-DNP (B) Schiff's reagent (C) NaHSO3 (D)All of these

Q.47 Which statement is incorrect.

(A) All monosaccharide whether aldose or Ketose are reducing sugars
(B) Lactose is present in milk
(C) Fehling solution and Tollen's reagent can reduce, reducing sugars
(D) Glucose is aldohexose

Q.48 In Fructofuranose –OH group of which carbon adds to >C=O group.

(A) C3 (B) C4 (C) C5 (D) C2

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Q.49 Hydrolysis of d-sucrose results in change of optical rotation because
(A) number of moles of d-Glucose are less than number of moles of l-Fructose produced during hydrolysis.
(B) Angle of rotation of l-Fructose is greater than AOR of d-Glucose
(C) Both glucose and fructose are levorotatory.
(D) Glucose is l and fructose is d and number of moles of glucose formed during reaction are greater
than number of moles of fructose.

[MULTIPLE CORRECT CHOICE TYPE]

Q.50 Carbohydrates may be :
(A) Sugars (B) Starch
(C) Polyhydroxy aldehyde/ ketones (D) Compounds that can be hydrolysed to sugar

Q.51 Select the correct statement :

(A) S-glyceraldehyde is also known as L-glyceraldehyde
(B) The configuration of the stereocenter most distant from the carbonyl group determines whether a
monosaccharide is D or L.
(C) Glucose and all naturally occurring sugars are D-sugars
(D) D-erythrose and D-threose are diastereomers.

Q.52 Select the incorrect statement.

(A) Monosaccharide are insoluble in organic solvents like diethyl ether.
(B) Anomers of a cyclic monosaccharides differ in the position of the OH group at the hemiacetal
carbon.
(C) D-ribose the OH group used to form the five membered furanose ring is located on C4.
(D) Aldopentoses and ketohexoses form pyranose rings in solution.

Q.53 All sixteen possible aldohexose have been isolated or synthesised. Only three occur in nature there are
(A) D-Glucose (B) D-Galatose (C) D-Mannose (D) L-Mannose

Q.54 Which is not correct about monosaccharides.

(A) Optically active polyhydroxy carbonyl compounds.
(B) Fructose & glucose can not be distinguish by Br2/H2O
(C) Glucose and mannose are anomers
(D) Fructose & glucose can be distinguish by Fehling solution

Q.55 Which of the following is diasaccharides ?

(A) Lactose (B) Sucrose (C) Cellulose (D) Maltose

Q.56 Select the correct statement.

(A) Proteins upon hydrolysis gives -amino acid only.
(B) Except glycine, all other naturally occuring -amino acids are optically active.
(C) In fibrous proteins polypeptide chains are held together by hydrogen and disulphide bonds.
(D) Fats upon hydrolysis gives -amino acids

Q.57 Select the correct statement.

(A) Coiling of polypeptide chain form fibrous protein.
(B) Quarternary structure of protein also exist.
(C) Lysine is an amino acid with basic side chain.

(D) The absolute configuration of H 3 N – CH(CH2OH)COO¯ (L-serine) is S.

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Q.58 Select the correct option.

(A) Isoelectric point is the pH at which an amino acid exists primarily in its neutral form.
(B) Isoelectric point is the average of pKa values of -COOH amino - NH3 groups [valid only for
neutral amino acid]
(C) Glycine is characterised by two pKa values.
(D) For neutral amino acid the concentration of zwitter ion is maximum at its isoelectric point.

Q.59 Amino acids are synthesised from.

(A) -Halo acids by reaction with NH3.
(B) Aldehydes by reaction with NH3 and cyanide ion followed by hydrolysis.
(C)Alkyl halides by reaction with the enolate anion derived from diethyl acetamidomalonate & hydrolysis.
(D) Alcohols by reaction with NH3 and CN¯ ion followed by hydroysis.

Q.60 Which of the following carbohydrates developes blue colour on treatment with iodine solution ?
(A) Glucose (B)Amylose (C) Starch (D) Fructose

Q.61 Which of the following are correct.

(A) Br2/H2O can be used to differentiate between aldose & ketose.
(B)All monosaccharides are reducing sugar
(C) Osazone formation destroys the configuration about C2 of an aldose but not effect the rest of
molecule
(D) Mono saccharides undergoes mutarotation.

Q.62 -D-(+) Glucopyranose is

(A) Cylic structure of glucose
(B) It is dextrorotatory because 'D' is placed before name of it
(C) it is anomer of -D-(+) Glucopyranose
(D) Pyranose naming is due to it's anlogy with pyrane

Q.63 Which statement(s) is / are correct.

(A) Glucose react with NH2–OH to form oxime.
(B) Pentaacetate of glucose does not react with NH2–OH
(C) HI can reduce glucose to n-Hexane
(D) Glucose reacts with HCN to form epimeric cyonohydrin.

Q.64 Glycosidic linkage is present in :

(A) sucrose (B) starch (C) glucose (D) fractose

Q.65 Free aldehyde group can be produced on hydrolysis of which compound(s) :

(A) maltose (B) sucrose (C) -D Glucose (D) Lactose

Q.66 Which is non reducing sugar :

(A) maltose (B) sucrose (C) Lactose (D)All of these

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Q.67 Which statement is correct regarding to following structure :

CH2OH CH2OH
HO O H O OH
H O H
OH H OH H
H H
H OH H OH
(A) It is lactose
(B) It is composed of -D-galactose and -D-glucose
(C) Left side unit is galactose and right side unit is glucose
(D) there is free anomeric OH

Q.68 Which statement are correct about following?

CH2OH CH2OH
H O H H O H
H H
OH H OH H
HO
O OH
H OH H OH
(A) It is sucrose
(B) Due to presence of heme acetal linkage it is non reducing sugar.
(C) Left side unit is -D-Glucose and rightside unit is -D-fructose.
(D) Due to absence of heme acetal linkage it can't show mutarotation.

[REASONING TYPE]
Q.69 Statement 1 : Bromine water changes glucose to gluconic acid.
Statement 2 : Bromine water acts as oxidising agent.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.

[PARAGRAPH TYPE]
Comprehension (70 to 72)
Consider the following reversible process for a reaction of D-glucose.
+
H
Y
CH 2OH O -D-isomer
HO H
CH 3OH
OH [X]
OH
HO
D-glucose Z
( or form) + -D-isomer
H

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Q.70 The structure of intermediate [X] is

CH 2OH O
CH 2OH O
HO OH
(A) (B) OH
OH
HO O
CH 2OH O CH 2OH O
HO HO
(C) (D) OH
O OH
HO HO

Q.71 Select the correct option.

CH2OH O CH 2OH O
HO HO
(A) Y is H (B) Z is OCH 3
OH OH
HO OMe HO H
CH2OH O
CH2OH O HO
H
(C) Y is OH (D) Z is OH
OH HO OCH3
H
OCH 3
Q.72 Which of the following is positional isomer of -D glucose?

CH2OH O
CH2OH O
HO
(A) OH (B) HO OCH3
HO OH
OCH3 HO

OH OH
CH2OH CH 2OH

(C) HO OH (D) HO OCH 3

OH OH
O O

[MATCH THE COLUMN]

Q.73 Match the column :
Column I Column II
(A) Sucrose (P) Two acetals

(B) Maltose (Q) No hemiacetal

(C) Lactose (R) 1,4'-glycosidic bond

(D) Cellulose (S) Hydrolysis product is glucose

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Q.74 Match the column :

Column I Column II
(Carbohydrate) (Properties)

(A) Starch (P) Mutarotation

(B) Sucrose (Q) Non reducing sugar

(C) Lactose (R) -glycosidic bond

(D) Maltose (S) -glycosidic bond

(T) Reducing sugar

(U) Hemiacetal

Q.75 Match the column :

Column I Column II
CH 2OH O
HO H
CH2OH O
(A) HO OH O OH (P) -glycoside bonds
OH
HO H
H OH
O H
CH 2 O OH
HO
(B) OH O (Q) Reducing sugar
OH OH
HO HOCH 2

CH 2OH O
HO H
HO
HO
O
HOCH 2 O
(C) (R) Forms enediol intermediate
HO
CH2OH
OH
HOCH2 HOCH2
O O
(D) HO O OH (S) -glycoside bond
HO OH HO OH

Q.76 Column I Column II

(Monosaccharide units on hydrolysis)
(A) Sucrose (P) -D Glucose

(B) Maltose (Q) -Fructose

(C) Lactose (R) -D-Glucose

(S) -D-Galactose

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Q.77 Column I Column II

(Compound) (Relationship)
(A) -D-Glucose and -D-Glucose (P) Functional group isomer

(B) Glucose and Mannose (Q) Anomer

(C) Glucose and Fructose (R) Epimer

(D) -D-Fructo Furanose and

-D-Fructo Furanose

Q.78 Column I Column II

(Compound) (Relative configuration)
CHO
HO H
(A) H OH (P) D
CH2OH
COOH
H OH
(B) HO H (Q) L
H OH
H OH
COOH
CHO
H OH
HO H
(C)
H OH
HO H
CH2OH
CH2OH
H OH
H OH
(D) HO H
H OH
CHO

Q.79 Column I Column II

(Carbohydrate) (Type)
(A) Glucose (P) Monosaccharide

(B) Sucrose (Q) Oligosaccharide

(C) Maltose (R) Polysaccharide

(D) Glycogen

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Q.80 Column I Column II

(Reactant) (Number of functional group in product)
HI
(A) Glucose (P) 2
Br2 water
(B) Glucose (Q) 3
HCN
(C) Glucose (R) 0
O O
CH3–C–O–C–CH3
(D) Glucose (excess)
HNO3
(E) Glucose

Q.81 Column I Column II

(Compounds) (Reagents used to differentiate)
(A) Glucose / Fructose (P) Br2 water

(B) Glucose / Sucrose (Q) Fehling's solution

(C) Fructose / Sucrose (R) Tollen's reagent

Q.82 Column I Column II

(Compound) (Characteristic property)
(A) Amylose (P) Polysaccharide

(B) Cellulose (Q) -Glycosidic linkage

(C) Maltose (R) -Glycosidic linkage

(D) Galactose (S) Hydrolysed to glucose

Q.83 Column I Column II

CH2OH
H O H
H
(A) OH H (P) -D(+)-Glucopyranose
HO OH
H OH
CH2OH
H O OH
H
(B) OH H (Q) -D(+)-Glucopyranose
HO H
H OH
HOH 2C O CH2OH

H HO
(C) H OH (R) -D(–)-Fructofuranose
OH H
HOH 2C O OH

H HO
(D) H CH2OH (S) -D(–)-Fructofuranose
OH H

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EXERCISE-2 (Subjective Questions)

Q.1 The pKa values for the three acidic group P,Q,R are 4.3, 9.7 and 2.2 respectively
(R) (P)
HOOC–CH –CH2–COOH
+
NH3
(Q)
Calculate the isoelectric point of the amino acid ?

Q.2 Number of 1° and 2° alcoholic group in sucrose are x and y value of x & y is

Q.3 Number of chiral centre in maltose are ________.

Q.4 Number of Chiral carbons in Fischer projection of D-Fructose in which are OH groups are in RHS are
__________.

Q.5 Number of chiral carbon in Haworth structure of -Glucopyranose having OH group downside are
________.

Q.6 Saccharic acid can be obtained by oxidation of glucose number of C, H and O present in Saccharic acid
are _________.

Q.7 D-Glucose, D-Mannose and D-Fructose form same osazone on reaction with phenyl hydrazine because
they have same confuguration at carbon x,y & z. Values of x, y and z are ____

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EXERCISE-3
SECTION-A
(JEE ADVANCED Previous Year's Questions)

Q.1 Give the structures of the products in each of the following reactions. [JEE 2000]
H
Sucrose A+ B

Q.2 Aspartame, an artificial sweetener, is a peptide and has the following structure: [JEE 2001]
CH 2 C 6 H 5
|
H 2 N CH CONH CH COOCH 3
|
CH 2 COOH
(i) Identify the four functional groups.
(ii) Write the zwitterionic structure.
(iii) Write the structures of the amino acids obtained from the hydrolysis of aspartame.
(iv) Which of the two amino acids is more hydrophobic?

Q.3 Followingtwo amino acids lysine and glutamine form dipeptide linkage. What are two possible dipeptides?
[JEE 2003]
NH2 NH2

H2N COOH + HOOC COOH

Q.4 The structure of D-Glucose is as follows- [JEE 2004]

CHO (a) Draw the structure of L -Glucose
H OH (b) Give the reaction of L - Glucose with Tollen's reagent.
HO H
H OH
H OH

HO

Q.5 Which of the following pairs give positive Tollen’s test? [JEE 2004]
(A) Glucose, sucrose (B) Glucose, fructose
(C) Hexanal, Acetophenone (D) Fructose, sucrose

Q.6 Two forms of D – glucopyranose, are called. [JEE 2005]

(A) Enantiomers (B)Anomers
(C) Epimers (D) Diastereomers

Q.7 When benzene sulphonic acid and p-nitrophenol are treated with NaHCO3, the gases released
respectively are [JEE 2006]
(A) SO2, NO2 (B) SO2, NO
(C) SO2, CO2 (D) CO2, CO2

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Q.8 Statement-1 : Glucose gives a reddish-brown precipitate with Fehling’s solution.

because [JEE 2007]
Statement-2 : Reaction of glucose with Fehling’s solution gives CuO and gluconic acid.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.

Q.9 Statement-1 : p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.
because [JEE 2007]
Statement-2 : o-Hydroxybenzoic acid has intramolecular hydrogen bonding.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.

Q.10 Match the chemical substances in Column I with type of polymers / type of bonds in Column II. Indicate
your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [JEE 2007]
Column I Column II
(A) Cellulose (P) Natural polymer

(B) Nylon-6, 6 (Q) Synthetic polymer

(C) Protein (R) amide linkage

(D) Sucrose (S) Glycoside linkage

Q.11 Cellulose upon acetylation with excess acetic anhydride / H2SO4 (catalytic) gives cellulose triacetate
whose structure is [JEE 2008]
AcO
H O O–
H
AcO
(A) OAc H
H O O H
AcO H
OAc H H OAc
O O
H H H
OAc H H OAc
–O H
H OAc
AcO
H O O–
H
AcO
OH H
H O O H
AcO H
OH H H OH
O O
H H H
(B)
OH H H OH
–O H
H OH
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AcO AcO AcO

H H O H H O H H
(C) H O O–
OAc H H
OAc H OAc H
–O O O H
H OAc H OAc H OAc
AcO AcO AcO
O H H O H
H H H O O–
(D) H H
H H H H
–O O H H
O H
OAc OAc OAc OAc OAc OAc

Q.12 The correct statement(s) about the following sugars X and Y is(are) [JEE 2009]
CH2OH
CH2OH O
CH2OH H H
O H
H H HOH2C O H O H HO
H H O OH
OH H O H
HO H HO CH2OH OH H OH H
HO H
H OH OH H
H OH

(X)
(Y)
(A) X is a reducing sugar and Y is a non-reducing sugar
(B) X is a non-reducing sugar and Y is a reducing sugar
(C) The glucosidic linkages in X and Y are and , respectively.
(D) The glucosidic linkages in X and Y are and , respectively.

Q.13 Among cellulose, poly vinyl chloride, nylon and natural rubber, the polymer in which the intermolecular
force of attraction is weakest in [JEE 2009]
(A) Nylon (B) Poly vinyl chloride
(C) Cellulose (D) Natural Rubber

Q.14 The correct statement about the following disaccharide is [JEE 2010]
CH2OH
O H CH2OH O
H H
H
OH H H HO
OH CH2OH
OCH2CH2O
H OH OH H
(a) (b)
(A) Ring (a) is pyranose with -glycosidic link.
(B) Ring (a) is furanose with -glycosidic link
(C) Ring (b) is furanose with -glycosidic link
(D) Ring (b) is pyranose with -glycosidic link

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Q.15 The following carbohydrate is [JEE 2011]

(A) a ketohexose (B) an aldohexose (C) an -furanose (D) an -pyranose

Q.16 A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and
phenylalanine. Glycine contributes 47.0 % to the total weight of the hydrolysed products. The number
of glycine units present in the decapeptide is [JEE 2011]

Q.17 ThesubstituentsR1 and R2 for nine peptides are listed in the table given below. How many of these
peptides are positively charged at pH = 7.0 ? [JEE 2012]

H 3 N C H CO NH C H CO NH C H CO NH C H CO O
| | | |
H R1 R2 H

Peptide R1 R2
I H H
II H CH 3
III CH 2 COOH H
IV CH 2CONH 2 (CH 2 ) 4 NH 2
V CH 2CONH 2 CH 2CONH 2
VI (CH 2 ) 4 NH 2 (CH 2 ) 4 NH 2
VII CH 2COOH CH 2CONH 2
VIII CH 2OH (CH 2 ) 4 NH 2
IX (CH 2 ) 4 NH 2 CH 3

Q.18 Atetrapeptide has –COOH group on alanine. This produces glycine (Gly), Valine (Val), PhenylAlanine
(Phe) andAlanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences
(primary structures) with –NH2 group attached to a chiral center is [JEEAdvance 2013]

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SECTION-B
(JEE Main Previous Year's Questions)

Q.1 RNA contains - [AIEEE-2002]

(A) Urasil, Ribose (B) Thiamine, Ribose
(C) Cytocine, Deoxyribose (D)Adenine, Deoxyribose
Q.2 Complete hydrolysis of cellulose gives – [AIEEE-2003]
(A) D-glucose (B) L-glucose (C) D-fructose (D) D-ribose
Q.3 The reason for double helical structure of DNA is operation of – [AIEEE-2003]
(A) Hydrogen bonding (B) Electrostatic attractions
(C) vander Waal’s forces (D) Dipole-dipole interaction
Q.4 Coordination compounds have great importance in biological systems, In this context which of the following
statements is incorrect ? [AIEEE-2004]
(A) Chlorophylls are green pigments in plants and contain calcium
(B) haemoglobin is the red pigment of blood and contains iron
(C) Cyanocobalamin is B12 and contains cobalt
(D) Carboxypeptidase–A is an enzyme and contains zinc
Q.5 Which base is present in RNA but not in DNA ? [AIEEE-2004]
(A) Uracil (B) Cytosine (C) Guanine (D) Thymine
Q.6 Insulin production and its action in human body are responsible for the level of diabetes. This compound
belongs to which of the following categories ? [AIEEE-2004]
(A)A co-enzyme (B)Ahormone (C)An enzyme (D)An antibiotic
Q.7 Identify the correct statement regarding enzymes : [AIEEE-2004]
(A) Enzymes are specific biological catalysts that can normally function at very high temperatures
(T~ 1000 K)
(B) Enzymes are normally heterogeneous catalysts that are very specific in their action
(C) Enzymes are specific biological catalysts that cannot be poisoned
(D) Enzymes are specific biological catalysts that possess well-defined active sites

Q.8 In both DNA and RNA, heterocylic base and phosphate ester linkages are at – [AIEEE-2005]
(A) C2' and C5' respectively of the sugar molecule
(B) C5' and C2' respectively of the sugar molecule
(C) C5' and C1' respectively of the sugar molecule
(D) C1' and C5' respectively of the sugar molecule
Q.9 The pyrimidine bases present in DNA are – [AIEEE 2006]
(A) cytosine and guanine (B) cytosine and thymine
(C) cytosine and uracil (D) cytosine and adenine
Q.10 The term anomers of glucose refers to – [AIEEE 2006]
(A) a mixture of (D)-glucose and (L)-glucose
(B) enantiomers of glucose
(C) isomers of glucose that differ in configuration at carbon one (C-1)
(D) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)

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Q.11 The secondary structure of a protein refers to – [AIEEE 2007]

(A) -helical backbone
(B) hydrophobic interactions
(C) sequence of -amino acids
(D) fixed configuration of the polypeptide backbone
Q.12 -D-(+)-glucose and -D-(+)-glucose are [AIEEE 2008]
(A) epimers (B) anomers (C) enantiomers (D) conformers
Q.13 The two functional groups present in a typical carbohydrate are : [AIEEE 2009]
(A) – OH and –COOH (B) – CHO and –COOH
(C) > C = O and –OH (D) – OH and –CHO

Q.14 Which of the following compounds can be detected by Molisch's test ? [AIEEE 2012]
(A)Amines (B) Primary alcohols (C) Nitro compounds (D) Sugars

Q.15 Which one of the following statement is correct? [AIEEE 2012]

(A)All amino acids except glycine are optically active.
(B)All amino acids except glutamic acid are optically active.
(C)All amino acids except lysine are optically active.
(D)All amino acids are optically active.

Q.16 Synthesis of each molecule of glucose in photosynthesis involves: [JEE Main 2013]
(A) 10 molecules of ATP (B) 8 molecules of ATP
(C) 6 molecules of ATP (D) 18 molecules of ATP

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EXERCISE-4

SECTION-A
(CBSE Previous Year's Questions)
Q.1 What are hormones ? State the function of the following hormones : [CBSE-2003, 3]
(i) Testosterone
(ii) Oxytocin

Q.2 How does DNA replicate ? Describe the mechanism of replication. How is the process responsible for
preservation of heredity ? [CBSE-2004, 5]

Q.3 Write the major classes in which the carbohydrates are divided depending upon whether these undergo
hydrolysis, and if so, on the number of products formed. [CBSE-2004, 2]

Q.4 (a) Write chemical equations for the reactions of glucose with [CBSE-2005, 2]
(i)Acetic anhydride and
(ii)Ammoniacal silver nitrate solution.
(b) What do you understand by replication by DNA ? How does DNA differ from RNA structurally ?

Q.5 (a) Define vitamins and state their classification. List two vitamins of each class. [CBSE-2005, 5]
(b) What are enzymes ? State the activity of an enzyme

Q.6 (a) Write the chemical reactions of glucose with (i) NH2OH and (ii) (CH3CO)2O. Also draw simple
Fischer projections of D-glucose and L-glucose.
(b) Name the food sources and the deficiency diseases caused due to lack of any two of vitamin A, C,
E and K. [CBSE-2005, 5]

Q.7 Describe briefly two important functions of nucleic acid. [CBSE-2005, 3]

Q.8 Can a copolymer be formed in both addition and condensation polymerisation ? Explain with examples.
[CBSE-2005, 2]

Q.9 What are nucleotides ? Name two classes of nitrogen containing bases found in nucleotides.
[CBSE-2005, 2]

Q.10 Describe how allergic response is produced in the body. Write the role of an antihistamine.
[CBSE-2005, 3]

Q.11 (a) What is digestion ? How does digestion of carbohydrate take place ? [CBSE-2005, 3]
(b) Differentiate between DNA and RNA. Give four differences.

Q.12 (a) What are essential and non-essential amino acids ? Give two examples of each.
(b) What are the two types of photosynthesis in green plants ? Give the basic equations of photosynthesis.
(c) Mention the two products of glycolysis. [CBSE-2006, 2]

Q.13 Give one example each for reducing and non-reducing sugars. [CBSE-2006, 1]

Q.14 (i) What changes occur during digestion of a protein in human body ?
(ii) Name the different types of RNA found in the cell. Mention their function [CBSE-2006, 2]

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Q.15 Name the purines present in DNA. [CBSE-2007, 1]

Q.16 (a) What are natural and synthetic polymers ?

(b) Define and classify vitamins. [CBSE-2007, 2]

Q.17 (a) Name the three major classes of carbohydrates and give an example of each of these classes.
(b)Answer the following :
(i) What type of linkage is responsible for the primary structure of propeins ?
(ii) Name the location where protein synthesis occurs in our body. [CBSE-2007, 5]

Q.18 Why do amines react as nucleophiles ? [CBSE-2007, 1]

Q.19 B-complex is an often prescribed vitamins. What is complex about it and what is its usefulness.
[CBSE-2007, 1]

Q.20 (a)Answer the following questions briefly :

(i) What are reducing sugars ?
(ii) What is meant by denaturation of a protein ?
(iii) How is oxygen replenished in out atmosphere ? [CBSE-2007, 3, 2]
(b) Define enzymes :

Q.21 Define the following terms in relation to proteins. [CBSE-2008, 1]

Peptide linkage

Q.22 Write two main functions of carbohydrate in plants. [CBSE-2008, 1]

Q.23 What happens when D-glucose is treated with the following reagents : [CBSE-2008, 3]
(i) HI (ii) Bromine water (iii) HNO3

Q.24 How are the vitamins classified ? Mention the chief sources of vitamins Aand C. [CBSE-2008, 2]

Q.25 Discuss the two ways in which drugs prevent attachment of natural substrate on active site of an enzyme.
[CBSE-2008, 2]

Q.26 What are amino acids ? On electrolysis in an acidic solution, the relevant amino ions migrate towards
the cathode while in alkaline solution they migrate towards the anode. Give reason.
[CBSE-2008, 3]

Q.27 What are essential and non-essential acids ? Give one example of each type. [CBSE-2008, 1]

Q.28 Mention the type of linkage responsible for the formation of the following : [CBSE-2008, 2]
(i) Primary structure of proteins
(ii) Cross-linking of polypeptide chains
(iii) -helix formation
(iv) -sheet structure

Q.29 Name two water soluble vitamin, their sources and the diseases caused due to their deficiency in diet.
[CBSE-2009, 2]

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Q.30 (a) What is the structural difference between a nucleoside and a nucleotide ? [CBSE-2009, 3]
(b) The two strands in DNA are not identical but are complementary.
Explain.
Q.31 What is the biological effect of denaturation of proteins ? [CBSE-2009, 1]
Q.32 Explain the following terms with one example in each case : [CBSE-2010, 3]
(i) Food preservation (ii) Enzymes (iii) Detergents
Q.33 Differentiate between fibrous proteins and globular proteins. What is meant by the denaturation of a
protein ? [CBSE-2010, 3]
Q.34 (a) What type of bonding helps in stabilising the -helix structure of proteins.
(b) Differentiate between globular and fibrous proteins. [CBSE-2010, 3]
Q.35 What are vitamin ? Deficiency of which vitamin causes [CBSE-2010, 2]
(i) Pernicious anaemia (ii) Convulsions

Q.36 What is essentially the difference between -form of glucose and -form of glucose ? Explain.
[CBSE-2011]
Q.37 Explain what is meant by (i) a peptide linkage, (ii) a glycosidic linkage. [CBSE-2011]
Q.38 Name the bases present in RNA. Which one of these is not present in DNA ? [CBSE-2011]
Q.39 Write the main structural difference between DNA and RNA of the four bases, name those which are
common to both DNA and RNA. [CBSE-2011, 2]

SECTION-B
(Potential Problems Based on CBSE)
Q.1 (a) Give reasons for the following :
(i) Glucose does not give 2, 4-DNA test and Schiff's test.
(ii)Amino acids have high melting points and are soluble in water.
(b) What is meant by the secondary structure of protein ?
Q.2 What are polysaccharides ? Name one of them. How is it important for us ?
Q.3 What do you understand by the statement : "ATP molecules are the currency of energy metabolism in a
cell?"
Q.4 Differentiate between RNA and DNA
Q.5 Gives reasons for the following :
(a) Enzyme catalysts are highly specific in their action.
(b) The path of light becomes visible when it is passed through As2S3 sol. in water.
(c) The enthalpy in case of chemisorption is usually high than that of physiosorption.
Q.6 State the significance of primary and secondary structures of protein.
Q.7 Describe the following :
(i) Denaturation of proteins (ii) Mutation in DNA.
Q.8 What is the importance of amino acids to us ?

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ANSWER KEY
EXERCISE 1
Q.1 D Q.2 A Q.3 D Q.4 B Q.5 A Q.6 A Q.7 B
Q.8 B Q.9 A Q.10 C Q.11 A Q.12 D Q.13 C Q.14 A
Q.15 D Q.16 B Q.17 A Q.18 A Q.19 C Q.20 A Q.21 C
Q.22 B Q.23 D Q.24 A Q.25 A Q.26 A Q.27 C Q.28 A
Q.29 B Q.30 B Q.31 D Q.32 C Q.33 B Q.34 B Q.35 A
Q.36 B Q.37 B Q.38 B Q.39 B Q.40 B Q.41 B Q.42 C
Q.43 A Q.44 D Q.45 C Q.46 D Q.47 C Q.48 C Q.49 B
Q.50 ABCD Q.51 ABCD Q.52 ABC Q.53 ABC
Q.54 BCD Q.55 ABD Q.56 ABC Q.57 BCD
Q.58 ABCD Q.59 ABC Q.60 BC
Q.61 ABCD Q.62 ACD Q.63 ABCD Q.64 AB
Q.65 ACD Q.66 AB Q.67 ABCD Q.68 AD
Q.69 A Q.70 A Q.71 D Q.72 C Q.73 (A) Q,S (B) S (C) R,S (D) R,S
Q.74 (A) Q,S (B) Q,S (C) P,R,T,U (D) P,S,T,U Q.75 (A) P,Q,R (B) P (C) P (D) P,Q,R
Q.76 (A) PQ (B) PR (C) RS Q.77 (A) QR (B) R (C) P (D) Q
Q.78 (A) P, (B) P (C) L (D) L Q.79 (A) P (B) Q (C) Q (D) R
Q.80 (A) R, (B) P (C) P (D) P (E) P Q.81 (A) P (B) PQR (C) QR
Q.82 (A) PQ (B) PS (C) S (D) S Q.83 (A) Q (B) P (C) R (D) S

EXERCISE 2
Q.1 3.25 Q.2 3&5 Q.3 10 Q.4 2 Q.5 3
Q.6 6108 Q.7 345

EXERCISE 3
SECTION-A
Q.2 (i)Amine, carboxylic acid,Amide, Ester

CHO COO
HO H HO H
+
H OH [Ag(NH 3) 2] H OH
Q.4 HO H HO H
HO H HO H
HO HO
L - Glucose
Q.5 B Q.6 B Q.7 D Q.8 C Q.9 D
Q.10 (A) P,S; (B) Q, R; (C) P, R; (D) S Q.11 A Q.12 BC Q.13 D Q.14 A
Q.15 B Q.16 6 Q.17 4 Q.18 4

SECTION-B
Q.1 A Q.2 A Q.3 A Q.4 A Q.5 A Q.6 B Q.7 D
Q.8 D Q.9 B Q.10 C Q.11 A Q.12 B Q.13 C Q.14 D
Q.15 A Q.16 D

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EXERCISE 4
SECTION-A
Q.1 Harmones are complex organic compounds which are produced in endocrine glands and are directly
secreted into blood stream. These control the various metabolic processes.
(i) Testosterone controls normal function of male sex organs.
(ii) Oxytocin produces milk in the mammory glands of animals.

Q.2 Replication is the process by which a single DNA molecule produces two identical copies of itself,
during cell division. During this process, the two strands of the double helix first separate and each
separated strand then serves as a template for the synthesis of new strand. Due to specificity of base-
pair, the sequence in the two new strands is complementary to the original. In other words, replication
leads to the production of two identical copies of DNA from a single parent DNA and copy of each then
passes on the new cells resulting from cell division. In this way, the hereditary effect are transmitted from
one cell to another.

T T

G C

A T

C G

T A

C G C G
A T A T
G C G C
C G C G
T A T A

Q.3 On the basis of hydrolysis, carbohydrates can be divided in three major classes :
(i) Monosaccharides : They cannot be hydrolyse into simpler molecules. They are further classified as
aldoses and ketose.
(ii) Oligosaccharides : These carbohydrates on hydrolysis give 2-10 monosaccharides. For example,
sucrose.
(iii) Polysaccharides : These are high molecular mass carbohydrates which give many molecules of
monosaccharide on hydrolysis. For example starch and cellulose.

Q.4 (a) (i) HOCH2 – (CHOH)4 – CHO + 5 (CH3 CO)2 O

Glucose Acetic anhydride
OHC – (CHOCOCH3)4 – CH2OOCH3
Pentaacetyl glucose
(ii) Glucose reduces ammonical silver nitrate solution (Tollen's reagent) to metallic silver.
HOCH2 – (CHOH)4 – CHO + Ag2.O
HOCH2 – (CHOH)4 – COOH + 2 Ag
Glucose acid

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(b) Replication is the process by which a single DNA molecule produces two identical copies of itself,
during cell division. During the process, two strands of the double-helix first separate and each separated
strand then serves as a template for the synthesis of new strand. Due to specificity of base pair, the
sequence in the two new strands is complimentary to the original .
The structural differences between DNA and RNA are
(i) DNA has dexyribose sugar while RNA has ribose sugar
(ii) DNA contains thymine and RNA has uracil
(iii) DNA is double stranded while RNA is single stranded

Q.5 (a) Vitamin can be defined as the essentials dietary factors required by an organisms in minute quantities
and whose absence causes specific deficiency diseases.
Vitamins are broadly classified into two types :
1. Fat soluble vitamin : These are oily substances not readily soluble in water. These include vitamins
A, D, E and K.
2. Water soluble vitamins : The vitamin B and C are water soluble and are stored in much smaller
amounts in the cells.
Vitamin H (Biotin) as an exception, it is neither soluble in water nor in fat.
(b) Enzymes are naturally occurring simple conjugate proteins acting as specific catalysts in all processes.
In contrast to ordinary chemical catalyst, it loses activity by pH or temperature change.
Enzymes are highly specific, i.e., a particular enzyme catalyses a specific reaction. For example ureas
attacks on urea. This specific action is due to active sites present in the enzyme molecule (E) that fits
into substrate (S) and forms E-S complex which changes into product P and E.
For example
NH4CONH2 + H2O Urase CO2 + 2 NH3
Urea

Q.6 (a) (i) CHO CH = N – OH

NH2OH
(CHOH)4 (CHOH)4

CH2OH CH2OH
Glucose Glucose monoxime

(CH3CO)2O
(ii) OHC – (CHOH) 4¯ CH 2OH
Glucose
OHC – (CHOCOCH 3) 4 – CH 2OOCCH 3
Pentacetate

Name of Vitamin Food Source Deficiency Disease

Vitamin-A Fish oil, Liver Hardening of cornea of eye

Vitamin-C Citrus fruits, Green vegetables Scurvy

(b)
Vitamin-E Wheat germ oil, Cotton Seed oil Steritity

Vitamin-K Cereals, Leafy vegetables Hemorrhagic condition

Q.7 Functions of nucleic acid : (i) RNA acts as an intermediatry compound in the synthesis of proteins from
the genetic codes of DNA. Molecules of RNA being specific amino acids to the ribosomes for
incorporation into the protein.
(ii) DNA contains the genetic information of a particular species and passes on the same by a process
named replication. It takes part in the synthesis of RNA by the process named transcription.

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Q.8 A copolymer is formed by the polymerisation of two or more than two different monomers.
Yes, a copolymer can be formed in both addition and condensation polymerisation. For example
(i) The addition polymerisation of styrene and butadiene forms a copolymer styrene-butadiene rubber
(addition polymerisation)
n CH2 = CH – CH = CH2 + n C6H5 CH = CH2
C6H5

( CH2 – CH = CH – CH2 – CH2 – CH ) n

(ii) The condensation polymerisation of hexamethylenediamine and adipic acid form a copolymer nylon-
66
HOOC (CH2)4 COOH + NH2 + (CH2)6 NH2 ( NH (CH2)4 NHCO (CH2)2 CO ) n

Q.9 Nucleotides : There are the monomeric units of a type of long polymers called nucleic acids. These are
compound of three components viz. nitrogen containing heterocylic base, a five carbon sugar and a
phosphoric acid moiety.
Nitrogen containing bases
Purines and pyridines

Q.10 Allergic response produced in the body :Allergy in a state of hypersensitivity characterized by difficult
respiration and skin rashes. The substances that cause allergy is called allergin. When these come in
contact with a cell to which a particular type of antibody is attached, antibody - antigen complex is
formed. The formation of the later leads to the release of allergy mediators which are responsible for
various allergy symptoms.
Antithistamine : These are the drugs used to treat allergy. These drugs act by blocking, one specific
type of allergy mediator called histamine.

Q.11 (a) Digestion : Digestion is a process in which complex nutrient molecules such as fats, proteins, starch,
carbohydrates etc. are converted into simple nutrient molecules by the action of enzyme secreted by the
wall of the digestive tract.
Digestion of carbohydrates : Carbohydrates are class of compounds which include polyhydroxy aldehydes,
polyhydroxy ketones and large polymeric molecules. In the digestive tract, the carbohydrate is ultimately
converted into monosaccharide (e.g. fructose, galactose and glucose) by the action of enzyme amylase,
secreted by saliva and the walls of intestine. Glucose is absorbed by the blood and get stored in liver
and muscles. The glucose burns in cells to liberate energy.

Q.12 (a) These are amino acids which are essential for maintenance of proper nitrogen balance in the body but
cannot be synthesised within living organisms. Out of 20 amino acids that make up protein , 10 are
synthesised by our bodies (non-essential amino acids) and other 10 amino acids known as essential
amino acids must be supplied in the diet through protein-rich diets like milk, fish eggs, meat, bean and
peas.
(b) Photosynthesis : It is a series of complex biochemical reaction involving two stages :
(i) Light reactions : These reactions occur in the presence of sunlight. The sunlight is absorbed by the
chlorophyll of plant cells. This energy is used to synthesis energy richATP molecules with the liberation
of oxygen.
hv
H2O + CO2 Chlorophyll
Organic molecules + O2 + ATP

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(ii) Dark reactions : These reactions takes place in the absence of light but immediately after light
period. The energy of ATP molecules is used in the conversion of CO2 of the atmosphere into glucose
and then into starch.

6CO 2 6H 2 O ATP C 6 H12 O 6 6CO 2

The synthesis of each molecule of glucose consumes 18 molecules of ATP.
(c) The two main products of glycolysis are pyruvate (a 3-carbon molecule) and ATP molecules.

Q.13 Reducing sugar : D-fructose

Nonreducing sugar : Sucrose

Q.14 (i) The digestion of protein begins in the stomach which contains the enzyme pepsin. The digestion
continues in the intestine in presence of typsin, dry motrypin and various other peptidases secreted by
pancreas and intestine. During digestion the proteins are converted into -amino acids which are
absorbed by blood.
(ii) Different types of RNA founds in cells are
(a) Messenger RNA : It carries the genetic code form DNA to ribosomes where protein is synthesised.
(b) Ribosomal RNA : It is present in ribosomes. It function is to provide the site for protein synthesis.
It helps in binding m-RNA to ribosomes during protein synthesis.
(c) Transfer RNA : It transfers amino acids from different parts of cytoplasm to the ribosomes during
protein synthesis.

Q.15 Adenine and Guanine.

Q.16 (a) Polymers are classified on the basis they are naturally occurring or synthetic. Some of the natural
polymers are starch, cellulose, proteins and nucleic acids etc. Synthetic polymers are man-made
polymers such as polyethylene, PVC, teflon, nylon etc.
(b) It has been observed that there are some organic compounds other than carbohydrates, proteins
and fats which are necessary for normal growth and maintenance of health. These specific organic
compounds are known as vitamins. They are classified into two broad types based on their solubility in
fats and water.
Vitamin A, D and K are fat soluble where as vitamin of B group such as B1, B6, B12 and vitamin C are
water soluble.

Q.17 (a) Based on structure, the carbohydrate have been classified into three main classes.
(i) Monosaccharide
Examples : Glucose, Fructose
(ii) Diasaccharides
Example : Maltose , Sucrose
(iii) Polysaccharides
Example : Starch, glycogen
(b)
(i) The primary structure of a protein determines its functions and biological activity.
(ii) Protein synthesis takes place in ribosomes.

Q.18 Because the electron pair of nitrogen can conjugate with the benzene ring giving carbon nitrogen bond
with double bond character.

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Q.19 B-complex is a group of vitamins contains vitamins B1, B2, B6, B12, biotin, folic acid, pantothenic acid
and nicotinic acid. It is required to release energy from food and to promote healthy skin and muscles. Its
deficiency causes beriberi and pernicious anaemia.

Q.20 (a) (i) Reducing sugars :All those carbohydrates which reduce Fehling's solution and Tollen's reagent
are called reducing sugars. All monosaccharide are reducing sugars in nature.
(ii) Denaturation of protein : The complex three-dimensional structure of proteins gets distributed by
changing the pH, temperature or by adding some salt. This disruption of the native structure of protein
is called denaturation of protein.
(iii) Oxygen is replenished in out atmosphere through photosynthesis in plants.
(b) Enzymes : Most of the chemical reactions which occur in living systems process at very high rates
even under mild conditions of temperature and pH. These reactions are catalysed by a group of
biomolecules called enzymes.

Q.21 Peptide Linkage : Proteins are condensation polymers of -amino acids in which the same or different
-amino acids are connected by peptide bonds. Chemically, a peptide bond is an amide linkage formed
between – COOH group of one -amino acid and NH2 group of the other -amino acid by loss of a
molecule of water.

Q.22 Main functions of carbohydrates in plants :

(i) Carbohydrate are used as storage molecules as starch in plants.
(ii) Cell wall of bacteria and plants is made up of cellulose

Q.23 (i) On prolonged heating with HI, D-glucose forms n-hexane.

CHO
HI,
(CHOH)4 CH3 – CH2 – CH2 – CH2 – CH2 – CH3
n–Hexane
CH2OH
D-Glucose
(ii) D-glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with bromine
water
CHO
Br2 water
(CHOH) 4 COOH

CH2OH (CHOH) 4
D-Glucose
CH2OH
Gluconic acid
(iii) On oxidation with nitric acid, D-Glucose yields a saccharic acid
CHO COOH
HNO 3
(CHOH)4 (CHOH)4
Oxidation
CH2OH COOH
D-Glucose D-saccharide acid
Q.24 Classification of Vitamins : Vitamins are classified into two groups depending upon their solubility in
water or fat.
(i) Fat soluble vitamins
(ii) Water soluble vitamins
Sources of Vitamin A: Fish liver oil, carrots, butter and milk.
Sources of Vitamin C : Citrus fruits, amla and green leafy vegetables.
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Q.25 The drug attachment of natural substrate on active site on an enzyme can be prevented by two ways :
(i) Catalytic action of enzymes
(ii) Drug enzyme interaction

Q.26 Amino acids :Amino acids contains amino (–NH2) and Carboxyl (–COOH) functional groups. Depending
upon the relative position of amino group with respect to carboxyl group, the amino acid can be classified
as , , , and so on.
In acidic medium , a-amino acids exist as cation (I) and thus migrate towards cathode under the
influence of an electric field.
On the other hand, in alkaline medium, a-amino acids exists as anions (II) and thus migrate towards
anode under the influence of an electric field.

Q.27 Essential amino acids : -amino acids which are needed for health and growth of human beings but
are not synthesised by the human body are called essential amino acids.

Examples : Valine, Leucine etc.

Non-Essential amino acids : -amino acids which are needed for health and growth of human being and
synthesised by the human body are called non-essential amino acids.

Example : Glycine, aspartic acid etc.

Q.28 Biomolecule Type of linkages

1. Primary structure of proteins Peptide bond or peptide linkage
2. Cross-linking of polypeptide chains Polypeptide linkage
3. -helix formation Hydrogen bonds
4. -sheet structure Intermolecular hydrogen bonds.

Q.29 Example of water soluble vitamin : B group vitamins and vitamin C.

Name of vitamins Sources Deficiency

Vitamin B1 Yeast, milk, green Beri-beri

vegetables and cereals
Vitamin C Citrus fruits, amla and Scurvy (Bleeding gums)
green leafy vegetable

Q.30 (a) Nucleotide is formed by condensation of a purine or pyrimidine base with Pentose sugar at position
1. When nucleoside is linked to phosphoric acid at 5' position of sugar moiety, we get a nucleotide.
(i) Nucleoside has two units : Pentose sugar and a base.
(ii) Nucleotide has three units : Phosphate group, pentose sugar and a base.
O
5' O 5' O
HO – H2C Base O – P – O – H2 C Base
4' H1' 4' H1'
H O H
H 3' 2' H H 3' 2' H
OH OH OH OH
Nucleoside Nucleotide
(b) DNA is a double helix in which the two strands of DNA are held by the hydrogen bonds between the
bases on the two strands. Thymine (T) pairs with adenine through two hydrogen bonds and cystosine
(C) pairs with guanine (G) through three hydrogen bonds.
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So opposite each adenine (A) on one strand there is always a thymine on the other strand and opposite
guanine there is cytosine.

Hence the two strands of DNA are complementary to each other.

Q.31 During denaturating , the protein molecule uncoils from an ordered and specific conformation into a
more random conformation. Denaturation does not change the primary structure of protein.

Q.32 (i) Food preservation : Food preservation are the substances used to prevent spoiloge of food due to
microbial growth during storage. The most common preservation used are table salt, vegetable oils and
sodium benzoate (C6H6COONa)
(ii) Enzymes : The proteins which perform the role of biological catalysts in the body are called
enzymes. Examples : E. coli.
(iii) Detergents : Detergents may be defined as ammonium, sulphate or hydrogen sulphate salts of long
chain hydrocarbons containing 12-18 carbon atoms.
Example : The more common detergents are the sodium salts of long chain sulphonic acids.

Q.33 Fibrous Protein Globular proteins

(i) When the polypeptide chains run parallel (i) when the chains of polypeptide coil
and are held together by hydrogen and around to give a spherical shape
disulphide bonds to give a fibre like
structure
(ii) They are generally insoluble in water (ii) They are usually soluble in water
Example : Keratin and myosin Example : Insulin and albumins

Q.34 (a) -Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen
bonds by twisting into a right handed screw (helix) with the –NH group of each amino acid residue
hydrogen bonded to the > C = O of an adjacent turn to the helix.
(b)
Globular proteins Fibrous proteins
(i) These proteins are cross linked These are linear condensation product
condensation product of basic
and acid amino acids
(ii) These are soluble in water or aqueous These are insoluble in common solvent
solution of acids, bases or salts. but soluble in strong acids and bases
(iii) Examples of globular proteins are Example are myosin in muscles, keratin
albumines in egg; all enzymes and in hair and fibroin in silk.
hormones

Q.35 Vitamin : The organic compounds which cannot be produced by the body and must be supplied in small
amounts in diet to perform specific biological functions for the normal health, growth and maintenance
body is called vitamins.
(i) Pernicious anemia is caused by the deficiency of vitamin B12
(ii) Convulsion is caused by the deficiency of vitamin B6

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O
1 1
H – C – OH 1 H–C–H
H–C
2 2 2
H OH H OH H OH
3 O 3 3 O
HO H HO H HO H
4 4 4
H OH H OH H OH
Q.36 5 5 5
H H OH H
6 6 6
CH2OH CH2OH CH2OH
-D–(+)–Glucose -D–(+)–Glucose

Glucose is found to exist in two different crystalline form which are named as and . The -form of
glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K
while the -form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at
371 K.

Q.37 (i) Disaccharide on hydrolysis with dilute acids or enzymes yield two molecules of either the same or
different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed
by the loss of a water molecule. Such a linkage between two monosaccharides units through oxygen
atom is called glycosidic linkage. Proteins are the polymers of a-amino acids and they are connected to
each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between
–COOH group and – NH2 group.

Q.38 DNA contains a five carbon sugar molecules called deoxyribose whereas RNA contains ribose. Both
DNA and RNA contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in
RNA. The structure of DNA is double strand whereas RNA is a single strand molecules. DNA is the
chemical basis of heredity and have the coded message for proteins to be synthesised in the cell.

Q.39 RNA : (Ribonucleic acids) contains the sugar D-ribose. The base thymine is never present in it. Its
molecule consists of a single strand. It is mainly formed in the cytoplasm in the cell. It does not self-
replicate.

SECTION-B
Q.1 (a) (i) In the cyclic structure of glucose-CHO group is not free as it forms a hemiacetal linkage with
–OH group at C-5.
(ii) The amino acids have high melting points and solubility in water due to zwitter ion (polar) structure
and strong intermolecular forces between them.
R R R

H2N – CH – COOH H3N – CH – COO¯

(b) Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They
are found to exist in two different types of structures viz. -helix and -pleated sheet structure.

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O N N N
C
C RCH RCH RCH
C O C O
H C O
O
H O H N HN
H N HCR
O HCR HCR
HO
C C
O N H O C O C H
H C N
O N H N
O RCH RCH RCH
H C
H
H N O C C O
H N
H N O HN
N
HCR HCR HCR
C C C

Q.2 Polysaccharides are the polymers in which more than six molecules of monosaccharides are joined
together by glycosidic bonds, with the loss of water molecules each time monosaccharides molecule is
added. For example starch which is a reserve food material of plant cells.

Q.3 ATP molecules are energy rich molecules. Their hydrolysis releases a large amount of energy. When
hydrolysis ofATP is coupled with an energy requiring reaction during metabolic changes in a cell, ATP
molecules are known as the currency of energy in a cell.

Q.4 RNA DNA

1. RNA (Ribonucleic acid) contains the DNA (Deoxyribonucleic acids) contains
sugar D-ribosome the sugar D-2-deoxyribose
2. The base thymine is never present in it The base uracil is never present
3. Its molecule consists of a single strand It is a double stranded molecule
4. It is mainly formed in the cytoplasm It is mainly formed in the nucleus of the
in the cell cells
5. It does not self-replicate It has the property of self replication

Q.5 (a) This is because each enzyme has a specific active site on which only a particular substrate can bind.
(b) This is because of Tyndal effect caused by the scattering of light by colloidal particles of As2 S3 sol.
(c) Chemisorption involves the formation of a chemical bond between adsorbent and adsorbate molecules
which involves high energy changes while in physisorption, the molecules of adsorbate and adsorbent
are held by weak Vander Wall's interaction.

Q.6 Primary structure : The sequence in which various amino acids are arranged in a protein is called
primary structure. The amino acid sequence of a protein determines its function and is critical of its
biological activity.

Q.7 (i) Denaturation of proteins : On heating or on treatment with minerals acids, the water solule globular
proteins undergo coagulation or precipitation with loss of biological activity to give water insoluble
fibrous proteins. This process is called denaturation of proteins.
(ii) Mutation :Amutation is a chemical change in the base sequence along the DNAstrand which leads
to the synthesis of proteins with changed sequence of amono acids. These change are caused by X-
rays, U.V. radiations, chemical agents or viruses. Generally these changes are repaired byspecial reenzymes
present in the cell. If the change is not repaired a mutation is said to have occurred.

Q.8 Amino acids are building block of peptides and proteins which are essential for proper growth and
development of our body. Deficiency of amino acids lead to diseases like Marasinus and Kwashiorkar.

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COLUMN : A COLUMN : B

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solve in first attempt questions

Exercise # 1

Exercise # 2

Exercise # 3 (A)

Exercise # 3 (B)

Exercise # 4 (A)

Exercise # 4 (B)

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