ORGANIC CHEMISTRY

Daily Practice Problems

CLASS : XIII (ALL) DPP. NO.-20
Time : 90 Min. Max. Marks : 54
CH 3
| conc.H SO
Q.1 CH 3CH 2 CH 2 CH 2 CCH 2 OH  24  Major product is: [2]

|
CH 3

CH 2
||
(A*) CH3CH2CH2CH2CH= (B) CH 3CH 2CH 2CH 2C
|
CH 2CH 3

(C) CH 3CH 2CH 2CH 2C  (D) CH 3CH 2 CH 2CH  C  CH 2CH 3

| |
CH 3 CH 3

LAH
Q.2  A PBr3 Alcoholic KOH
dry ether
 B     C

What are A to C ? [3]

[Sol. (A) (B) (C) ]

Q.3 Assign A to D in the given reaction sequence [4]

( i ) CH I ( excess ) ( i ) CH I ( excess )
 3   A  3   B
( ii ) AgOH ,  ( ii ) AgOH , 

[Sol. A= B = MeOH ]

Q.4 How will you convert cyclohexane into benzene in four steps. [3]

Br
[Sol. 2  alc
KOH NBS
 alc
KOH ]
h excess

Q.5 PhOCH2Cl + Ph3P + tBuO—K+  A (C25H21P) [3]

A + ethyl methyl ketone  Ph3P = O + B (C11H14O)
B + dilute aqueous acid  C (C5H10O)
What are A to C?
O
||
[Sol. A = Ph3P = CHOPh B= C = C 2 H 5  CH  C  H ]
|
CH 3
Q.6 Complete the following reactions.

(a) + PPh3  A + B
(b) ClCH2COMe Ph 3P
 EtOH
 A EtONa   B [4]
[Sol. (a) A = CH2 = CHMe + Ph3P = O = B


(b) A = (Ph 3 P CH 2 COMe)Cl
B = CHC–Me ]

Q.7 Give structure of A to J. [9]

reduction
  A (C9H17ON, an alcohol)

A + heat  B (C9H15N)
B + CH3I + Ag2O  C (C10H19ON)
C + heat  D (C10H17N)
D + CH3I + Ag2O  E (C11H21ON)
E + heat  F (C8H10)
F + Br2  G (C8H10Br2) [Thermodynamically controlled pot]
G + Me2NH  H (C12H22N2)
H + MeI + Ag2O  I (C14H30O2N2)
I + heat  J (C8H8)

[Sol. (A) (B) (C) (D)

(E) (F) (G) (H)

(I) ]

Q.8(a)Complete the following reactions.

EtOLi
(i) PhCH=CHCH2PPh3+Cl— A PhCHO
  B + C

(ii) Ph3P + (i) ClCH2COMe  D Na

2CO 3
 E PhCHO
  F.. [6]

(b) Rate law for substitution reaction of 2-bromobutnae with OH in 75% ethanol & 25% water at 30°C is
rate = 3.20 × 10–5 (2-bromobutane) (OH–) + 1.5 × 10–6 (2-bromobutane) what percent of reaction
take place by SN2 when [OH–] = 1.00 m.
[Sol. (a) (i) A = PhCH = CHCH2= PPh3
B = Ph–CH=CH–CH=CH–Ph
C = Ph3PO
 

(ii) D = (Ph 3 P CH 2 COMe)Cl
E = Ph3P = CHCOMe
F = Ph–CH=CHCOMe
S
N2
(b) % by S N 2 = × 100
S 2 S 1
N N

3.20  105[2  bromobutane][1.00]  100

=
3.20  10 5[2  bromobutane][1.00]  1.5 10  6 [2  Bromobutane]

3.20 10 5
= 5 × 100 = 96% ]
3.20 10  0.15 10 5


Q.9 
 Which of the following product cannot be obtained by this reaction. [2]
Ei

(A) (B) (C) (D*)

Q.10 CH2 = C = O (

i ) Br2
 C4H8O [2]
(ii ) CH 3MgBr
( 2 equi )

(A*) (B) (C) (D) All of these

Br2
Q.11  HOOC  CH  CH  COOH
 (X) [2]
CCl 4 | |
Br Br

Br
2  HOOC  CH  CH  COOH (Y)
CCl 4 | |
Br Br
The correct statement with respect to above reactions are
(A*) The reaction is stereospecific
(B*) (X) is erythro and (Y) is threo isomer
(C) (X) is threo and (Y) is erythro isomer
(D) each gives mixture of (X) and (Y)

( i ) Mg / ether
Q.12 Which is B in the sequence HBr
 A    
B [2]
(ii ) CO 2 / H 3 O

(A) (B*) (C) (D)

Q.13 Give mechanism for given reaction: [4]

H  / HOH
 

-Terpeniol

[Ans. H  HOH / H  ]
  

Q.14 Draw the structures of stable configuration obtained after acidic hydration of the following unsaturated
compounds: (exclude rearranged products) [4]

(I) (II) (III) (IV)

[Ans. (I) (II) (III) (IV) ]

Q.15 Formulate mechanisms for the following transformations. [4]

CH N
2
2 

[Sol.

]
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-21
Q.1 Classical carbocation contain:
(A) a lone pair (B) a vaccant orbital (C) three bond pairs (D*) B and C both

Conc . H SO
2
Q.2    4  A, A is:

(A) (B*) (C) Both (D) None

Q.3 Which of the following is most acidic


(A) (B) CH 3  NH 3 (C) (D*)

Q.4 The decreasing order of reactivity towards electrophilic addition is:

(A) III > II > II > I (B) IV > I > II > III (C) II > III > I > IV (D*) II > III > IV > I

CH 3
| H 
Q.5(a) CH 3  C  OH  ? (b) CH 3  CH  CH  CH 3 H ?  ?  ?
| Heat
| | Heat ( Major ) ( Minor ) (Minor )
CH 3 OH CH 3

(c) H ?  ?
Heat ( Major ) (Minor )

conc.H SO AlCl
(d)  24  (e) 3  C
HCl, 

Q.6 + HBr 

This involves 1,4 addition of HBr followed by enolisation of keto group. Explain the mechanism.
Q.7 Arrange the following in increasing order of stability.

(a) , , , , (regarding stability of free radical)

 
(b) CH 2  C H 2 , C H 2  CH  CH 2 , , (regarding to stability of free radical)

Q.8 Supply the structure & type of intermediate species designated by

(a)   ? + H2O
+ H+  (b) CH3CH2 – N = N – CH2CH3 
  ? + N2


(c) CH3CHI2 + Zn  ? + ZnI2 (d) CH3C  CH + NaNH2  ? N a + NH3
(e) CH3CH2Cl + AlCl3  AlCl4– + ?

Q.9 Give major organic product in each case.

 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DATE : 06/10/2006 DPP. NO.-22
Q.1 As many as nine different Diels–Alder products can be obtained by heating a mixture of 1,3-butadiene
and 2-methyl-1,3-butadiene. Identify the products.

Q.2 Dr. Harry Potter treated 1,3-cyclohexadiene with Br2 and obtained two products (ignoring stereoisomers).
Her lab partner treated 1,3-cyclohexadiene with HBr and was surprised to find that only one product
was formed (ignoring stereoisomers). Account for these results.

Q.3 How would the following substitutents affect the rate of a Diels–Alder reaction?
(a) an electron-donating substituent in the diene
(b) an electron-donating substituent in the dienophile
(c) an electron-withdrawing substituent in the diene
[Ans. (a) increases, (b) decreases, (c) decreases]

Q.4 Give two sets of a conjugated diene and a dienophile that could be used to prepare the following
compound.

Q.5 Which of the following conjugated diene would not react with a dienophile in a Diels–Alder reaction?

(a) (b) (c) (d)

(e) (f) [Ans. a, d]

Q.6 What diene and what dienophile should be used to synthesize the following compounds?

(a) (b) (c) (d)

(e) (f)

Q.7
(a) How many linear dienes are there with molecular formula C6H10? (Ignore cis–trans isomers)
(b) How many of these are conjugated dienes?
(c) How many of these are isolated dienes? [Ans. 6, 2, 2]
Q.8
(a) Which dienophile is more reactive in a Diels–Alder reaction?
O O
|| ||
(i) CH 2  CHCH or CH 2  CHCH 2CH

O
||
(ii) CH 2  CHCH or CH2=CHCH3
(b) Which diene is more reactive in a Diels–Alder reaction?
CH2=CHCH=CHOCH3 or CH2 = CHCH = CHCH2OCH3
O O
|| ||
[Ans.(a)(i) CH 2  CHCH (ii) CH 2  CHCH , (b) CH2=CHCH=CHOCH3]

Q.9
(a) Give all possible products of the following reaction:

+ Br2 

(b) How many stereoisomers of each product could be obtained?

[Ans.

stereoisomers 2 2 4 ]

Q.10 The following equilibrium is driven to the right if the reaction is carried out in the presence of maleic
anhydride. What is the function of maleic anhydride?

[Ans. Maleic anhydride react with cyclopentadiene and shift equilibrium toward right]

Q.11 Give the product or products that would be obtained from each of the following reactions:

(a) + CH2= CH – CH = CH2 

CH 3
|

(b) + CH 2  CH  C  CH 2 

O
||

(c) + CH 2  CHCCH 3 
 
(d) + CH2 = CHCH2Cl 

Q.12 Explain why the following are not optically active:

(a) The product obtained from the reaction of 1,3-butadiene with cis-1,2-dichloroethene.
(b) The product obtained from the reaction of 1,3-butadiene with trans-1,2-dichloroethene.
[Ans. (a) meso, (b) racemic]

Q.13 Write mechanism for following reaction.

Br
(a) 2 
H 2O

(b) + 


Br
(c) 2

CCl 4

Q.14
(a) Give the products obtained from the reaction of one mole of HBr with one mole of 1,3,5-hexatriene.
(b) Which product(s) will predominate if the reaction is under kinetic control?
(c) Which product(s) will predominate if the reaction is under thermodynamic control?
[Ans. 3 products, CH3–CH=CH–CH=CH–CH2–Br (TC)]

Q.15 How could the following compounds be synthesized using a Diels – Alder reaction?

(a) (b)

(c) (d)
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DATE : 09/10/2006 DPP. NO.-23
Q.1 How many litres of methane would be produced when 0.6 gm of CH3MgBr is treated with excess of
C4H9NH2
(A) 0.8 litre (B) 0.08 litre (C*) 0.112 litre (D) 1.12 litre
[Sol.  119 gm MeMgBr gives 22.4 lit
22.4
 0.6 gm will give  0.6 = 0.112 lit. ]
119

( i ) CH ONH
3
Q.2 2CH3 MgBr ( 2 
ii ) H 

(A) CH3 –O–NH–CH3 (B) CH3–NH–CH3

(C*) CH3 –NH2 (D) CH3 –O–CH3
[Sol. CH3 MgBr + CH3 ONH2  CH4 + CH3 ONHMgCl


H3O
CH3 OH + CH3 –NH2   CH3NHMgCl + CH3OMgBr ]

H 3O Mg HCHO
Q.3 + MeMgCl   A HBr
 B  C  
D
Et 2O H 3O

D is :

(A) (B) (C*) (D)

O
||
Q.4 C 2 H 5 O  C  OC 2 H 5 2
CH 3MgBr
  A. Product A formed
(A) is ethyl acetate
(B) further react with CH3MgBr/H2O+ to give acetone
(C*) further react with CH3MgBr/H2O+ to give t-butyl alcohol
(D) A & B are correct

Q.5 Br–CH2–CC–CH2–Br Mg

 BrMg–CH2–CC–CH2–MgBr
( excess)
Et 2O

Product
The major product is:
(A) Br–Mg–CH2–CC–CH2–Br (B) Cyclobutyne
(C) —(CH 2  C  C  CH 2—
)n (D*) CH2 = C=C=CH2
Q.6 The number of moles of grignard reagent consumed per mole of the compound
is:
 (A*) 4 (B) 2 (C) 3 (D) 1

AgNO
Q.7 Ph  C(OH )  C(I)Me 2  3  ?
| 
Me
Major product is
OAg
|
(A) Ph C(OH)  C( NO3 )Me 2 (B) Ph C  C( NO 3 )Me 2
| |
Me Me

(C) Ph CO  CMe 3 (D*) Ph C (Me2) C O Me

CH MgBr (1 eq .)
Q.8 3   ?

The product is:

(A) (B) (C) (D*)

Q.9 Which of the following reacts with grignard reagent?

(A) Acinitro form of CH3NO2 (B) enol form of acetyl acetone
(C) Methane (D) Both (A) & (B)
CH MgBr
Q.10 CH 3CCH 2CH 2CH 2Cl 3 A, A is

||
O
CH 3
|
(A) CH 3CCH 2CH 2CH 2Cl (B) CH 3CCH 2CH 2CH 2CH 3
| ||
OH O

(C*) (D)

O O
|| ||
Q.11 CH 3CCH 2 CH 2 COCH 2 CH 3 (
i ) CH 3MgBr ( one mol )
     A, A formed in this reaction is
( ii ) H 3O 

OH O O O
| || || ||
(A) CH 3CCH 2CH 2COCH 2CH 3 (B) CH 3CCH 2CH 2CCH 3
|
CH 3
 CH 3 CH 3
| |
(C*) (D) CH 3CCH 2CH 2CCH 3
| |
OH OH

O
||

Q.12 CH3MgBr + CH2=CH  C  H H Product (1 : 4 addition). It is
3O

OH
|
(A) CH 2  CH  C  H (B) CH 2CH  CH  CH 3
| |
CH 3 OH
(C*) CH3CH2CH2CHO (D) none

Q.13  1
. Mg / ether
    Product (s)

2. CH 3CHCH2 CH 3. H 3O
| ||
OH O

Select the product from the following

I: II : CH 3CHCH 2CH III :

| ||
OH O
(A) III (B) I, III (C*) I, II (D) II, III

Q.14 On conversion into grignard followed by treatment with ethanol, how many alkyl halides (excluding
stereoisomers) would yield 2-methyl butane.
(A) 2 (B) 3 (C*) 4 (D) 5
Q.15 Propose mechanism for following conversion.

( i ) CH MgBr
 3 
( ii ) H 2O

Q.16 (A) + xCH3MgBr  xCH4

C4 H10 O4
an alcohol (A) 20 gm when reacts with RMgx 14.6 l of CH4 will formed at STP? Find value of x & give
structure formula.

Q.17 Two different grignard reagents (X) and (Y) produce

PhCH2C(CH3)2OH on reaction with P and Q respectively. Give structure of X, Y, P and Q.

Q.18 Identify (A) to (B) :

Q.19 Addition of phenyl magnesium bromide to 4-tert-butyl cyclohexanone gives two isomeric tertiary alcohol
as product. Both alcohol yield same alkene when subjected to acid-catalyzed dehydration. Suggest a
reasonable structure for these two alcohol.

4-tert-butyl cyclohexanone

Q.20 Neither of these methods of making pentan-1,4 diol will work. Explain why not – what will happen
instead?
(i) CHO–CH2–CH2–CH2–OH MeMgBr
  CH 3  CH  CH 2  CH 2  CH 2
followed by H  | |
OH OH
CH CHO
(ii) Br–CH2–CH2–CH2–OH Mg
 MgBr–CH2–CH2–CH2–OH 3

Et 2O followed by H

CH 3  CH  CH 2  CH 2  CH 2
| |
OH OH
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-24
OBJECTIVE
Q1. What is the correct order of nitration of the following compounds?

C6 H5 CH3 C6H6 C6 D 6 C6 T6 C6 H5 Br C6 H5 N R3 C6 H5 NMe2
a b c d e f g
(A) g > a > b > c > d > e > f (B) g > b > c > d > e > a > f
(C*) g > a > b = c = d > e > f (D) g > a > b > c = d > e > f
Q2. What is the correct order of o/p ratio when E+ attacks the following system?
PhF PhCl PhBr PhI
a b c d
(A*) a < b < c < d (B) a = b = c = d (C) d < c < b < a (D) d < b < a < c
Q3. Types of reaction o/p ratio
Chlorination x
Nitration y
Bromination z
sulfonation w
which of the following order is correct regarding o/p ratio?
(A) x > y > w> z (B*) x > y > z > w (C) y > x > z > w (D) x > z > y > w
Q4. What is the correct order of EAS of the following species?

a b c d e
(A) e > c > d > b > a (B) e > b > c > d > a
(C*) b > e > c > d > a (D) b = e > c > d = a
Q5. What is the order of reactivity of following hetero cyclic compound towards electrophile ?

I II III
(A) I > II > III (B) III > II > I (C) I > III > II (D*) III > I > II
Q6. Which of the following ions/ groups are o–p directing ?
(A) –CO2 – (B) –CH = CH – C HO (C) –CMe3 (D*) All

Q7. + PhCH = 3  Product (85%). Product is

AlCl


(A) (B) (C*) (D) None

Q8. In which of the following reaction tertiary butyl benzene is formed ?

(A ) + Me3 C–OH + BF3  (B) + Me2 C = CH2 + H2 SO4 

(C) + Me2 CH – CH2 Cl + AlCl3 (hot)  (D*) All

Q9. + + H2 SO4  Product

Product is
(A*) Phenolphthalein indicator (B) Fluoroscein dye
(C) Methyl orange indicator (D) Butter yellow dye
Q10. What is the crrect order of EAS of the following compounds?

(a) (b) (c) (d)

(A*) c > d > b > a (B) d > b > c > a (C) a > b > d > c (D) none
SUBJECTIVE

NBS
 A KCN
 B H2O
, HCl
 C SOCl 2 , AlCl 3
 D Zn
(
Hg ), HCl
 E Pd
Q.1 C
   F

What are A to F ?
1,3, 5 trimethyl benzene
 4  B A
CH3CO2H AlPO  CH3CO2H + D Zn Hg ) conc . HCl
Q.2  C     
AlCl
(    E
3

What are B to E ?

) Zn
( Hg )
Q.3 A(C7H6O) (gives +ve test with tollens reagent)KCN
B(C14H12O2  C (which decolorises Br2 water)
 conc. HCl
dil. HNO3
. KOH , 2. H 
D (C14H10O2) 1    E (C14H12O3)
What are A to E?

H3O PhH
Q.4 HCl
  A KCN
CH 2O
 B   C SOCl
2  D  E
ZnCl 2 AlCl 3
What are A to E?
HBr
Q.5 PhCH=CH–CH3 + Br2 (Fe)  A  B
peroxide
 NaNH 2
Q.6 A   B
n propyl bromide
H2 + Lindlar’s Catalyst
HBr
C D
+H+ +
HBr + Peroxide
1,1 diphenyl ethane
E
What are A to E?
A(C13H10NOBr)conc . HCl HCl 3  H 2SO 4
 B(C7H6O2) + CNaNO  DCu
(I ) bromide
Q.7   2   EHNO
   F
Displace CO2 only
from Na2CO3 solution one pdt.

Q.8 Conversion : 

Q.9 Conversion : 

Q.10 Conversion :  1-phenyl napthalene

Q.11 Phenol is converted into nitrophenol in presence of mixture of HNO2 & HNO3. Here HNO3 having
dual purpose. Explain with mechanism.
Q.12 When 1 methoxy 2 phenyl ethane is nitrated with nitrating mixture (HNO3 & H2SO4), p isomer is
formed as major product whereas with N2O5 results in formation of 69% o & 28% p-isomer. Explain.
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DATE : 13/01/2007 DPP. NO.-25
OBJECTIVE

Q1. +NaNO2 + HCl 

This reaction is example of

(A) Intermolecular C–N coupling (B) Intramolecular C–N coupling
(C) Intermolecular N–N coupling (D*) Intramolecular N–N coupling
Q2. For chlorination of benzene which of the following reagent should be used?
(A) BrCl (B) ICl (C*) HOCl + H+ (D) All
Q3. Which of the following statement is not correct regarding desulfonation ?
(A) All steps are reversible (B) it exhibits kinetic isotope effect
(C) Phenol when reacts with SO3 at low temperature o–product is formed & at high temperature
p–product is formed as mojor product.
(D*) none
Br2
Q4. 4  B(C8H6O4) 
A(C9H12) KMnO Fe
 C(C H BrO ) (product only). A is
8 5 4

(A) (B) (C*) (D)

Q5. What is the correct order of EAS of the following compounds ?

PhCH3 PhCH2CO2Et PhCH(CO2Et)2 PhC(CO2Et)3
I II III IV
(A*) I > II > III > IV (B) IV > III > II > I (C) I = II = III = IV (D) none
Q6. What is the order of EAS of the following compounds?
Terephthalic acid Toluene p Toluic acid p Xylene
I II III IV
(A) II > III > I > IV (B) I > III > II > IV (C) IV > I > II > III (D*) IV > II > III > I
Q7. What is the order of EAS of the following compounds?
Mesitylene mXylene p Xylene Toluene
I II III IV
  
PhCH2 N Me3 PhCH2CH2 N Me3 Ph N Me3
V VI VII
(A) I > III > II > IV > V > VI > VII (B) I > II > III > IV > VI > V > VII
(C*) I > II > III > IV > V > VI > VII (D) none

Q8. 1
. PhNMeCHO
 P. P is
2 . POCl 3

(A) (B) (C*) (D)

Q9. + + H+  P is

(A*) (B) (C) (D) none

Q10. Compound A (C7H8O) is insoluble in water, dilute HCl & aqueous NaHCO3, it dissolves in dilute
NaOH. When A is treated with Br2 water it is converted into a compound C7H5OBr3 rapidly. The
structure of A is

(A) (B) (C*) (D)

SUBJECTIVE

Q1. (i) + Me3C – COCl + AlCl3  Product

(ii) + CH2Cl2 + AlCl3 (excess)  Product

Q2. The action of Me3C–CH2Cl /GaCl3 on benzene is found to yield 100% PhCMe2CH2Me whereas in
presence of FeCl3 product is 100% PhCH2CMe3 . Explain.

Q3. Aromatic bromination catalysed by lewis acid thallium acetate l (CH3CO2)3 gives only para-isomer.
Explain.

Q4. Give a mechanism for DDT preparation from chlorobenzene & chloral in presence of H2SO4.

Q5. When aniline is subjected to Fridel Craft Alkylation in presence of catalytic amount of AlCl3 alkylation
does not occur while in presence of excess AlCl3 small amount of meta product is formed.

Q6. Phenol reacts with Br2 water to form 2, 4, 6 tribromophenol but in presence of Br2 in CS2 medium it
form either o- bromo phenol or p- bromo phenol. Give mechanism.

Q7.
SO 3
0
 160
 C

80 C SO3

Explain in terms of stability of the product with energy profile.

Q8. Phenyl triphenyl methyl ether when on treatment with acid gives p-triphenyl methyl phenol. Explain.

Q9. m-t-butyl phenol reacts with Cl2 to give trichloro derivative, with bromine gives dibromo derivative, &
with I2 to give monoiodo derivative. Explain.

Q10. Accounts for the following facts

(i) N, N dimethyl aniline reacts with HNO2 & so does the corresponding 2, 6 dimethyl derivative but the
rate of reaction of latter is much less w.r.t. former.

(ii) Ph N 2 Cl– couples with PhNMe2 but not with 2, 6 Me2 C6H3NMe2.
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-26
Q.1 Consider the reaction

OH 
+  X

here X is

(A*) (B) (C) (D) None

Q.2 Indicate the correct choice among the following.

(A) N, N, 2, 6-tetramethyl aniline is less basic than N, N–dimethylaniline

(B) C6H5 – NH – C6H5 reacts with nitrous acid to give


H
(C) Ph – O – CPh3 rearranges
   (major product)
to

(D*) N, N-dimethyl picramide is a stronger base than picramide

Q.3 In the given reaction:

C H ONa
2 5
   (P)
C 2 H 5OH

(P) is:

(A*) (B) (C) (D)

Q.4 Which one of the following compounds is most reactive for ArSN2 reaction ?

(A) (B) (C*) (D)

Q.5 In the given reaction.

i ) KOH / 
(  (P)
( ii ) CH I
3

(A) (B) (C*) (D)

Q.6 Which one of the following aryl amines undergoes diazotisation most readily?

(A) NO2 NH2 (B) Cl NH2

(C*) CH3O NH2 (D) CH3 NH2

Q.7 In the given reaction:

HNO H O / H NaNO / conc . HCl H PO

3  (A) 2  (B)  2  
 (C) 3 
2  (D)

(D) is:

(A*) (B)

(C) (D)

Q.8 The final product of the given reaction sequence

( NH ) S NaNO
  
4 2
  2
 H2O
,Cu 2O

HCl Cu ( NO 3 ) 2
0 5 C
is:

(A) (B*) (C) (D)

Q.9 In the given reaction:

(P)

(P) will be

(A) (B)

(C*) mixture of (A) and (B) (D)

Q.10 Compound A & B both having molecular weight 104.


A  C C & D having molecular weight 172.

B  D C shows geometrical isomerism but D cannot. What is the structure of A to D?

Q.11 (Subsituted electrophilic aromatic) S.E.Ar in most favourable at position

Q.12 Complete the following equations

H PO
(a) Resorcinol + MeCO2H 3 
4 


2ICl H
(b)  (c) + A 

Q.13 Which of following is in-capable to show Iodoform test

OH
|
(A) (B) Ph  CH  CH  CH  CH 3

(C*) (D)
 OH O
| || HO 
Q.14 CH 3  CH  CH 2  C  H  (A)
(Retro aldol)

Na CO ( B)
3HCHO + A 23  (82%)
40 C
Product (B) of above reaction is
CH 2  OH CHO
| |
(A) HO  CH 2  C  CH 2OH (B) HO  CH 2
 C  CHO
| |
CH 2  OH CH 2  OH

CH 2OH CHO
| |
(C*) HO  CH 2
 C  CH 2OH (D) HO  CH 2
 C  CH 2  CH 2  OH
| |
CHO CH 2  OH
Q.15 Arrange the following carbonyl compounds in decreasing order of their reactivity in nucleophilic addition
reaction.

(A) b > c > a > d (B*) b > a > d > c (C) c > b > a > d (D) c > a > d > b

Q.16 Identify (X) of the following reaction:

Caro 's acid (X)

 


(A*) (B)

(C) (D)

Q.17 Identify the product obtained in the following reaction:

Br
2  (A)
CHCl / 0C
3
 (A) (B)

(C*) (D)

Q.18 In the given reaction:

O
||
ZnCl
Cl  C  Cl + 2C6H– 2  (X)

O
||
(A) Cl  C (B)

O
||
(C) CH 3  C (D)

Q.19 Final product of the given reaction:

Br
2  (A) Mg
 /
ether
 (B) (C)
Fe

(C) is:

(A) (B) (C) (D)

Q.20 The compound that gives a lactone on heating is

(A) pentaedioic acid (B*) 4-hydroxypentanoic acid
(C) 4-aminopentanoic acid (D) 2-hydroxypentanoic acid
Q.21 Point out the wrong formulation.
AlCl
(A) C6H5 – Br + (CH3– CO)2 O 3 

AlCl
(B) + CH3 – COCl 3 
(C*) + CH3 – COCl AlCl
3 

AlCl
(D) C6H5 – NO2 + EtCl 3  no reaction
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DATE : 15/01/2007 DPP. NO.-27
Q.1 Give structure and names of the principal organic products expected from mononitration of
(a) o-nitrotoluene (b) m-dibromobenzene
(c) m-dinitrobenzene (d) o-cresol
(e) p-cresol (f) m-nitrotoluene
(g) p-xylene (p-C6H4(CH3)2) (h) terephthalic acid (p-C6H4(COOH)2)
(i) anilinium hydrogen sulfate (C6H5NH3+ HSO4—)
(j) m-cresol (m-CH3C6H4OH)
(k) p-nitroacetanilide (p-O2NC6H4NHCOCH3)

Q.2 For each of the following compounds, indicate which ring you would expect to be attacked in nitration,
and give structures of the principal product

Q.3 Conversion :

(i) to Aniline (ii) CH2 = CH–CH = CH2 

Q.4 An aromatic compound having the molecular formula C13H9OCl A reacts with hydroxylamine
hydrochloride forming a compound B. A dry ethereal solution of B on treatment with PCl5 at 0° forms a
compound C, which may be hydrolysed to equal molar quantities of benzoic acid and p-chloro-aniline.
What structure would you assign to A? Explain the reactions involved.

Q.5 Identify A – H.
H ,H O —
2
A    B HO
 C

C5 H8 O3

H ,H O SOCl —
2
D    E HO excess CH OH
   F 2  G 3 H .
 ( i ) I 2 excess
( ii ) H 

H is methyl propionate.
O O O OH O
|| || 
H ,H2O
| | 
| | |
[Sol. CH 3CCH 2COCH 3    CH 3CCH 3 HO
 CH 3CCH 2CCH 3

|
CH 3
(A) (B) (C)
C5H8O3
 O O O O O
|| || 
H ,H2O | |  | | ||
CH 3CCHCOCH 3    CH 3CCH 2CH 3 HO
 excess
 CH 3CH 2COH SOCl
 2 CH 3 CH 2 CCl
 I 2 excess
|
CH 3
(D) (E) (F) (G)

O
||
CH 3CH 2COCH 3
(H) ]

KOH Br / H 2O
Q.6 HNO
 2  HNO 3

SOCl
 A 2  B  C 2   D Sn
/ 
HCl
 E
DMF H 2O

( i ) NaH , ( ii ) MeI
G     F
( iii ) NaOH / H 2O
'G' is used for the preparation of "Antileukemia drug" 'G' is

(A) G = (B) G =

(C*) G = (D) G =

Q.7 The correct pK a a order of following compounds in aqueous solution.

  
NH 4 N H 3Me N H 2 Me 2 N HMe3
(I) (II) (III) (IV)
(A) I < II < III < IV (B) III > II > I > IV (C*) III > II > IV > I (D) II > IV > III > I
Q.8 In the given reaction
Ac O H 2O Cl
2
C6H5 – NH2  2 Y 
 X   Z
Pyridine
the product (Z) will be:
(A) o-chloroaniline (B) 2,4-dichloroaniline
(C) 2, 4, 6-trichloroaniline (D*) mixture of o and p-chloroaniline
Q.9 A rrange the following compounds in order of decreasing reactivity towards electrophilic aromatic
substitution reactions and select the correct answer from the codes.
(I) m-chloroanisole (II) p-chloroanisole (III) anisole (IV) N-methylaniline
Codes:
(A*) IV > III > I > II (B) IV > I > III > II (C) IV > II > III > I (D) IV > III > II > I
Q.10 In which compound aromatic electrophilic substitution takes place at ortho and/ or para position?

(A) C6H5 – B(OH)2 (B) C6H5 – N(CH 3 ) 2 (C*) C6H5 – C6H5 (D) C6H5 – NO2

Q.11 In the given reaction:

C6H6 + X H2SO
4  C6H5

(X) is:

(A) (B) OH (C*) Both (A) and (B) (D) Cl

Q.12 In the given reaction:

Anhy AlCl
3 
A + B   


(A) and CH3 (B) and

(C) and (D*) Both (A) and (B)

Q.13 Give structures and names of the principal products expected from the ring monobromination of each of
the following compounds. In each case, tell whether bromination will occur faster or slower than with
benzene itself.
(a) acetanilide (C6H5NHCOCH3) (b) iodobenzene
(c) sec-butylbenzene (d) N-methylaniline (C6H5NHCH3)
(e) ethyl benzoate (C6H5COOC2H5) (f) acetophenone (C6H5COCH3)
(g) ethyl phenyl ether (C6H5OC2H5) (h) diphenylmethane (C6H5CH2C6H5)
(i) benzonitrile (C6H5CN) (j) benzotrifluoride (C6H5CF3)
(k) biphenyl (C6H5–C6H5)
Q.14 Arrange the compounds of each set in order of reactivity towards electrophilic substitution. Indicate in
each set which would yield the highest percentage of meta isomer, and which would yield the lowest.
(a) C6H5N(CH3)3+, C6H5CH2N(CH3)3+, C6H5CH2CH2N(CH3)3+, C6H5CH2CH2CH2N(CH3)3+,
(b) C6H5NO2 , C6H5CH2NO2 , C6H5CH2CH2NO2
(c) C6H5CH3 , C6H5CH2COOC2H5 , C6H5CH(COOC2H5)2 , C6H5C(COOC2H5)3


Q.15 3 methyl phenol + A (C10H13O2Cl) OH
 B (C10H12O2).

B Me CHNH
2 2  C (C13H21NO2) Toliprolol. What are A to C ?

Q.16 Explain the following changes and deduce the nature of compounds A to F.

Q.17 Complete the reaction :

SOCl 
H
(i) RMgBr + C*O 2  A  B 2  C Ag
 2O
 D H
2O
E
CH N 2 2


H
(ii) PhCH=NR MeI
 ?  A + B

H 2O

(iii) R2C(OH) CH2CR2NH2 HNO 2  A + B

(A) give positive Iodoform test when R = – CH3
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-28

Q.1 Conversion : 

Q.2 Conversion : 

Q.3 Conversion : Ph  CH  C  Ph 1

. NaBrO3 , 2. NaOH
   
| ||
OH O


Q.4  ?

Q.5 An aromatic compound A (C6H5O2N) by clemmenson reduction produces B (C6H7N) which reacts
with sodium nitrite in presence of mineral acid to give C. Compound C when treated with
hypophosphorous acid gives D and this reacts with equimolar amount of bromine in presence of iron
catalyst to give E (C6H5Br). Reaction of acetic anhydride on compound B produces F. Compounds E
and F when refluxed with potassium carbonate in an inert solvent in presence of copper powder gives G
(C12H11N). Identify A, B, C, D, E, F and G. Give reactions.
Q.6 Compound A (C7H12O4) is an ester of a dicarboxylic acid and undergoes hydrolysis to give a product
which gives positive iodoform test. Compound B (C7H6O) gives oxime and hydrazone when reacts with
hydroxyl amine and hydrazine respectively, and it is synthesized by passing a mixture of carbon monoxide
and hydrogen chloride in benzene in presence of catalysts. A and B gives a condensation reaction in
presence of a base, Pyridine to give a compound C of formula (C14H16O4) which on hydrolysis and
subsequent heating produces cinnamic acid. Give structures of A, B and C. Give reactions.
Q.7 A naturally occurring compound C10H12O2 (A) is soluble in aqueous NaOH but not in NaHCO3.
Treatment of (A) (i) with methyl sulphate and aqueous NaOH gives compound (B) (C11H14O2) ;
(ii) with hot concentrated base gives compound (C) C10H12O2. Compound (B) is insoluble in aqueous
NaOH and readily decolourises dilute KMnO4 and Br2/CCl4. Treatment of (B) with hot concentrated
base gives (D) C11H14O2. Ozonolysis of (C) gives a compound that is isomeric with vanillin. Ozonolysis
of (D) gives a compound that identical with the one obtained from the treatment of vanillin with methyl
sulphate. What are structures of (A), (B), (C) and (D) ?
Q.8 A compound A (C13H11NO) on hydrolysis produces B (C7H6O2) and C (C6H7N). Compound B when
heated with sodalime produces benzene, while compound C on heating with chloroform and alkali produces
phenyl isocyanide of characteristic bad odour. Compound A is formed by treating the ethereal solution of
compound D (C13H11ON) with phosphorous pentachloride. Identify A, B, C and D. Give reactions.
Q.9 A compound A (C9H10O) on treatment with NaOH and iodine gave iodoform and a compound B
(C8H8O2) which on treatment with SOCl2 and then with NH3 gave a compound C (C8H9NO). The
compound C was treated with NaOH and Br2 to give a compound D (C7H9N) which was first treated
with cold dil. HCl and NaNO2 and then with a solution of KI to give p-iodotoluene. What are the
structures of A, B, C and D?
Q.10 On treatment with bromine, certain diarylmethanols (I) are converted into a 50 : 50 mixture of aryl
bromide (II) and aldehyde (III)
+ Br2  +
(I) (II) (III)
Whether G is –NO2, –H, –Br, or –CH3, bromine appears only in the ring containing the –OCH3 group.
The rate of reaction is affected moderately by the nature of G, decreasing along the series :
G = – CH3 > –H > –Br > – NO2.
Outline all steps in the most likely mechanism for this reaction. Show how your mechanism accounts for
each of the above facts.
 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-29
Q.1 End products of the following sequence of reaction is
CH CN
NBS
 Mg
 /
ether
  3

H3O

(A) (B) (C) (D)

Q.2 1-octene NBS

 ?
Major product in the above reaction is
(A*) 3-bromo-1-octene (B) 2-bromo octane
(C) 1-bromo octane (D) 1-bromo-2-octene

anhy
 . HI
 A
Q.3
conc
 . HI
 B
A & B are respectively:

(A) & (B) &

(C) Both (D*) Both

Q.4 OH
 Olefinic product.


Identify major product

(A*) (B) (C) (D) None of these

Q.5 + HCO2 Et EtONa

  (X)

Identify unknown (X) in above reaction

(A) (B) (C*) (D)

Q.6 Guess the product
CH3CH2CONH2 PCl
5  ?

(A*) CH3CH2–CN (B) CH3CH2COCl (C) CH3CCl2CONH2 (D) CH3CH2CONHCl

Q.7 Correct order of yield of Hofmann alkene in following reaction will be

CH 3CH 2CHCH 3 X may be F, Cl, Br or I
|
X
(A*) F > Cl > Br > I (B) I > Be > Cl > F
(C) Cl > F > Br > I (D) I > Br > F > Cl

 ( i ) CH X
Q.8 OH
 (A) (
i ) CH 3X 3
 (B)    (C).
 
( ii ) OH  ,  ( ii ) OH , 

Identify (C)

(A) (B) (C*) (D)

Q.9 EtOH
 products. Expected products are:

(I) (II) (III)

Select correct alternates

(A) I, II & III (B) I, III (C*) II, III (D) I, II

CH COOK
Q.10 CH COCl
3 3 
 A  B

(A) (B*) (C) (D) None of these


Q.1 
 Which of the following product cannot be obtained by this reaction.
Ei

(A) (B) (C) (D*)

 ORGANIC CHEMISTRY
Daily Practice Problems
CLASS : XIII (ALL) DPP. NO.-30
Q.1 Compare the two methods shown for the preparation of carboxylic acids:
Mg (i ) CO 2
Method 1: RBr   RMgBr    RCO2H
diethyl ether (ii ) H3O 

H O , HCl
Method 2 : RBr NaCN
 RCN 2  RCO2H
heat
Which of the following statements correctly describes this conversion?



(A) Both method 1 and method 2 are appropriate for carrying out this conversion.
(B) Neither method 1 nor method 2 is appropriate for carrying out this conversion.
(C*) Method 1 will work well, but method 2 is not appropriate.
(D) Method 2 will work well, but method 1 is not appropriate.
Q.2 The final product Z in the following reaction is:

( CH ) COK
NBS
 [ ]  
3 3
 [ ] Z
CCl4

(A) (B) (C*) (D)

Q.3 What are the most likely products of the reaction shown below?

3 H O
 

(A) & (B*) &

(C) & (D) &

Q.4 Identify the product (Y) in the following reaction sequence:

(i ) AlCl H PO 4 
+ 3  (X) Zn
(Hg ) / H
 (Y) 3 


(ii ) H 3O

(A*) (B)

(C) (D)
Q.5 Identify the final product.
HNO MnO 
2 5 ( i ) SOCl PO
PhCH3 3  P–O2N–C6H4CH3 
4
 (X)  2
 (Y)  (Z)
H 2SO 4 H ( ii ) NH 3
(A) P–O2N–C6H4–CO2H (B*) P–O2N–C6H4–CN
(C) P–O2N–C6H4–CONH2 (D) P–O2N–C6H4–CH2CN

Q.6 The halogen compound most likely to undergo nucleophilic substitution is in which compound

(A) (B) (C) (D*)

Q.7 'A' is a higher phenol and 'B' is an aromatic carboxylic acid. Separation of a mixture of 'A' and 'B' can be
carried out easily by having a solution of
(A) NaOH (B) Na/ether (C) Lime (D*) NaHCO3

Br
Q.8 2 substitution take place at the position.

Fe

(A) 1 (B) 2 (C*) 3 (D) both (A) & (B)

O
|| ( i ) aq.HCl
Q.9 + CH 3  C  Cl A 
 B   C

Cl2 , FeCl3 ( ii ) OH

Product 'C' is

(A) (B*) (C) (D)

Q.10

—
(i ) I  OH Br  H 2O
A  2  B excess
  2  C
( ii ) H 3O 

'C' form white precipitate compound 'C' is:

(A) (B) (C*) (D)

Q.11 NaOH
 A. A is:
excess

(A) (B*) (C) (D)

Q.12 Arrange these compounds in order of increasing SN2 reaction rate:

(A) III < I < II < IV (B) III < II < I < IV (C*) IV < III < I < II (D) III <IV < I < II

Q.13 2-bromopentane on reaction with alcoholic KOH give(s)

(A) Pent-1-ene
(B) cis and trans pent-2-ene but trans is in major yield
(C) cis and trans pent-2-ene but cis isomer is in major yield
(D*) pent-1-ene, cis pent-2-ene and trans-pent-2-ene but trans pent-2-ene is in major yield

Q.14 Match List-I with List-II (no. of products in -elimination) and select the correct answer.
List-I List-II

(a) (i) Three

(b) (ii) Zero

(c) (iii) One

(d) (iv) Two

(a) (b) (c) (d) (a) (b) (c) (d)

(A) (i) (ii) (iv) (iii) (B*) (iv) (iii) (i) (ii)
(C) (iv) (iii) (ii) (i) (D) (i) (iii) (iv) (ii)


Q.15 (A) C9H10 O2 H3O (B) + (C)
(positive lodoform) (1, 4-dihydroxy benzene)

O
||
[Ans. (A) , (B) = CH 3  C  CH 3 ]