1. Chemistry : General Organic Chemistry & Isomerism
SECTION - I
SUBJECTIVE TYPE PROBLEMS
Problem 1 : Write the IUPAC name of the following compounds
CH _ C = C _ CH _ CH _ C CH
3
Cl CH3CH C H
3 2 5
Solution : 6-Chloro-3-ethyl-4,5-dimethyl-5-hepten-1-yne
Problem 2 : Name the following compounds as per the IUPAC rules
(1) (1)
Solution : (1) 2-Hexen-4-yne
(2) pent-1-ene-3-yne or 1-penten-3-yne
Problem 3 : Write down the structure for the given names
(1) 3,3,4,4-tetramethyl heptane (2) 3-ethyl-2-methylhexane
Solution : (1) CH2
6
CH3
7
CH2
5
C
4
CH3
CH3
C
3
CH3
CH3
CH2
2
CH3
1
(2) CH2
6
CH2
5
CH2
4
CH
3
CH2
2
CH3
1
CH3 C2H5
3, 3, 4, 4-tetramethyl heptane
Problem 4 : Write the correct order of stability of the given carbocation.
(1)  
6 5 3
C H C

(2)
2 2
CH CH CH

 
(3)  
6 5 2
C H CH

(4) 6 5 2
C H CH


(5) 3
CH

(6)
2
CH CH


Solution :    
6 5 6 5 6 5 2 2 2
3 2
C H C C H CH C H CH CH CH CH
   
       3 2
CH CH CH
 
 
Problem 5 : Arrange the following in their decreasing order of acidity and explain.
CHF3
, CHCl3
,
Solution : 3 3
3 3
CHCl CCl H
CHF CF H




 
 
3
CCl

is more stable than 3
CF

because of dispersion of lone pair on carbon to the vacant d-
orbital in chlorine. that is why 3
CHCl can loose H+
more easily.
.

2. Chemistry : General Organic Chemistry & Isomerism
Problem 6 : Which of the ion is most stable and why ?
I CH2 Br CH2 Cl CH2
,
,
Solution : The most stable carbanion is 2
ClCH

, due to the maximum – I effect of chlorine.
Problem 7 : The Singlet Carbenes are given as CBr2
, CF2
, CCl2
. Arrange them in their decreasing
order of stability.
Solution : CF2
> CCl2
> CBr2
This is because of  
2p 2p
   back bonding in 2
CF , 
2p 3p
   back bonding in 2
CCl
and  
2p 4p
   back bonding in 2
CBr .
Problem 8 : Which of the amine (in gaseous state) is most basic and why ?
CH3
NH2
, (CH3
)2
NH, F – CH2
NH2
, CH3
– CH2
NH2
Solution : (CH3
)2
NH is the most basic amongs the given amines because the + I effect of CH3
groups
increases the availability of electron on nitrogen.
Problem 9 : Write the resonance structures of CH2
= CH – CHO and arrange them in order of
decreasing stability.
Solution : CH = CH __ CH = O
2 :
:
I
CH CH = CH O
2
_ _ :
:
II
+
:
: CH CH = CH O
2
_ _ +
:
:
III
Structure I is most stable since each C and O atom has an octet of electron and neither of
these carries a charge.
Stucutres II and III both involved separation of charge and hence both are less stable than
structure I.
However, structure II is more stable than structure III since it carries a –ve charge on the
more electronegative O atom and +ve charge on the less electronegative C atom while in
structure III, the more electronegative O atom carries the +ve charge while the less
electronegative C atom carries –ve charge. Thus the decreasing order of stability is : I > II
> III
Problem 10 : Identify the electrophile and nucleophile amongst the given species.
NH3
, BeCl2
, BCl3
, 2
CF , CH3
–
, SO3
, H2
O
Solution : Electrophiles - BeCl2
,BCl3
, SO3
, 2
CF
Nucleophile - NH3
, H2
O, CH3
–

3. Chemistry : General Organic Chemistry & Isomerism
SECTION - II
SINGLE CHOICE PROBLEMS
Problem 1 : Which among the following compounds will show tautomerism
(a) CH3
– CH = NH (b) 3 3
O
||
CH C CH
 
(c) 3 2
3
CH CH NO
|
CH
  (d) All of these
Solution : Each is containing the  -H with respect to functional group.
Problem 2 : In which compound, cis-trans nomenclature cannot be used ?
(a) CH3
– CH = CH – CH3
(b) CH3
– CH = CH – COOH
(c) C =
= C
CH3
C H5
2
Cl
Br
(d) C6
H5
– CH = CH – CHO
Solution : For this compound E-Z nomenclature will be used.
Problem 3 : Which among the following compound will have meso form :
(a) CH2
OH – CHOH – CHOH – CHO (b) CH2
OH – CHOH – CHOH – COOH
(c) CH2
OH – (CHOH)2
– CH2
OH (d) C6
H5
– CHCl – CHOH – CH3
Solution : Due to the plane of symmetry.
Problem 4 : Which among the following compounds will be dissymmetric but not asymmetric :
(a)
3
COOH
|
H C OH
|
H C OH
|
CH
 
 
(b) 3
2 5
OH
|
CH C COOH
|
C H
 
(c)
COOH
|
H C Br
|
H C Br
|
COOH
 
 
(d) C C C
 
3
CH
H
3
CH
H
Solution : (d)
Problem 5 : Number of configurational isomers of the compound
2 2
* * *
CH OH CHOH CHOH CHOH CH OH
   
(a) 8 (b) 2
(c) 4 (d) 6
Solution : (c)

4. Chemistry : General Organic Chemistry & Isomerism
Problem 6 : How many geometrical isomers are possible for given compound
C6
H5
– CH = CH – CH = CH – COOH
(a) 3 (b) 4
(c) 2 (d) 1
Solution : Due to the presence two double bonds with respect to the different groups.
Problem 7 : Meso tartaric acid is optically inactive due to the presence of
(a) Two chiral carbons (b) Molecular asymmetry
(c) Internal compensation (d) External compensation
Solution :
H
_
C
_
OH
COOH
H
_
C
_
OH
COOH
Problem 8 : Angle strain in which compound is maximum
(a) Propane (b) Cyclopropane
(c) n-butane (d) Cyclobutane
Solution : In cyclopropane the bond angle ( CCC
 ) is 60o
, which creates the strain.
Problem 9 : Optical acitivity is measured by
(a) Polarimeter (b) Refracto meter
(c) Both (d) Ordinary reaction
Solution : (a)
Problem 10 : In which of the following compounds carbon-carbon single bond length will be
maximum?
(a) CH3
– CH2
– CH3
(b) CH3
– CH = CH – CH3
(c) 3 3
3 3
CH C C CH
| |
CH CH
   (d) CH2
= CH2
Solution : In the alkene due to the hyperconjugation the bond length decreases for the C – C.

5. Chemistry : General Organic Chemistry & Isomerism
SECTION - III
MULTIPLE CHOICE PROBLEMS
Problem 1: Which of the given will show hyperconjugation?
(a) 3
3
CH —CH —CH C—
|
|
CH
 (b) 3
CH — CH C—
|

(c)
3
3
3
CH
|
CH — C — CH C—
|
|
CH
 (d) 3 2
CH — CH — CH C—
|

Solution : (a, b, d)
Problem 2: What products can be expected in the following reaction?
1
N
2 5
3
S
3 2 C H OH
3
CH
|
CH —C —CH Br Products
|
CH


(a)
3
3 3
CH
|
CH —C CH — CH
 (b)
3
3 2 3
2 5
CH
|
CH —C —CH CH
|
OC H
(c) both are correct (d) none is correct
Solution : (c)
Problem 3:
OCH3
Br
2
NaNH
Product



Product (P) and reaction (R) are :
(a)
OCH3
NH2
, cine substitution (b)
OCH3
NH2
,eliminationaddition
(c)
OCH3
NH2
, elimination addition (d)
OCH3
NH2
, cine substitution
Solution : (a, b)

6. Chemistry : General Organic Chemistry & Isomerism
Problem 4: Which of the following is not found in alkenes?
(a) Geometrical isomerism (b) Metamerism
(c) Position isomerism (d) Chain isomerism
Solution : (b)
Problem 5: How many functional isomers are present for C2
H4
O2
?
(a) 3 (b) 2
(c) 5 (d) 4
Solution : (b)
Problem 6: How many optically active isomers of tartaric acid are possible?
CH(OH)COOH
|
CH(OH)COOH
(a) 3 (b) 2
(c) 4 (d) 5
Solution : (b)
Problem 7: Which compounds can show tautomerism?
(a) HCN (b) CH3
CH2
NO2
(c) O (d) C
O
CH3
Solution : (a, b, c, d)
Problem 8: The compounds
C
H3 H
H
CH3
CH3
H
H CH3
CH3
H
H
CH3
(a) diastereomers (b) geometrical isomers
(c) enantiomers (d) same structure
Solution : (c)
Problem 9: ThetotalnumberofalkylradicalspossiblefortheformulaC4H9 is
(a) 3 (b) 4
(c) 5 (d) 6
Solution : (b)
Problem 10: How many optically active stereoisomers are possible for butane-2, 3-diol?
(a) 1 (b) 2
(c) 3 (d) 4
Solution : (b)

7. Chemistry : General Organic Chemistry & Isomerism
MISCELLANEOUS PROBLEMS
SECTION - IV
COMPREHENSION TYPE PROBLEMS
Write up – I
Kinetics of E2 reaction is second order. The reaction is first order with respect to the substrate and first
order with respect to the base. In this reaction, product formation takes place by Hofmann and Saytzeff
rule. In this reaction both leaving groups should be anti periplanar.
Problem 1: Which one of the following pairs is not correctly matched?
(a) 3 2 3 2 5
Br
|
CH — CH —CH — CH / C H O; Saytzeff

(b) 3 2 3 2 5
F
|
CH — CH —CH —CH /C H O;Saytzeff

(c) 3 2 3 3 3
Br
|
CH — CH —CH —CH /(CH ) CO; Hofmann

(d) 3 2 3 3
3
CH — CH —CH — N(CH ) OH; Hofmann
|
CH
 
Solution : (b)
Problem 2: In the given reaction :
alk. KOH/
3 2 2
Br
|
CH — CH —CH —CH CH [X]

 

[X] will be :
(a) CH3
—CH2
—CH=C=
=CH2
(b) CH2
=
=CH—CH2
—CH=
=CH2
(c) CH3
—CH=
=CH—CH=
=CH2
(d) CH3
—CH2
—CH2
—CCH
Solution : (c)

8. Chemistry : General Organic Chemistry & Isomerism
Problem 3: Consider the following alkyl halides.
(1) CH3
—CH2
—CH2
—CH2
—Br (2) 3 2 3
Br
|
CH — CH —C —CH
(3) 3 2 3
3
Br
|
CH —C— CH — CH
|
CH
Arrange these compounds in decreasing order of reactivity for E2 reactions and select the
correct answer from the codes given below :
Codes :
(a) 1 > 2 > 3 (b) 3 > 2 > 1
(c) 2 > 1 > 3 (d) 3 > 1 > 2
Solution : (b)
Write up – II
Type of isomerism occuring by the migration of atom (generally acidic hydrogen) and the movement of
a double bond is called tautomerism. Tautomers are true isomers. The most common tautomerism is
keto-enol tautomerism. This is an example of 1, 3-tautomerism. The keto form is more stable than the
enol form. However some enols are more stable than keto form. Enol from is more stable than the keto
form either due to resonance or due to hydrogen bonding or due to the nature of solvent.
Problem 4: Which of the following compounds does not exist in the enol form?
(a)
O
O
(b)
O
NH
(c) Both (a) and (b) (d) 3 2 2 5
O
||
CH —C —CH — COOC H
Solution : (b)

9. Chemistry : General Organic Chemistry & Isomerism
Problem 5: Which of the compounds exist mainly in the enol form?
(a)
O
O
(b)
O
(c)
O
O O
(d) all of these
Problem 6: Which of the compound will show tautomerism?
(a) 3 3
O
||
CH —C —CH (b)
O
(c)
O
CH3
(d) all of these
Solution : (b)
MATCHING TYPE PROBLEM
7. Column (I) Column (II)
(P) cis-2-kulene + Br2
(A) Free radical
(Q) trans-2-butene + Br2
(B) Ionic reaction
(R) Cu2
=Cu2
(Cu2
–Cu2
)n
(C) Resolvable product
(S) C6
H5
CH3
+ Cl2
 C6
H5
CH2
Cl (D) Non-resolvable product
Sol.
(P) (Q) (R) (S)
(a) A,C B, D A A
(b) B, C B, D B C, D
(c) A, C B, C B C, D
(d) B, C B, D A A

10. Chemistry : General Organic Chemistry & Isomerism
ASSERTION-REASON TYPE PROBLEMS
The question given below consist of an ASSERTION and the REASON. Use the following key for the
appropriate answers
(a) If both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(b) If both Assertion and Reason are correct and Reason is not the correct explanation for
Assertion
(c) If Assertion is correct but Reason is not correct.
(d) If Assertion is incorrect but Reason is correct.
8. Assertion : Alkyl free radical is more stable than vinyl free radical.
Reason : Allyl free radical is stabilized by resonance.
Solution : (a)
9. Assertion : meso-Tartaric acid is optically inactive.
Reason : meso-Tartaric acid contains two asymetric carbon atoms.
Solution : (b)
10.Assertion : p-Xylene has zero dipole moment.
Reason : In p-Xylene all the carbon atoms are in sp2-hybrid state.
Solution : (c)