Ncert Exercises & Solutions

10.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement:

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,
1-Bromo-3-methylbutane.

(i)

$C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} - B r C H_{3} \overset{B r}{\overset{|}{C}} H_{2} C H_{2} C H C H_{3} C H_{3} C H_{2} {C_{|}^{|}}_{C H_{3}}^{B r} C H_{3}$
$1 - B r o m o p e n \tan e (1 °) 2 - B r o m o p e n \tan e (2 °) 2 - B r o m o - 2 - m e t h y l b u \tan e (3 °)$

An S $_{N}$ 2 reaction involves the approaching of the nucleophile to the carbon atom to which
the leaving group is attached. When the nucleophile is sterically hindered, then the
reactivity towards S $_{N}$ 2 displacement decreases. Due to the presence of substituents,
hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards S $_{N}$
2 displacement is:
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

(ii) $C H_{3} B r B r$
$| | |$
$C H_{3} C H C H_{2} C H_{2} B r C H_{3} C C H_{2} C H_{3} C H_{3} C H C H C H_{3}$
$1 - B r o m o - 3 - 2 - B r o m o - 2 2 - B r o m o - 3$
$m e t h y l b u \tan e - M e t h y l b u \tan e - m e t h y l b u \tan e$
$(1 °) (3 °) (2 °)$

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing
order of reactivity towards SN2 displacement is 3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their
reactivity towards S $_{N}$ 2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane [2-
Bromo-3-methylbutane is incorrectly given in NCERT]

iii) $C H_{3} C H_{3}$
$| |$
$C H_{3} - C H_{2} - C H - C H_{2} - B r C H_{3} - C - C H_{2} - B r$
$|$
$C H_{3}$
$l - B r o m o - 2 - m e t h y l b u \tan e l - B r o m o - 2, 2 - d i m e t h y l p r o p a n e$

The steric hindrance to the nucleophile in the S $_{N}$ 2 mechanism increases with a decrease in
the distance of the substituents from the atom containing the leaving group. Further, the
steric hindrance increases with an increase in the number of substituents. Therefore, the
increasing order of steric hindrances in the given compounds is as below:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards S $_{N}$ 2 displacement
is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane
< 1-Bromobutane

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions