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NCERT Class XII Chemistry
Chapter - Haloalkanes and Haloarenes
Questions with Solutions

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Question : 16
Total: 31
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, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane,1-Bromo-2-methyl butane, 1-Bromo-3-methylbutane.
Solution:  
SN2 reaction proceeds via the formation of transition state where the carbon atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to form, the steric interactions have to be minimum. Therefore, the most favourable substratesfor SN2 reactions are 1° alkyl halides followed by 2° and 3° alkyl halide. Order of reactivity towards SN2 : 1° > 2° > 3° > aryl halide.
Based on this, the order will be
(i)
H3C(CH2)3CH2Br
1−Bromopentane
>
CH3
Br
|
C
H(CH2)2CH3
2−Bromopentane
>
CH3
Br
|
C
|
CH3
(CH2)2CH3
2−Bromo−2−methylpentane
(ii)
H3C
CH3
|
C
HCH2CH2Br
1−Bromo−3−methylbutane
>
CH3
C
|
Br
H
CH3
|
C
HCH3
2−Bromo−3−methylbutane
>
CH3
CH3
|
C
|
Br
CH2CH3
2−Bromo−2−methylbutane
(iii)
CH3(CH2)CH2Br
(I)1−Bromobutane
>
(CH3)2CHCH2CH2Br
(II)1−Bromo−3−methylbutane
>
CH3CH2
CH3
|
C
HCH2Br
(III)1−Bromo−2−methylbutane
>
CH3
CH3
|
C
|
CH3
CH2Br
(IV)1−Bromo−2,2−dimethylpropane
In example (iii), although all the given alkyl halides are 1° but the steric hindrance around the carbon bearing the –Br atom decides the order of reactivity. More the number of bulky groups around this carbon lower will be its reactivity towards SN2.
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