Question -
Answer -
(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-methylbutane, 1-Bromo-3-methylbutane.
Answer
(i)
An SN2 reaction involves theapproaching of the nucleophile to the carbon atom to which the leaving group isattached. When the nucleophile is sterically hindered, then the reactivitytowards SN2 displacement decreases. Due to the presence ofsubstituents, hindrance to the approaching nucleophile increases in thefollowing order.
1-Bromopentane < 2-bromopentane <2-Bromo-2-methylbutane
Hence, the increasing order of reactivitytowards SN2 displacementis:
2-Bromo-2-methylbutane < 2-Bromopentane <1-Bromopentane
(ii)
Since steric hindrance in alkyl halidesincreases in the order of 1° < 2° < 3°, the increasing order ofreactivity towards SN2 displacement is
3° < 2° < 1°.
Hence, the given set of compounds can bearranged in the increasing order of their reactivity towards SN2 displacementas:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane <1-Bromo-3-methylbutane
[2-Bromo-3-methylbutane is incorrectly given in NCERT]
(iii)
The steric hindrance to the nucleophile inthe SN2 mechanism increases with a decrease in the distance ofthe substituents from the atom containing the leaving group. Further, thesteric hindrance increases with an increase in the number of substituents.Therefore, the increasing order of steric hindrances in the given compounds isas below:
1-Bromobutane < 1-Bromo-3-methylbutane <1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity ofthe given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane <1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane