Question

Arrange the compounds of each set in order of reactivity towards S_N² 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.

Answer 1

Since, due to steric reasons, the order of reactivity in SN2 reactions follows the order : 1⁰ > 2⁰ >3^0, therefore, order of reactivity of the given alkyl bromide is m1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane.

Since, due to steric reasons, the order of reactivity of alkyl halides in SN2 reaction follows the order: n1⁰> 2⁰> 3^0, therefore, the order of reactivity of the given alkyl bromides is 1-Bromo-3-methylbutane (1⁰) > 2-Bromo-3-methylbutane (2⁰) > 2-Bromo-2-methylbutane (3⁰ ).

Since, in case of 1⁰ alkyl halides, steric hindrance increases in the order: n-alkyl halides, alkyl halide with a substituent at position other than the -β - position, one substituent at the β -position, two substituents at the β-position, therefore, the reactivity decreases in the same order. Thus, the reactivity of the given alkyl bromides decreases in the order:
1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane.