Question

Which is the correct decreasing order of reactivity towards $S{N}_2$ displacement of the following compounds?

• A. 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethy-1-propane> 1-Bromo-3-methylbutane> 1-Bromobutane
• B. 1-Bromo-2,2-dimethylpropane > 1-Bromobutane > 1- Bromo-2-methylbutane > 1-Bromo-3-methylbutane
• C. 1-Bromobutane > 1-Bromo-2-methylbutane> 1-Bromo2,2-dimethylpropane > 1-Bromo-3-methylbutane
• D. 1 -Bromobutane > 1 -Bromo-3-methylbutane > 1 -Bromo2- methylbutane > 1-Bromo-2,2-dimethylpr

Answer 1

Answer: (D)

Since in case of $1^{\circ }$ alkyl halides, steric hindrance increases in the order : $n$ -alkyl halides, alkyl halide with a substituent at any position other than the $\beta$ -position, one substituent at the $\beta$ -position, two substituents at the $\beta$ -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-2methylbutane $>1$ -Bromo-2,2-dimethylpropane.