Q 11-11-41 E-d - Organic Chemistry (Mcmurry)

Question

Which compound in each of the following pairs will react faster in an S_N2 reaction with OH^- ?

(d) ${\mathrm{CH}}_2=\mathrm{CHBr} \mathrm{or} {\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{Br}$

Step-by-Step Solution

Answer

${\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{Br}$

S N 2 reaction

It is a nucleophilic substitution reaction in which an old bond is broken and new bond is formed both processes occur simultaneously. Hence, it’s a single-step reaction.

The two reacting species are involved in the rate determining steps.

Nucleophile

Nucleophiles are negatively charged ions or the neutral species having the lone pair of electrons that are ready for bonding.

Species react faster with OH -

If the species is a better leaving group then it react faster with the nucleophile 

 ${\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{Br}$is a better leaving group because of vicinal halide whereas ${\mathrm{CH}}_2=\mathrm{CHBr}$ is unreactive in the substitution reaction

${\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{Br} \mathrm{show} \mathrm{fast} \mathrm{reaction} \mathrm{tha} {\mathrm{CH}}_2=\mathrm{CHBr}$