Question

Benzyl bromide

(C6H5CH2Br)\left( \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { CH } _ { 2 } \mathrm { Br } \right)

reacts rapidly with

CH3OH\mathrm { CH } _ { 3 } \mathrm { OH }

to afford benzyl methyl ether

(C6H5CH2OCH3).\left( \mathrm { C } _ { 6 } \mathrm { H } _ { 5 } \mathrm { CH } _ { 2 } \mathrm { OCH } _ { 3 } \right).

Draw a stepwise mechanism for the reaction, and explain why this

11 ^ { \circ }

alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an

SN1S_N1

mechanism. Would you expect the para-substituted benzylic halides

CH3OC6H4CH2Br and O2NC6H4CH2Br\mathrm { CH } _ { 3 } \mathrm { OC } _ { 6 } \mathrm { H } _ { 4 } \mathrm { CH } _ { 2 } \mathrm { Br } \text { and } \mathrm { O } _ { 2 } \mathrm { NC } _ { 6 } \mathrm { H } _ { 4 } \mathrm { CH } _ { 2 } \mathrm { Br }

to each be more or less reactive than benzyl bromide in this reaction? Explain your reasoning.

Solution

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Answered 1 year ago
Answered 1 year ago
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In the first step, there happens departure of bromide ion via S$_N$1 pathway to form carbocation which in the second step gets nucleophilically attacks by methanol to form ether compound having a positive charge.

The above formed compound in the final step undergoes deprotonation to form benzyl methyl ether.

Below image shows the mechanism for the same :-

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