Benzyl bromide

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

reacts rapidly with

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

to afford benzyl methyl ether

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

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

$$1 ^ { \circ }$$

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

$$S_N1$$

mechanism. Would you expect the para-substituted benzylic halides

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

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