Benzyl bromide (C6H5CH2Br) reacts rapidly with CH3OH to afford benzyl methyl ether (C6H5CH2OCH3). Draw a stepwise mechanism for the reaction, and explain why this 1° alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an SN1 mechanism. Would you expect the para-substituted benzylic halides CH3OC6H4CH2Br and O2NC6H4CH2Br to each be more or less reactive than benzyl bromide in this reaction? Explain your reasoning.

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Asked Dec 29, 2019
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Benzyl bromide (C6H5CH2Br) reacts rapidly with CH3OH to afford benzyl methyl ether (C6H5CH2OCH3). Draw a stepwise mechanism for the reaction, and explain why this 1° alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an SN1 mechanism. Would you expect the para-substituted benzylic halides CH3OC6H4CH2Br and O2NC6H4CH2Br to each be more or less reactive than benzyl bromide in this reaction? Explain your reasoning.

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Expert Answer

Step 1

The first step involves the attack of Br ion via SN1 pathway to form the carbocation. The next step will involve the attack of nucleophile to for...

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