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(A) The reaction of $2$-bromo-$3$-methylbutane with methanol $(CH_3OH)$ proceeds via an $S_N1$ mechanism.First,the $Br^-$ ion leaves to form a seconda

Vedclass · Accepted Answer

$CH_3-C(CH_3)(OCH_3)-CH_2-CH_3$

Vedclass · Answer

(A) The reaction of $2$-bromo-$3$-methylbutane with methanol $(CH_3OH)$ proceeds via an $S_N1$ mechanism.First,the $Br^-$ ion leaves to form a secondary carbocation $CH_3-CH(CH_3)-C^+H-CH_3$.This carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation $CH_3-C^+(CH_3)-CH_2-CH_3$.Finally,the nucleophile $CH_3OH$ attacks the tertiary carbocation,followed by deprotonation,to yield $2$-methoxy-$2$-methylbutane $(CH_3-C(CH_3)(OCH_3)-CH_2-CH_3)$ as the major product.

The major product of the following reaction is $CH_3-CH(CH_3)-CH(Br)-CH_3 \xrightarrow{CH_3OH}$

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