CH_3-C(CH_3)_2-Cl xrightarrow{NaCN} A xrighta | vedclass

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Question

(B) Tertiary alkyl halides like $CH_3-C(CH_3)_2-Cl$ undergo elimination $(E2)$ when treated with $NaCN$ (which acts as a base) to form isobutylene $(C

Vedclass · Accepted Answer

$CH_3-C(CH_3)_2-OH$

Vedclass · Answer

(B) Tertiary alkyl halides like $CH_3-C(CH_3)_2-Cl$ undergo elimination $(E2)$ when treated with $NaCN$ (which acts as a base) to form isobutylene $(CH_2=C(CH_3)_2)$.Subsequent hydration with $dil. H_2SO_4$ follows Markovnikov's rule to yield tert-butyl alcohol $(CH_3-C(CH_3)_2-OH)$.Reaction:$CH_3-C(CH_3)_2-Cl$ $\xrightarrow{NaCN} CH_2=C(CH_3)_2 (A)$ $\xrightarrow{dil. H_2SO_4} CH_3-C(CH_3)_2-OH (B)$

$CH_3-C(CH_3)_2-Cl$ $\xrightarrow{NaCN} A$ $\xrightarrow{dil. H_2SO_4} B$. Compound $B$ is:

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