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${CH}_{3} OH \overset{HI}{\to} A \overset{KCN}{\to} B \to_{∆}^{H_{3} O^{\oplus}} C \overset{{LiAlH}_{4}, H_{2} O}{\to} D$
D is

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CH₃OH

CH₃ - O - CH₃

CH₃-CH₂-OH

CH₃COOH

answer is C.

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Detailed Solution

$A = {CH}_{3} I, B = {CH}_{3} CN, C = {CH}_{3} COOH, D = {CH}_{3} {CH}_{2} OH$

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