What are A,B,C in the following reaction?
$\begin{array}{l} {CH}_{3} COOH \overset{{LiAlH}_{4} / {(C_{2} H_{5})}_{2} O}{\to} A \\ {CH}_{3} COONa \to_{Δ}^{NaOH / CaO} B \\ {CH}_{3} COOH \overset{P_{2} O_{5} / Δ}{\to} C \end{array}$

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$A - C H_{3} C H_{2} O H; B - C H_{4}; C - {(C H_{3} C O)}_{2} O$

$A - C H_{3} C H O; B - C H_{3} C O C H_{3}; C - C H_{3} O C H_{3}$

$A - C H_{3} C H_{2} O H; B - C H_{3} C H_{3}; C - {(C H_{3} C O)}_{2} O$

$A - C H_{3} C H O; B - C H_{3} C H_{3}; C - {(C H_{3} C O)}_{2} O$

answer is B.

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Detailed Solution

$C H_{3} - C O O H \to_{{(C_{2} H_{5})}_{2} O}^{L i A l H_{4}} \underset{A}{C H_{3} - C H_{2} - O H}$

${CH}_{3} - COONa \to_{Δ}^{NaOH - CaO} \underset{B}{{CH}_{4}}$

$C H_{3} - C O O H \to_{}^{P_{2} O_{5} / ∆} \underset{C}{C H_{3} - C O - O - C O - C H_{3}}$

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