A nitrile X is treated with LiAlH_{4} to obta | vedclass

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(C) $1$. The nitrile $X$ is $CH_{3}CN$ (acetonitrile or ethanenitrile).$2$. Reduction of $CH_{3}CN$ with $LiAlH_{4}$ gives $CH_{3}CH_{2}NH_{2}$ (ethan

Vedclass · Accepted Answer

$(CH_{3}COO^{-})(CH_{3}CH_{2}NH_{3}^{+})$

Vedclass · Answer

(C) $1$. The nitrile $X$ is $CH_{3}CN$ (acetonitrile or ethanenitrile).$2$. Reduction of $CH_{3}CN$ with $LiAlH_{4}$ gives $CH_{3}CH_{2}NH_{2}$ (ethanamine),which is $Y$ $(C_{2}H_{7}N)$.$3$. Acidic hydrolysis of $CH_{3}CN$ gives $CH_{3}COOH$ (ethanoic acid),which is $Z$.$4$. When $Y$ (an amine,$CH_{3}CH_{2}NH_{2}$) and $Z$ (a carboxylic acid,$CH_{3}COOH$) are mixed,an acid-base reaction occurs to form a salt: $CH_{3}COOH + CH_{3}CH_{2}NH_{2} \rightarrow CH_{3}COO^{-}CH_{3}CH_{2}NH_{3}^{+}$.$5$. Thus,the product is $(CH_{3}COO^{-})(CH_{3}CH_{2}NH_{3}^{+})$.

$A$ nitrile $X$ is treated with $LiAlH_{4}$ to obtain compound $Y$ $(C_{2}H_{7}N)$. In a separate reaction,$X$ is hydrolyzed in an acidic medium to obtain $Z$. The product obtained after mixing $Y$ and $Z$ will be:

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