A nitrile X is treated with LiAlH_4 to obtain compound Y (C_2H_7N). In a separate reaction X is hydrolyzed in an acid medium to obtain

Q: A nitrile $X$ is treated with $LiAlH_4$ to obtain compound $Y (C_2H_7N).$ In a separate reaction $X$ is hydrolyzed in an acid medium to obtain $Z$. The product obtained after mixing $Y$ and $Z$ will be

c $\mathrm{X} \frac{\mathrm{LiAlH}_{4}}{\text { Reduction }} \mathrm{C}_{2} \mathrm{H}_{7} \mathrm{N}($ Amine $=\mathrm{Y})$ $\mathrm{X}$ is $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{N}$ (Cyanomethane) $\mathrm{X} \stackrel{\mathrm{H}}{\longrightarrow} \mathrm{Z}$ $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{N} \stackrel{\mathrm{H}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOH}$ $\mathrm{Z}=\mathrm{CH}_{3} \mathrm{COOH}$ Product after mixing $\mathrm{Y}$ and $\mathrm{Z}$ is $\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)+\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}\right)$ (Substituted Ammonium Carboxylate Salt)

SECTION - A [CHEMISTY - MCQ]

GSEB - English Medium

A nitrile $X$ is treated with $LiAlH_4$ to obtain compound $Y (C_2H_7N).$ In a separate reaction $X$ is hydrolyzed in an acid medium to obtain $Z$. The product obtained after mixing $Y$ and $Z$ will be

A$CH_3 CONHCH_2CH_3$
B$CH_3 CH_2CONHCH_2CH_3$
C$(CH_3COO^-)(CH_3CH_2NH_3^+)$
D$(CH_3CH_2COO^- )(CH_3 NH^+_2)$

Difficult

STD 11 Science
JEE
STD 12 - 9. Amines
Chemistry

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