Let’s break down the reaction sequence step by step:

Identification of the starting compound (C₃H₅N):

A three‐carbon nitrile with the formula $$\mathrm{C_3H_5N}$$ is propionitrile (also called propanenitrile), which has the structure

$$\mathrm{CH_3CH_2CN}.$$

The degree of unsaturation, calculated by

$$\frac{(2 \times 3 + 2 + 1 - 5)}{2} = 2,$$

is consistent with the presence of a triple bond in the –CN group.

Hydrolysis to form compound X:

Nitriles hydrolyze to carboxylic acids. Thus, hydrolyzing $$\mathrm{CH_3CH_2CN}$$ converts the –CN group into –COOH, giving propionic acid:

$$\mathrm{CH_3CH_2CN \xrightarrow{H_2O/H^+} CH_3CH_2COOH.}$$

This carboxylic acid is compound X.

Formation of compound Y using Cl₂/P:

When a carboxylic acid is treated with halogens (in the presence of phosphorus, as in the Hell–Volhard–Zelinsky reaction), halogenation occurs at the alpha carbon (the carbon adjacent to the carboxyl group).

For propionic acid, the alpha carbon is the middle carbon. Halogenation replaces one of its hydrogens to give 2‑chloropropanoic acid:

$$\mathrm{CH_3CH_2COOH \xrightarrow{Cl_2/P} CH_3CHClCOOH.}$$

This acid is compound Y.

Conversion of compound Y to compound Z with NH₃:

The alpha‑chloro acid (2‑chloropropanoic acid) can undergo nucleophilic substitution when reacted with ammonia, where the chlorine is replaced by an amino group:

$$\mathrm{CH_3CHClCOOH + NH_3 \to CH_3CH(NH_2)COOH.}$$

The product is 2‑aminopropanoic acid, which is the amino acid alanine, and is compound Z.

To summarize, the sequences are:

$$\mathrm{CH_3CH_2CN \xrightarrow{hydrolysis} CH_3CH_2COOH\, (X)}$$

$$\mathrm{CH_3CH_2COOH \xrightarrow{Cl_2/P} CH_3CHClCOOH\, (Y)}$$

$$\mathrm{CH_3CHClCOOH \xrightarrow{NH_3} CH_3CH(NH_2)COOH\, (Z)}$$

Thus, the final compound Z is 2‑aminopropanoic acid.

Answer: Option A.

Compounds Containing Nitrogen

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