$C{H}_3-COOH\stackrel{C{l}_2\left(Onemole\right)\mathrm{Re}d\text{\hspace{0.17em}\hspace{0.17em}}P}{\to }X\stackrel{alc.KCN}{\to }Y$ $\stackrel{H_{2}O/H^{+}}{\to }Z;$identify ‘Z’ in The above reaction

Solution:

Step 1: Formation of X

Reaction: CH₃-COOH + Cl₂/Red P → ClCH₂-COOH

This is the Hell-Volhard-Zelinsky (HVZ) reaction. Acetic acid reacts with chlorine in the presence of red phosphorus, where one α-hydrogen is replaced by chlorine.

X = ClCH₂-COOH (Chloroacetic acid)

Step 2: Formation of Y

Reaction: ClCH₂-COOH + alc. KCN → NCCH₂-COOH

Chloroacetic acid undergoes nucleophilic substitution with alcoholic potassium cyanide. The chlorine atom is replaced by the cyanide group (-CN).

Y = NCCH₂-COOH (Cyanoacetic acid)

Step 3: Formation of Z

Reaction: NCCH₂-COOH + H₃O⁺/H⁺ → HOOC-CH₂-COOH

Cyanoacetic acid undergoes acid hydrolysis. The cyanide group (-CN) is hydrolyzed to a carboxylic acid group (-COOH).

Z = HOOC-CH₂-COOH (Malonic acid)

Final Answer:

Z is Malonic acid (HOOC-CH₂-COOH) or Propanedioic acid.