In the given reaction: CH_3-CHCl-COOH xrighta | vedclass

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Question

(A) In this reaction,the chlorine atom at the $\alpha$-position of the carboxylic acid is substituted by an amino group $(-NH_2)$ via nucleophilic sub

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$\alpha$-Amino acid

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(A) In this reaction,the chlorine atom at the $\alpha$-position of the carboxylic acid is substituted by an amino group $(-NH_2)$ via nucleophilic substitution using ammonia $(NH_3)$. $CH_3-CHCl-COOH + NH_3 \rightarrow CH_3-CH(NH_2)-COOH + HCl$ The resulting product $[X]$ is $2$-aminopropanoic acid (Alanine),which is an $\alpha$-amino acid.

In the given reaction: $CH_3-CHCl-COOH \xrightarrow[(ii) H_2O]{(i) NH_3 (1 \text{ mole})} [X]$. Product $[X]$ will be:

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