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Question: The final product (B) in the reaction is; \(C{{H}_{3}}-C{{H}_{2}}-COOH\xrightarrow[\Delta ]{Br/re...

The final product (B) in the reaction is;
CH3−CH2−COOH→ΔBr/red P(A)→ Alc.. NH3(B)C{{H}_{3}}-C{{H}_{2}}-COOH\xrightarrow[\Delta ]{Br/red\text{ }P}(A)\xrightarrow{\text{ Alc.}\text{. }N{{H}_{3}}}(B)
(1) Alanine
(2 Pyruvic acid
(3) Citric acid
(4) Lactic acid

Explanation

Solution

When the given acid i.e. propanoic acid is treated with the bromine in the presence of the small amount of red phosphorous it results in the formation of the alpha compound i.e. alpha-halo carboxylic acid and when this compound is made to react with the ammonia, the resulting compound consists of the −NH2-N{{H}_{2}}group in place of the halogen atom. Now you can easily answer the given statement accordingly. Solve it.

Complete answer:
When the given compound which consists of a carboxylic group as the functional group i.e. acid reacts with the bromine in the presence of the small amount of red phosphorus, the compounds in which the alpha-hydrogen atoms get replaced by the halogen atoms. The chemical reaction takes place as;
CH3−CH2−COOH propanoic acid →ΔBr/red PCH3−CH(Br)−COOH Bromopropanoic acid (A)  \begin{matrix} C{{H}_{3}}-C{{H}_{2}}-COOH \\\ propanoic\ acid \\\ \end{matrix}\xrightarrow[\Delta ]{Br/red\text{ }P}\begin{matrix} C{{H}_{3}}-CH(Br)-COOH \\\ \begin{aligned} & Bromopropanoic\text{ }acid \\\ & (A) \\\ \end{aligned} \\\ \end{matrix}
When this alpha halo carboxylic acid i.e. Bromo propanoic acid is made to undergo reaction with the alcoholic ammonia, halogen i.e. bromine is replaced with the −NH2-N{{H}_{2}} group and thus, results in the formation of alanine. The chemical reaction takes place as;
CH3−CH(Br)−COOH Bromopropanoic acid (A)  →Alc. NH3CH3−CH(NH2)−COOH Alanine \begin{matrix} C{{H}_{3}}-CH(Br)-COOH \\\ \begin{aligned} & Bromopropanoic\text{ }acid \\\ & (A) \\\ \end{aligned} \\\ \end{matrix}\xrightarrow{Alc.\text{ N}{{\text{H}}_{3}}}\begin{matrix} C{{H}_{3}}-CH(N{{H}_{2}})-COOH \\\ Alanine \\\ \end{matrix}
Thus, the overall reaction occur as;
CH3−CH2−COOH→ΔBr/red PCH3−CH(Br)−COOH→ Alc. NH3CH3−CH(NH2)−COOHC{{H}_{3}}-C{{H}_{2}}-COOH\xrightarrow[\Delta ]{Br/red\text{ }P}C{{H}_{3}}-CH(Br)-COOH\xrightarrow{\text{ Alc}\text{. }N{{H}_{3}}}C{{H}_{3}}-CH(N{{H}_{2}})-COOH
So, the end product (B) in the reaction is:- alanine.

So, the correct answer is “Option 1”.

Note:
The reaction of the carboxylic acid with the halogen in the presence of the red phosphorous resulting in the formation of alpha halo carboxylic acid is known as the Hell Volhard-Zelinsky reaction and this reaction is given by only those acids which consists of the alpha hydrogen atoms.