In the equation, $C{{H}_{3}}COOH+C{{l}_{2}}\xrightarrow[-HCl]{\text{Red P}}A$, the compound A is:A.CH$_3$CH$_2$ClB.ClCH$_2$COOHC.CH$_3$ClD.CH$_3$COCl

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Hint: The chemical reaction of halogenation of carboxylic acid in the presence of Red P takes place at $\alpha$- carbon atom to form a mono form of acetic acid attached to a halogen. Now, we can find the product.

-Firstly, let us discuss the chemical reaction. It is a HVZ reaction i.e. Hell-Volhard-Zelinsky reaction.
-As we know, in the HVZ reaction, the carboxylic acid reacts with halogen, and a catalytic amount of phosphorus is added, leading to the -halogenation of carboxylic acids.
- The halogen that gives this reaction is generally Chlorine (Cl) and Bromine (Br).
-Now, we are given with the carboxylic acid i.e. acetic acid and halogen used is chlorine.
-Thus, when the acetic acid reacts with chlorine in the presence of Red P, $\alpha$-H atom is replaced with Cl atom.
-It leads to the formation of monochloroacetic acid, its chemical formula is ClCH$_2$COOH.
-Therefore, the chemical reaction can be written as:

$C{{H}_{3}}COOH+C{{l}_{2}}\xrightarrow[-HCl]{\text{Red P}}ClC{{H}_{2}}COOH$

The compound A is ClCH$_2$COOH.

So, the correct option is B.

The reaction temperature is maintained at above 373K in Hell-Volhard-Zelinsky reaction. Also, chlorination and bromination is only accomplished through this reaction at the $\alpha$-carbon. It should be noted that fluorination and iodination is not executed through these reaction conditions.