A) \[PC{{l}_{5}}\]
B) \[C{{l}_{2}}\]
C) \[SOC{{l}_{2}}\]
D) \[PC{{l}_{3}}\]
Correct Answer: B
Solution :
\[PC{{l}_{3}},\,PC{{l}_{5}}\] and \[SOC{{l}_{2}}\] are used to replace \[-OH\] group of an alcohol or an acid by \[-Cl\]group \[R-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-OH+PC{{l}_{5}}\xrightarrow{{}}R-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-Cl\] \[+\,POC{{l}_{3}}+HCl\] \[R-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-OH+SOC{{l}_{2}}\to R-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-Cl\] \[+\,S{{O}_{2}}+HCl\] when acid reacts with \[C{{l}_{2}}\] in presence of red phosphorus, \[\alpha \]-chloroacid is obtained. (Hall-Vohlard Zelinsky reaction). \[C{{H}_{3}}C{{H}_{2}}COOH+C{{l}_{2}}\xrightarrow[\Delta ]{red\,\,S}\] \[\underset{\alpha -chloropropionic\text{ }add}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,HCOOH+HCl}}\,\]
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