Question

The best method for the conversion of an alcohol into alkyl chloride is by treating the alcohol with:
A. $PC{l_3}$
B. $PC{l_5}$
C. $SOC{l_2}$ in the presence of pyridine
D.Dry HCl in the presence of anhydrous $ZnC{l_2}$

Answer 1

For formation of alkyl halides from alcohols, we need to perform a substitution reaction to replace the hydroxyl group with the halide group. Many a times, the alkyl halide products that are formed are either in the gaseous state or are mixed with other liquid products formed because of the reaction.

Complete step by step answer:
Before we move forward with the solution of the given question, let us first discuss some important basic concepts.
Because of the above-mentioned issue, the actual extractable amount of alkyl halides actually reduces and also, we need to perform extra procedures to isolate it. This is usually not a desirable method. Hence, a reaction where the alkyl halide can be directly extracted without any extra processes can be considered as the best method. Let us discuss the conversion of alcohol into alkyl chloride using the compounds given in the options.
$PC{l_3}$ : alcohols react with $PC{l_3}$ to give alkyl chloride along with phosphorous acid. Both these products are obtained in the same solution.
$R - OH + PC{l_3} \to R - Cl + {H_3}P{O_3}$
$PC{l_5}$ : alcohols react with $PC{l_5}$ to give alkyl chloride along with phosphorus oxychloride and hydrochloric acid. Again, al l these products are formed in the same liquid mixture.
$R - OH + PC{l_5} \to R - Cl + POC{l_3} + HCl$
$SOC{l_2}$ : Alcohols react with $SOC{l_2}$ in the presence of pyridine to give alkyl chloride along with sulphur dioxide and hydrochloric acid. Except the alkyl chloride, all the other products are gaseous. Hence, they are released from the system leaving just alkyl chloride to be collected in the liquid form. Hence, this is the best method to extract alkyl halide from alcohol.
$R - OH + SOC{l_2} \to R - Cl + S{O_2} + HCl$

Hence, Option C is the correct option.

Note:
When alcohol is reacted with $SOC{l_2}$ in the presence of pyridine to form alkyl chloride, the alkyl chloride formed will have the same stereochemistry as of the hydroxyl group in the alcohol. This is a very important feature of using $SOC{l_2}$ .