$C H_{3} \underset{l a u r y l a l c o h o l}{{(C H_{2})}_{10}} C H_{2} O H \overset{H_{2} S O_{4}}{\to} A \overset{N a O H (a q)}{\to} C H_{3} \underset{\begin{array}{l} sodium laurylsuphate \\ (anionic detergent) \end{array}}{{(C H_{2})}_{10} C H_{2} O S {\bar{O}}_{3} \overset{+}{N a}}$
Here, in the above reaction the intermediate (A) formed after treating Lauryl alcohol and with sulphuric acid gives

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lauryl sulphate

hydrogen sulphate

lauryl hydrogen sulphate

sodium sulphate

answer is C.

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Detailed Solution

$C H_{3} \underset{l a u r y l a l c o h o l}{{(C H_{2})}_{10}} C H_{2} O H \overset{H_{2} S O_{4}}{\to} A C H_{3} \underset{\begin{array}{l} lauryl hydrogen sulphate \end{array}}{{(C H_{2})}_{10} C H_{2} O S O_{3} \overset{}{H}} \overset{N a O H (a q)}{\to} C H_{3} \underset{\begin{array}{l} sodium laurylsulphate \\ (anionic detergent) \end{array}}{{(C H_{2})}_{10} C H_{2} O S {\bar{O}}_{3} \overset{+}{N a}}$