Tertiary butyl alcohol gives tertiary butyl chloride on treatment with:

conc.

H C l

/ anhydrous

Z n C l_{2}

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K C N

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N a O C l

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C l_{2}

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Solution

The correct option is B conc. $H C l$ / anhydrous $Z n C l_{2}$
Alcohols (secondary and tertiary) on reaction with conc. $H C l$ and dry anhydrous $Z n C l_{2}$ forms corresponding alkyl halide via $S_{N} 1$ mechanism. It also a chemical test to distinguish between primary, secondary and tertiary alcohols and is called the Lucas test.
Option A is correct.

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