An organic compound (A) (molecular formula \(\mathrm{C_6H_{12}O_2}\)) was hydrolysed with dil. \(\mathrm{H_2SO_4}\) to give a carboxylic acid (B) and an alcohol (C). 'C' gives white turbidity immediately when treated with anhydrous \(\mathrm{ZnCl_2}\) and conc. HCl. The organic compound (A) is :
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The Correct Option is A
Solution and Explanation
Step 2: Lucas test: tertiary alcohols give turbidity immediately, secondary alcohols after about 5 minutes, primary alcohols give no turbidity at room temperature. So C is a tertiary alcohol.
Step 3: Check each ester (all are \(\mathrm{C_6H_{12}O_2}\)): option (1) tert-butyl acetate gives \(\mathrm{CH_3COOH}\) + \((\mathrm{CH_3})_3\mathrm{COH}\) (tertiary); option (2) sec-butyl acetate gives a secondary alcohol; option (3) isobutyl acetate gives a primary alcohol; option (4) propyl propanoate gives n-propanol (primary).
Step 4: Only tert-butyl acetate gives a tertiary alcohol, which shows immediate turbidity (the tertiary carbocation forms fastest by SN1 and \(\mathrm{Cl^-}\) attacks to give insoluble \((\mathrm{CH_3})_3\mathrm{CCl}\)).
Answer: (1) tert-butyl acetate, \(\mathrm{CH_3COOC(CH_3)_3}\)
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