Question

Consider the following sequence of reactions:



The major product $P$ is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

First, dehydrate the tertiary alcohol using hot copper to get an alkene. Then, perform an acid-catalyzed addition of benzoic acid to that alkene to form an ester following Markovnikov's logic.

Answer 1

The Correct Option is C

This question involves a multi-step organic transformation starting from tert-butyl alcohol.

Step 1: Reaction of tert-butyl alcohol with $Cu$ at 573 K
Primary and secondary alcohols are typically oxidized to aldehydes and ketones respectively when passed over heated copper at 573 K. However, tertiary alcohols like tert-butyl alcohol ($(CH_3)_3C-OH$) lack an $\alpha$-hydrogen required for oxidation. Instead, they undergo dehydration (loss of water) at these high temperatures to form an alkene. The product formed is 2-methylpropene (also known as isobutylene):
$$(CH_3)_3C-OH \xrightarrow{Cu, 573 K} (CH_3)_2C=CH_2 + H_2O$$

Step 2: Reaction with Benzoic Acid ($PhCOOH$) in the presence of $H^+$
The resulting alkene, isobutylene, is then reacted with benzoic acid under acidic conditions. This is an acid-catalyzed addition of a carboxylic acid to an alkene, which follows Markovnikov's rule. The catalyst $H^+$ adds to the terminal carbon of the double bond to create the more stable tertiary carbocation:
$$(CH_3)_2C=CH_2 + H^+ \rightarrow (CH_3)_3C^+$$
Next, the benzoic acid molecule acts as a nucleophile. Its carboxyl oxygen attacks the carbocation, followed by the loss of a proton to restore the catalyst and form the final ester product:
$$(CH_3)_3C^+ + PhCOOH \rightarrow Ph-COO-C(CH_3)_3 + H^+$$
The final product $P$ is tert-butyl benzoate. This corresponds to the structure shown in option 3.