Major Product of Alcohol Oxidation and Esterification

To determine the major product $P$ of the given reaction sequence, let us analyze each step systematically:

Step 1: Reaction of 2-methylpropan-2-ol with $\text{Cu} / 573\text{ K}$

The starting material is $tert$-butanol (2-methylpropan-2-ol), a tertiary alcohol: $(CH_3)_3\text{C}-\text{OH}$

When alcohols are passed over heated copper at $573\text{ K}$:

• Primary alcohols undergo dehydrogenation to form aldehydes.
• Secondary alcohols undergo dehydrogenation to form ketones.
• Tertiary alcohols, lacking an $\alpha$-hydrogen atom, undergo dehydration to yield an alkene.

$\text{(CH}_3)_3\text{C-OH} \xrightarrow{\text{Cu} / 573\text{ K}} (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O}$

Thus, the intermediate formed in step (i) is 2-methylpropene (isobutene).

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Step 2: Acid-catalyzed reaction with Benzoic Acid ($\text{H}^+, \text{PhCOOH}$)

In the presence of an acid catalyst ($\text{H}^+$):

1. Carbocation Formation: Protonation of the alkene double bond occurs according to Markovnikov's rule, generating a highly stable tertiary carbocation ($tert$-butyl carbocation): $(\text{CH}_3)_2\text{C=CH}_2 + \text{H}^+ \rightleftharpoons (\text{CH}_3)_3\text{C}^+$

2. Nucleophilic Attack: The oxygen atom of the hydroxyl group in benzoic acid ($\text{PhCOOH}$) attacks the $tert$-butyl carbocation: $\text{PhCOOH} + (\text{CH}_3)_3\text{C}^+ \longrightarrow \text{Ph-C}(=\text{O})-\overset{+}{\text{O}}\text{H}-\text{C}(\text{CH}_3)_3$

3. Deprotonation: Loss of a proton ($\text{H}^+$) yields the ester $tert$-butyl benzoate: $\text{Ph-C}(=\text{O})-\overset{+}{\text{O}}\text{H}-\text{C}(\text{CH}_3)_3 \xrightarrow{-\text{H}^+} \text{Ph-C}(=\text{O})-\text{O}-\text{C}(\text{CH}_3)_3$

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Conclusion:

The major product $P$ formed in the sequence is $tert$-butyl benzoate ($\text{Ph}-\text{COO}-\text{C}(\text{CH}_3)_3$).

Correct Answer: Option B