JEE Main 2026 · 6 Apr · Shift 1 · Chemistry Q67

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ChemistrySingle correct+4 / -1Question 67 of 75

Consider the following sequence of reactions

The major product P is:

tert-Butyl alcohol (drawn skeletally) treated with (i) Cu/573 K, (ii) H⁺, PhCOOH gives P.

4-tert-Butylbenzoic acid: COOH and tert-butyl para on a benzene ring.

3-tert-Butylbenzoic acid: COOH and tert-butyl meta on a benzene ring.

2-Methylprop-2-enyl (methallyl) benzoate: PhC(=O)O-CH₂-C(CH₃)=CH₂.

Answer and worked solution

Official answer: B

Step (i). Cu at 573 K does not dehydrogenate a 3° alcohol; it dehydrates it. tert-Butyl alcohol gives 2-methylpropene.

Step (ii). In acid, the alkene gives the tert-butyl cation. This alkylates benzoic acid at the position the $\ce{-COOH}$ group directs to, which is meta.

P: 3-tert-butylbenzoic acid.

(B)

Topics:Alcohols Aromatic hydrocarbons and electrophilic substitution

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JEE Main 2026 · 6 Apr · Shift 1, Chemistry question 67