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Friday, 9 October

JEE Main 2026 · ChemistryMultiple choiceOtherHardMulti-step

JEE Main 6 April 2026, Shift 1, Chemistry Q67

Question 67 of 75 in this shift, Chemistry question 17 of 25, Section A.

Consider the following sequence of reactions [skeletal structure: a central carbon with a vertical line up and down, a line to the left, and OH on the right, i.e. tert-butyl alcohol] →(ii) H+,PhCOOH(i) Cu / 573K\xrightarrow[\text{(ii) } \mathrm{H^+, PhCOOH}]{\text{(i) Cu / 573K}} P. The major product P is:

The figure, in words

Stem shows a skeletal tertiary alcohol (central carbon with three methyl groups and OH) treated with (i) Cu/573 K, (ii) H+, PhCOOH. Options are drawn structures.
  1. (1)[figure: benzene ring with a tert-butyl group at the lower-left vertex and COOH at the upper-right vertex (para)]
  2. (2)[figure: benzene ring with a tert-butyl group at the lower-left vertex and COOH at the lower-right vertex]Official answer
  3. (3)[figure: tert-butyl benzoate, C6H5COOC(CH3)3\mathrm{C_6H_5COOC(CH_3)_3}]
  4. (4)[figure: 2-methylprop-2-enyl benzoate, C6H5COOCH2C(CH3)=CH2\mathrm{C_6H_5COOCH_2C(CH_3)=CH_2}]

Official answer

Option 2

NTA final key.

Chapter
Alcohols, Phenols and Ethers
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.