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JEE Advanced 2025 Paper 1 · Chemistry · Alcohols, Phenols, Ethers and Haloalkanes · medium

3,3-Dimethylbutan-2-ol is heated with concentrated H2SO4 and the major alkene M is isolated.

One or more options correct

3,3-Dimethylbutan-2-ol is heated with concentrated H2SO4\mathrm{H_2SO_4} and the major alkene M\mathrm{M} is isolated. Choose the correct statement(s).
  1. A. M\mathrm{M} is the tetrasubstituted alkene 2,3-dimethylbut-2-ene
  2. B. Ozonolysis of M\mathrm{M} followed by reductive work-up (Zn/H2O\mathrm{Zn/H_2O}) gives acetone as the only organic product
  3. C. Hydroboration followed by oxidative work-up (BH3⋅THF\mathrm{BH_3\cdot THF}; H2O2/OH−\mathrm{H_2O_2/OH^-}) of M\mathrm{M} gives an optically active alcohol
  4. D. The first-formed secondary carbocation rearranges to a tertiary carbocation by a 1,2-hydride shift
Show answer and solution

Answer: A, B

Protonation of the alcohol and loss of water gives the secondary cation CH3 ⁣− ⁣CH+ ⁣− ⁣C(CH3)3\mathrm{CH_3\!-\!CH^+\!-\!C(CH_3)_3}. A 1,2-shift of a methyl group (not a hydride, because the adjacent quaternary carbon bears no hydrogen) gives the tertiary cation (CH3)2C+ ⁣− ⁣CH(CH3)2\mathrm{(CH_3)_2C^+\!-\!CH(CH_3)_2}. Loss of a proton by the Zaitsev rule gives mainly 2,3-dimethylbut-2-ene, tetrasubstituted, so A is correct and D is wrong (it is a methyl shift). Ozonolysis cleaves the symmetric double bond (CH3)2C=C(CH3)2\mathrm{(CH_3)_2C=C(CH_3)_2} into two molecules of (CH3)2C=O\mathrm{(CH_3)_2C=O}, so B is correct. Hydroboration–oxidation adds H\mathrm{H} and OH\mathrm{OH} across the double bond to give (CH3)2CH ⁣− ⁣C(OH)(CH3)2\mathrm{(CH_3)_2CH\!-\!C(OH)(CH_3)_2}, 2,3-dimethylbutan-2-ol, which has no stereocentre and is achiral, so C is wrong. Answer: A, B.
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