MHTCET 10 May 2024 Shift 2 Paper

Concept:A tertiary allylic alcohol must have the $-OH$ group on a carbon that is directly attached to a $C=C$ double-bond carbon, and that carbon must be bonded to three carbon groups with no hydrogen on it.Explanation:For option A: $\mathrm{HOCH_2{-}CH{=}CH_2}$ is prop-2-en-1-ol, where the $-OH$ carbon is primary.For option B: $\mathrm{CH_3{-}CH(OH){-}CH{=}CH_2}$ is but-3-en-2-ol, where the $-OH$ carbon is secondary.For option C: $\mathrm{HOCH_2{-}C(CH_3){=}CH_2}$ is 2-methylprop-2-en-1-ol, where the $-OH$ carbon is primary.For option D: $\mathrm{CH_2{=}CH{-}C(OH)(CH_3)_2}$ is 2-methylbut-3-en-2-ol. The carbon bearing $-OH$ is attached to two methyl groups and the vinyl group, so it has no H and is tertiary. It is adjacent to the double bond, hence allylic.Answer:Option D: 2-methylbut-3-en-2-ol