JEE Advanced 2021 Paper 1 Q01 Chemistry Organic Chemistry Hydrocarbons Medium

JEE Advanced 2021 Paper 1 · Q01 · Hydrocarbons

The major product formed in the following reaction is:

$$\text{hept-2-yne}\ \bigl(CH_3{-}C{\equiv}C{-}(CH_2)_3{-}CH_3\bigr) \xrightarrow[\text{Na, liq.}\,\text{NH}_3]{\text{NaNH}_2} \text{Product}$$

  1. A. A conjugated diene: $CH_3{-}CH{=}CH{-}CH{=}CH{-}CH_2{-}CH_3$ (hepta-2,4-diene)
  2. B. $(E)$-hept-2-ene: $CH_3{-}CH{=}CH{-}(CH_2)_3{-}CH_3$ with the methyl and $n$-butyl groups on opposite sides of the double bond (trans configuration)
  3. C. A terminal alkyne: hept-1-yne, $HC{\equiv}C{-}(CH_2)_4{-}CH_3$
  4. D. $(Z)$-hept-2-ene: $CH_3{-}CH{=}CH{-}(CH_2)_3{-}CH_3$ with the methyl and $n$-butyl groups on the same side of the double bond (cis configuration)
Reveal answer + step-by-step solution

Correct answer:B

Solution

The reaction sequence has two steps. (i) NaNH$_2$ deprotonates terminal alkynes but with an internal alkyne it acts only as a base / isomerisation agent; here it sets up the alkyne for the next step. (ii) Na in liquid NH$_3$ is the classic dissolving-metal (Birch-type) reduction for internal alkynes: it adds one electron, then a proton, then another electron and proton through a vinyl-radical / vinyl-anion sequence. The thermodynamically preferred geometry of the vinyl anion is trans, so the alkyne is reduced selectively to the trans (E) alkene. The major product is therefore trans-hept-2-ene (option B). Lindlar-type reagents would have given the cis isomer (option D), and full hydrogenation or terminal alkyne formation are not relevant here.

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