JEE Main 2022 · 25 Jul · Shift 1 · Chemistry Q89

← JEE Main 2022 · 25 Jul · Shift 1

ChemistryInteger+4 / -1Question 89 of 90

The number of sp $^3$ hybridised carbons in an acyclic neutral compound with molecular formula $\mathrm{C_4H_5N}$ is ______.

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Answer and worked solution

Official answer: 0

Isomers. Acyclic $\mathrm{C_4H_5N}$ (3 degrees of unsaturation) includes allyl cyanide $\mathrm{CH_2{=}CHCH_2CN}$ with one sp³ carbon and $\mathrm{CH_2{=}CH{-}CH{=}CH{-}}$ type imines with none; NTA's key accepts 0 or 1.

Answer: 1

Topics:Hybridisation Structural isomerism

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JEE Main 2022 · 25 Jul · Shift 1, Chemistry question 89