JEE Main 2022 Chemistry question

The number of hybridised carbons in an acyclic neutral compound with molecular formula is .

Answer: 1

Solution

For the degree of unsaturation is (C + 1) − (H + X − N)/2 = 5 − (5 − 1)/2 = 3 double-bond equivalents. An acyclic compound with 3 DBE must

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Question ID VK-024802