Dehydrohalogenation of neopentyl bromide, (CH_3)_3C–CH_2Br, with alcoholic KOH gives mostly

JEE Main Chemistry question

Dehydrohalogenation of neopentyl bromide, (CH)C–CHBr, with alcoholic KOH gives mostly

(A)2-Methylbut-1-ene
(B)2,2-Dimethylbut-1-ene
(C)2-Methylbut-2-eneCorrect
(D)But-2-ene

Answer: (C)

Solution

Neopentyl bromide has no -hydrogen available (its only neighbouring carbon is fully quaternary), so simple E2 elimination is impossible. Under ionising conditions, the resulting primary

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-142732