JEE Main Chemistry question

Which of the following compounds reacts with HBr obeying Markownikoff's rule (requires a genuinely unsymmetrical alkene)? [J & K CET 2003]

(A)CH=CH (symmetrical — no regiochemical choice)
(B)a *cis*- or *trans*-disubstituted but still symmetrical alkene (e.g. (E)- or (Z)-3-hexene type, identical groups both ends)
(C)2-methylbut-2-ene, (CH)C=CH–CH (genuinely unsymmetrical, trisubstituted)Correct
(D)a tetrasubstituted, fully symmetrical alkene (e.g. 2,3-dimethyl-2-butene)

Answer: (C)

Solution

Markovnikov's rule describes which carbon of an *unsymmetrical* alkene the H of H–X attaches to; it is only meaningfully tested (and gives a clear regiochemical preference) when the two alkene carbons bear different substituents. A trisubstituted alkene such as 2-methylbut-2-ene is genuinely unsymmetrical: HBr adds

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-142226