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Chapter 16: PROBLEM 16.42 (page 635)

Question: Ignoring stereoisomers, draw all products that form by addition of HBr to (E)-hexa-1,3,5-triene.

Short Answer

Expert verified

HBr is a strong acid and undergoes reaction by donating an acidic proton of its own and thus forming a carbocationic center in the reacting compound.

Step by step solution

01

HBr

HBr is a strong acid and undergoes reaction by donating an acidic proton of its own and thus forming a carbocationic center in the reacting compound.

02

Carbocations

The bond electrons from the C=C are donated to the acidic hydrogen atom of the hydrogen bromide acid, and thus a carbocation is formed. This addition of the hydrogen from HBr to the double bond takes place according to Markovnikoff’s rule.

03

Product of the given reaction

The product of the attack of C=C on the hydrogen of HBr is resonance stabilized. Thus, 1,2-, 1,4- and 1,6-addition products of the reaction are obtained.

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