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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鈥檚 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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Most popular questions from this chapter

Question: Draw additional resonance structures for each ion.

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Question: a. Draw resonance structures for the carbocation formed after loss of a leaving group from phytyl diphosphate. b. Draw the two-step mechanism for Friedel鈥揅rafts alkylation of 1,2-dihydroxynaphthoic acid with this carbocation to form X.

Question: Rank the following compounds in order of increasing stability.

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Question: Draw a stepwise mechanism for the biological conversion of linalyl diphosphate to limonene.

Question: a. Draw the two isomeric dienes formed when CH2=CHCH2CHClCHCH32is treated with an alkoxide base.

b. Explain why the major product formed in this reaction does not contain the more highly substituted alkene.

See all solutions

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