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Chapter 11: Q11-71E (page 350)

The reaction of HBr with (R)-3-methyl-3-hexanol leads to racemic 3-Bromo-3-methyl hexane. Explain.

Short Answer

Expert verified

Formation of carbocation

Formation of the desired product

Step by step solution

01

Tertiary alcohol

Whenever a tertiary alcohol is treated with a strong acid, a carbocation intermediate is formed.

Formation of carbocation

This carbocation is planar due to the empty p-orbital of the carbocation carbon. Moreover, the tetrahedral chirality that was present in the starting material was lost.

02

Explanation

The chiral starting material goes through an achiral intermediate, such as a carbocation, and the stereochemical information is destroyed. Hence, treatment of 3 -methyl -3-hexanol with HBr forms a carbocation intermediate because 3-methyl-3-hexanol is a tertiary alcohol. This intermediate loses all stereochemical information so the product is racemized.

Formation of the desired product

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Most popular questions from this chapter

Among the many examples of reactions that occur with incomplete racemization, the optically pure tosylate of 2,2-dimethyl-1-phenyl-1-propan([α]D= 230.3) gives the corresponding acetate ([α]D= 15.3) when heated inacetic acid. If complete inversion had occurred, the optically pure acetatewould have had[α]D= 153.6. What percentage racemization and what percentageinversion occurred in this reaction?

Which compound in each of the following pairs will react faster in an SN2reaction with OH-?

(c)(CH3)3CClorCH3Cl

Assign R or S configuration to the following molecule, write the product you would expect from SN2reaction with NaCN, and assign R or S configuration to the product (green = Cl):

How might you prepare each of the following molecules using a nucleophilic substitution reaction at some step?

c.CH3CH2CH2CH2CN

Which compound in each of the following pairs will react faster in an SN2reaction with OH-?

(d)CH2=CHBrorCH2=CHCH2Br
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