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Chapter 16: Problem 353

A naive graduate student attempted the preparation of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDBrCH}_{3}\) from \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDOHCH}_{3}\) by heating the deuterio-alcohol with \(\mathrm{HBr}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\). He obtained a product having the correct boiling point, but a careful examination of the spectral properties by his research director showed that the product was a mixture of \(\mathrm{CH}_{3} \mathrm{CHDCHBrCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDBrCH}_{3}\). What happened?

Short Answer

Expert verified
The graduate student obtained a mixture of CH3CHDCHBrCH3 and CH3CH2CDBrCH3 because the carbocation intermediate formed during the reaction was prone to rearrangement via hydride shift, resulting in the formation of a more stable carbocation. This then reacted with the bromide ion, leading to the formation of both the expected product and the rearranged product.

Step by step solution

01

Identify the given compounds

We have the following compounds: 1. CH3CH2CDOHCH3 - starting compound (deuterio-alcohol) 2. CH3CH2CDBrCH3 - expected product 3. CH3CHDCHBrCH3 - actual product obtained (part of the mixture) 4. HBr and H2SO4 - reactants used
02

Identify the initial reaction

The reaction that the student attempts is the substitution of the -OD group in the starting compound (CH3CH2CDOHCH3) with a -Br group from the HBr/H2SO4 reagents.
03

Formation of a carbocation intermediate

As the reaction proceeds, a carbocation intermediate is formed. This occurs due to the protonation of the deuterio-alcohol by the strong acid H2SO4, followed by the loss of the deuterium (D)+ ion: \[CH3CH2CDOHCH3 + H^+ \rightarrow CH3CH2CD(OH2)CH3^+\] \[CH3CH2CD(OH2)CH3^+ \rightarrow CH3CH2C^+CH3 + D^+\]
04

Reaction with HBr

The carbocation intermediate then reacts with the bromide ion (Br-) from HBr, which leads to the formation of two products: \[CH3CH2C^+CH3 + Br^- \rightarrow CH3CH2CDBrCH3\] \[CH3CH2C^+CH3 \rightarrow CH3CH^+CHBrCH3\] The formation of the second carbocation can arise from the hydride shift in the carbocation, which occurs to achieve a more stable carbocation intermediate (as it forms a more substituted, tertiary carbocation).
05

Formation of both CH3CHDCHBrCH3 and CH3CH2CDBrCH3

Since the carbocation intermediate can rearrange via hydride shift, both CH3CHDCHBrCH3 and CH3CH2CDBrCH3 products can be obtained.
06

Concluding the explanation

The graduate student obtained a mixture of CH3CHDCHBrCH3 and CH3CH2CDBrCH3 because the carbocation intermediate formed during the reaction was prone to rearrangement. This resulted in the formation of a more stable carbocation, which then reacted with the bromide ion. Therefore, both the expected product and the rearranged product were formed.

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Most popular questions from this chapter

In the esterification of an acid with an alcohol, how could you distinguish between \(\mathrm{C}-\mathrm{O}\) and \(\mathrm{O}-\mathrm{H}\) cleavage of the alcohol using heavy oxygen \(\left({ }^{18} \mathrm{O}\right)\) as a tracer?

What are alcohols and how are they classified?

(a) From the reaction of either \((2 \mathrm{R}, 3 \mathrm{~S})\) or \(\left(2 \mathrm{~S}, 3 \mathrm{R}_{7}\right)-\) 3 -bromo-2-butanol with HBr, the product is meso-2, 3 -dibromobutane. Show how this product is formed. Is the intermediate bromonium ion chiral? (b) Contrast the reaction of \((2 \mathrm{R}, 3 \mathrm{~S})-3\) -bromo \(-2\) -butanol with HBr and the reaction of the same substrate with sodium ethoxide.

You prepare sec-butyl tosylate from alcohol of \([\alpha]+6.9^{\circ}\). On hydrolysis with aqueous base, this ester gives sec-buty1 alcohol of \([\alpha]-6.9^{\circ}\). Without knowing the configuration or optical purity of the starting alcohol, what (if anything) can you say about the stereochemistry of the hydrolysis step?

Sodium metal was added to tert-butyl alcohol and allowed to react. When the metal was consumed, ethyl bromide was added to the resulting mixture. Work-up of the reaction mixture yielded a compound of formula \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\). In a similar experiment, sodium metal was allowed to react with ethanol. When tert-butyl bromide was added, a gas was evolved, and work-up of the remaining mixture gave ethanol as the only organic material. (a) Write equations for all reactions, (b) What familiar reaction type is involved in each case? (c) Why did the reactions take different courses?
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