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Chapter 4: Problem 124

Treatment of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{MgBr}\) with \(\mathrm{I}_{2}\) gives \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2}\) I in good yield whereas, \(\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) on reaction with HBr gives mainly a rearranged product namely \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}\) and not \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Br}\). Why?

Short Answer

Expert verified
Answer: The first reaction involves a Grignard reaction and no carbocation formation, so rearrangement does not occur, giving \((CH_3)_3CCH_2 I\) in good yield. In the second reaction, a carbocation intermediate forms and rearranges via a 1,2-hydride shift, leading to the formation of the more stable product \((CH_3)_2CBrCH_2 CH_3\).

Step by step solution

01

Treatment of \((CH_3)_3CCH_2 MgBr\) with \(I_2\) to form \((CH_3)_3CCH_2I\)

The reaction is a Grignard reaction, in which \((CH_3)_3CCH_2MgBr\) acts as a nucleophilic organometallic species and \(I_2\) acts as an electrophile. The mechanism involves the nucleophilic attack of the Grignard reagent on \(I_2\) followed by the release of \(MgBrI\). Since the nucleophilic center is the carbon bonded to \(MgBr\), no carbocation is formed and no rearrangement is possible. The reaction proceeds to form the desired product \((CH_3)_3 CCH_2I\) in good yield.
02

Reaction of \((CH_2)_3 CCH_2OH\) with HBr to form rearranged product, \((CH_3)_2CBrCH_2 CH_3\)

This reaction involves the protonation of the hydroxyl group (OH) by HBr, forming a tertiary alcohol. A tertiary carbocation forms as a result of the loss of water (H₂O). Since carbocations are prone to rearrangement, a 1,2-hydride shift occurs so that a more stable secondary carbocation is formed. Finally, the Br⁻ ion attacks the carbocation to form the main product \((CH_3)_2CBrCH_2 CH_3\).
03

Explaining the differences in the two reactions

While the first reaction involves a Grignard reaction and no carbocation formation, the second reaction proceeds through a carbocation intermediate that rearranges via a 1,2-hydride shift. The rearrangement will not happen in the Grignard reaction, leading to the formation of \((CH_3)_3CCH_2I\) without rearrangement. In the reaction of \((CH_2)_3 CCH_2 OH\) with HBr, the rearrangement occurs due to the higher stability of the secondary carbocation, leading to the main product \((CH_3)_2CBrCH_2 CH_3\).

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Most popular questions from this chapter

IUPAC name of allyl chloride is (a) 2 -chloropropene (b) 3 -chloropropene (c) 1 -chloropropene (d) 1 -(chloromethyl)ethane

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