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Chapter 11: Q15P (page 559)

Neopentyl alcohol, (CH3)3CCH2OH, reacts with concentrated HBr to give 2-bromo-2-methylbutane, a rearranged product. Propose a mechanism for the formation of this product.

Short Answer

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Neopentyl alcohol(CH3)3CCH2OHis a primary alcohol, but the bulky group has hinderedit. The backside attack is not possible, so instead of the SN2 mechanism SN1followedto get the desired rearranged product.

The reaction forms a rearranged stable carbocation from primary carbocation to tertiary carbocation.

Step by step solution

01

Step 1: Rearranged Product

Neopentyl alcohol(CH3)3CCH2OHis a primary alcohol, but the bulky group has hinderedit. The backside attack is not possible, so instead of the SN2 mechanism SN1followedto get the desired rearranged product.

The reaction forms a rearranged stable carbocation from primary carbocation to tertiary carbocation.

02

 Mechanism for the formation of 2-bromo-2-methylbutane

Protonation of Neopentyl alcoholconverts the hydroxy group to a good leaving group.

Water leaves form a primary carbocation rearranged through methyl shift, thus forming a stable tertiary carbocation.

Bromide ion attacks the carbocation to form the product 2-bromo-2-methylbutane

Mechanism showing the formation of 2-bromo-2-methylbutane

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Most popular questions from this chapter

Show the alcohol and the acid chloride that combine to make the following esters.

(a)

(b)

(c)

(d)

Two products are observed in the following reaction.

(a) Suggest a mechanism to explain how these two products are formed.

(b) Your mechanism for part (a) should be different from the usual mechanism of the reaction of SOCl2 with alcohols. Explain why the reaction follows a different mechanism in

this case.

(a) The reaction of butan-2-ol with concentrated aqueous HBr goes with partial racemization, giving more inversion than retention of configuration. Propose a mechanism that accounts for racemization with excess inversion.

(b)Under the same conditions, an optically active sample of trans-2-bromocyclopentanol reacts with concentrated aqueous HBr to give an optically inactive product, (racemic) trans-1,2-dibromocyclopentane. Propose a mechanism to show how this reaction goes with apparently complete retention of configuration, yet with racemization. (Hint: Draw out the mechanism of the reaction of cyclopentene with in water to give the starting material, trans-2- bromocyclopentanol. Consider how parts of this mechanism might be involved in the reaction with HBr.)

The following reaction involves a starting material with a double bond and a hydroxyl group, yet its mechanism resembles a pinacol rearrangement. Propose a mechanism, and point out the part of your mechanism that resembles a pinacol rearrangement.

Under normal circumstances, tertiary alcohols are not oxidized. However, when the tertiary alcohol is allylic, it can undergo a migration of the double bond (called an allylic shift) and subsequent oxidation of the alcohol. A particularly effective reagent for this reaction is Bobbitt’s reagent, similar to TEMPO used in many oxidations. (M. Shibuya et al., J. Org. Chem., 2008, 73, 4750.)

Show the expected product when each of these 3° allylic alcohols is oxidized by Bobbitt’s reagent.

(a)

(b)

(c)

(d)
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