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Chapter 6: Problem 7

The acid-catalyzed hydration of 3,3-dimethyl-1-butene gives 2,3 -dimethyl-2-butanol as the major product. Propose a mechanism for the formation of this alcohol.

Short Answer

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Question: Propose a mechanism for the acid-catalyzed hydration of 3,3-dimethyl-1-butene, which gives 2,3-dimethyl-2-butanol as the major product. Answer: The mechanism for the acid-catalyzed hydration of 3,3-dimethyl-1-butene involves the following steps: 1. Protonate the alkene - Formation of 2,3-dimethyl-2-butyl carbocation. 2. Carbocation rearrangement (if necessary) - Not needed in this case. 3. Attack by water - Formation of 2,3-dimethyl-2-butyl oxonium ion. 4. Deprotonation to form the alcohol - Formation of 2,3-dimethyl-2-butanol as the major product.

Step by step solution

01

Identify the reactants and reagents

In this reaction, we have the following reactants and reagents: - Alkene: 3,3-dimethyl-1-butene - Acid: Acid catalyst (typically H2SO4 or H3PO4) - Water Our goal is to propose a mechanism that leads to the formation of the major product, 2,3-dimethyl-2-butanol.
02

Protonate the alkene

The first step in the acid-catalyzed hydration of an alkene is the protonation of the alkene, which forms a carbocation. A lone pair from the alkene's pi bond will attack the acidic hydrogen (H+) from the acid catalyst, forming a new sigma bond with the hydrogen and leaving a positive charge (carbocation) on the other carbon that was part of the pi bond. In our case: 3,3-dimethyl-1-butene + H+ -> 2,3-dimethyl-2-butyl carbocation
03

Carbocation rearrangement (if necessary)

Carbocations can rearrange to form more stable species. In the current case, the initially formed carbocation is already a secondary carbocation which is stabilized by hyperconjugation, so a rearrangement is not needed.
04

Attack by water

The carbocation will now be attacked by a molecule of water, which acts as a nucleophile. The oxygen atom with a lone pair in the water molecule will form a bond with the positively charged carbon atom of the carbocation. This step forms an oxonium ion: 2,3-dimethyl-2-butyl carbocation + H2O -> 2,3-dimethyl-2-butyl oxonium ion
05

Deprotonation to form the alcohol

The final step is the deprotonation of the oxonium ion to form the alcohol product. Another water molecule or another base present in the reaction mixture can take a proton from the oxonium ion, resulting in the major product, 2,3-dimethyl-2-butanol: 2,3-dimethyl-2-butyl oxonium ion + H2O/Base -> 2,3-dimethyl-2-butanol + H3O+/BH+ This completes the proposed mechanism for the formation of 2,3-dimethyl-2-butanol from the acid-catalyzed hydration of 3,3-dimethyl-1-butene.

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Most popular questions from this chapter

Treating 4-penten- \(1-\mathrm{ol}\) with bromine in water forms a cyclic bromoether.

As we have seen in this chapter, carbon-carbon double bonds are electron-rich regions and are attacked by electrophiles (for example, HBr); they are not attacked by nucleophiles (for example, diethylamine). However, when the carbon-carbon double bond has a carbonyl group adjacent to it, the double bond reacts readily with nucleophiles by nucleophilic addition (Section \(19.8\) ). Account for the fact that nucleophiles add to a carbon-carbon double bond adjacent to a carbonyl group, and account for the regiochemistry of the reaction.

The 2-propenyl cation appears to be a primary carbocation, and yet it is considerably more stable than a \(1^{\circ}\) carbocation such as the 1 -propyl cation. $$ \begin{array}{lc} \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}^{+} & \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^{+} \\ \text {2-Propenyl cation } & \text { 1-Propyl cation } \end{array} $$ How would you account for the differences in the stability of the two carbocation?

Draw a structural formula of an alkene that undergoes acid-catalyzed hydration to give each alcohol as the major product (more than one alkene may give each alcohol as the major product). (a) S-Hexanol (b) 1-Methylcyclobutanol (c) 2-Methyl-2-butanol (d) 2-Propanol

Draw the alternative chair conformations for the product formed by the addition of bromine to 4-tert-butylcyclohexene. The Gibbs free energy differences between equatorial and axial substituents on a cyclohexane ring are \(21 \mathrm{~kJ}\) ( \(4.9 \mathrm{kcal}) / \mathrm{mol}\) for tert-butyl and \(2.0-2.6 \mathrm{~kJ}(0.48-0.62 \mathrm{kcal}) / \mathrm{mol}\) for bromine. Estimate the relative percentages of the alternative chair conformations you drew in the first part of this problem.
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