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Chapter 2: Problem 121

When 3-methyl-2-butanol reacts with HBr, the expected product is 2-bromo-3-methyl-butane. But the actual product formed is 2 -bromo-2-methylbutane. Account for this.

Short Answer

Expert verified
Answer: The actual product, 2-bromo-2-methylbutane, is formed due to the carbocation rearrangement in the reaction mechanism. After protonation and elimination, a secondary carbocation is initially formed, which then undergoes a hydride shift to form a more stable tertiary carbocation. The bromide ion then attacks the tertiary carbocation to form the final product, 2-bromo-2-methylbutane.

Step by step solution

01

Identify the reactants

The reactants in this reaction are 3-methyl-2-butanol (also known as 3-methyl-2-butanol) and hydrobromic acid (HBr). 3-methyl-2-butanol has the following structure: ``` CH3 | CH3--C--OH | CH-CH3 ```
02

Protonation of the alcohol group

The first step in this reaction involves the protonation of the alcohol (OH) group on the 3-methyl-2-butanol by the strong acid, HBr. This results in the formation of a more stable oxonium ion (a positively charged oxygen): ``` CH3 | CH3--C--O-H | CH-CH3 + ```
03

Formation of the carbocation

Once the alcohol group is protonated, a water molecule (H2O) is eliminated, and a carbocation (a positive charge on the carbon) intermediate is formed. In this case, a secondary carbocation is initially formed: ``` CH3 | CH3--C+ | CH-CH3 ```
04

Carbocation rearrangement

The secondary carbocation can undergo a hydride shift to form a more stable tertiary carbocation. A hydrogen atom from the neighboring carbon moves to the positively charged carbon, forming the following intermediate: ``` CH3 | CH3--C-CH3 + | CH2-CH3 ```
05

Nucleophilic attack by bromide ion

In the final step, the bromide ion (Br-) from HBr acts as a nucleophile and attacks the positively charged carbon in the tertiary carbocation. This results in the formation of the actual product, 2-bromo-2-methylbutane: ``` CH3 | CH3--C-CH3 Br | CH2-CH3 ``` Hence, due to the carbocation rearrangement in the reaction mechanism, the actual product formed is 2-bromo-2-methylbutane instead of the expected 2-bromo-3-methylbutane.

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Most popular questions from this chapter

An alkene on mild oxidation with dil. aqueous \(\mathrm{KMnO}_{4}\) gave a product which on further treatment with \(\mathrm{HIO}_{4}\) gave hexane- 2,5 -dione. The alkene is (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) Cyclohexane (d) 1, 2 dimethyl cyclobutene

What are the products formed when buta-1, 3 -diene is treated with HBr in \(1: 1\) molar proportion? Explain the formation of the products.

Directions: This section contains 3 paragraphs. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I The three main classes of hydrocarbons are alkanes, alkenes and alkynes. Alkanes having cyclic structure are called cycloalkanes. They exhibit chain and ring chain isomerism. Alkenes and alkynes are more reactive than alkanes. A certain hydrocarbon has the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{8} .\) Which of the following is not a structural possibility for this hydrocarbon? (a) a cycloalkane (b) contains a ring and a double bond (c) contains two double bond and no ring (d) an alkyne

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