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Chapter 4: Problem 4

When 3 -methyl butan-2-ol is heated with HBr, the major product formed is 2-bromo-2-methyl butane. Explain the formation of the product from the mechanism of the reaction

Short Answer

Expert verified
Answer: The mechanism for the formation of 2-bromo-2-methylbutane from the reaction of 3-methylbutan-2-ol with HBr involves nucleophilic substitution, including protonation of the alcohol, formation of a carbocation, and a nucleophilic attack by the bromide ion.

Step by step solution

01

Identify the starting materials and reagents

In this reaction, we have the following starting materials and reagents: - 3-methylbutan-2-ol which is the alcohol - HBr which is the hydrogen halide
02

Protonation of the alcohol

Since -OH is a poor leaving group, the first step in the reaction is protonation of the alcohol to form a good leaving group. The lone pair on the oxygen in alcohol attacks the hydrogen atom in HBr, forming a bond. This results in the formation of a water molecule (H2O) which is a better leaving group and a positively charged bromine ion (Br-). 3-methylbutan-2-ol + HBr -> 3-methylbutan-2-ol-H^+ + Br^-
03

Formation of the carbocation

The next step is the formation of the carbocation. The water molecule, as a good leaving group, departs from the protonated alcohol, and a positively charged carbocation is formed at the 2nd carbon in the 3-methylbutane molecule. 3-methylbutan-2-ol-H^+ -> 3-methylbutan-2-ylium + H2O
04

Nucleophilic attack by bromide ion

The carbocation formed in step 3 acts as an electrophile and is attacked by the negatively charged bromine ion (Br-) from step 1. The nucleophilic attack occurs at the positively charged carbon, forming a bond with the bromine and yielding our final product, 2-bromo-2-methylbutane. 3-methylbutan-2-ylium + Br^- -> 2-bromo-2-methylbutane In conclusion, the formation of 2-bromo-2-methylbutane from the reaction of 3-methylbutan-2-ol with HBr involves a nucleophilic substitution mechanism that includes the protonation of the alcohol, the formation of a carbocation, and the nucleophilic attack by the bromide ion.

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Most popular questions from this chapter

But-1-ene \(\frac{\left(\text { i) } \mathrm{Hg}(\mathrm{OAc})_{2} / \mathrm{THF}-\mathrm{H}_{2} \mathrm{O}\right.}{\longrightarrow}\) (A). The product (A) is (a) \(\mathrm{n}\) -butyl alcohol (b) iso-butyl alcohol (c) s-butyl alcohol (d) but-1-en-3-ol

The correct order of \(\mathrm{S}_{\mathrm{N}} 2\) reactivity among the following is

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \frac{(\text { i }) \mathrm{A}_{\mathrm{B}} \mathrm{CN}}{\left(\text { ii } \mathrm{H}_{2} / \mathrm{Ni}\right.}{\longrightarrow}\) product. The product is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\)

Reactions of Grignard reagents Organic compounds which contain atleast one carbon-metal bond are called organometallic compounds. Grignard reagents are very reactive organometallic compounds which are of great synthetic utility. They are used for the synthesis of alcohols, aldehydes, ketones, carboxylic acids, etc. They are however decomposed by compounds which have active hydrogen.

1,3 -diiodopropane on reaction with sodium in ether gives (a) \(\mathrm{IH}_{2} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)
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