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Chapter 4: Problem 63

n-butyl bromide can be prepared by treating \(n\) -butyl alcohol with \(\mathrm{KBr}\) and conc. \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) This process cannot be applied to prepare tertiary butyl bromide because (a) in tertiary alcohol, C - OH bond cleaves easily. (b) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dehydrates tertiary alcohol to form an alkene. (c) tertiary alcohol does not react with \(\mathrm{KBr}\). (d) in primary alcohol, C - OH bond cleaves easily.

Short Answer

Expert verified
Answer: The reaction does not work for tertiary alcohols because concentrated H2SO4 dehydrates tertiary alcohol to form an alkene instead of proceeding with the nucleophilic substitution with bromide ion.

Step by step solution

01

Understand the process for primary alcohols

When n-butyl alcohol (a primary alcohol) is treated with \(\mathrm{KBr}\) and concentrated \(\mathrm{H}_{2}\mathrm{SO}_{4}\), n-butyl bromide is formed. The \(\mathrm{H}_{2}\mathrm{SO}_{4}\) acts as an acid to protonate the alcohol, forming a good leaving group (water) that is then replaced by bromide ion generated from the reaction of \(\mathrm{KBr}\) with the acid. The reaction proceeds via a nucleophilic substitution mechanism (SN2).
02

Recognize the problem for tertiary alcohols

In the case of tertiary alcohols, the reaction with concentrated \(\mathrm{H}_{2}\mathrm{SO}_{4}\) leads to dehydration instead of substitution with bromide ion. Dehydration of tertiary alcohol proceeds via an E1 mechanism, forming an alkene as the final product.
03

Select the correct option

Based on the understanding of the reaction process for primary and tertiary alcohols, we can conclude that the right answer is (b), which states that the concentrated \(\mathrm{H}_{2}\mathrm{SO}_{4}\) dehydrates tertiary alcohol to form an alkene instead of proceeding with the nucleophilic substitution with bromide ion.

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Most popular questions from this chapter

Write equations for the following reactions and identify the product formed in each case (i) chlorobenzene + sodium \(\stackrel{\text { ether }}{\longrightarrow}\) (ii) chlorobenzene \(+\) chloral \(\stackrel{\text { con. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)

Which of the following is the correct order of ease of displacement of groups in substitution reactions? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}\)

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?

The number of possible isomeric compounds that can be formed by substitution of two hydrogen atoms of propane by two chlorine atoms is (a) 2 (b) 3 (c) 4 (d) 5

(a) \(\mathrm{n}\) -butyl alcohol (b) iso-butyl alcohol (c) sec-butyl alcohol (d) but-1-en-3-ol
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