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Chapter 4: Problem 100

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene

Short Answer

Expert verified
Answer: The product B formed in the given reaction is 2-phenyl-3-methylbutan-2-ol.

Step by step solution

01

Formation of Grignard reagent (A)

In the first step, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr}\) reacts with \(\mathrm{Mg}\) to form a Grignard reagent. The magnesium inserts itself between the carbon and bromine atoms. The Grignard reagent (compound A) formed is: \((\mathrm{CH}_{3})_{2} \mathrm{CH}\mathrm{MgBr}\).
02

Nucleophilic addition of Grignard reagent to a carbonyl compound

In the second step, Grignard reagent (A) \((\mathrm{CH}_{3})_{2} \mathrm{CH}\mathrm{MgBr}\) undergoes a nucleophilic addition reaction with a carbonyl compound, which is \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}\). The Grignard reagent acts as the nucleophile and attacks the electrophilic carbonyl carbon, leading to the formation of an alkoxide anion.
03

Protonation of alkoxide anion to form alcohol (B)

Finally, the alkoxide intermediate is protonated to form compound B, an alcohol. The hydrogen atom in this step comes from the reaction between Grignard reagent and a carbonyl compound. The alcohol product (B) is generated: \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}(\mathrm{CH}_{3})_{2} \mathrm{CH}\mathrm{OH}\).
04

Identifying the correct structure and name for product B

We now need to identify the correct structure and name for product B from the given options. The structure found in Step 3 is \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}(\mathrm{CH}_{3})_{2} \mathrm{CH}\mathrm{OH}\). This corresponds to the structure: \(\mathrm{CH}_{3}\mathrm{C}(\mathrm{CH}_{3})\mathrm{CH}\mathrm{C}( \mathrm{OH})\mathrm{C}_{6} \mathrm{H}_{5}\), which has the IUPAC name 2-phenyl-3-methylbutan-2-ol. Thus, option (a) is the correct answer. In conclusion, the product B in the given reaction is 2 -phenyl- 3 -methyl butan- 2 -ol.

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Most popular questions from this chapter

Equal quantities of \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{Cl}\) are treated with iodide ion in acetone. Which reacts faster? (a) both react at almost equal rate (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) reacts faster (c) \(\mathrm{CH}_{3} \mathrm{Cl}\) reacts faster (d) \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Cl}\) doest not reacts at all

A dextrorotatory alkyl halide is hydrolysed under \(\mathrm{S}_{\mathrm{N}} 2\) conditions to form the corresponding alcohol. The resulting alcohol (a) will be levorotary (b) will be dextrorotatory (c) will be optically inactive due to racemization (d) may be dextro or laevorotatory

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?

Biphenyl is formed when (a) bromobenzene is heated with Na in ether. (b) iodobenzene is heated with copper powder in a sealed tube. (c) chlorobenzene is heated with benzene in presence of anhydrous aluminium chloride. (d) phenyl magnesium bromide is treated with bromobenzene in the presence of a small amount of \(\mathrm{CoCl}_{2}\).

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
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