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Chapter 4: Problem 92

Methyl ethanoate \(\frac{\text { (i) } \mathrm{CH}_{3} \mathrm{MgCl}(\text { excess })}{\text { (ii)Hydrolysis }} \mathrm{x}\). The product \(\mathrm{x}\) in the above reaction is

Short Answer

Expert verified
Answer: The final product 'x' is tert-butanol (\(\mathrm{CH}_{3} \mathrm{C}(\mathrm{CH}_{3})_{2} \mathrm{OH}\)).

Step by step solution

01

Identify the functional group in the starting compound.

Methyl ethanoate is an ester (\(\mathrm{CH}_{3}\mathrm{COOCH}_{3}\)) which has the general formula \(\mathrm{R-COO-R'}\). In this case, both R and R' are methyl groups (CH₃).
02

Understand the reactivity of Grignard reagents with esters

Grignard reagents (\(\mathrm{RMgX}\)) are strong nucleophiles that can react with electrophiles like carbonyls. In this case, the ester (Methyl ethanoate) has a carbonyl group that will react with Grignard reagent (\(\mathrm{CH}_{3} \mathrm{MgCl}\)).
03

Perform the first reaction - Grignard reagent with Methyl ethanoate

Treat Methyl ethanoate with an excess of Grignard reagent, which will result in a nucleophilic attack on the carbonyl carbon. This reaction will lead to the formation of a ketone after the initial O-alkylation and substitution, and further nucleophilic attack by Grignard reagent because the reaction is carried out in excess which ultimately forms tertiary alcohol. The intermediate product would be \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{CH}_{3})_{2} \mathrm{OH}\) (tert-butanol).
04

Perform the second reaction - Hydrolysis

Hydrolysis is the reaction between a compound and water, resulting in the breaking of a bond in the compound and the formation of two new products. In our case, since tert-butanol isn't subject to hydrolysis, there will be no change in the product.
05

Determine the final product

After going through Grignard reaction and hydrolysis, the final product 'x' is \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{CH}_{3})_{2} \mathrm{OH}\) (tert-butanol).

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Most popular questions from this chapter

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?

[A] \(\frac{\text { (i) } \mathrm{B}_{2} \mathrm{H}_{6} \text { /ether }}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{OH}^{-}}\) Isobutyl alcohol. The compound \([\mathrm{A}]\) in the above reaction is (a) But-1-ene (b) But-2-ene (c) 2 -Methyl propene (d) 2, 2-Dimethyl but \(-1\) -ene

Directions: Each question contains Statement- 1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement- 1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{Br}\) undergoes hydrolysis faster than \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) and Statement 2 The carbocation formed by ionization of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2} \mathrm{Br}\) is stabilized by resonance.

The correct nucleophilicity order for \(\mathrm{S}_{\mathrm{N}} 2\) reaction in protic solvents is (a) \(\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{I}^{-}>\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{OH}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{CN}^{-}>\mathrm{I}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{I}^{-}\) (d) \(\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{I}^{-}\)

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 \(\mathrm{CoCl}_{2}\).
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