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In which case, the intermediate involved is incorrect? (a) Pinacol - pinacolone rearrangement - carbocation (b) Hofmann's bromamide reaction \(-\) Nitrene (c) Aldol condensation - carbocation (d) Anti Markonikoff's addition of HBr to propene Free radical.

Short Answer

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(c) Aldol condensation - carbocation is incorrect.

Step by step solution

01

Understand the Reactions

First, we need to understand the key intermediates involved in each of the given reactions. For Pinacol-Pinacolone rearrangement, the intermediate is a carbocation. In Hofmann's bromamide reaction, the intermediate is a nitrene. For Aldol condensation, the correct intermediate should be an enolate ion, while Anti Markovnikov addition of HBr involves a free radical as its intermediate.
02

Identify the Incorrect Intermediate

Compare the reaction names with their intermediates: (a) Pinacol - pinacolone rearrangement involves a carbocation, which is correct. (b) Hofmann's reaction involves a nitrene, which is correct. (c) Aldol condensation should involve an enolate ion instead of a carbocation, which is incorrect. (d) Anti Markovnikov addition of HBr does indeed involve a free radical, which is correct.
03

Conclusion

From the analysis, the intermediate associated with Aldol condensation is listed incorrectly as a carbocation. The correct intermediate should be an enolate ion. Therefore, option (c) is the incorrect pairing of reaction and intermediate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pinacol-Pinacolone rearrangement
The Pinacol-Pinacolone rearrangement is a fascinating reaction in organic chemistry where a vicinal diol (pinacol) is converted to a ketone (pinacolone). This process is driven by the rearrangement of a carbocation intermediate.

In the first step of this reaction, acid catalysis leads to the protonation of one of the hydroxyl groups in the pinacol. This causes water to leave, forming a carbocation. The stability and rearrangement of this carbocation are crucial.

  • Carbocations are prone to rearrangements to form more stable structures.
  • In pinacolone formation, a methyl group shifts to stabilize the positive charge.
  • This rearrangement finalizes with the deprotonation, creating the ketone structure of pinacolone.

This reaction is an excellent example of how carbocation rearrangements can dramatically alter the molecular structure, transforming an alcohol into a ketone through a series of mechanistic steps.
Hofmann's bromamide reaction
Hofmann's bromamide reaction is a chemical reaction used to convert an amide into a primary amine with one fewer carbon atom. This reaction uses nitrenes as key intermediates, which are pivotal in the rearrangement during the reaction process.

The reaction proceeds by treating the amide with bromine in an alkaline solution. Here's a simplified breakdown of the process:

  • Amide reacts with bromine, forming an N-bromoamide.
  • A base then deprotonates this, forming a nitrene intermediate.
  • Nitrenes, having a very reactive electron configuration, migrate with the removal of a -CO group.
  • This migration results in the formation of an isocyanate intermediate, which finally hydrolyzes to a primary amine and carbon dioxide.

The nitrene intermediate is essential, as it allows the conversion of the carbon-nitrogen bond of the amide into an amine structure, reducing carbon count by one.
Aldol condensation
Aldol condensation is a fundamental reaction in organic chemistry where an enolate ion acts as an intermediate, not a carbocation. This process is known for forming carbon-carbon bonds, crucial for building complex organic compounds.

In an aldol reaction, an enolate ion, which is a deprotonated aldehyde or ketone, attacks another carbonyl compound. This reaction sequence can be summarized as:

  • The enolate ion forms by deprotonation of the α-carbon of the aldehyde or ketone.
  • This nucleophilic enolate then attacks the electrophilic carbonyl carbon of another molecule.
  • This addition results in a β-hydroxy carbonyl compound known as an aldol.
  • With heat, dehydration occurs, yielding an α,β-unsaturated carbonyl compound.

Aldol condensation exemplifies the power of enolate ions in forming new organic frameworks, emphasizing the incorrect association of this process with carbocations.
Anti Markovnikov addition
Anti Markovnikov addition is an intriguing concept in organic chemistry, especially notable in the free radical addition of HBr to alkenes like propene.

This type of addition takes place under radical conditions, differing markedly from the regular Markovnikov rule by reversing the regioselectivity. Let's dive deeper into how this occurs:

  • Initiation begins with the formation of bromine radicals, often through the influence of light or a radical initiator.
  • Propagation involves radical attack on the less-substituted carbon of the alkene, adhering to the anti-Markovnikov rule.
  • The intermediate free radical formed subsequently attracts a hydrogen atom to form the alkyl bromide product.

Anti Markovnikov addition highlights the impact of reaction conditions on regioselectivity, utilizing the unique nature of free radicals to achieve results that traditional ionic pathways do not.

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Most popular questions from this chapter

Which of the following reaction does not involve a carbocation as intermediate? (a) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \quad \mathrm{AlBr}_{3} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{Br}_{2} \longrightarrow \mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr} \mathrm{H}^{+}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{H}_{2} \mathrm{O}\) (d) Both (b) and (c)

The order of effectiveness of o- and p- directing groups is (a) \(\mathrm{OH}>\mathrm{NR}_{2}>\mathrm{NH} \cdot \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{CH}_{3}\) (b) \(\mathrm{OH}>\mathrm{Cl}>\mathrm{CH}_{3}>\mathrm{NR}_{2}>\mathrm{NH} . \mathrm{COCH}_{3}\) (c) \(\mathrm{OH}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}>\mathrm{CH}_{3}\) (d) \(\mathrm{OH}>\mathrm{CH}_{3}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}\)

The increasing order of the rate of HCN addition to compounds \(\mathrm{A}-\mathrm{D}\) is (a) \(\mathrm{HCHO}\) (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (c) \(\mathrm{PhCOCH}_{3}\) (d) \(\mathrm{PhCOPh}\) (a) \(\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}\) (b) \(\mathrm{d}<\mathrm{b}<\mathrm{c}<\mathrm{a}\) (c) \(\mathrm{d}<\mathrm{c}<\mathrm{b}<\mathrm{a}\) (d) \(\mathrm{c}<\mathrm{d}<\mathrm{b}<\mathrm{a}\)

Which of the following undergoes nucleophillic substitution exclusively by \(\mathrm{SN}^{1}\) mechanism? (a) ethyl chloride (b) isopropyl chloride (c) chlorobenzene (d) benzyl chloride

In which cases, free radicals can be formed by hemolytic fission? C=O hv (a) \(\mathrm{CH}_{3} \mathrm{CCH}_{3}\) (b) \(\mathrm{R}-\ddot{\mathrm{N}}=\ddot{\mathrm{N}}-\mathrm{R}\) hv (c) O=CCc1ccccc1 (d) in all cases

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