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Chapter 24: Problem 71

4-Methylpyridine reacts with benzaldehyde \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\right)\) in the presence of base to form A. (a) Draw a stepwise mechanism for this reaction. (b) Would you expect 2-methylpyridine or 3-methylpyridine to undergo a similar type of condensation reaction? Explain why or why not.

Short Answer

Expert verified
4-Methylpyridine reacts through nucleophilic attack, not 2- or 3-methyl due to hindered nitrogen lone pair availability.

Step by step solution

01

Understanding the Reaction Components

In this reaction, 4-methylpyridine acts as a nucleophile while benzaldehyde \( (\text{C}_6 ext{H}_5 ext{CHO}) \) is the electrophile. A base is present to deprotonate and activate the pyridine for nucleophilic attack on the carbonyl carbon of benzaldehyde.
02

Deprotonation of 4-Methylpyridine

The base removes a proton from the 4-methylpyridine, specifically from the nitrogen atom, forming an anionic species. This increases the nucleophilicity of the nitrogen, making it ready to attack the carbonyl carbon of the benzaldehyde.
03

Nucleophilic Attack

The anionic nitrogen of 4-methylpyridine attacks the electrophilic carbonyl carbon of benzaldehyde, resulting in the formation of a tetrahedral intermediate. This step breaks the \( \text{C}=\text{O} \) double bond temporarily.
04

Formation of Intermediate A

The tetrahedral intermediate rearranges to re-establish the carbon-oxygen double bond, ejecting a hydroxide ion (since this reaction occurs under basic conditions). The product formed is intermediate A, a substituted aldol type product.
05

Consideration of Alternatives - 2- and 3-Methylpyridine

In the case of 2-methylpyridine or 3-methylpyridine, sterics and electronics play a role. Sterically, 2-methylpyridine, with the methyl group adjacent to the nucleophilic nitrogen, faces more hindered nucleophilic attacks, reducing its reactivity. 3-methylpyridine is less hindered than 2-methyl, but both are typically less reactive than 4-methylpyridine due to the electronic effect of the methyl substituent affecting nitrogen lone pair availability. Thus, they are less likely to undergo a similar condensation to form aldol products with benzaldehyde.

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Key Concepts

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

4-Methylpyridine
4-Methylpyridine is an aromatic heterocyclic compound. It consists of a pyridine ring, which is a six-membered ring with five carbon atoms and one nitrogen atom. The methyl group is attached at the fourth position relative to the nitrogen atom. This specific positioning influences its chemical behavior in nucleophilic reactions.
  • The nitrogen atom in pyridine has a lone pair of electrons, making it a good site for nucleophilic attack. This lone pair is less hindered in 4-methylpyridine compared to other isomers like 2-methylpyridine, where the methyl group is closer to the nitrogen.
  • Electronegative nitrogen can act as a base, making 4-methylpyridine more basic than benzene. This explains why it can be easily deprotonated by a base in a condensation reaction, enhancing its nucleophilicity.
  • The position of the methyl group in 4-methylpyridine allows for less steric hindrance during nucleophilic attacks, contributing to its reactivity in comparison to its 2- or 3-methyl counterparts.
Condensation Reactions
Condensation reactions are a key concept in organic chemistry, often involving the formation of a new bond while losing a small molecule such as water or alcohol. In the context of 4-methylpyridine and benzaldehyde, condensation involves these two larger molecules forming a bond with the elimination of a hydroxide ion due to basic conditions.
  • The reaction begins with the activation of a nucleophile, here the paired nitrogen of 4-methylpyridine, to enhance its attacking potential.
  • The carbonyl carbon of benzaldehyde is electrophilic, thus susceptible to nucleophilic attack by the activated nitrogen, leading to the formation of a complex intermediate.
  • This is a type of aldol reaction, modified due to the use of nitrogen instead of a carbon atom as the nucleophile, resulting in a substituted product different from traditional carbonyl-vinyl type aldols.
Condensation reactions illustrate how molecular components can combine through relatively simple mechanisms, pivotal in synthetic pathways for constructing more complex organic molecules.
Mechanism of Reaction
Understanding the mechanism of reaction between 4-methylpyridine and benzaldehyde helps to appreciate how chemical transformations occur at a molecular level. Let's break it down:
  • Initiation: The base deprotonates 4-methylpyridine. Here, the nitrogen atom loses a proton, forming a more reactive anionic species.
  • Nucleophilic Attack: The electron-rich nitrogen then attacks the positively charged carbon atom in the carbonyl group of benzaldehyde, forming a tetrahedral intermediate.
  • Intermediate Formation: The initial reaction leads to a temporary breakdown of the carbonyl double bond, which is then re-established as the intermediate collapses, ejecting a hydroxide ion. This forms the final product known as intermediate A.
Comparing this sequence to other methylpyridine isomers outlines the influence of steric and electronic effects, which play crucial roles in determining the overall feasibility and pathway of the condensation reaction.

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Most popular questions from this chapter

In theory, the intramolecular aldol reaction of 6-oxoheptanal could yield the three compounds shown. It turns out, though, that 1-acetylcyclopentene is by far the major product. Why are the other two compounds formed in only minor amounts? Draw a stepwise mechanism to show how all three products are formed.

Devise a synthesis of each compound from \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}\), benzene, and alcohols having \(\leq 2 \mathrm{C}^{\prime} \mathrm{s}\). You may also use any required organic or inorganic reagents.

What four compounds are formed from the crossed Claisen reaction of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}\) ?

Draw the product formed from an aldol reaction with the given starting material(s) using \(\mathrm{OH}, \mathrm{H}_{2} \mathrm{O}\)a. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO}\) only b. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO}+\mathrm{CH}_{2}=\mathrm{O}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) d. \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{C}=\mathrm{O}\) only e. \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{CH}_{2}=\mathrm{O}\)

Draw the organic products formed when butanal \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\right)\) is treated with each reagent. a. \(\mathrm{OH}, \mathrm{H}_{2} \mathrm{O}\) f. \(\mathrm{NaBH}_{4}, \mathrm{CH}_{3} \mathrm{OH}\) k. \(\mathrm{CrO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}\) b. \({ }^{-} \mathrm{OH}, \mathrm{CH}_{2}=\mathrm{O}, \mathrm{H}_{2} \mathrm{O}\) g. \(\mathrm{H}_{2}, \mathrm{Pd}-\mathrm{C}\) L. \(\mathrm{Br}_{2}, \mathrm{CH}_{3} \mathrm{COOH}\) c. \([1] \angle D A ;[2] \mathrm{CH}_{3} \mathrm{CHO} ;[3] \mathrm{H}_{2} \mathrm{O}\) h. \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{TsOH}\) m. \(\mathrm{Ph}_{3} \mathrm{P}=\mathrm{CH}_{2}\) d. \(\mathrm{CH}_{2}\left(\mathrm{CO}_{2} \mathrm{Et}\right)_{2}, \mathrm{NaOEt}, \mathrm{EtOH}\) i. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), mild acid n. \(\mathrm{NaCN}, \mathrm{HCl}\) e. \([1] \mathrm{CH}_{3} \mathrm{Li} ;[2] \mathrm{H}_{2} \mathrm{O}\) 隆. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\), mild acid o. \([1] \mathrm{LDA} ;[2] \mathrm{CH}_{3} \mathrm{I}\)
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