Question

Ph-CHO can be participated in (A) Iodoform reaction (B) Aldol reaction (C) Cannizaro's reaction (D) Tischenko reaction

Short answer

Ph-CHO (benzaldehyde) can participate in (B) Aldol reaction, as it has an α-hydrogen atom and can undergo a condensation reaction, and (D) Tishchenko reaction, under specific conditions with an appropriate catalyst. It cannot participate in (A) Iodoform reaction due to the lack of a methyl group directly bonded to the carbonyl group, and (C) Cannizaro's reaction because it has an α-hydrogen atom.

Step-by-step solution

Understand the Iodoform reaction

The Iodoform reaction occurs with aldehydes and ketones that have a methyl group directly bonded to the carbonyl group. In this reaction, the methyl group is oxidized to generate iodoform (CHI_3). Since Ph-CHO does not have a methyl group directly bonded to the carbonyl group, it does not undergo the Iodoform reaction.

Understand the Aldol reaction

The Aldol reaction is an organic reaction that involves the condensation of two carbonyl compounds, typically an aldehyde and a ketone. The reaction occurs in the presence of a base or an acid as a catalyst. For an aldehyde to participate in an Aldol reaction, it should have an α-hydrogen atom that can be deprotonated. Ph-CHO, or benzaldehyde, has an α-hydrogen atom, which makes it capable of participating in the Aldol reaction.

Understand the Cannizaro's reaction

Cannizaro's reaction is a disproportionation reaction that occurs between two aldehyde molecules, facilitated by a base. One aldehyde molecule is oxidized to a carboxylic acid, while the second aldehyde is reduced to an alcohol. A key aspect of this reaction is that the aldehyde must not have any α-hydrogen atoms. Since Ph-CHO has an α-hydrogen atom, it cannot participate in Cannizaro's reaction.

Understand the Tishchenko reaction

Tishchenko reaction is a disproportionation reaction involving two aldehyde molecules, facilitated by a metal alkoxide catalyst, typically aluminum alkoxide. In this reaction, one aldehyde molecule is oxidized to an ester, while the second aldehyde molecule is reduced to an alcohol. Benzaldehyde can participate in the Tishchenko reaction under specific conditions, using an appropriate catalyst.

Answer Conclusion

Based on the above explanations, Ph-CHO (benzaldehyde) can participate in: (A) Iodoform reaction - No, because it lacks a methyl group directly bonded to the carbonyl group. (B) Aldol reaction - Yes, because it has an α-hydrogen atom, and it can undergo a condensation reaction. (C) Cannizaro's reaction - No, because it has an α-hydrogen atom. (D) Tishchenko reaction - Yes, under specific conditions with an appropriate catalyst. Thus, the correct answer is (B) Aldol reaction and (D) Tischenko reaction.

Key concepts

Aldol Reaction

The Aldol reaction is a significant transformation in organic chemistry, allowing the formation of carbon-carbon bonds. It involves the condensation of two carbonyl compounds, typically aldehydes and ketones, in the presence of a base or acid catalyst. During this reaction, the participating molecule must have at least one α-hydrogen atom. This α-hydrogen is essential since it can be deprotonated to form an enolate ion, which then attacks another carbonyl group.

Here are the steps for Aldol reaction with benzaldehyde (Ph-CHO):

• Deprotonation: The base deprotonates the α-hydrogen, forming an enolate ion.
• Nucleophilic Attack: This enolate ion attacks the carbonyl carbon of another molecule.
• Formation of Aldol Product: The aldol product is formed, which contains a β-hydroxy aldehyde or ketone.

Once the aldol is formed, it can undergo further reactions, such as dehydration, to form α,β-unsaturated carbonyl compounds. It is important to note that since benzaldehyde does contain an α-hydrogen atom, it can participate in the Aldol reaction.

Tishchenko Reaction

The Tishchenko reaction is an interesting process in organic chemistry, often confused with Cannizaro's reaction, since both involve disproportionation. In the Tishchenko reaction, two aldehyde molecules undergo disproportionation to form an ester and an alcohol.

One key difference from Cannizaro's reaction is the role of the catalyst. Tishchenko reaction requires a metal alkoxide catalyst, commonly aluminum alkoxide. This catalyst specifically facilitates the transfer of a hydride ion, causing one aldehyde to be reduced to an alcohol while the other is oxidized to form an ester.

An overview of the Tishchenko reaction:

• Aldehyde Mix: Two equivalents of an aldehyde are combined.
• Hydride Transfer: A hydride ion is transferred from one aldehyde molecule to another.
• Formation of Ester and Alcohol: An ester and an alcohol are simultaneously produced.

Benzaldehyde, under the right conditions and with the proper catalyst, can engage in the Tishchenko reaction, making it able to form these products.

Cannizaro's Reaction

Cannizaro's reaction is another notable reaction involving aldehydes but stands out because it does not require an α-hydrogen. This reaction is a base-catalyzed disproportionation process where two molecules of an aldehyde yield an alcohol and a carboxylic acid.

The steps of the Cannizaro's reaction include:

• Base Catalysis: A strong base deprotonates the non-α hydrogen, triggering the reaction.
• Formation of Intermediate: The base attack forms an alkoxide intermediate.
• Disproportionation: This intermediate causes one aldehyde to be reduced and another to be oxidized.

Benzaldehyde, having an α-hydrogen atom, does not typically undergo Cannizaro's reaction. This is an essential consideration when predicting which reactions benzaldehyde can participate in.