Aldehydes, Ketones and Carboxylic Acids

This is Wolff-Kishner reduction :

Let's analyze the given statements in detail:

Statement I: Acidity of $\alpha$-hydrogens of aldehydes and ketones is responsible for Aldol reaction.

This statement is correct. The presence of acidic $\alpha$-hydrogens in aldehydes and ketones makes it possible for these compounds to undergo deprotonation, forming enolates. These enolates can then attack the carbonyl carbon of another aldehyde or ketone, leading to the formation of the Aldol product. Therefore, the acidity of the $\alpha$-hydrogens is crucial for the Aldol reaction.

Statement II: Reaction between benzaldehyde and ethanal will NOT give Cross-Aldol product.

This statement is incorrect. Benzaldehyde (which lacks an $\alpha$-hydrogen) can indeed react with ethanal (which has an $\alpha$-hydrogen) in a Cross-Aldol reaction. In a Cross-Aldol reaction, one molecule (ethanal) forms an enolate and reacts with another molecule (benzaldehyde), resulting in the formation of a Cross-Aldol product.

Based on these explanations, the most appropriate answer is:

Option B

Statement I is correct but Statement II is incorrect.

Clemmensen reduction will convert

To predict the boiling point of these compounds, we need to consider the types of intermolecular forces present in each molecule. These intermolecular forces include van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

The compounds given are:

• Option A : Butanol ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}$) - a primary alcohol


• Option B : Butane ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$) - an alkane


• Option C : Butyraldehyde ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}$) - an aldehyde


• Option D : Diethyl ether ($\mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5$) - an ether

Now let's analyze each one:

Option A: Butanol is capable of hydrogen bonding because it has an $\mathrm{-OH}$ group. Hydrogen bonding is the strongest intermolecular force among the forces affecting these molecules. Therefore, butanol will have the highest boiling point among the compounds without considering the others.

Option B: Butane only has van der Waals forces because it is a non-polar molecule. Van der Waals forces are the weakest intermolecular forces, resulting in a lower boiling point compared to molecules that can hydrogen bond or have permanent dipoles.

Option C: Butyraldehyde can form dipole-dipole interactions because of its polar carbonyl ($\mathrm{C=O}$) group, but it cannot form hydrogen bonds with itself since there is no $\mathrm{OH}$ group. Dipole-dipole interactions are stronger than van der Waals forces but weaker than hydrogen bonds. As a result, butyraldehyde will have a higher boiling point than butane but a lower boiling point than butanol.

Option D: Diethyl ether has an oxygen atom, which makes it polar and enables dipole-dipole interactions. However, ethers cannot form hydrogen bonds with themselves, so diethyl ether will have a higher boiling point than butane but lower than butanol.

Considering the intermolecular forces, the compound with the highest boiling point is the one that can form hydrogen bonds, which in this list is the primary alcohol, butanol.

Therefore, the compound with the highest boiling point is:

Option A : Butanol ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}$).

Meta-chlorobenzaldehyde will undergo

Cannizzaro reaction with $50 \% \mathrm{~KOH}$ to give $\mathrm{m}-$ chlorobenzoate ion and m-chlorobenzyl alcohol.

It is known as rosenmund reduction that is the partial reduction of acid chloride to aldehyde.

Acidity of an acid depends upon the stability of it's conjugate base.

H_C > H_D > H_A > H_B

H_C is hydrogen of carboxylic acid.

H_D removal will lead to stable carbanion.

DIBAL-H reduces both the cyanides and esters to aldehydes.

Na₂Cr₂O₇, H₂SO₄/H₂O is the strongest oxidising agent and it will oxidise 1° alcohol into acids.