Aldehydes, Ketones and Carboxylic Acids

The rate of nucleophilic addition decreased due to steric crowding around carbonyl carbon \& increased by electron withdrawing group if the steric crowding is same hence the reactivity towards nucleophilic addition will be

$\mathrm{k}_{\mathrm{eq}}=2280$ is for HCHO

$\mathrm{k}_{\mathrm{eq}}=2000$ is for chloral

Both data is given in clayden and warren book.

$\mathrm{k}_{\mathrm{eq}}>1$ because HCHO and chloral are more electrophilic.

This is an example of Tollen's reaction i.e. multiple cross aldol followed by cross Cannizaro reaction

Note : In question about molecule "P" is not clarified, weather it is alcohol or alkene and as in question language rearrangement product is asking hence according to question language ans. is either (2) or (4). As alkene also undergoes rearrangement in presence of acid but option (2) also fulfil all conditions.

Due to NGP effect of phenyl ring Nucleophilic substitution of Br will occurs.

The reaction involved here is the Tollen's test, which detects aldehydes and alpha-hydroxy ketones, as they are able to reduce the ammoniacal silver nitrate solution to metallic silver, producing a mirror-like coating on the glassware. This test specifically distinguishes these reducing sugars from non-reducing sugars and is often used to identify the presence of an aldehyde functional group.

Let's analyze each option:

A. Formic acid - Formic acid ($HCOOH$) is a carboxylic acid but importantly, it can also reduce silver ions to metallic silver because it can act as a reducing agent just like aldehydes. Thus, it will give a positive Tollen's test.

B. Formaldehyde ($HCHO$) is the simplest aldehyde, and aldehydes are known to give positive results with the Tollen's test since they can be oxidized to carboxylic acids, reducing the silver ions in the process to metallic silver.

C. Benzaldehyde ($C_6H_5CHO$) is an aromatic aldehyde and, similar to formaldehyde, it can be oxidized to an acid, thereby reducing the silver ions to silver in the Tollen's test, giving a positive result.

D. Acetone ($CH_3COCH_3$) is a ketone. Generally, ketones do not undergo oxidation easily and do not reduce ammoniacal silver nitrate, and thus, do not give a silver mirror with the Tollen's reagent.

Therefore, the compounds that will give a silver mirror with ammoniacal silver nitrate (positive Tollen's test) are Formic acid, Formaldehyde, and Benzaldehyde, but not Acetone.

The correct answer from the options given is therefore:

Option C: A, B, and C only.

is known as Gattermann-Koch reaction.

Given reaction is HVZ reaction.

The major product (P) formed in the given reaction is 2-hydroxy propanoic acid.

Formaldehyde is more reactive due to least hindrance.

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.