Aldehydes, Ketones and Carboxylic Acids

Esterification of carboxylic acid with an alcohol is nucleophilic acyl substitution and presence of electron withdrawing group in the carboxylic acid increases the rate of esterification reaction.

Statement (I) is correct as monocarboxylic acids with even number of carbon atoms show better packing efficiency in the solid state, statement (II) is also correct as the solubility of carboxylic acids decreases with an increase in molar mass due to increase in the hydrophobic portion with an increase in the number of carbon atoms.

As for Reason R, it is false.

A molecule with 2 chiral carbons is not necessarily optically active.

It may be if the molecule is not superimposable on its mirror image (i.e., it's chiral).

But if the molecule is superimposable on its mirror image (i.e., it's achiral), then it will not be optically active.

This can occur if the molecule has a plane of symmetry, in which case it is a meso compound.

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 ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}$).

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.

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.

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

To determine the correct match between Item I and Item II, let's analyze each item and its corresponding match : 1.

Benzaldehyde: Benzaldehyde is an aromatic aldehyde, which is typically considered as an adsorbate in chromatography.

An adsorbate is a substance that is adsorbed on a surface.

2.

Alumina: Alumina (Aluminum oxide) is well known for its use as an adsorbent in chromatography processes.

An adsorbent is a material that adsorbs another substance.

3.

Acetonitrile: Acetonitrile is an organic solvent often used as a mobile phase in chromatography.

The mobile phase is the phase that moves in chromatography, helping to carry the sample through the stationary phase.

Now, let's match the items : Item - I Item - II (A) Benzaldehyde (R) Adsorbate (B) Alumina (Q) Adsorbent (C) Acetonitrile (P) Mobile phase Therefore, the correct match is : (A) $\to$ (R); (B) $\to$ (Q); (C) $\to$ (P) This corresponds to Option B.

In Formaldehyde (HCHO) both side of the C atom H presents but in Acetaldehyde (CH3CHO) one side of C atom -CH3 group and other side H atom presents.

So Formaldehyde is less crowded then it is more reactive.

CH3OH is 1o alcohol and tBuOH is 3o alcohol.

As we know on 3o alcohol steric hindrance is very high compare to 1o alcohol and due to steric hindrance reactivity of 3o alcohol is less compare to 1o alcohol.

.tg .tg S.No.

Name of Reaction Acetaldehyde Image Acetone Image 1 Iodoform reaction $\oplus$ve $\oplus$ve 2 Cannizaro

ve

ve 3 Aldol reaction $\oplus$ve $\oplus$ve 4 Tollen's test $\oplus$ve

ve 5 Clemmensen reduction $\oplus$ve $\oplus$ve Ans. (2) A, C and E only.