Question

Select the correct options :

• A. Boiling point, (P >Q)
• B. Melting point, (Q >P)
• C. Water solubility, (P <Q)
• D. Acid Strength, (Q<P)

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The given reaction is the Reimer-Tiemann reaction. Phenol reacts with chloroform ($\text{CHCl}_3$) and sodium hydroxide ($\text{NaOH}$) followed by acidification ($\text{H}^+$) to form ortho- and para-hydroxybenzaldehydes. The electrophile in this reaction is dichlorocarbene (:$\text{CCl}_2$).

The products formed are:

1. P: Salicylaldehyde (2-hydroxybenzaldehyde) SMILES: O=Cc1ccccc1O This is the ortho isomer. It is formed as the major product due to intramolecular hydrogen bonding (chelation) in the transition state. The problem states (P > Q) % yield, confirming P as the ortho isomer.

2. Q: p-hydroxybenzaldehyde (4-hydroxybenzaldehyde) SMILES: O=Cc1ccc(O)cc1 This is the para isomer.

Now let's compare the properties of P and Q:

1. Boiling Point:

• P (Salicylaldehyde): The hydroxyl (-OH) group and the aldehyde (-CHO) group are ortho to each other, allowing for intramolecular hydrogen bonding. This reduces the extent of intermolecular hydrogen bonding.

• Q (p-hydroxybenzaldehyde): The -OH and -CHO groups are para to each other, so intramolecular hydrogen bonding is not possible. It forms extensive intermolecular hydrogen bonds.

Stronger intermolecular forces lead to higher boiling points. Therefore, Q has a higher boiling point than P. Conclusion: Boiling point (Q > P). Option (A) states (P > Q), which is incorrect.

2. Melting Point:

• Melting point depends on the strength of intermolecular forces and the efficiency of crystal packing.

• Q (p-hydroxybenzaldehyde): Its more symmetrical structure allows for more efficient packing in the crystal lattice compared to P. Combined with stronger intermolecular hydrogen bonding, this leads to a higher melting point for Q.

• P (Salicylaldehyde): Less symmetrical and forms intramolecular hydrogen bonds, which reduces overall intermolecular forces available for crystal lattice formation. Conclusion: Melting point (Q > P). Option (B) states (Q > P), which is correct.

3. Water Solubility:

• Water solubility depends on the ability to form hydrogen bonds with water molecules.

• P (Salicylaldehyde): The intramolecular hydrogen bonding in P makes the -OH proton less available for forming hydrogen bonds with water molecules.

• Q (p-hydroxybenzaldehyde): Both the -OH group and the aldehyde oxygen are fully available to form hydrogen bonds with water molecules, as there is no intramolecular hydrogen bonding. Therefore, Q is more soluble in water than P. Conclusion: Water solubility (Q > P) or (P < Q). Option (C) states (P < Q), which is correct.

4. Acid Strength:

• The acid strength of phenols is increased by electron-withdrawing groups. The -CHO group is an electron-withdrawing group by both inductive (-I) and resonance (-M) effects.

• P (Salicylaldehyde): The -CHO group is at the ortho position. The -I effect is stronger at the ortho position due to proximity. The -M effect is also significant. While intramolecular hydrogen bonding slightly stabilizes the neutral molecule, the dominant effect of the ortho-CHO group is to stabilize the phenoxide ion, making it more acidic.

• Q (p-hydroxybenzaldehyde): The -CHO group is at the para position. The -M effect is strong, but the -I effect is weaker at the para position compared to ortho.

A lower pKa value indicates higher acid strength. Conclusion: Acid Strength (P > Q) or (Q < P). Option (D) states (Q < P), which is correct.