Compound P is neutral, Q gives effervescence with NaHCO$$_3$$ while R reacts with Hinsberg's reagent to give solid soluble in NaOH. Compound P is

Step 1: Calculating the Double Bond Equivalent (DBE)

The Double Bond Equivalent (also known as the Index of Hydrogen Deficiency) tells us the total number of rings and/or $$\pi$$-bonds present in the molecule. The general formula for a compound with the molecular formula $$\text{C}_\text{c}\text{H}_\text{h}\text{O}_\text{o}\text{N}_\text{n}$$ is:

$$\text{DBE} = \text{c} + 1 - \frac{\text{h} - \text{n}}{2}$$

Substituting the values from the given molecular formula ($$\text{C}_{14}\text{H}_{13}\text{ON}$$):

$$\text{DBE} = 14 + 1 - \frac{13 - 1}{2}$$

$$\text{DBE} = 15 - 6 = 9$$

A DBE value of 9 strongly indicates the presence of:

• Two benzene rings: Each benzene ring accounts for 4 structural units (1 ring + 3 double bonds), totaling $$2 \times 4 = 8$$ units of unsaturation.
• One carbonyl or amide group: The remaining 1 unit of unsaturation ($$9 - 8 = 1$$) corresponds to a double bond, specifically a carbonyl/amide carbon-oxygen double bond ($$\text{C=O}$$).

Step 2: Deducing the Functional Groups from Chemical Tests

1. Neutral Nature of Compound P:

   Compound P is described as neutral. Since it contains a nitrogen atom and an oxygen atom but lacks acidic or basic traits, it points directly toward a substituted amide group ($$\text{--CONH--}$$) rather than a free amine or carboxylic acid.




2. Acidic Hydrolysis of Amides:

   When an amide is heated with dilute hydrochloric acid ($$\text{HCl}$$), it undergoes cleavage into a carboxylic acid and an amine salt:

   $$\text{R--CONH--R'} \xrightarrow{\text{HCl, }\Delta} \text{R--COOH (Residue Q)} + \text{R'--NH}_3^\oplus\text{Cl}^\ominus \xrightarrow{\text{NaOH}} \text{R'--NH}_2\text{ (Oily Liquid R)}$$



3. Analysis of Residue Q:

   Residue Q gives brisk effervescence when treated with sodium bicarbonate ($$\text{NaHCO}_3$$), liberating carbon dioxide ($$\text{CO}_2$$) gas. This test is a hallmark confirmation for the presence of a carboxylic acid group ($$\text{--COOH}$$).




4. Analysis of Oily Liquid R (Hinsberg Test):

   Oily liquid R reacts with Hinsberg's reagent (benzenesulfonyl chloride, $$\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$$) to produce a solid sulfonamide compound that dissolves readily in aqueous sodium hydroxide ($$\text{NaOH}$$). This specific solubility profile proves that R must be a primary amine ($1^\circ\text{ amine}$ marketing as $$\text{R'--NH}_2$$), as its sulfonamide derivative still contains an acidic hydrogen on the nitrogen atom.

Step 3: Reconstructing Compound P

Since the hydrolysis fragments are a carboxylic acid and a primary amine, the parent structure P is a secondary amide linking two aromatic rings:

$$\text{4-methyl-N-phenylbenzamide} \quad (\text{p-CH}_3\text{--C}_6\text{H}_4\text{--CONH--C}_6\text{H}_5)$$

Let's verify the atom count for this structure:

• Carbons: 7 from the 4-methylbenzoyl part + 6 from the aniline ring = 14 carbons.
• Hydrogens: 7 from the tolyl fragment + 1 from the amide nitrogen + 5 from the phenyl ring = 13 hydrogens.
• Heteroatoms: 1 Oxygen and 1 Nitrogen.

The total formula is precisely $$\text{C}_{14}\text{H}_{13}\text{ON}$$, fitting every physical parameter and chemical constraint perfectly.

Conclusion:

The combination of a calculated DBE of 9, its neutral nature, and the formation of a carboxylic acid along with a primary amine upon acid hydrolysis identifies compound P as N-(4-methylphenyl)benzamide / 4-methyl-N-phenylbenzamide.

Answer: The option displaying the structure with a 4-methylphenyl ring on one side and a phenyl group attached to the amide nitrogen ($$\text{p-CH}_3\text{--C}_6\text{H}_4\text{--CONH--C}_6\text{H}_5$$).