To solve this sequence of reactions, we work backward from compound (R).
1. Identifying Compound (R): The molecular formula given is C6H7N. This formula corresponds to Aniline (C6H5NH2). Aniline is a primary aromatic amine.
2. Identifying Compound (Q): Compound (R) is formed by heating (Q) with Br2 and KOH. This is the classic Hoffmann Bromamide Degradation reaction. In this reaction, an amide is converted into a primary amine with one fewer carbon atom. Since (R) is Aniline (C6), (Q) must be Benzamide (C6H5CONH2).
3. Identifying Compound (P): Compound (Q) is formed by treating (P) with aqueous ammonia under hot conditions. Carboxylic acids react with ammonia to form ammonium salts, which upon heating dehydrate to form amides. Thus, (P) must be Benzoic acid (C6H5COOH).
Thus, P = Benzoic acid, Q = Benzamide, and R = Aniline.
Organic chemistry in JEE Main often focuses on named reactions and functional group interconversions. This specific problem integrates two major pillars: the chemistry of carboxylic acid derivatives and the synthesis of amines. Understanding these mechanisms is vital for cracking multi-step synthesis questions.
1. Preparation of Amides from Carboxylic Acids
The reaction of a carboxylic acid with ammonia doesn't initially produce an amide. Instead, an acid-base reaction occurs to form an ammonium carboxylate salt. For example, Benzoic acid reacts with aqueous ammonia to form Ammonium benzoate. When this salt is heated strongly, a water molecule is lost (dehydration), resulting in the formation of Benzamide. This is a standard laboratory method for amide synthesis. In industrial setups, acid chlorides (R-COCl) are often preferred as they react more vigorously with ammonia to yield amides without requiring intense heat.
2. Hoffmann Bromamide Degradation Reaction
This is one of the most important named reactions in the JEE syllabus. It is used specifically to descend a homologous series—meaning it reduces the carbon chain length by one. The general reaction is:
\[ \text{R-CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O} \]
The mechanism involves the formation of an N-bromoamide, followed by the migration of the alkyl or aryl group to the nitrogen atom, leading to an isocyanate intermediate. The hydrolysis of this isocyanate eventually yields the primary amine. A key point to remember for exams: only primary amides undergo this reaction. Secondary and tertiary amides do not form amines via this pathway.
3. Physical and Chemical Properties of Aniline (R)
Aniline (C6H5NH2) is the simplest aromatic amine. Its molecular formula C6H7N shows a high degree of unsaturation (Degree of Unsaturation = 4, indicating a benzene ring). Aniline is a weak base because the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring through resonance. This makes the electrons less available for protonation compared to aliphatic amines like methylamine. In JEE Advanced, you might be asked to compare the basic strength of aniline with substituted anilines (using +I/-I and +M/-M effects).
4. Analysis of Incorrect Options
- Option A: Toluic acid has a methyl group on the benzene ring (C7H8O2). The final product (R) would have been methylaniline (C7H9N), which contradicts the given formula C6H7N.
- Option C: Similar to Option A, 4-methylbenzamide would yield 4-methylaniline, which has 7 carbons.
- Option D: Phenylethanoic acid (C6H5CH2COOH) has 8 carbons. Hoffmann degradation of its amide would yield benzylamine (C6H5CH2NH2), which has 7 carbons.
5. Strategic Importance for JEE
In the JEE exam, when you see "Br2 + KOH" (or NaOH), your mind should immediately think of Hoffmann Bromamide or Haloform reaction. If the starting material is an amide, it’s Hoffmann. If the product has one less carbon, it confirms the suspicion. Learning to "read" reagents is faster than trying to derive every mechanism during the exam. Furthermore, knowing the molecular formulas of common compounds like Benzene (C6H6), Aniline (C6H7N), and Phenol (C6H6O) saves precious seconds.
6. Common Mistakes to Avoid
- Carbon Count: Many students forget that Hoffmann degradation removes the carbonyl carbon as a carbonate. Always subtract one carbon from the amide to get the amine.
- Reagent Confusion: Do not confuse NH3/heat with LiAlH4. While NH3/heat turns acid to amide, LiAlH4 would reduce the acid all the way to a primary alcohol (Benzyl alcohol), and if used on an amide, it would reduce it to an amine without losing a carbon atom.
- Formula Mapping: C6H7N is only possible if the benzene ring is intact. Always check if the formula fits the aryl-amine structure.
7. Related Shortcut Tricks
If a question gives a formula like CnH2n-5N, it usually indicates an aromatic amine. For \(n=6\), it is Aniline. For \(n=7\), it is Toluidine or Benzylamine. Keeping these general formulas in mind helps in quickly eliminating options in the MCQ format of JEE Main.