Q.Write resonance structures of CH₂=CH–CHO. Indicate relative stability of the contributing structures.

Concept & Intuition

This is a classic organic synthesis puzzle. The first step uses chromia (Cr₂O₃) at high temperature and pressure—a typical condition for the dehydrogenation of an alkylbenzene to an aromatic aldehyde or acid. But here the product (A) is then treated with alkaline KMnO₄ followed by acid, which is a strong oxidation that converts any alkyl side‑chain (or aldehyde) into a carboxylic acid. So (B) must be a benzoic acid derivative. Finally, nitration with conc. HNO₃/H₂SO₄ introduces a nitro group. The key is to identify the starting material from the given options: the final compound (C) is aromatic with two substituents D and E. Working backwards, the only combination that fits the oxidation and nitration pattern is –COOH and –NO₂.

Step‑by‑Step Reasoning

1. Identify the starting material

   The first arrow shows a structure (not drawn here, but typical in such problems) being passed over Cr₂O₃ at 773 K and 10–20 atm. This is the dehydrogenation of an alkylbenzene (e.g., toluene) to benzaldehyde or benzoic acid. In fact, Cr₂O₃ at high temperature often gives the aldehyde, but the exact product (A) is not yet fully oxidised.

2. Oxidation to (B)

   Step (i) KMnO₄/OH⁻ followed by (ii) H₃O⁺ is a vigorous oxidation that converts any alkyl group (–CH₃) or aldehyde (–CHO) directly to a carboxylic acid (–COOH). So (B) must be benzoic acid (or a substituted benzoic acid if the starting material already had a substituent). Since the starting material is a simple aromatic hydrocarbon (likely toluene), (B) is unsubstituted benzoic acid.

3. Nitration to (C) …

The key here is to understand what pKa tells us about acidity. A lower pKa means a stronger acid — the molecule more readily donates its proton. The reason given compares the stability of the conjugate bases (phenoxide vs. benzoate) via resonance. Let’s check both statements carefully.

1. Check the Assertion (A):

   The problem states: pKa of phenol is 4.19 and that of benzoic acid is 10.

   In reality, the pKa of phenol is about 10, and the pKa of benzoic acid is about 4.2.

   So the assertion has the numbers reversed — phenol is the weaker acid, benzoic acid is the stronger one.

   Therefore, Assertion (A) is false.

2. Check the Reason (R):

   The reason says: Phenoxide ion is stabilised by non-equivalent resonance structures whereas benzoate ion by two equivalent resonance structures.

   This is chemically correct:

   • In the phenoxide ion, the negative charge can be delocalised into the ring, but the resonance structures are not all equivalent (some place the charge on carbon, which is less stable than on oxygen).
   • In the benzoate ion, the two major resonance structures place the negative charge equally on the two oxygen atoms — they are equivalent, giving extra stability.
   • Greater stabilisation of the conjugate base means a stronger acid. Benzoate is more stabilised than phenoxide, so benzoic acid (pKa ~4.2) is stronger than phenol (pKa ~10). …

The key to this question is recognizing the structural features of the molecule. The given compound contains a nitrobenzene ring, a dichloroacetamide group (NHCOCHCl₂), and a chain with two hydroxyl groups and a primary alcohol. This exact arrangement is the hallmark of chloramphenicol, a well-known antibiotic.

1. Identify the functional groups: The molecule has:

   • A para-nitrophenyl group (O₂N–C₆H₄–).
   • An amide linkage (–NHCO–) attached to a dichloromethyl group (–CHCl₂).
   • A three-carbon chain with two hydroxyl groups (–OH) and a terminal –CH₂OH.

2. Recall the known drug structure: Chloramphenicol is a natural antibiotic (originally from Streptomyces venezuelae) with the systematic name: 2,2-dichloro-N-[(1R,2R)-1,3-dihydroxy-1-(4-nitrophenyl)propan-2-yl]acetamide. Its structure matches exactly: a p-nitrophenyl ring, a dichloroacetamide, and a dihydroxypropyl side chain.

3. Eliminate other options:

   • (B) Antiseptic: Antiseptics (e.g., phenol, iodine) are simpler, non-specific germicides; chloramphenicol is a specific systemic antibiotic.
   • (C) Analgesic: Pain relievers like aspirin or paracetamol lack the nitro and dichloroacetamide groups.
   • (D) Tranquilizer: Sedatives (e.g., diazepam) have different ring systems (benzodiazepines), not this structure. …

The concept first

When you meet an aldehyde and a strong base, the first question is always: does it have an $\alpha$-hydrogen?

• With an $\alpha$-H (e.g. ethanal), the base pulls it off to make an enolate → aldol condensation.
• Without an $\alpha$-H, no enolate is possible. With concentrated alkali the aldehyde has only one escape route: it oxidises one of its own molecules and reduces another. That self-oxidation–reduction is the Cannizzaro reaction.

In benzaldehyde the carbon attached to $-$CHO is an aromatic ring carbon carrying no hydrogen on an $sp^3$ centre, so there is no $\alpha$-hydrogen — Cannizzaro is guaranteed.

Step-by-step mechanism

Step 1 — Hydroxide attacks the carbonyl.

A tetrahedral alkoxide intermediate forms.

Step 2 — Hydride transfer. This intermediate (or its doubly deprotonated dianion, which is the better hydride donor) collapses: the C$-$H bond breaks and the hydrogen leaves with its bonding pair as $\text{H}^-$, attacking the carbonyl carbon of a second benzaldehyde molecule.

Step 3 — Two different fates.

• The molecule that gave away the hydride becomes benzoic acid, immediately deprotonated by the alkali to $\text{C}_6\text{H}_5\text{COO}^-\text{Na}^+$ — the oxidation product (printed in the question). …

Concept and Intuition

In organic chemistry, the concept of "overlapping p-orbitals" is central to understanding the stability and reactivity of molecules, particularly those with double bonds and cyclic structures. When we talk about overlapping p-orbitals in a cyclic system, we are referring to the formation of a continuous $\pi$-electron cloud above and below the plane of the ring. This continuous overlap is a prerequisite for aromaticity, a special stability found in certain cyclic, planar, fully conjugated systems.

For a cyclic system to have continuous overlap of p-orbitals, two main conditions must be met:

1. Each atom in the ring must be $sp^2$ or $sp$ hybridized. This ensures that each atom has at least one unhybridized p-orbital available. In most aromatic systems, the atoms are $sp^2$ hybridized.
2. These p-orbitals must be aligned parallel to each other. This allows for effective side-by-side overlap, forming a delocalized $\pi$-system.

The number of overlapping p-orbitals is simply the count of atoms in the ring that contribute an unhybridized p-orbital to this continuous $\pi$-system. In the case of a carbocation, if the carbon bearing the positive charge is part of a conjugated system, it will be $sp^2$ hybridized and contribute an empty p-orbital to the overall delocalization.

Step-by-Step Solution

1. Identify the structure of cycloheptatrienyl cation: The name "cycloheptatrienyl cation" indicates a 7-membered carbon ring ("cyclohept-") containing three double bonds ("-triene") and a positive charge ("-yl cation"). The structure can be drawn as a 7-membered ring with three double bonds and one carbon atom bearing a positive charge.

(This is a simplified representation; imagine a heptagon with alternating double bonds and a positive charge on one carbon.)

2. Determine the hybridization of each carbon atom in the ring:

* Carbons involved in double bonds (e.g., C1, C2, C3, C4, C5, C6) are $sp^2$ hybridized. Each $sp^2$ carbon has one unhybridized p-orbital.

* The carbon bearing the positive charge (C7) is a carbocation. Carbocations are typically $sp^2$ hybridized, with the positive charge residing in an empty p-orbital. This empty p-orbital is crucial for conjugation and delocalization.

Since all 7 carbon atoms in the ring are either part of a double bond or bear a positive charge, they are all $sp^2$ hybridized.

3. Count the number of p-orbitals involved in the cyclic overlap: …