Q.An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

The problem presents a multi‑step sequence starting from benzonitrile. The trick is to track the functional‑group transformations carefully, because the reagents in the first step are not the usual acidic hydrolysis of a nitrile. Let’s unpack each stage.

Concept & Intuition

• SnCl₂/HCl in ether is a selective reducing agent for nitriles: it converts –C≡N to –CH=NH (an imine), which then hydrolyses to an aldehyde (–CHO) upon aqueous work‑up. This is the Stephen reduction.
• The resulting aldehyde (benzaldehyde) then reacts with HCN to form a cyanohydrin, which upon hydrolysis gives an α‑hydroxy acid.
• Finally, treating two moles of the aldehyde with concentrated NaOH triggers a Cannizzaro reaction (since benzaldehyde has no α‑hydrogen), producing one molecule of benzyl alcohol and one molecule of sodium benzoate.

Step‑by‑step reasoning

1. Stephen reduction of benzonitrile Benzonitrile (C₆H₅–C≡N) reacts with SnCl₂/HCl in dry ether. The SnCl₂ reduces the triple bond to an imine:

On adding water (H₃O⁺), the imine hydrolyses to an aldehyde:

So the product [X] is benzaldehyde.

2. Formation of [P₁] – cyanohydrin then α‑hydroxy acid Benzaldehyde reacts with HCN (in the presence of a trace of base) to give the cyanohydrin:

Acidic hydrolysis of the cyanohydrin converts the –CN group to –COOH:

This product is α‑hydroxy phenylacetic acid (also called mandelic acid). Hence [P₁] is α‑hydroxy phenylacetic acid.

3. Formation of [P₂] – Cannizzaro reaction of benzaldehyde …

We start by understanding the clues. Compound Y reduces Tollen’s reagent — that means Y is an aldehyde (or an α-hydroxy ketone, but here it’s an aldehyde). Y also undergoes the Cannizzaro reaction, which is a disproportionation of an aldehyde that has no α-hydrogen atoms (i.e., the carbon next to the –CHO group has no hydrogen). So Y must be an aromatic aldehyde like benzaldehyde or a substituted benzaldehyde.

Next: On rigorous oxidation with acidic KMnO₄, Y gives phthalic acid. Phthalic acid is benzene-1,2-dicarboxylic acid. That means the original aldehyde group in Y must be on a benzene ring that also has another carbon-containing substituent at the ortho position, which gets oxidized to a –COOH group. So Y is an ortho-substituted benzaldehyde, where the substituent is something that can be oxidized to –COOH (like –CH₃, –CH₂OH, –CHO, etc.).

Now, Y comes from X by reaction with H₂ / Pd–BaSO₄. This is the Lindlar catalyst — it reduces an acyl chloride (–COCl) to an aldehyde (–CHO) without over-reducing it to an alcohol. So X must be an acyl chloride. Therefore, X has a –COCl group on the benzene ring, and the other ortho substituent is something that, after reduction of –COCl to –CHO, gives Y, which then oxidizes to phthalic acid.

Let’s check the options:

1. Option (A): Benzene with –OH, –CH₂CH₃, and –Cl. No –COCl group. So reduction with H₂/Pd–BaSO₄ would not give an aldehyde. Eliminated.

2. Option (B): Benzene with –COCl and –CH₃ (ortho to each other).

   • Reduction: –COCl → –CHO, giving ortho-methylbenzaldehyde.
   • This aldehyde has no α-hydrogen (the –CHO carbon is attached directly to the ring), so it undergoes Cannizzaro reaction.
   • It also reduces Tollen’s reagent (aldehyde). …

Concept & Intuition

The Cannizzaro reaction is a disproportionation of an aldehyde (without an α‑hydrogen) into a carboxylic acid and an alcohol, under strong base. The key condition: no α‑hydrogen (i.e., no H on the carbon next to the –CHO group). Why? Because if an α‑hydrogen exists, the base will instead deprotonate that position, leading to an enolate and then an aldol reaction — not Cannizzaro. So to identify which compounds will not undergo Cannizzaro, we check each for the presence of α‑hydrogens. Also, the compound must be an aldehyde — alcohols don’t undergo Cannizzaro at all.

Step‑by‑step reasoning

1. Compound A: 2‑Chlorobutanal

   Structure: CH₃–CH₂–CH(Cl)–CHO

   The carbon next to the –CHO (the α‑carbon) has a hydrogen (the carbon is CH(Cl)–). That hydrogen is an α‑hydrogen. Therefore, under basic conditions, this aldehyde will form an enolate and undergo aldol condensation, not Cannizzaro.

   → Will NOT undergo Cannizzaro.

2. Compound B: 2,2‑Dimethylpropanal (pivalaldehyde)

   Structure: (CH₃)₃C–CHO

   The α‑carbon is fully substituted with three methyl groups — no hydrogen attached. Hence no α‑hydrogen. This is a classic example of an aldehyde that undergoes Cannizzaro reaction.

   → Will undergo Cannizzaro.

3. Compound C: Benzaldehyde

   Structure: C₆H₅–CHO

   The α‑carbon is part of the aromatic ring; there is no α‑hydrogen (the carbon adjacent to –CHO is a quaternary aromatic carbon). Benzaldehyde is the textbook example of an aldehyde that undergoes Cannizzaro. …

Concept

The Cannizzaro reaction disproportionates an aldehyde that has no $\alpha$-hydrogen (aromatic aldehydes qualify) using a strong base: one molecule is oxidised to a carboxylate and another reduced to an alcohol. The phrase "undergoes self-oxidation and reduction" is the signature of this reaction, and it fixes what the two reagents must accomplish.

Solution

Tracing the scheme:

1. Cu$^{2+}$/OH$^-$ step: one $-\mathrm{CHO}$ of the aromatic dialdehyde is oxidised to a carboxylate.
2. C$_2$H$_5$OH/H$^+$ step: Fischer esterification converts that acid into an ethyl ester, leaving one aldehyde and one ester group.
3. Reagent I: to regenerate the second aldehyde needed for Cannizzaro, the ester must be reduced only as far as the aldehyde. DIBAL-H at low temperature followed by aqueous work-up is the classic reagent for ester $\rightarrow$ aldehyde.
   • Zn/Hg-HCl (Clemmensen) and NH$_2$NH$_2$/KOH (Wolff-Kishner) reduce all the way to $-\mathrm{CH_2}-$; H$_2$/Pd-BaSO$_4$ (Lindlar) is for alkynes. None fit. …

Both p-methoxybenzaldehyde and methanal (HCHO) lack α-hydrogens, so with concentrated NaOH they undergo a crossed Cannizzaro (disproportionation) reaction.

Formaldehyde is the strongest reducing agent among aldehydes, so it is preferentially oxidised to formate:

$HCHO \rightarrow HCOO^-Na^+ \text{ (sodium formate)}$ …

The key here is to recognise what functional group chemistry is at play. An aldehyde has a carbonyl group ($\ce{C=O}$) that is electrophilic at the carbon. An alcohol has an $\ce{-OH}$ group that can act as a nucleophile. When they react, the oxygen of the alcohol attacks the carbonyl carbon, leading to addition — not substitution or elimination. This is fundamentally different from ester formation (which requires a carboxylic acid, not an aldehyde) or glycol formation (which requires two $\ce{-OH}$ groups on adjacent carbons, typically from diols).

Let’s walk through the reaction step by step.

1. Nucleophilic addition of one alcohol molecule The lone pair on the alcohol oxygen attacks the electrophilic carbonyl carbon of the aldehyde. The $\pi$ bond of $\ce{C=O}$ breaks, and the oxygen picks up a proton (from the alcohol or from the medium). This gives a hemiacetal — a molecule with both an $\ce{-OH}$ and an $\ce{-OR}$ group on the same carbon. For a generic aldehyde $\ce{RCHO}$ and alcohol $\ce{R'OH}$:

2. Further reaction with a second alcohol molecule The hemiacetal is still reactive. Under acidic conditions, the $\ce{-OH}$ group can be protonated and leave as water, generating a carbocation-like intermediate. A second molecule of alcohol then attacks, replacing the $\ce{-OH}$ with another $\ce{-OR'}$ group. The final product is an acetal:

3. Why the other options are wrong …

Step 1 — Interpret the three tests.

• Insoluble in $NaHCO_3$: 'A' is not a carboxylic acid (only acids stronger than carbonic acid dissolve in $NaHCO_3$ with effervescence).
• Dissolves in NaOH: 'A' is acidic enough to form a salt with a strong base — the hallmark of a phenol ($pK_a\approx10$), whose phenoxide is resonance-stabilised.
• Characteristic colour with neutral $FeCl_3$: the classic confirmatory test for a phenolic $-OH$ (a violet/blue coloured iron–phenoxide complex).

So 'A' contains an $-OH$ directly attached to the ring. This rules out (A) benzyl alcohol, whose $-OH$ is on a side chain: it is neutral, does not dissolve in NaOH, and gives no colour with $FeCl_3$.

Step 2 — Use the molecular formula.

$C_7H_8O$ with a phenolic OH $\Rightarrow$ a cresol (methylphenol). The remaining candidates are the o-, m- and p-isomers, (C), (B) and (D).

Step 3 — Use the bromine-water product, $C_7H_5OBr_3$.

Going from $C_7H_8O$ to $C_7H_5OBr_3$ means three ring hydrogens have been replaced by three Br atoms. Bromine water brominates a phenol only at the positions ortho and para to the $-OH$ (these are the positions activated by the strongly electron-donating $-OH$ through resonance). So 'A' must have three free o/p positions relative to its OH.

Number the ring with $-OH$ at C-1; the activated positions are C-2, C-4 and C-6. …