896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation

NCERT Exemplar · Q49

Q. 896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H2SO4 and HgSO4 gives a ketone 'B' having molecular formula C6H12O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.

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Concept understanding — Cannizzaro Reaction

Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions

These two reactions are often grouped together in textbooks because they both involve carbonyl compounds (C=O), but they do entirely different things. Let's take them one at a time.

Clemmensen Reduction

Intuition first. Imagine you have a ketone or aldehyde — a molecule with a C=O group. You want to remove that oxygen entirely and replace the C=O with two hydrogen atoms, turning it into a simple hydrocarbon chain. That's a reduction (adding hydrogen, removing oxygen). The Clemmensen reduction is a brute-force way to do this using a strongly acidic, reducing environment.

The precise reaction:

A ketone or aldehyde is heated with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The C=O group is reduced to a CH₂ group.

\ce{R-C(=O)-R' + 4[H] ->[Zn(Hg), HCl, heat] R-CH2-R' + H2O}

\text{Aldehyde or Ketone} \xrightarrow{\text{Zn(Hg), conc. HCl, } \Delta} \text{Hydrocarbon}

Key points for exams:

Works only for ketones and aldehydes that are stable in strong acid.
Does not work for compounds that get destroyed by conc. HCl (e.g., acid-sensitive groups like esters, nitriles).
The mechanism is complex and not usually tested in detail — just know it's a reductive removal of C=O.
The product is always a saturated hydrocarbon (alkane).

Watch out

Clemmensen reduction cannot reduce carboxylic acids, esters, or amides. Only aldehydes and ketones.

Example:

Acetophenone ( $\ce{C6H5-CO-CH3}$ ) → Ethylbenzene ( $\ce{C6H5-CH2-CH3}$ )

Cannizzaro Reaction

Intuition first. This is a disproportionation reaction — one molecule of aldehyde gets oxidised (to a carboxylic acid) while another gets reduced (to an alcohol). It happens only with aldehydes that have no alpha-hydrogen atoms (i.e., the carbon next to the C=O has no H). Why? Because if there were alpha-hydrogens, the aldehyde would undergo aldol condensation instead.

The precise reaction:

An aldehyde without $\alpha$ -hydrogen is treated with concentrated aqueous or alcoholic base (NaOH/KOH). Two molecules of aldehyde react: one becomes a carboxylate salt, the other becomes a primary alcohol.

\ce{2 R-CHO + OH- -> R-COO- + R-CH2OH}

After acidification, the carboxylate salt gives the carboxylic acid.

2\,\text{HCHO} \xrightarrow{\text{conc. NaOH}} \text{HCOONa} + \text{CH}_3\text{OH}

Key points for exams:

Only works for aldehydes with no $\alpha$ -hydrogen: formaldehyde, benzaldehyde, trimethylacetaldehyde, etc.
The base must be concentrated (dilute base won't work).
Formaldehyde is the most common example — it gives formic acid (as formate) and methanol.
Crossed Cannizzaro: When formaldehyde is mixed with another aldehyde (like benzaldehyde), formaldehyde is always the one that gets oxidised (to formate), and the other aldehyde gets reduced (to alcohol). This is because formaldehyde is the strongest reducing agent among aldehydes.

Tip

In a crossed Cannizzaro, formaldehyde always becomes the carboxylate. The other aldehyde becomes the alcohol. This is a common exam question.

Example: …

Why this formula?

Cannizzaro Reaction: Why the Key Formulas Hold

The Cannizzaro reaction is a disproportionation reaction of aldehydes (without α-hydrogens) in the presence of a strong base. Let's build the understanding from the ground up.

1. What Happens in the Reaction?

An aldehyde (like formaldehyde or benzaldehyde) reacts with concentrated base to give:

One molecule is oxidized to a carboxylic acid (or its salt)
Another molecule is reduced to a primary alcohol

General equation (for two identical aldehydes):

2 \text{RCHO} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{RCH}_2\text{OH}

2. Why Does Disproportionation Occur?

The Key Insight: No α-Hydrogen

Aldehydes with α-hydrogens undergo aldol condensation instead.
Without α-hydrogens, the only available reaction path is hydride transfer.

The Mechanism (Step-by-Step Reasoning)

Nucleophilic attack: $\text{OH}^-$ attacks the carbonyl carbon of one aldehyde molecule.
- Forms a tetrahedral intermediate (a gem-diolate).
Hydride shift: The intermediate acts as a hydride donor ( $\text{H}^-$ ) to a second aldehyde molecule.
- This is the rate-determining step.
- The hydride comes from the C–H bond of the intermediate (not from the OH).
Products:
- The donor aldehyde becomes a carboxylate ion (oxidized).
- The acceptor aldehyde becomes an alkoxide ion (reduced).
Protonation: In workup, the carboxylate gives the acid, and the alkoxide gives the alcohol.

3. The Key Formula(e) and Their Derivation

Formula 1: Stoichiometry

2 \text{RCHO} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{RCH}_2\text{OH}

Why this holds:

One aldehyde loses a hydride ( $\text{H}^-$ ) → gains an oxygen → oxidation state increases by 2.
The other aldehyde gains a hydride → oxidation state decreases by 2.
The base ( $\text{OH}^-$ ) is consumed stoichiometrically (one per two aldehydes).

Formula 2: Oxidation State Change

For an aldehyde carbon (carbonyl carbon):

In $\text{RCHO}$ : oxidation state = +1
In $\text{RCOO}^-$ : oxidation state = +3 (gain of +2)
In $\text{RCH}_2\text{OH}$ : oxidation state = -1 (loss of -2)

Net change: $+2$ (oxidation) + $(-2)$ (reduction) = 0 — consistent with disproportionation.

Formula 3: Rate Law (for the hydride transfer step)

\text{Rate} = k[\text{aldehyde}]^2[\text{OH}^-]

Why:

First aldehyde reacts with $\text{OH}^-$ to form the hydride donor (first order in each).
Second aldehyde accepts the hydride (first order in aldehyde).
Overall: second order in aldehyde, first order in base.

4. Why Only Certain Aldehydes Work?

Condition: Aldehyde must have no α-hydrogen atoms.

Examples: $\text{HCHO}$ (formaldehyde), $\text{C}_6\text{H}_5\text{CHO}$ (benzaldehyde), $\text{(CH}_3\text{)}_3\text{CCHO}$ (pivalaldehyde). …

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The hydrocarbon A is an alkyne (C₆H₁₀) whose empirical formula (CH₁.₆₆) and molar mass (~82 g/mol) give the molecular formula C₆H₁₀. Hydrogenation to 2‑methylpentane and hydration to a methyl ketone (positive iodoform test) identify A as 4‑methylpent‑1‑yne.

1. Finding the molecular formula of A

First, the vapour data: 896 mL at STP corresponds to

$\frac{896}{22400} = 0.04$ moles.

Mass of this sample = 3.28 g, so the molar mass is

$M = \frac{3.28}{0.04} = 82\ \text{g mol}^{-1}$ .

Now the percentage composition:

Carbon: 87.80 % → in 100 g, mass of C = 87.80 g → moles of C = $\frac{87.80}{12} = 7.317$

Hydrogen: 12.19 % → in 100 g, mass of H = 12.19 g → moles of H = $\frac{12.19}{1} = 12.19$

Divide by the smaller number (7.317) to get the simplest ratio:

C : H = $1 : 1.666$ → multiply by 3 → C₃H₅ as the empirical formula.

Empirical formula mass = $3 \times 12 + 5 \times 1 = 41$  g mol⁻¹.

Since molar mass = 82 g mol⁻¹, the molecular formula is twice the empirical: C₆H₁₀.

Tip

C₆H₁₀ has a degree of unsaturation (DoU) = $\frac{2 \times 6 + 2 - 10}{2} = 2$ . Two degrees of unsaturation means either two double bonds, one triple bond, or a ring plus one double bond. The hydration reaction (next step) will tell us which.

2. Hydrogenation gives 2‑methylpentane

Hydrogenation of A with H₂/Ni (or Pd/C) adds H₂ across all multiple bonds, yielding a saturated alkane. The product is 2‑methylpentane, whose carbon skeleton is:

   C
   |
C—C—C—C

That is, a five‑carbon straight chain with a methyl branch on carbon 2. So the carbon skeleton of A must be exactly this: 2‑methylpentane skeleton.

Since A has two degrees of unsaturation and the skeleton is fixed, A must be an alkyne (one triple bond) or a diene. The hydration reaction will decide.

3. Hydration gives a ketone B (C₆H₁₂O) that gives a positive iodoform test

Hydration of an alkyne in the presence of H₂SO₄ and HgSO₄ follows Markovnikov’s rule: the OH ends up on the more substituted carbon of the triple bond, and the product tautomerises to a carbonyl compound.

If the triple bond is terminal (‑C≡CH), hydration gives a methyl ketone (‑COCH₃).
If the triple bond is internal, hydration gives a mixture or a different ketone.

The ketone B has formula C₆H₁₂O (one oxygen, so it’s a saturated ketone). A positive iodoform test means B contains a CH₃‑CO‑ group (methyl ketone). Therefore, A must have a terminal triple bond.

So A is a terminal alkyne with the 2‑methylpentane skeleton. The only place to put a terminal triple bond is at the end of the chain:

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
(ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_2 / \mathrm{HCl} / \text { ether }}[\mathrm{X}] \end{aligned} $(A)$ \left[\mathrm{P}_1\right] \text { Phenol and }\left[\mathrm{P}_2\right] \text { Sodium benzoate } $(B)$ \left[\mathrm{P}_1\right] \text { Benzoyl chloride. and }\left[\mathrm{P}_2\right] \text { Acetophenone } $(C)$ \left[P_1\right] \text { Acetophenone. } \quad \text { and } \quad\left[P_2\right] \text { Benzoic acid } $(D)$ \left[\mathrm{P}_1\right] \alpha \text {-Hydroxy phenylacetic acid and }\left[\mathrm{P}_2\right] \text { Benzyl alcohol } $$
›Reveal solutionSolution
The key is to recognise that the first step (SnCl₂/HCl/ether) reduces the nitrile to an imine that hydrolyses to an aldehyde (benzaldehyde), not a carboxylic acid; the second step (HCN then hydrolysis) gives a cyanohydrin that hydrolyses to an α‑hydroxy acid; the third step (2 moles of X + 50% NaOH) is a Cannizzaro reaction of the aldehyde, yielding benzyl alcohol and sodium benzoate. The correct option is (D).

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

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

$\ce{C6H5-C#N ->[SnCl2/HCl] C6H5-CH=NH2+ Cl-}$
On adding water (H₃O⁺), the imine hydrolyses to an aldehyde:
$\ce{C6H5-CH=NH2+ ->[H3O+] C6H5-CHO + NH4+}$
So the product [X] is benzaldehyde.

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

$\ce{C6H5-CHO + HCN -> C6H5-CH(OH)-CN}$
Acidic hydrolysis of the cyanohydrin converts the –CN group to –COOH:
$\ce{C6H5-CH(OH)-CN ->[H3O+] C6H5-CH(OH)-COOH}$
This product is α‑hydroxy phenylacetic acid (also called mandelic acid). Hence [P₁] is α‑hydroxy phenylacetic acid.

Formation of [P₂] – Cannizzaro reaction of benzaldehyde …
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COMEDK 2025Set 2025-M1 markMCQ
$[\mathrm{X}]$
›Reveal solutionSolution
The key is to work backwards from the final product (phthalic acid) and the reactions of compound Y (reduces Tollen’s reagent, undergoes Cannizzaro reaction) to deduce that Y is an aromatic aldehyde with no α-hydrogen, and X is its acyl chloride precursor. The correct option is (B).

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:

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

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). …
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COMEDK 2024Set 2024-E1 markMCQ
$Q. Given are the names of 4 compounds. Two of these compounds will not undergo Cannizzaro's reaction. Identify the two. A. 2-Chlorobutanal B. 2,2-Dimethylpropanal C. Benzaldehyde D. 2-Phenyl ethanol (A) A & D (B) D & C (C) C & B (D) A & B$
›Reveal solutionSolution
Cannizzaro reaction requires an aldehyde with no alpha‑hydrogen; 2‑chlorobutanal has an alpha‑hydrogen (so it undergoes aldol instead) and 2‑phenyl ethanol is not an aldehyde at all — so the two that will not undergo Cannizzaro are A and D, making the correct option (A).

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

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.

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.

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. …
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COMEDK 2024Set 2024-M1 markMCQ
$\text { Reagent I: DIBAL- } \mathrm{H} / \mathrm{H}_2 \mathrm{O} \quad \text { Reagent II: conc. } \mathrm{KOH} / \text { heat }$
›Reveal solutionSolution
The final "self oxidation & reduction" step is a Cannizzaro reaction, so Reagent I must stop the reduction at the aldehyde (DIBAL-H $/\mathrm{H_2O}$ ) and Reagent II must be the strong base that drives the disproportionation (conc. $\mathrm{KOH}$ /heat). The correct option is (A).

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:

Cu $^{2+}$ /OH $^-$ step: one $-\mathrm{CHO}$ of the aromatic dialdehyde is oxidised to a carboxylate.

C $_2$ H $_5$ OH/H $^+$ step: Fischer esterification converts that acid into an ethyl ester, leaving one aldehyde and one ester group.

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. …
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COMEDK 2023Set 2023-E1 markMCQ
$50 \%$
›Reveal solutionSolution

A crossed Cannizzaro reaction: methanal is oxidised to sodium formate and p-methoxybenzaldehyde is reduced to p-methoxy benzyl alcohol.

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:

$HCHO \rightarrow HCOO^-Na^+ \text{ (sodium formate)}$ …
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KCET 2022Set B-31 markMCQ
$Q. The general name of the compound formed by the reaction between aldehyde and alcohol is (A) Glycol (B) Acetate (C) Ester (D) Acetal$
›Reveal solutionSolution
The reaction between an aldehyde and an alcohol forms an acetal (or hemiacetal) via nucleophilic addition, not an ester or glycol. The correct option is (D).

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.

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 $\ce{R'OH}$ :

$\ce{RCHO + R'OH -> RCH(OH)(OR')}$

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:

$\ce{RCH(OH)(OR') + R'OH -> RCH(OR')2 + H2O}$

Why the other options are wrong …
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KCET 2021Set B-21 markMCQ
$\mathrm{C_7H_8O}$
›Reveal solutionSolution
The tests identify a cresol; the tri-bromo product then fixes the isomer, because only m-cresol leaves all three o/p positions of the $-OH$ free.

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. …
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←Previous questionQ48 — An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alken…→Next questionQ50 — An unsaturated hydrocarbon 'A' adds two molecules of H2 and on reducti…

1. Finding the molecular formula of A

2. Hydrogenation gives 2‑methylpentane

3. Hydration gives a ketone B (C₆H₁₂O) that gives a positive iodoform test

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