(a) (i) An organic compound (X) has the molecular formula C₅H₁₀O. Draw structures for (X) if it : (I) does not

Question

Q. (a)

(i) An organic compound (X) has the molecular formula

C_5H_{10}O

. Draw structures for (X) if it : (I) does not give Tollen's test but gives a positive iodoform test. (II) does not give Tollen's test and iodoform test but undergoes Aldol condensation. (III) undergoes Cannizzaro's reaction.

(ii) Show how each of the following compounds can be converted to benzoic acid : (I) Acetophenone (II) Ethyl benzene

(OR)

(b) Answer the following questions :

(i) Draw structure of the 2, 4-dinitrophenyl hydrazone derivative of benzaldehyde.

(ii) Arrange the following in increasing order of their reactivity towards HCN : Di-tert. butyl ketone, Acetaldehyde, Acetone

(iii) Give a simple chemical test to distinguish between benzoic acid and ethyl benzoate.

(iv) Write the name of the reagent to convert Ethanenitrile to Ethanal.

(v) Draw the structure of 'X' in the following reaction : Cyclohexanol (

C_6H_{11}OH

)

\xrightarrow{CrO_3}

'X'

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Concept understanding — Cannizzaro Reaction

The Cannizzaro Reaction: An Intuition

Imagine two identical molecules of an aldehyde meeting in a strongly basic solution. Normally, aldehydes with a hydrogen on the carbon next to the carbonyl (the $\alpha$ -carbon) undergo aldol condensation. But what if that $\alpha$ -carbon has no hydrogen at all? The molecule cannot do the usual reaction. Instead, something remarkable happens: one aldehyde molecule gets reduced to an alcohol, while the other gets oxidized to a carboxylic acid (which, in base, exists as its salt). One molecule gives away electrons; the other accepts them. This is a disproportionation — a single species (the aldehyde) acts as both the oxidising and the reducing agent.

Note

The reaction requires concentrated alkali (typically NaOH or KOH). Dilute base will not work.

The Precise Statement

The Cannizzaro reaction is the base-catalysed disproportionation of an aldehyde that lacks an $\alpha$ -hydrogen atom (i.e., the carbon adjacent to the $-\text{CHO}$ group has no hydrogen attached). In the presence of concentrated aqueous or alcoholic alkali, two molecules of such an aldehyde yield one molecule of a primary alcohol and one molecule of the salt of a carboxylic acid.

The general equation (using benzaldehyde as the classic example):

2\,\text{C}_6\text{H}_5\text{CHO} + \text{NaOH} \xrightarrow{\text{conc.}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa}

Benzaldehyde gives benzyl alcohol and sodium benzoate.

2\,\text{RCHO} \xrightarrow[\text{conc. alkali}]{\text{no }\alpha\text{-H}} \text{RCH}_2\text{OH} + \text{RCOO}^- \text{M}^+

Why "No $\alpha$ -Hydrogen" Matters

The key is the mechanism. The first step is the attack of hydroxide ion ( $\text{OH}^-$ ) on the carbonyl carbon. This forms a tetrahedral intermediate. If an $\alpha$ -hydrogen were present, this intermediate would lose water and form an enolate — leading to aldol condensation. Without that hydrogen, the intermediate cannot do that. Instead, it transfers a hydride ion ( $\text{H}^-$ ) to a second molecule of aldehyde. That second molecule gets reduced to the alkoxide (which later picks up a proton to become the alcohol), while the first molecule becomes the carboxylate.

Watch out

A common mistake: thinking the reaction works for any aldehyde. It does not. If the aldehyde has even one $\alpha$ -hydrogen, the aldol pathway dominates. Only aldehydes like formaldehyde ( $\text{HCHO}$ ), benzaldehyde ( $\text{C}_6\text{H}_5\text{CHO}$ ), and trimethylacetaldehyde ( $(\text{CH}_3)_3\text{CCHO}$ ) undergo the Cannizzaro reaction.

Crossed Cannizzaro Reaction …

Part (a): $C_5H_{10}O$ = pentan-2-one (I), pentan-3-one (II), 2,2-dimethylpropanal (III); acetophenone->benzoic acid by haloform, ethylbenzene->benzoic acid by hot $KMnO_4/H^+$ . Part (b): DNP-hydrazone of $PhCHO$ ; HCN reactivity di-tert-butyl ketone < acetone < acetaldehyde; $NaHCO_3$ distinguishes benzoic acid; DIBAL-H gives ethanal from ethanenitrile; cyclohexanol $+CrO_3$ -> cyclohexanone.

(i) Deducing the $C_5H_{10}O$ isomers. One degree of unsaturation => a carbonyl (aldehyde/ketone).

(I) No Tollens' test => not an aldehyde; positive iodoform => a $CH_3CO{-}$ (methyl ketone). The fit is pentan-2-one, $CH_3{-}CO{-}CH_2CH_2CH_3$ .
(II) No Tollens' (not an aldehyde) and no iodoform (no $CH_3CO$ ) but undergoes aldol (needs $\alpha$ -H). The only $C_5$ ketone with $\alpha$ -H and no methyl-ketone group is pentan-3-one, $CH_3CH_2{-}CO{-}CH_2CH_3$ . (Note: pentanal would give a Tollens' test, so it is ruled out.)
(III) Cannizzaro is given only by aldehydes with no $\alpha$ -hydrogen. Here that is 2,2-dimethylpropanal (pivalaldehyde), $(CH_3)_3C{-}CHO$ - its $\alpha$ -carbon is quaternary.

(ii) Conversions to benzoic acid.

(I) Acetophenone -> benzoic acid (haloform/iodoform oxidation of the methyl ketone):

C_6H_5COCH_3 + 3NaOI \to C_6H_5COONa + CHI_3 + 2NaOH

then $C_6H_5COONa + HCl \to C_6H_5COOH$ .

(II) Ethylbenzene -> benzoic acid (the whole benzylic side chain is oxidised to $-COOH$ as long as a benzylic H exists):

C_6H_5CH_2CH_3 \xrightarrow[\Delta]{KMnO_4/OH^-,\ \text{then } H^+} C_6H_5COOH

Three C5H10O carbonyl isomers: pentan-2-one giving a positive iodoform test, pentan-3-one undergoing aldol condensation, and 2,2-dimethylpropanal undergoing Cannizzaro reaction

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Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions) …

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

Let's build this from first principles — understanding why nucleophilic addition happens, not just memorising the steps.

1. The Core Problem: Why Does Addition Happen at All?

A carbonyl group ( $\ce{C=O}$ ) has a polarised double bond:

Oxygen is more electronegative than carbon → it pulls electron density toward itself.
This creates a partial positive charge on carbon ( $\delta^+$ ) and a partial negative charge on oxygen ( $\delta^-$ ).

\ce{C^{\delta+}=O^{\delta-}}

Key insight: The carbon is electron-deficient — it wants electrons. A nucleophile (Nu⁻) is electron-rich — it wants to give electrons. This is a natural match.

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

The nucleophile donates its lone pair to the electrophilic carbonyl carbon.

\ce{Nu^- + C=O -> [Nu-C-O]^-}

Why this happens:

The $\pi$ bond between C and O breaks — the electrons move entirely to oxygen.
Oxygen now has a full negative charge (alkoxide ion).
The carbon changes from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

This step is slow because the $\pi$ bond must break — it requires energy.

Step 2: Protonation (Fast)

The negatively charged oxygen picks up a proton ( $\ce{H^+}$ ) from the solvent or acid.

\ce{[Nu-C-O]^- + H^+ -> Nu-C-OH}

Why this happens:

The alkoxide ion is a strong base — it wants to neutralise its charge.
Protonation gives a stable neutral alcohol product.

3. The Key Formula: Rate Law Derivation

For a general nucleophilic addition:

\ce{Nu^- + R2C=O ->[k] products}

The rate law comes from the slow step (Step 1):

\text{Rate} = k[\text{Nu}^-][\ce{R2C=O}]

Why this form?

The reaction is bimolecular — two species must collide with correct orientation.
Doubling either concentration doubles the rate (first order in each).
This is second order overall.

Exam tip: This is why nucleophilic addition is often called addition-elimination when followed by loss of a leaving group (like in acyl substitution), but here it's just addition.

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

The intermediate is tetrahedral ( $sp^3$ hybridised carbon):

Bond angles: ~109.5°
Four groups around carbon: Nu, R, R', O⁻

Why it's unstable:

The negative charge on oxygen is high-energy.
The tetrahedral geometry is sterically crowded (especially with bulky R groups).
The intermediate collapses quickly — either back to starting materials or forward to product. …

2,4-Dinitrophenylhydrazone of benzaldehyde. Benzaldehyde condenses with 2,4-DNPH ( $H_2N{-}NH{-}C_6H_3(NO_2)_2$ ) losing water:
Structure of the 2,4-dinitrophenylhydrazone of benzaldehyde, with a CH=N-NH bridge joining the benzene ring to the 2,4-dinitrophenyl ring
$C_6H_5CHO + H_2N{-}NH{-}C_6H_3(NO_2)_2 \to C_6H_5CH{=}N{-}NH{-}C_6H_3(NO_2)_2 + H_2O$ (NO $_2$ groups at positions 2 and 4 of the hydrazine ring.)
Reactivity toward HCN. Nucleophilic addition is faster with less steric hindrance and a more electrophilic (less +I) carbonyl carbon. Acetaldehyde (one small H + one $CH_3$ ) is most reactive; acetone (two $CH_3$ ) is intermediate; di-tert-butyl ketone (two bulky $t$ -Bu groups) is least reactive. Increasing order: di-tert-butyl ketone < acetone < acetaldehyde. …

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

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When chloroform reacts with acetone then which of the following is formed ?$
(a) Ethylene dichloride
(b) Mesitylene
(c) Chloretone
(d) Chloral
›Reveal solutionSolution

Chloroform adds across the carbonyl of acetone to give chloretone, 1,1,1-trichloro-2-methyl-2-propanol.

Chloroform (CHCl3) in the presence of a base loses a proton and its CCl3 carbanion adds to the carbonyl carbon of acetone. The addition product is chloretone (also written chlorbutol), a well-known hypnotic/preservative.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Name the reaction in which aldehydes with no alpha-hydrogen atom undergo self-oxidation and reduction reaction in presence of conc. alkali.$
›Reveal solutionSolution

The Cannizzaro reaction is a base-mediated disproportionation unique to aldehydes with no alpha-hydrogen, where one molecule is reduced to an alcohol and another is oxidised to a carboxylate salt.

Aldehydes that have no alpha-hydrogen (i.e., no hydrogen atom on the carbon adjacent to the -CHO group, such as benzaldehyde, formaldehyde, or trimethylacetaldehyde) cannot undergo the usual base-catalysed aldol condensation, since that requires removing an alpha-hydrogen to form an enolate.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Contrary to electrophilic addition reactions observed in alkenes, the aldehydes and ketones undergo nucleophilic addition reactions.$
›Reveal solutionSolution

True - the polar C=O of aldehydes/ketones is attacked by nucleophiles.

In alkenes the C=C double bond is electron-rich, so it attracts electrophiles (electrophilic addition). In aldehydes and ketones the carbonyl C=O bond is polar: oxygen is electronegative and pulls electrons, leaving the carbonyl carbon partially positive (electron-deficient). Therefore th …

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CBSE 2025Set 56/5/11 markMCQ
$CH_3{-}CH(CH_3){-}CHO$
›Reveal solutionSolution

The Cannizzaro reaction requires an aldehyde with no alpha-hydrogen atoms (i.e., not enolizable). Among the given options, only $(CH_3)_3C{-}CHO$ (pivalaldehyde) has no $\alpha$ -H, so it alone undergoes the reaction. The correct option is (B).

Why the Cannizzaro reaction happens — and when it doesn’t

The Cannizzaro reaction is a disproportionation of an aldehyde in concentrated base: one molecule is reduced to a primary alcohol, the other is oxidised to a carboxylate salt. But this reaction only works if the aldehyde cannot form an enolate. Why? Because if there is even one hydrogen on the carbon next to the carbonyl (the $\alpha$ -carbon), the base will preferentially pull that hydrogen off, leading to aldol condensation instead. So the key condition is: no $\alpha$ -hydrogen atoms.
$R{-}CHO$
Now let’s examine each option.

1. Option (A): $CH_3{-}CH(CH_3){-}CHO$

Draw the structure: the carbonyl carbon is at the end. The $\alpha$ -carbon is the one directly attached to the CHO group — that’s the carbon bearing the $CH_3$ and $CH(CH_3)$ groups. Count its hydrogens: it has one hydrogen (since it’s a tertiary carbon with one H). That one $\alpha$ -H makes this aldehyde enolizable. In strong base, that H will be abstracted, and the resulting enolate will undergo aldol condensation — not Cannizzaro.

Watch out

A common mistake: thinking that a branched aldehyde automatically has no $\alpha$ -H. Check the $\alpha$ -carbon carefully — even one H is enough to block the Cannizzaro path.

2. Option (B): $(CH_3)_3C{-}CHO$ …
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CBSE 2025Set ANNUAL1 mark
$Q. Passage: Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former (i.e. than one alkyl group does). A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from sp2 to sp3 in this process and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product.$
(b) What product is formed when CH3CHO reacts with NaHSO3? Give chemical equation.
›Reveal solutionSolution

Bisulfite ion adds across the carbonyl of acetaldehyde to give a crystalline addition compound.

Acetaldehyde undergoes nucleophilic addition with saturated sodium bisulphite solution: the bisulphite ion ( $HSO_3^-$ ) acts as the nucleophile, attacking the carbonyl carbon and forming a tetrahedral addition compound, which is a white crystalline solid (the 'bisulphite addition product'):

$CH_3CHO + NaHSO_3 \to CH_3CH(OH)SO_3Na$

…

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CBSE 2025Set A1 mark
$Q. Write True or False: Ketones containing carbonyl group.$
›Reveal solutionSolution

By definition, a ketone is a carbonyl compound in which the C=O group is bonded to two carbon (alkyl/aryl) groups.

The carbonyl group (a carbon doubly bonded to oxygen, >C=O) is the functional group common to aldehydes, ketones, and carboxylic acids. In a ketone, this carbonyl carbon is attached to two other carbon atoms (R–CO–R′), unlike an aldehyde, where the carbonyl carbon is attached to at least one hydr …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following aldehydes undergo Cannizzaro reaction -$
(a) CH3-CHO
(b) CH3-CH2-CHO
(c) (CH3)2-CH-CHO
(d) C6H5-CHO
›Reveal solutionSolution
The Cannizzaro (base-catalysed self oxidation-reduction/disproportionation) reaction occurs only with aldehydes that lack an alpha-hydrogen atom.

An aldehyde with alpha-hydrogens instead undergoes base-catalysed aldol condensation (since the alpha-H is acidic enough to be removed by strong base, forming an enolate that attacks another aldehyde molecule), NOT the Cannizzaro reaction.

Checking each option:

CH3-CHO (acetaldehyde): has alpha-H (on the CH3 carbon) - undergoes aldol, not Cannizzaro.

CH3-CH2-CHO (propanal): has alpha-H - undergoes aldol.

(CH3)2-CH-CHO (isobutyraldehyde): has alpha-H (on the CH carbon) - undergoes aldol. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Cannizzaro's reaction is not given by -$
(a) 1-methylcyclohexane-1-carbaldehyde (a cyclohexane ring drawn with both a -CHO group and a -CH3 group attached to the same ring carbon, i.e. no alpha-hydrogen on the carbonyl carbon)
(b) Benzaldehyde (a benzene ring drawn with a -CHO group attached, no alpha-hydrogen)
(c) HCHO (formaldehyde, printed as a plain chemical formula, no ring drawn)
(d) $CH_3CHO$ (acetaldehyde, printed as a plain chemical formula, no ring drawn)
›Reveal solutionSolution
Cannizzaro's reaction needs an aldehyde with NO alpha-hydrogen; acetaldehyde has alpha-hydrogens and undergoes a different reaction (aldol condensation) instead.

Cannizzaro's reaction (base-mediated disproportionation into one molecule of alcohol and one of the carboxylate salt) occurs only for aldehydes that have no $\alpha$ -hydrogen — because if $\alpha$ -hydrogens were present, the far faster aldol condensation pathway would dominate instead.

Checking each option for $\alpha$ -hydrogens:

(a) 1-methylcyclohexane-1-carbaldehyde: the carbon bearing $-CHO$ has no H on it (fully substituted, quaternary-type carbon holding both $-CHO$ and $-CH_3$ plus two ring bonds) → no $\alpha$ -H → gives Cannizzaro.

(b) Benzaldehyde: the carbonyl carbon is attached directly to the aromatic ring, with no adjacent sp³ carbon bearing H → no $\alpha$ -H → gives Cannizzaro (the classic textbook example). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not a characteristic of carbonyl compounds?$
(a) They have a polarized C=O bond.
(b) They undergo nucleophilic addition reactions.
(c) They show geometric isomerism.
(d) They can be reduced to alcohol.
›Reveal solutionSolution

Geometric (cis-trans) isomerism about a C=O needs two distinguishable groups on BOTH ends of the double bond, but the oxygen end carries only a lone pair on a single atom, so plain aldehydes/ketones cannot show it.

Carbonyl compounds genuinely have a polarized C=O bond (a), readily undergo nucleophilic addition at the electrophilic carbonyl carbon (b), and can be reduced to alcohols (d) — all true. But geometric (cis–trans) isomerism requires restricted rotation about a double bond WITH two different substituents on each doubly-bonded atom; in a simple aldehyde/ketone (>C=O), the oxygen end carries only a lone pair (not tw …

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CBSE 2024Set 56/1/11 markMCQ
$Q. The formation of cyanohydrin from an aldehyde is an example of: (A) nucleophilic addition (B) electrophilic addition (C) nucleophilic substitution (D) electrophilic substitution$
›Reveal solutionSolution
Cyanohydrin formation involves the cyanide ion ( $\ce{CN^-}$ ) attacking the electrophilic carbonyl carbon of an aldehyde — a textbook case of nucleophilic addition. The answer is (A).

Why this is nucleophilic addition

The carbonyl group ( $\ce{C=O}$ ) in aldehydes is polarized: oxygen is more electronegative than carbon, so the carbon carries a partial positive charge ( $\delta^+$ ) and becomes electron-deficient. This makes it a prime target for nucleophiles — species that are electron-rich and "love" positive centers.

When we treat an aldehyde with a source of cyanide ion (typically $\ce{HCN}$ or $\ce{NaCN}$ ), the $\ce{CN^-}$ acts as a nucleophile. It donates its electron pair to the carbonyl carbon, and the $\pi$ -bond of the carbonyl breaks, with both electrons moving onto the oxygen. The result? A new $\ce{C-CN}$ bond forms, and we add two groups across the original double bond — the hallmark of an addition reaction.

The key distinction: nothing leaves the molecule. In substitution reactions, one group replaces another; here, we're simply adding to the existing structure.

Step-by-step mechanism

Generation of the nucleophile In aqueous or alcoholic medium, $\ce{HCN}$ dissociates (or $\ce{NaCN}$ provides) the cyanide ion:

$\ce{HCN <=> H^+ + CN^-}$
The $\ce{CN^-}$ is a strong nucleophile with a lone pair on carbon.

Nucleophilic attack on the carbonyl carbon The cyanide ion attacks the electrophilic carbonyl carbon of the aldehyde:

$\ce{R-CHO + CN^- -> R-CH(O^-)-CN}$
The $\pi$ -electrons of the $\ce{C=O}$ bond shift entirely onto oxygen, forming an alkoxide intermediate ( $\ce{O^-}$ ).

Protonation of the alkoxide The negatively charged oxygen picks up a proton from the medium (from $\ce{HCN}$ , water, or the solvent):

$\ce{R-CH(O^-)-CN + H^+ -> R-CH(OH)-CN}$
This gives the final cyanohydrin, which contains both a hydroxyl group ( $\ce{-OH}$ ) and a nitrile group ( $\ce{-CN}$ ) on the same carbon. …
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CBSE 2024Set 56/3/11 markMCQ
$\xrightarrow{KCN}$
›Reveal solutionSolution
Cyanide ion is a strong nucleophile and displaces only the reactive benzylic chlorine by $S_N2$ ; the aromatic (aryl) C–Cl is inert under these conditions. The major product is 4-chlorobenzyl cyanide — option (C).

The molecule 4-chlorobenzyl chloride, $Cl\text{–}C_6H_4\text{–}CH_2\text{–}Cl$ , has two very different C–Cl bonds, and the whole question turns on telling them apart.

The benzylic C–Cl is highly reactive. The $-CH_2Cl$ carbon is a primary, benzylic position. Its $S_N2$ transition state is stabilised by the adjacent aromatic ring, so cyanide readily displaces this chlorine:

$Cl\text{–}C_6H_4\text{–}CH_2Cl + CN^- \rightarrow Cl\text{–}C_6H_4\text{–}CH_2CN + Cl^-$

The aryl C–Cl is essentially inert. In an aryl chloride the C–Cl carbon is $sp^2$ and the bond has partial double-bond character from resonance with the ring, so it is short and strong. There is no $S_N2$ at an aromatic carbon, and an aryl cation is far too unstable for $S_N1$ . Nucleophilic aromatic substitution ( $S_NAr$ ) would need strong electron-withdrawing groups (e.g. $-NO_2$ ) ortho/para to the chlorine to stabilise the Meisenheimer intermediate; a weakly withdrawing $-CH_2CN$ group does not provide that, so the ring chlorine survives.

Result. Only the benzylic chlorine is replaced, giving 4-chlorobenzyl cyanide — a benzene ring carrying $-CH_2CN$ with the ring $-Cl$ still para to it. …
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CBSE 2024Set A11 markMCQ
$Q. Nucleophilic attack on carbonyl carbon atom changes its hybridization from :$
(a) $sp$ to $sp^2$
(b) $sp^2$ to $sp^3$
(c) $sp^3$ to $sp^2$
(d) $sp$ to $sp^3$
›Reveal solutionSolution

Nucleophilic addition changes the carbonyl carbon from $sp^2$ to $sp^3$ — option (b).

In a carbonyl group $>\!\text{C}=\text{O}$ the carbon is $sp^2$ hybridised and planar (trigonal, ~120°). When a nucleophile attacks the electrophilic carbonyl carbon, the C=O π bond breaks and a new σ bond forms; the carbon now has four …
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Q.(a)

Concept understanding — Cannizzaro Reaction

The Cannizzaro Reaction: An Intuition

The Precise Statement

Why "No α\alphaα-Hydrogen" Matters

Crossed Cannizzaro Reaction …

Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

The Precise Statement

Why It Happens – The Driving Force

A Concrete Example – Addition of HCN to a Ketone

What Makes a Good Nucleophile?

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

1. The Core Problem: Why Does Addition Happen at All?

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

Step 2: Protonation (Fast)

3. The Key Formula: Rate Law Derivation

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

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Why "No $\alpha$ -Hydrogen" Matters

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