(a) (i) Write the major product(s) in the following reactions : (1) Ethylbenzene (CH₃CH₂C₆H₅) xrightarrowKMnO₄

Question

Q. (a)

(i) Write the major product(s) in the following reactions :

(1) Ethylbenzene (

CH_3CH_2C_6H_5

)

\xrightarrow{KMnO_4, KOH}

\xrightarrow{H^+}

(2) Benzaldehyde (

C_6H_5CHO

)

+ CH_3COCH_3

\xrightarrow{dil\,NaOH}

(3) Benzoic acid (

C_6H_5COOH

)

\xrightarrow{Br_2/FeBr_3}

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :

(1) Acetophenone (

C_6H_5COCH_3

) and Propiophenone (

C_6H_5COCH_2CH_3

)

(2) Pentanal (

CH_3CH_2CH_2CH_2CHO

) and Pentan-3-one (

CH_3CH_2COCH_2CH_3

)

(OR)

(b)

(i) Give reasons for the following :

(1) In semicarbazide, only one

-NH_2

group is involved in the formation of semicarbazone.

(2) Acetaldehyde is more reactive than acetone towards addition of HCN.

(ii)

(1) Arrange the following in decreasing order of their acidic strength :

CH_3COOH

O_2N - CH_2 - COOH

HCOOH

(2) Name the reagent in the following reaction :

CH_3 - CH = CH - CH_2 - CN \xrightarrow{?} CH_3 - CH = CH - CH_2 - CHO

(iii) Write the reaction involved in Hell-Volhard-Zelinsky reaction.

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Part (a)Concept understanding — Aldol Condensation

Aldol Condensation – From Intuition to Precision

Imagine you have two identical aldehyde molecules. Each has a carbon–oxygen double bond (the carbonyl) that is electron-hungry — the oxygen pulls electron density toward itself, leaving the carbonyl carbon slightly positive. Now look at the carbon next to the carbonyl (the $\alpha$ -carbon). The hydrogens attached to it are unusually acidic, because if you remove one, the negative charge that forms can be stabilised by resonance with the carbonyl group.

What if you could make one molecule act as an "electrophile" (electron-poor, at its carbonyl carbon) and the other as a "nucleophile" (electron-rich, at its $\alpha$ -carbon)? That is exactly what aldol condensation does. The two molecules join together, forming a new carbon–carbon bond.

Note

The name "aldol" comes from aldehyde + alcohol — the initial product has both functional groups.

The Mechanism in Two Stages

Stage 1: The Aldol Addition (the "aldol" part)

Under base catalysis (typically dilute NaOH or KOH), the base abstracts an $\alpha$ -hydrogen from one molecule of the aldehyde or ketone. This generates an enolate ion — a carbanion that is resonance-stabilised.

The enolate then attacks the carbonyl carbon of a second, unreacted molecule. The result is a $\beta$ -hydroxy carbonyl compound — an "aldol" (if starting from an aldehyde) or a "ketol" (if starting from a ketone).

2 \text{CH}_3\text{CHO} \xrightarrow{\text{OH}^-} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

Stage 2: Dehydration (the "condensation" part)

The $\beta$ -hydroxy carbonyl compound now has an $\alpha$ -hydrogen and a $\beta$ -hydroxyl group. Under the reaction conditions (often mild heat or slightly stronger base), a molecule of water is eliminated. This creates a conjugated $\alpha,\beta$ -unsaturated carbonyl compound — a much more stable product because the double bond is in conjugation with the carbonyl.

\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH}=\text{CHCHO} + \text{H}_2\text{O}

The overall process — addition followed by dehydration — is called aldol condensation.

The Precise Statement

Aldol condensation is a base-catalysed reaction in which two molecules of an aldehyde or ketone, each possessing at least one $\alpha$ -hydrogen, combine to form a $\beta$ -hydroxy carbonyl compound (the aldol addition product), which then undergoes dehydration to yield an $\alpha,\beta$ -unsaturated carbonyl compound.

Watch out

A common mistake: students think "condensation" means the reaction stops at the $\beta$ -hydroxy stage. In fact, the term "condensation" here refers to the loss of a small molecule (water) — the dehydration step is essential to the full condensation. If no dehydration occurs, the reaction is simply an aldol addition.

Key Conditions and Limitations …

Part (b)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. …

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Part (a): oxidation of ethylbenzene → benzoic acid; crossed aldol → benzalacetone; bromination of benzoic acid → 3-bromobenzoic acid; iodoform distinguishes acetophenone, Tollens'/Fehling's distinguishes pentanal.

Part (b): semicarbazide uses only its terminal $-NH_2$ ; acetaldehyde > acetone toward HCN; acidity $O_2N CH_2COOH > HCOOH > CH_3COOH$ ; reagent = DIBAL-H; HVZ gives $\alpha$ -bromo acid.

Part (a)

(i) Major products

Ethylbenzene + KMnO $_4$ /KOH, then H $^+$ . Hot alkaline $KMnO_4$ oxidises any benzylic side chain (that has a benzylic H) completely to $-COOH$ . First step gives potassium benzoate; acidification gives benzoic acid, $C_6H_5COOH$ (not phenylacetic acid).
Benzaldehyde + acetone, dil. NaOH. Benzaldehyde has no $\alpha$ -H; acetone's enolate attacks it, and the $\beta$ -hydroxyketone dehydrates to the $\alpha,\beta$ -unsaturated ketone benzalacetone, $C_6H_5CH=CHCOCH_3$ (4-phenylbut-3-en-2-one).
Benzoic acid + Br $_2$ /FeBr $_3$ . $-COOH$ is deactivating and meta-directing, so bromination gives 3-bromobenzoic acid.

(ii) Distinguishing tests

Acetophenone vs propiophenone — iodoform test. Acetophenone ( $C_6H_5COCH_3$ ) is a methyl ketone → yellow $CHI_3$ precipitate with $I_2/NaOH$ . Propiophenone ( $C_6H_5COCH_2CH_3$ ) has no $CH_3CO-$ group → negative. …

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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 33 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 markMCQ
$Q. Aldol condensation does not occur between$
(a) two different aldehydes
(b) two different ketones
(c) an aldehyde and a ketone
(d) an aldehyde and an ester
›Reveal solutionSolution
Aldol condensation is defined between aldehydes/ketones bearing $\alpha$ -hydrogens; esters instead undergo the distinctly-named Claisen condensation, so the aldehyde–ester combination falls outside 'aldol condensation'.

(a) Two different aldehydes: a crossed aldol condensation is possible — e.g. base generates the enolate of one aldehyde, which attacks the carbonyl carbon of the other, followed by dehydration to the $\alpha,\beta$ -unsaturated carbonyl product. This does occur (though it can give a mixture of products).

(b) Two different ketones: a crossed aldol condensation between ketones can also occur, though it is often slower and lower-yielding (ketones are less electrophilic and more sterically hindered than aldehydes), but it is still classified as an aldol-type reaction.

(c) An aldehyde and a ketone: a crossed aldol (Claisen–Schmidt-type when aromatic) reaction readily occurs — typically the aldehyde (lacking $\alpha$ -H if aromatic, or simply more electrophilic) acts as the electrophile and the ketone's enolate as the nucleophile. …
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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 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 D1 markMCQ
$Q. Which of the following would give Aldol condensation reaction?$
(a) CCl3CHO
(b) CH3-C(CH3)(CH3)-CHO
(c) CH3CHO
(d) HCHO
›Reveal solutionSolution
Aldol condensation needs an alpha-H; only CH3CHO has alpha-hydrogens.

Aldol condensation occurs only with aldehydes/ketones that possess at least one alpha-hydrogen (a hydrogen on the carbon next to the carbonyl group).

CCl3CHO: alpha-carbon carries three Cl, no alpha-H.

(CH3)3C-CHO (2,2-dimethylpropanal): the alpha-carbon is quaternary, no alpha-H. …
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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 mark
$Q. Name the reagent which is used to convert aldehydes or ketones having alpha-hydrogen into beta-hydroxy aldehydes or beta-hydroxy ketones.$
›Reveal solutionSolution

Aldehydes/ketones with an alpha-hydrogen undergo base-catalysed aldol addition with dilute NaOH, giving a beta-hydroxy carbonyl product.

When an aldehyde or ketone possessing at least one alpha-hydrogen is treated with dilute aqueous sodium hydroxide (or another dilute alkali), the base removes an alpha-hydrogen to form an enolate, which then attacks the carbonyl carbon of a second carbonyl molecule. The result is a beta-hydroxy aldehyde (aldol) or beta-hydroxy ketone (ketol) - t …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Acetaldehyde undergoes aldol-condensation with NaOH (dil). Reason (R): Aldehydes which do not contain α-hydrogen undergo aldol-condensation.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
›Reveal solutionSolution

Acetaldehyde does undergo aldol condensation with dilute NaOH (true), but the stated reason is backwards — aldol condensation requires aldehydes WITH $\alpha$ -hydrogens, not without them.

Assertion: Acetaldehyde ( $CH_3CHO$ ) has $\alpha$ -hydrogens (on the methyl carbon adjacent to the carbonyl) and readily undergoes base-catalyzed (dil. NaOH) aldol condensation to give 3-hydroxybutanal, which on further heating dehydrates to crotonaldehyde. This is TRUE.

…

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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 2025Set ANNUAL1 markMCQ
$Q. The product formed in Aldol condensation is ........................ .$
(a) an α, β unsaturated ester
(b) an α-hydroxy aldehyde or ketone
(c) a β-hydroxy acid
(d) a β-hydroxy aldehyde or ketone.
›Reveal solutionSolution

Base-catalysed self-addition of an aldehyde/ketone with an α-hydrogen first gives a β-hydroxy carbonyl compound (the aldol); loss of water on heating gives the α,β-unsaturated product.

In Aldol condensation, a base (e.g. dilute NaOH) removes an α-hydrogen from one molecule of aldehyde/ketone to form a carbanion (enolate), which then attacks the carbonyl carbon of a second molecule. The direct product of this nucleophilic addition — before any dehydration — is a β-hydroxy aldehyde (an 'aldol') or β-hydroxy ketone (a 'ketol'), because the newly formed –OH group is on the carbon β to the original c …

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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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Q.(a)

Part (a)

(i) Major products

(ii) Distinguishing tests

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