(A) (a) How will you convert the following : (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii)

Question

Q. (A)

(a) How will you convert the following :

(i) Propanone to Propene

(ii) Benzoic acid to Benzaldehyde

(iii) Benzene to m-Nitroacetophenone

(b)

(i) Arrange the following compounds in increasing order of their reactivity towards HCN : Propanone

(CH_3)_2C{=}O

, Di-tert-butyl ketone

((CH_3)_3C)_2C{=}O

, Ethanal

CH_3CHO

(ii) Identify the compounds, which would undergo Aldol condensation : Methanal, HCHO Benzaldehyde,

C_6H_5CHO

Ethanal,

CH_3CHO

Cyclohexanone (

C_6H_{10}O

, drawn)

(OR)

(B)

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

(i) Acetophenone and Benzophenone

(ii) Propanal and Propanone

(iii) Pentan-2-one and Pentan-3-one

(b)

(i) Which of the following acids is stronger and why ?

CH_2FCOOH

CH_3COOH

(ii) Arrange the following compounds in the increasing order of their boiling points :

CH_3OCH_3

CH_3CH_2CH_3

CH_3CH_2OH

CH_3CHO

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

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Part (a)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. …

Part (b)Concept understanding — Tollens Test

The Tollens Test: The Silver Mirror That Spots an Aldehyde

Imagine you have two unlabelled bottles — one contains an aldehyde, the other a ketone. They look identical, smell similar, and both are carbonyl compounds. How do you tell them apart in a single, dramatic step? The Tollens test gives you a visible, unmistakable answer: a beautiful silver mirror coating the inside of your test tube.

The Intuition: Why Aldehydes Are Special

The key difference lies in the carbonyl carbon. In an aldehyde, that carbon is bonded to at least one hydrogen atom. In a ketone, it is bonded to two carbon groups. That hydrogen on the aldehyde is not just any hydrogen — it is weakly acidic and, more importantly, easily oxidised. The aldehyde can lose this hydrogen and become a carboxylic acid, donating electrons in the process.

Tollens reagent is a solution of silver ions ( $\text{Ag}^+$ ) held in a complex with ammonia. Silver ions are excellent oxidising agents — they want to gain electrons and become neutral silver metal. When an aldehyde meets these silver ions, the aldehyde gets oxidised (loses its hydrogen) and the silver ions get reduced (gain electrons), forming metallic silver.

Tip

Think of it as a trade: the aldehyde gives away electrons (gets oxidised), and the silver ions accept those electrons (get reduced). The silver atoms then clump together on the glass surface, forming that characteristic mirror.

The Precise Statement

Tollens test is a qualitative chemical test that distinguishes aldehydes from ketones. When an aldehyde is warmed with ammoniacal silver nitrate (Tollens reagent), the aldehyde is oxidised to a carboxylate ion, and the silver(I) ions are reduced to metallic silver, which deposits as a shiny mirror on the walls of the test tube. Ketones do not react under these mild conditions.

\text{RCHO} + 2[\text{Ag(NH}_3\text{)}_2]^+ + 3\text{OH}^- \longrightarrow \text{RCOO}^- + 2\text{Ag} \downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O}

How the Reagent Is Made

You cannot just use silver nitrate directly — the silver ions would precipitate as silver oxide in the alkaline conditions needed for the reaction. So you first dissolve silver nitrate in water, add sodium hydroxide to get a brown precipitate of silver oxide, and then add just enough ammonia to redissolve that precipitate. The result is a clear solution containing the complex ion $[\text{Ag(NH}_3\text{)}_2]^+$ .

Watch out

Never store Tollens reagent. It forms explosive silver nitride ( $\text{Ag}_3\text{N}$ ) on standing. Prepare it fresh, use it immediately, and destroy any leftover by adding dilute acid.

The Procedure in the Lab

Clean a test tube thoroughly — any grease will ruin the mirror.
Add a few drops of the unknown compound.
Add about 2 mL of freshly prepared Tollens reagent.
Warm the mixture gently in a water bath (do not boil). …

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Part (a): NaBH $_4$ /dehydration (propanone->propene), SOCl $_2$ +Rosenmund (benzoic acid->benzaldehyde), F-C acylation+nitration (benzene->m-nitroacetophenone); HCN reactivity di-tert-butyl ketone < propanone < ethanal; aldol given by ethanal and cyclohexanone. Part (b): iodoform/Tollens' distinguish the three pairs; $CH_2FCOOH>CH_3COOH$ ; boiling points $C_3H_8<CH_3OCH_3<CH_3CHO<C_2H_5OH$ .

Part (a)

(a) Conversions

(i) Propanone -> propene. Reduce the ketone to propan-2-ol ( $NaBH_4$ ), then acid-dehydrate:

CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3 \xrightarrow{\text{conc. }H_2SO_4,\ \Delta} CH_3CH{=}CH_2

(ii) Benzoic acid -> benzaldehyde. Direct reduction over-reduces to the alcohol, so make the acid chloride, then use the selective Rosenmund reduction (Pd poisoned with $BaSO_4$ ):

C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{H_2,\ Pd\text{-}BaSO_4} C_6H_5CHO

(iii) Benzene -> m-nitroacetophenone. Acylate first because $-COCH_3$ is a meta-director; then nitrate:

C_6H_6 \xrightarrow{CH_3COCl,\ \text{anhyd. }AlCl_3} C_6H_5COCH_3 \xrightarrow{\text{conc. }HNO_3/H_2SO_4} m\text{-}NO_2C_6H_4COCH_3

(b)(i) Reactivity toward HCN. Nucleophilic addition is faster with less steric bulk and a more electrophilic carbonyl: di-tert-butyl ketone < propanone < ethanal. …

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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 35 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
$Q. Which of the following gives silver mirror with Tollen's reagent?$
(a) Formic acid
(b) Acetic acid
(c) Acetone
(d) Ethyl alcohol
›Reveal solutionSolution

Formic acid is unusual among carboxylic acids in giving Tollens' test because it structurally contains an aldehyde-like H–C=O unit.

Tollens' reagent ( $[Ag(NH_3)_2]^+OH^-$ , ammoniacal silver nitrate) is reduced to metallic silver (depositing as a bright 'silver mirror' on the test tube) by compounds that can be oxidised easily, classically aldehydes ( $R-CHO \rightarrow R-COOH$ , with $Ag^+ \rightarrow Ag$ ).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Tollen's reagent is used to distinguish between$
(a) Alcohol and ether
(b) Aldehydes and ketones
(c) Carboxylic acid and esters
(d) 1 degree, 2 degree and 3 degree amines
›Reveal solutionSolution

Tollen's test distinguishes aldehydes from ketones because only aldehydes are easily oxidised, reducing the diamminesilver(I) complex to metallic silver (a bright mirror).

Tollen's reagent is an ammoniacal solution of silver nitrate, containing the complex ion [Ag(NH3)2]+. When warmed gently with an aldehyde, the aldehyde is oxidised to a carboxylate ion, while Ag+ is reduced to metallic silver, which deposits as a shiny 'silver mirror' on the test-tube wall:

RCHO + 2[Ag(NH3)2]+ + 3OH- -> RCOO- + 2Ag(s) + 4NH3 + 2H2O

…

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CBSE 2026Set ANNUAL1 mark
$Q. Among aldehydes or ketones which gives positive Tollen's Test?$
›Reveal solutionSolution

Tollens' reagent (ammoniacal silver nitrate) oxidizes aldehydes to carboxylate ions while itself being reduced to metallic silver (silver mirror); ketones do not respond (with the exception of α-hydroxy ketones).

Tollens' reagent, $[Ag(NH_3)_2]^+$ , is a mild oxidizing agent. Aldehydes are readily oxidized (they have an oxidizable C–H bond on the carbonyl carbon):

$RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag\downarrow + 4NH_3 + 2H_2O$

…

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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 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
$Q. Which of the following is not a characteristic of carbonyl compounds?$
(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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CBSE 2024Set FZ1 markMCQ
$Q. Which of the following compound is identified by Tollen's Reagent?$
(a) Alcohol
(b) Aldehyde
(c) Ketone
(d) Carboxylic acid
›Reveal solutionSolution

Tollen's reagent gives a silver mirror only with aldehydes → option (b).

Concept. Tollen's reagent is a mild oxidising agent, the diamminesilver(I) ion $[Ag(NH_3)_2]^+$ . Aldehydes are readily oxidised to carboxylate; in doing so they reduce $Ag^+$ to metallic silver, which deposits as a bright silver mirror.

Reaction.

$RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow RCOO^- + 2Ag\downarrow + 4NH_3 + 2H_2O$ …

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Q.(A)

Part (a)

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