The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ____________. ![Toluene ring-chlorination options -- benzyl chloride, o-chlorotoluene, and p-chlorotoluene -- each drawn as a real benzene ring matching the NCERT Exemplar page](/figures/haloalkanes-and-haloarenes/diagram-6-q4656-toluene-chlorination.svg) (iv) Mixture of (ii) and (iii)

The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ______. (iv)

NCERT Exemplar · Q20

Q. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ____________.

Toluene ring-chlorination options -- benzyl chloride, o-chlorotoluene, and p-chlorotoluene -- each drawn as a real benzene ring matching the NCERT Exemplar page — Figure

(iv) Mixture of

(ii) and (iii)

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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. …

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Toluene undergoes electrophilic aromatic substitution with chlorine in the presence of a Lewis acid (Fe/FeCl₃) and in the absence of light. The methyl group is an ortho/para director, so the products are o-chlorotoluene and p-chlorotoluene — a mixture. The correct option is (iv).

The key here is to recognise that the reaction conditions — chlorine gas, iron (which generates FeCl₃ in situ), and no light — are the classic setup for electrophilic aromatic substitution (EAS) on an aromatic ring.

If light were present, you’d get free-radical substitution at the benzylic position (giving benzyl chloride). But without light, the iron catalyst activates chlorine to form an electrophile, and the reaction proceeds on the ring.

Toluene has a methyl group attached to the benzene ring. The methyl group is an activating group and an ortho/para director in EAS reactions. This is because the methyl group donates electron density into the ring via hyperconjugation and the inductive effect, stabilising the intermediate carbocation (arenium ion) when substitution occurs at the ortho or para positions.

So, when chlorine attacks the ring, it will preferentially go to the ortho and para positions relative to the methyl group. Both ortho and para products are formed, and they are not separated under typical reaction conditions — you get a mixture.

Watch out

A very common mistake is to think that the absence of light automatically means side-chain chlorination. That’s wrong — side-chain chlorination requires light or heat (free-radical conditions). Without light, the iron catalyst directs the reaction to the ring.

Tip

Remember the mnemonic: No light, ring site; with light, side-chain site. For toluene:

Fe/Cl₂, dark → ring substitution (o/p mixture)
Cl₂, hv or heat → benzylic substitution (C₆H₅CH₂Cl)

Let’s walk through the reasoning step by step:

Identify the reaction type.

Given: toluene + Cl₂, in presence of iron (Fe), in absence of light.

Iron reacts with Cl₂ to form FeCl₃, a Lewis acid. This is the classic catalyst for electrophilic aromatic chlorination. No light means no free radicals are generated, so the mechanism is ionic, not radical.
Formation of the electrophile.

FeCl₃ polarises the Cl–Cl bond, making one chlorine more electrophilic:

\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^\delta+ \cdots \text{FeCl}_4^-

The actual attacking species is effectively $\text{Cl}^+$ (or a chlorine atom with strong positive character).

Attack on the aromatic ring. …

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Here’s a breakdown of the common mistakes students make on this question and how to avoid each.

Common Mistake #1: Confusing the catalyst/condition with free radical chlorination

The mistake:

Students see “chlorine” and “toluene” and immediately think of free radical substitution (which gives benzyl chloride, $\mathrm{C_6H_5CH_2Cl}$ ). They pick option (i).

Why it happens:

In school, the reaction of toluene with $\mathrm{Cl_2}$ in sunlight or UV light (or at high temperature) is drilled as:

\mathrm{C_6H_5CH_3 + Cl_2 \xrightarrow{h\nu} C_6H_5CH_2Cl + HCl}

So the presence of chlorine + toluene triggers that memory.

How to avoid:

Read the condition carefully. The question says:

“in the presence of iron and in the absence of light”
Iron ( $\mathrm{Fe}$ ) acts as a Lewis acid (forms $\mathrm{FeCl_3}$ in situ) — this is an electrophilic aromatic substitution catalyst, not a free radical initiator.
Rule of thumb:
- Light / heat → free radical → side chain chlorination (benzyl chloride)
- $\mathrm{Fe}$ / $\mathrm{FeCl_3}$ / dark → electrophilic substitution → ring chlorination

Common Mistake #2: Forgetting that the methyl group is ortho/para directing

The mistake:

Students correctly identify that ring substitution occurs, but they pick only one isomer — either (ii) or (iii) — instead of the mixture.

Why it happens:

They remember that $\mathrm{-CH_3}$ is an activating group, but forget that it directs both ortho and para positions.

How to avoid:

Recall the directing effect:
- Alkyl groups (like $\mathrm{-CH_3}$ ) are ortho/para directing (due to hyperconjugation and +I effect).
- So chlorination of toluene in the presence of $\mathrm{FeCl_3}$ gives both ortho and para products.
The correct answer is therefore a mixture of o-chlorotoluene and p-chlorotoluene → option (iv).

Common Mistake #3: Ignoring the “absence of light” clause

The mistake: …

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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 17 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
$Q. Write True or False: Contrary to electrophilic addition reactions observed in alkenes, the aldehydes and ketones undergo nucleophilic addition reactions.$
›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 D1 markMCQ
$Q. Chloretone is formed when chloroform reacts with$
(a) Formaldehyde
(b) Acetaldehyde
(c) Acetone
(d) Benzaldehyde
›Reveal solutionSolution

Chloroform + acetone -> chloretone.

Chloroform (CHCl3) adds across the carbonyl group of acetone (in presence of base) to give chloretone, chemically 1,1,1-trichloro-2-methyl-2-propanol, (CH3)2C(OH)CCl3, which is used as a hypnotic/sedative:

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Change occurs in hybridisation state of carbonyl carbon in nucleophilic addition reaction is -$
(a) sp2 to sp
(b) sp to sp2
(c) sp2 to sp3
(d) sp3 to sp2
›Reveal solutionSolution

Nucleophilic addition converts the planar, trigonal carbonyl carbon into a tetrahedral carbon, so its hybridisation changes from sp2 to sp3.

In an aldehyde/ketone, the carbonyl carbon is sp2 hybridised: it forms three sigma bonds (to O and to two other groups) that lie in one plane, plus a pi bond to oxygen, with bond angles close to 120 degrees. …

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CBSE 2023Set 56/2/11 markMCQ
$HCHO$
›Reveal solutionSolution
Nucleophilic addition to carbonyls is controlled by steric hindrance and electronic effects; formaldehyde ( $HCHO$ ) has the least steric hindrance and strongest electrophilic carbon, making it the most reactive. The correct option is (A).

Nucleophilic addition to a carbonyl group ( $\ce{C=O}$ ) is the fundamental reaction of aldehydes and ketones. A nucleophile attacks the electrophilic carbonyl carbon, and the key question is: what makes that carbon more or less attractive to an incoming nucleophile? Two factors dominate: steric hindrance around the carbonyl carbon, and electronic effects (inductive and hyperconjugative) that stabilise or destabilise the carbonyl.

Let’s examine each compound in the list.

Steric hindrance

The nucleophile must physically approach the carbonyl carbon. If bulky groups are attached to it, they block the path.

In $HCHO$ (formaldehyde), the carbonyl carbon is attached to two hydrogen atoms — the smallest possible substituents. There is almost no steric barrier.

In $CH_3CHO$ (acetaldehyde), one hydrogen is replaced by a methyl group. That methyl is larger than hydrogen, so approach is slightly hindered.

In $CH_3COCH_3$ (acetone), both substituents are methyl groups. The carbonyl carbon is now flanked by two bulky groups, making it significantly more crowded.

In $CH_3COC_2H_5$ (butanone, or ethyl methyl ketone), one methyl is replaced by an ethyl group — even bulkier. This is the most sterically hindered of the four.

So purely on steric grounds, reactivity order is: $HCHO > CH_3CHO > CH_3COCH_3 > CH_3COC_2H_5$ .

Electronic effects

Alkyl groups are electron-donating via the inductive effect (+I) and hyperconjugation. They push electron density toward the carbonyl carbon, making it less electrophilic (less positive). More alkyl groups = more electron donation = lower reactivity toward nucleophiles.

$HCHO$ has zero alkyl groups — the carbonyl carbon is the most electron-deficient.

$CH_3CHO$ has one alkyl group — slightly less electrophilic.

$CH_3COCH_3$ and $CH_3COC_2H_5$ each have two alkyl groups, so they are the least electrophilic. (The ethyl group is slightly more electron-donating than methyl, but the difference is small; both are much less reactive than $HCHO$ .) …
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CBSE 2023Set 56/3/11 markMCQ
$HCHO$
›Reveal solutionSolution
The reactivity of carbonyl compounds toward nucleophilic addition is governed by steric hindrance and electronic effects. Formaldehyde (I) is the most reactive, followed by acetaldehyde (II), and then acetone (III). The correct order is I > II > III, which corresponds to option (B).

Nucleophilic addition to a carbonyl group is one of the most fundamental reactions in organic chemistry. The carbonyl carbon is electrophilic because oxygen is more electronegative and pulls electron density away, leaving the carbon partially positive. A nucleophile attacks this carbon, forming a tetrahedral intermediate.

But why do different carbonyl compounds react at different rates? Two factors matter here: steric hindrance and electronic effects.

Steric hindrance: The nucleophile must physically approach the carbonyl carbon. If the carbon is surrounded by bulky groups, the approach is blocked, and the reaction slows down. Formaldehyde has two small hydrogen atoms attached to the carbonyl carbon — almost no hindrance. Acetaldehyde has one methyl group and one hydrogen — moderate hindrance. Acetone has two methyl groups — maximum hindrance among these three.

Electronic effects: Alkyl groups are electron-donating (through hyperconjugation and inductive effect). More alkyl groups attached to the carbonyl carbon mean more electron density pushed toward that carbon, making it less electrophilic (less positive). This also slows down nucleophilic attack. Formaldehyde has no alkyl groups, so its carbonyl carbon is the most electrophilic. Acetaldehyde has one methyl group, so it's less electrophilic. Acetone has two methyl groups, making it the least electrophilic.

Both factors — steric and electronic — work in the same direction here. So the order is clear.

Let's walk through it step by step.

Identify the carbonyl compounds and their substituents.

(I) HCHO: formaldehyde — two H atoms on the carbonyl carbon.

(II) CH₃CHO: acetaldehyde — one CH₃ and one H.

(III) CH₃COCH₃: acetone — two CH₃ groups.

Consider steric hindrance.

The nucleophile must approach the carbonyl carbon from above or below the plane. In formaldehyde, the two H atoms are tiny — no obstruction. In acetaldehyde, the methyl group is larger than H, so it partially blocks one side. In acetone, two methyl groups crowd the carbon from both sides, making approach difficult.

So steric hindrance increases: I < II < III.

Since more hindrance means slower reaction, reactivity due to sterics: I > II > III.

Consider electronic effects.

Methyl groups donate electrons via hyperconjugation and the inductive effect. More electron donation makes the carbonyl carbon less δ⁺, so less attractive to nucleophiles.

Formaldehyde: no donation → most δ⁺. …
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Q.The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ____________.

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)

Concept: Electrophilic Aromatic Substitution (EAS) vs. Free Radical Substitution

Method: Electrophilic Aromatic Substitution (EAS) — Chlorination of Toluene

Steps:

Common Mistake #1: Confusing the catalyst/condition with free radical chlorination

Common Mistake #2: Forgetting that the methyl group is ortho/para directing

Common Mistake #3: Ignoring the “absence of light” clause

Why this is nucleophilic addition

Step-by-step mechanism

Q.The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ____________.

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