Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile

NCERT Exemplar · Q13

Q. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction. H3C—HC=CH2 + H^+ ⟶ ?

(i) 2° Carbanion

(ii) 1° Carbocation

(iii) 2° Carbocation

(iv) 1° Carbanion

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Concept understanding — Inductive Effect

Inductive Effect on Acidity – From Intuition to Precision

Imagine you are holding a rope tied to a heavy box. If you pull the rope, the box moves toward you. Now imagine the rope is made of rubber bands — the pull still reaches the box, but it gets weaker the farther away you are. That is exactly how the inductive effect works inside a molecule.

The Core Intuition

An acid donates a proton ( $H^+$ ). After it does, the remaining part (the conjugate base) carries a negative charge. The stability of that negative charge determines how willing the molecule is to give up the proton. More stable conjugate base → stronger acid.

Now, some atoms or groups are electron-withdrawing — they pull electron density toward themselves through the sigma bonds. If such a group is attached near the acidic proton, it pulls some electron density away from the negative charge on the conjugate base. That spreads out (delocalises) the negative charge, making the conjugate base more stable. The acid becomes stronger.

Conversely, electron-donating groups push electron density toward the negative charge, concentrating it and making the conjugate base less stable. The acid becomes weaker.

Note

The inductive effect operates through sigma bonds only. It does not involve pi bonds or resonance. It is a permanent, through-bond polarisation.

The Precise Statement

Inductive effect on acidity: The acidity of a compound increases with the presence of electron-withdrawing groups (EWGs) near the acidic site, and decreases with electron-donating groups (EDGs). The effect is strongest when the group is closest to the acidic proton, and diminishes rapidly with distance.

Mathematically, for a series of substituted carboxylic acids:

\text{R-COOH} \quad \text{where R = substituent}

The acid dissociation constant $K_a$ changes as:

If R is electron-withdrawing (e.g., $-\text{Cl}$ , $-\text{NO}_2$ , $-\text{CF}_3$ ): $K_a$ increases → stronger acid.
If R is electron-donating (e.g., $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ ): $K_a$ decreases → weaker acid.

Why Distance Matters

The inductive effect falls off with distance because sigma bonds are localised. Each bond attenuates the effect by roughly a factor of 2–3. For example, compare:

Compound	$pK_a$	Explanation
$\text{CH}_3\text{COOH}$	4.76	Reference (no EWG)
$\text{ClCH}_2\text{COOH}$	2.86	Cl withdraws through one bond
$\text{Cl}_2\text{CHCOOH}$	1.29	Two Cl atoms, stronger withdrawal
$\text{Cl}_3\text{CCOOH}$	0.65	Three Cl atoms, strongest withdrawal
$\text{CH}_3\text{CH}_2\text{COOH}$	4.87	Ethyl group is electron-donating (slightly weaker acid)

Notice: $\text{ClCH}_2\text{COOH}$ is about 100 times stronger than acetic acid ( $\Delta pK_a \approx 1.9$ ). But if the Cl is moved further away:

Compound	$pK_a$
$\text{ClCH}_2\text{CH}_2\text{COOH}$	4.08
$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{COOH}$	4.52

The effect fades as the chlorine moves farther from the carboxyl group. …

Why this formula?

Inductive Effect: Why the Key Ideas Hold

The inductive effect is a fundamental concept in organic chemistry that explains how electron density shifts along a sigma ( $\sigma$ ) bond due to differences in electronegativity. Let's break down why the key principles work — not just what they are.

1. The Core Idea: Polarization of $\sigma$ Bonds

What happens?

When two atoms with different electronegativities form a $\sigma$ bond, the bonding electrons are not shared equally. The more electronegative atom pulls electron density toward itself.

Why does this happen?

Electronegativity is a measure of an atom's ability to attract shared electrons.
The $\sigma$ bond is a region of high electron density between the nuclei.
The more electronegative atom's nucleus exerts a stronger electrostatic pull on these electrons.
Result: The bond becomes polarized — one end becomes slightly negative ( $\delta^-$ ), the other slightly positive ( $\delta^+$ ).

Key formula (conceptual):

$\delta^- \leftarrow \text{Atom A} \rightarrow \delta^+$

where A is more electronegative than B.

2. Why the Effect Transmits Along a Chain

The puzzle:

If the inductive effect is about a single bond, how does it affect atoms several bonds away?

The reasoning:

The $\delta^+$ on the less electronegative atom creates a partial positive charge.
This partial charge polarizes the next $\sigma$ bond in the chain.
The effect is relayed through successive bonds, like a chain of dominoes.

Why does it weaken with distance?

Each bond acts as a dielectric medium — it partially screens the charge.
The electrostatic influence falls off with distance according to Coulomb's law:

F \propto \frac{q_1 q_2}{r^2}

In a molecular chain, the effective distance $r$ increases, so the induced dipole in each subsequent bond is smaller.

Key result: Inductive effect is significant only up to 3–4 bonds away.

3. The Quantitative Measure: Inductive Effect Constant ( $\sigma_I$ )

What is $\sigma_I$ ?

It's a Hammett-type constant that quantifies the electron-withdrawing or electron-donating power of a substituent through sigma bonds only.

Why does it have this form?

The inductive effect is additive — each substituent contributes independently.
For a substituent X attached to a carbon chain:

\sigma_I = \log \left( \frac{K_a(\text{X-CH}_2\text{COOH})}{K_a(\text{CH}_3\text{COOH})} \right)

where $K_a$ is the acid dissociation constant.

Why use acid dissociation?

The carboxyl group ( $-\text{COOH}$ ) is a sensitive probe.
An electron-withdrawing group (EWG) stabilizes the conjugate base ( $\text{R-COO}^-$ ) by dispersing its negative charge.
This increases $K_a$ (stronger acid).
An electron-donating group (EDG) destabilizes the conjugate base, decreasing $K_a$ .

Key formula:

$\sigma_I > 0$ for EWGs (e.g., $-\text{Cl}$ , $-\text{NO}_2$ )

$\sigma_I < 0$ for EDGs (e.g., $-\text{CH}_3$ , $-\text{C(CH}_3)_3$ )

4. Why Inductive Effect is Not Resonance

Common confusion:

Students often mix inductive and resonance effects.

The critical difference:

Property	Inductive Effect	Resonance Effect
Electron movement	Through $\sigma$ bonds only	Through $\pi$ bonds or lone pairs
Distance dependence	Dies off after 3–4 bonds	Can transmit over long distances in conjugated systems
Permanent or temporary	Permanent polarization	Can be temporary (delocalization)

Why this matters for exam problems:

In alkyl halides, the inductive effect of $-\text{Cl}$ explains the $\delta^+$ on carbon.
In benzene derivatives, the combined inductive and resonance effects determine reactivity.

--- …

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The first step of electrophilic addition to an alkene forms the more stable carbocation intermediate. For propene, the proton adds to the less substituted carbon, giving a 2° carbocation.

The question asks about the intermediate formed in the first step of an electrophilic addition reaction. The reaction shown is:

\ce{H3C-HC=CH2 + H+ -> ?}

This is propene reacting with a proton (the electrophile). Let’s understand why the answer is what it is.

The electrophile attacks the π bond.

The double bond in propene is electron-rich. The $\ce{H+}$ (a strong electrophile) seeks out this electron density. In the first step, the $\ce{H+}$ bonds to one of the two carbon atoms of the double bond, using a pair of π electrons to form a new $\ce{C-H}$ σ bond.
Which carbon gets the proton?

The key is Markovnikov’s rule: the proton adds to the carbon that already has more hydrogen atoms (the less substituted carbon). This is because the resulting carbocation must be as stable as possible.

In propene, $\ce{H3C-HC=CH2}$ , the two sp² carbons are:
- $\ce{CH2}$ (terminal, 1° carbon) — has two hydrogens.
- $\ce{HC}$ (internal, 2° carbon) — has one hydrogen.
The proton adds to the terminal $\ce{CH2}$ carbon. This leaves the positive charge on the internal carbon.
Identify the intermediate.

After the proton attaches to $\ce{CH2}$ , the other carbon becomes a carbocation:

\ce{H3C-HC+-CH3}

The positive charge is on the carbon that is bonded to two other carbons (the $\ce{HC}$ carbon is now $\ce{HC+}$ ). That makes it a secondary (2°) carbocation.

Why not a 1° carbocation? If the proton added to the internal carbon instead, the positive charge would land on the terminal carbon: $\ce{H3C-H2C+-CH2}$ …

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

KCET 2025Set D-41 markMCQ

Q. Which of the following has most acidic Hydrogen? (A) Propanoic acid (B) Dichloroacetic acid (C) Trichloroacetic acid (D) Chloroacetic acid

›Reveal solution

Compare the $-I$ (electron-withdrawing) effect of the substituents: more chlorines ⇒ better dispersal of the carboxylate's negative charge ⇒ more stable anion ⇒ stronger acid. Three Cl wins.

Step 1 — The principle: acidity = anion stability.

A carboxylic acid ionises as

\mathrm{RCOOH \rightleftharpoons RCOO^- + H^+}

The more stable the carboxylate anion $\mathrm{RCOO^-}$ , the further this equilibrium lies to the right, the more readily the H $^+$ leaves — and the stronger the acid (lower $\mathrm{p}K_a$ ). So the question reduces to: whose anion is most stabilised?

Step 2 — How the substituent stabilises the anion.

Chlorine is highly electronegative and exerts a $-I$ (inductive electron-withdrawing) effect. It pulls electron density along the $\sigma$ framework, away from the $\mathrm{COO^-}$ group. This disperses the concentrated negative charge on the carboxylate, which lowers its energy and stabilises it.

By contrast, an alkyl group is electron-releasing ( $+I$ ). It pushes electron density towards the carboxylate, intensifying the negative charge and destabilising the anion — making the acid weaker.

Step 3 — Rank the four compounds by their substituents.

Compound	Structure	Groups attached to the $\alpha$ -carbon	Net effect
Propanoic acid	$\mathrm{CH_3CH_2COOH}$	an ethyl group	$+I$ — destabilises the anion
Chloroacetic acid	$\mathrm{ClCH_2COOH}$	one Cl	$-I$ (weak)
Dichloroacetic acid	$\mathrm{Cl_2CHCOOH}$	two Cl	$-I$ (stronger)
Trichloroacetic acid	$\mathrm{Cl_3CCOOH}$	three Cl	$-I$ (strongest)

The $-I$ effect of several identical groups is cumulative — each additional chlorine withdraws more density and disperses the charge further.

Step 4 — The resulting acidity order. …

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KCET 2021Set B-21 markMCQ

>

›Reveal solution

Boiling point tracks the total hydrogen-bond network: water forms 4 H-bonds per molecule, HF only 2 (though stronger individually), $\mathrm{NH_3}$ 2 weak ones — giving $\mathrm{H_2O > HF > NH_3}$ .

1. The concept — why these three are anomalous at all.

$\mathrm{H_2O}$ , HF and $\mathrm{NH_3}$ all boil far above their group congeners ( $\mathrm{H_2S}$ , HCl, $\mathrm{PH_3}$ ) because H bonded to the highly electronegative F, O or N gives intermolecular hydrogen bonding. Breaking that network is what costs energy at the boiling point.

2. The deciding factor — number of H-bonds per molecule, not just strength.

An H-bond needs an H donor (an H attached to F/O/N) and an acceptor (a lone pair).

Molecule	H atoms (donors)	Lone pairs (acceptors)	H-bonds per molecule (limited by the smaller)
$\mathrm{H_2O}$	2	2	4 (2 donated + 2 accepted) — a perfectly matched 3-D network
HF	1	3	2 (1 donated + 1 accepted) — donors are the bottleneck; forms zig-zag chains
$\mathrm{NH_3}$	3	1	2 (1 donated + 1 accepted) — acceptors are the bottleneck

Water is the only one whose donors and acceptors are balanced, so it builds an extensive three-dimensional network — the largest total H-bonding energy, hence the highest boiling point (373 K).

3. HF vs $\mathrm{NH_3}$ . …

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COMEDK 2021Set 2021-B1 markMCQ
$Q. Identify which one of the following statements correctly describes the Inductive effect seen in Organic compounds. (A) This effect moves over the entire length of the Carbon chain. (B) It is a permanent effect which operates only in saturated compounds with polar bonds. (C) Inductive effect involves displacement of π electrons. (D) This effect causes a slight drift of π electrons towards the more electronegative atom.$
›Reveal solutionSolution
The inductive effect is a permanent polarization of σ-bonds due to electronegativity differences — described by option (B).

Key features of the inductive effect:

It is a permanent effect (unlike the temporary electromeric effect).

It arises from polar σ-bonds due to electronegativity differences.

It is transmitted through the σ-framework and weakens rapidly with distance, becoming negligible after ~3 carbons — so it does NOT operate over the entire chain length.

It involves σ electrons, not π electrons.

Evaluating options:

(A) "moves over the entire length of the carbon chain" — false; it dies out quickly. …
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KCET 2020Set A-11 markMCQ
$\text{HCOOH}$
›Reveal solutionSolution
Boiling point depends on the strength of intermolecular forces. Dimethyl ether ( $\text{CH}_3\text{OCH}_3$ ) has only weak dipole-dipole forces and no hydrogen bonding, giving it the lowest boiling point among the four compounds. The correct option is (D).

The key to comparing boiling points is understanding the intermolecular forces at play. Boiling requires overcoming the attractions between molecules in the liquid phase. Stronger forces mean a higher boiling point; weaker forces mean a lower one.

The four molecules are:

HCOOH (formic acid) — a carboxylic acid.

CH $_3$ CH $_2$ OH (ethanol) — an alcohol.

CH $_3$ CH $_2$ NH $_2$ (ethylamine) — a primary amine.

CH $_3$ OCH $_3$ (dimethyl ether) — an ether.

Let’s examine each.

Hydrogen bonding is the strongest intermolecular force among these. For a molecule to form hydrogen bonds, it must have a hydrogen atom bonded to a highly electronegative atom (N, O, or F) and have a lone pair on another electronegative atom to accept the hydrogen bond.

Formic acid has an O–H group and a C=O group — it can both donate and accept hydrogen bonds. Additionally, carboxylic acids form strong dimers (two molecules held together by two hydrogen bonds), which raises the boiling point significantly.

Ethanol has an O–H group — it can donate and accept hydrogen bonds.

Ethylamine has an N–H group — it can donate and accept hydrogen bonds, though N–H···N bonds are slightly weaker than O–H···O bonds because nitrogen is less electronegative than oxygen.

Dimethyl ether has no O–H or N–H bond. It has an oxygen atom with lone pairs, so it can accept hydrogen bonds from other molecules, but it cannot donate them. Therefore, it cannot form a network of hydrogen bonds with itself.

Without hydrogen bonding, the dominant forces are dipole-dipole interactions and London dispersion forces.

Dimethyl ether is a polar molecule (the C–O–C bond angle gives a net dipole), so it has dipole-dipole forces. But these are much weaker than hydrogen bonds.

Its molar mass (46 g/mol) is similar to ethanol (46 g/mol) and ethylamine (45 g/mol), so dispersion forces are comparable. The deciding factor is the absence of hydrogen bonding in the ether. …
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KCET 2019Set A-11 markMCQ
$Q. Match the following acids with their pKa values : Acid pKa a. Phenol i. 16 b. p-Nitrophenol ii. 0.78 c. Ethanol iii. 10 d. Picric acid iv. 7.1 a b c d (A) iii iv i ii (B) iii i iv ii (C) ii i ii iv (D) iv ii iii i$
›Reveal solutionSolution
The pKa of an acid tells you how acidic it is — the lower the pKa, the stronger the acid. Phenol (pKa ≈ 10), p‑nitrophenol (pKa ≈ 7.1), ethanol (pKa ≈ 16), and picric acid (pKa ≈ 0.78) match to options iii, iv, i, ii respectively, which corresponds to choice (A).

The key to matching pKa values is understanding what pKa actually measures. pKa is the negative logarithm of the acid dissociation constant Ka. A smaller pKa means a stronger acid — the equilibrium lies further to the right, favouring deprotonation. So when you see a list of acids, rank them by how readily they lose a proton, then assign the lowest pKa to the strongest acid.

Let’s think about each acid in turn.

Phenol ( $\ce{C6H5OH}$ ) is a weak acid. The phenoxide ion is stabilised by resonance into the aromatic ring, but the oxygen still carries a partial negative charge. Its pKa is around 10. That’s option iii.

p‑Nitrophenol has a nitro group ( $\ce{-NO2}$ ) at the para position. The nitro group is strongly electron‑withdrawing — it pulls electron density away from the O–H bond, making the proton easier to lose. It also stabilises the conjugate base by resonance. So p‑nitrophenol is a stronger acid than phenol. Its pKa is about 7.1, which is option iv.

Ethanol ( $\ce{CH3CH2OH}$ ) is a very weak acid. The alkoxide ion is not stabilised by resonance at all — it’s just a bare negative charge on oxygen. Ethanol is actually less acidic than water (pKa ≈ 15.7). Its pKa is around 16, which is option i. …
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Q.Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction. H3C—HC=CH2 + H^+ ⟶ ?

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Q.Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
H3C—HC=CH2 + H^+ ⟶ ?

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