JEE Mains Chemistry · Organic Chemistry - Some Basic Principles and Techniques

Electronic Effects, Resonance and Acidity

Electrons in a molecule shift through σ bonds (the inductive effect), through π systems (resonance), on demand when a reagent attacks (the electromeric effect) and from C–H bonds into a neighbouring empty or π orbital (hyperconjugation); these shifts rank resonance structures and set the strength of acids and bases.

All Organic Chemistry - Some Basic Principles and Techniques notes JEE Mains Chemistry strategy

Why this matters

Twenty-four PYQs, one of them asking for a number, and four from 2026. Ten name or compare the four electronic effects: which ones are permanent, the order of electron-withdrawing groups, the number of hyperconjugating hydrogens. Nine rank resonance structures, or use resonance to explain a dipole moment or a bond length. Five rank acids, conjugate bases or the acidity of marked hydrogens.

Concept 1 of 3: Inductive, resonance, electromeric and hyperconjugation effects

There are four ways electrons shift inside a molecule. Three of them are permanent features of the molecule: the inductive effect, resonance and hyperconjugation. Only the electromeric effect is temporary: it appears when a reagent approaches a multiple bond and disappears when the reagent is removed.

Definition

−I order: $\mathrm{-NO_2 > -CN > -COOH > -F > -Cl > -Br > -I}$ . +I groups are alkyl groups: $\mathrm{(CH_3)_3C{-} > (CH_3)_2CH{-} > CH_3CH_2{-} > CH_3{-}}$ . The inductive effect weakens quickly with distance.
−R groups have a multiple bond to a more electronegative atom: $\mathrm{-NO_2}$ , –CN, –CHO, –COOH, >C=O. +R groups carry a lone pair on the atom joined to the π system: –OH, –OR, $\mathrm{-NH_2}$ , –X.
Electromeric effect: +E when the π electrons move towards the atom the reagent attacks ( $\mathrm{H^+}$ adding to a C=C); −E when they move away from the atom the reagent attacks ( $\mathrm{CN^-}$ adding to the carbon of C=O). When it opposes the inductive effect, the electromeric effect wins.
Hyperconjugation: a C–H σ bond on the carbon next to a carbocation, a radical or a C=C overlaps the empty p orbital or the π orbital. The number of hyperconjugating hydrogens equals the number of α-hydrogens: $\mathrm{CH_3^+}$ has none, $\mathrm{CH_3CH_2^+}$ three and $\mathrm{(CH_3)_3C^+}$ nine.
More alkyl groups on a C=C give more hyperconjugation, a more stable alkene and a LOWER heat of hydrogenation.

Effect	Electrons move through	Permanent or temporary	Typical example
Inductive (I)	σ bonds, weakening with distance	Permanent	Cl pulls electrons along the chain in $\mathrm{ClCH_2COOH}$
Resonance (R or M)	π bonds and lone pairs on adjacent atoms	Permanent	$\mathrm{-NH_2}$ pushes its lone pair into the ring of aniline
Electromeric (E)	One π bond, shifted completely to one atom	Temporary; only while the reagent is present	The C=O of propanone as $\mathrm{CN^-}$ attacks The only one of the four that disappears when the reagent is taken away.
Hyperconjugation	A C–H σ bond into an adjacent empty p or π orbital	Permanent	The three C–H bonds of the $\mathrm{CH_3}$ group stabilise the C=C of propene

Resonance and the electromeric effect need a π system; the inductive effect needs only σ bonds.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q38Moderate

Example 1 · Organic Chemistry - Some Basic Principles and Techniques · Electronic Effects, Resonance and Acidity

Increasing order of electron withdrawing power of following functional groups is : (A) -CN (B) -COOH (C)

- {NO}_{2}

(D) -I

Hyperconjugation is a permanent effect

Hyperconjugation needs no reagent: it is present in the ground state of every molecule with an α-C–H next to an empty or π orbital. A statement calling it temporary is false.

H⁺ shows a +E effect, not −E

When

\mathrm{H^+}

attacks a C=C, the π electrons move towards the carbon it bonds to. That is the +E effect. The −E effect goes with a nucleophile such as

\mathrm{CN^-}

Iodine is the weakest −I group of the halogens

The −I effect of a halogen follows its electronegativity, so –I withdraws least. It also withdraws less than –COOH, –CN and

\mathrm{-NO_2}

Drill 9 more on inductive, resonance, electromeric and hyperconjugation effects

Concept 2 of 3: Rules for the stability of resonance structures

Resonance structures are drawings of one molecule that differ only in where π electrons and lone pairs sit; the atoms stay where they are. The real molecule is a blend of them, weighted towards the most stable drawings. So ranking the structures is a checklist, applied in order.

Definition

Only electrons move. No atom moves, and the number of unpaired electrons stays the same.
No second-period atom may exceed an octet: never five bonds to C, N or O.
Resonance energy = energy of the most stable contributing structure − energy of the actual molecule. The actual molecule is always lower in energy than any single structure.
Conjugation (a C=C next to a C=O) lets charge separate along the chain. This raises the dipole moment, and it gives the single bond between the two double bonds partial double-bond character, so that bond is SHORTER.
A conjugated diketone has its two C=O groups joined through a C=C: O=C–C=C–C=O, as in p-benzoquinone.

Rule	More stable contributor	Less stable contributor
Neutral beats charge-separated	$\mathrm{CH_2{=}CH{-}Cl}$	$\mathrm{^-CH_2{-}CH{=}\overset{+}{Cl}}$
Every atom with a complete octet	$\mathrm{CH_3{-}\overset{+}{O}{=}CH_2}$	$\mathrm{CH_3{-}O{-}\overset{+}{C}H_2}$ (carbon with a sextet)
Negative charge on the more electronegative atom	$\mathrm{CH_2{=}CH{-}O^-}$	$\mathrm{^-CH_2{-}CH{=}O}$
Opposite charges close, like charges apart	Unlike charges on neighbouring atoms	Like charges on neighbouring atoms (the worst case)
No atom beyond an octet	Nitrogen with four bonds and a + charge, as in $\mathrm{-\overset{+}{N}({=}O)O^-}$	Nitrogen with five bonds: not a valid structure A 'resonance structure' with five bonds to N or C is simply wrong, however it is charged.

Apply the rules from the top; the first rule that separates two structures decides.

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The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q43Moderate

Example 2 · Organic Chemistry - Some Basic Principles and Techniques · Electronic Effects, Resonance and Acidity

Given below are two statements. Statement I : The dipole moment of

\mathrm{CH_{3}-CH=CH-CHO}

is greater than that of

\mathrm{CH_{3}-CH_{2}-CH_{2}-CHO}

. Statement II : The

\mathrm{C_{1}-C_{2}}

bond length of

\mathrm{CH_{3}-CH=CH-CHO}

is greater than the

\mathrm{C_{1}-C_{2}}

bond length of

\mathrm{CH_{3}-CH_{2}-CH_{2}-CHO}

. In the light of the above statements, choose the correct answer from the options given below:

Five bonds to nitrogen is never allowed

In a nitro group the nitrogen has four bonds and a positive charge. A drawing that gives it five bonds breaks the octet rule and is not a resonance structure at all.

Charge separation costs stability

Among valid structures, the one without separated charges is the most stable. A charge-separated structure contributes less, even though it explains the dipole moment.

Conjugation shortens the single bond

In a conjugated enal, resonance gives the C–C bond between C=C and C=O some double-bond character. That bond is shorter than the same bond in the saturated aldehyde, not longer.

Drill 8 more on rules for the stability of resonance structures

Concept 3 of 3: Acid strength from conjugate-base stability

An acid is only as strong as its anion is stable. Anything that spreads or holds the negative charge makes the acid stronger: an electronegative atom, resonance, more s-character, an electron-withdrawing group. And the more stable the anion, the weaker it is as a base.

Definition

A stronger acid has a more stable conjugate base, and that conjugate base is a weaker base.
Order of common acids: sulphonic acid > carboxylic acid > phenol > water > alcohol > terminal alkyne. Their conjugate bases run the other way: $\mathrm{RC{\equiv}C^- > RO^- > OH^- > C_6H_5O^- > RCOO^- > RSO_3^-}$ .
s-character: an sp C–H (50% s) is more acidic than an sp² C–H (33%), which is more acidic than an sp³ C–H (25%).
Resonance: a C–H next to a C=O (an α-hydrogen), or on a benzylic carbon, is far more acidic than an ordinary alkane C–H; next to two C=O groups it is more acidic still.
Inductive effect: −I groups raise acidity ( $\mathrm{ClCH_2COOH > CH_3COOH}$ ); +I alkyl groups lower it, so a tertiary C–H is the least acidic sp³ C–H.

Acid strength and conjugate base

\mathrm{HA \rightleftharpoons H^+ + A^-}\qquad \text{more stable } \mathrm{A^-} \;\Rightarrow\; \text{larger } K_a,\ \text{smaller } \mathrm{p}K_a,\ \text{weaker base } \mathrm{A^-}

Worked example

Arrange ethanoic acid, phenol, ethanol and ethyne in decreasing order of acid strength.

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The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q34Moderate

Example 3 · Organic Chemistry - Some Basic Principles and Techniques · Electronic Effects, Resonance and Acidity

What will be the decreasing order of basic strength of the following conjugate bases?

\ ^{-}OH,R\overline{O},CH_{3}CO\overline{O},C\overline{l}

Basicity runs opposite to acidity

The anion of the strongest acid is the weakest base. Chloride, from the strong acid HCl, is a far weaker base than ethanoate, which is weaker than hydroxide.

An alkoxide is a stronger base than hydroxide

An alcohol is a slightly weaker acid than water because the +I alkyl group destabilises the alkoxide. So

\mathrm{RO^-}

is a stronger base than

\mathrm{OH^-}

Rank C–H acidity by s-character first

A C–H on an sp carbon is more acidic than one on an sp² carbon, whatever the size of the molecule. Only among sp³ C–H bonds do resonance and the inductive effect decide.

Drill 4 more on acid strength from conjugate-base stability

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Acid strength from conjugate-base stability
Acid strength and conjugate base
$\mathrm{HA \rightleftharpoons H^+ + A^-}\qquad \text{more stable } \mathrm{A^-} \;\Rightarrow\; \text{larger } K_a,\ \text{smaller } \mathrm{p}K_a,\ \text{weaker base } \mathrm{A^-}$

Reference tables (2)

Inductive, resonance, electromeric and hyperconjugation effects4 rows

Effect	Electrons move through	Permanent or temporary	Typical example
Inductive (I)	σ bonds, weakening with distance	Permanent	Cl pulls electrons along the chain in $\mathrm{ClCH_2COOH}$
Resonance (R or M)	π bonds and lone pairs on adjacent atoms	Permanent	$\mathrm{-NH_2}$ pushes its lone pair into the ring of aniline
Electromeric (E)	One π bond, shifted completely to one atom	Temporary; only while the reagent is present	The C=O of propanone as $\mathrm{CN^-}$ attacks The only one of the four that disappears when the reagent is taken away.
Hyperconjugation	A C–H σ bond into an adjacent empty p or π orbital	Permanent	The three C–H bonds of the $\mathrm{CH_3}$ group stabilise the C=C of propene

Resonance and the electromeric effect need a π system; the inductive effect needs only σ bonds.

Rules for the stability of resonance structures5 rows

Rule	More stable contributor	Less stable contributor
Neutral beats charge-separated	$\mathrm{CH_2{=}CH{-}Cl}$	$\mathrm{^-CH_2{-}CH{=}\overset{+}{Cl}}$
Every atom with a complete octet	$\mathrm{CH_3{-}\overset{+}{O}{=}CH_2}$	$\mathrm{CH_3{-}O{-}\overset{+}{C}H_2}$ (carbon with a sextet)
Negative charge on the more electronegative atom	$\mathrm{CH_2{=}CH{-}O^-}$	$\mathrm{^-CH_2{-}CH{=}O}$
Opposite charges close, like charges apart	Unlike charges on neighbouring atoms	Like charges on neighbouring atoms (the worst case)
No atom beyond an octet	Nitrogen with four bonds and a + charge, as in $\mathrm{-\overset{+}{N}({=}O)O^-}$	Nitrogen with five bonds: not a valid structure A 'resonance structure' with five bonds to N or C is simply wrong, however it is charged.

Apply the rules from the top; the first rule that separates two structures decides.

Watch out for (9)

Hyperconjugation is a permanent effect
→ Inductive, resonance, electromeric and hyperconjugation effects
H⁺ shows a +E effect, not −E
→ Inductive, resonance, electromeric and hyperconjugation effects
Iodine is the weakest −I group of the halogens
→ Inductive, resonance, electromeric and hyperconjugation effects
Five bonds to nitrogen is never allowed
→ Rules for the stability of resonance structures
Charge separation costs stability
→ Rules for the stability of resonance structures
Conjugation shortens the single bond
→ Rules for the stability of resonance structures
Basicity runs opposite to acidity
→ Acid strength from conjugate-base stability
An alkoxide is a stronger base than hydroxide
→ Acid strength from conjugate-base stability
Rank C–H acidity by s-character first
→ Acid strength from conjugate-base stability

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