Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain. (A) CH3COOH and

NCERT Exemplar · Q49

Q. Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain. (A) CH3COOH and (B) CH3COO^-

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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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Resonance stabilisation is measured by the number and quality of equivalent contributing structures. The acetate ion (B) has two identical resonance forms, giving it far greater stabilisation than neutral acetic acid (A), where the contributing structures are unequal in energy.

Resonance stabilisation is not about how many structures you can draw — it is about how much those structures lower the energy of the real molecule. The key idea is equivalence. When two or more resonance contributors are identical in energy, the actual molecule sits exactly halfway between them, and that delocalisation gives maximum stabilisation. When the contributors are unequal, the real structure leans heavily toward the more stable one, and the stabilisation is much smaller.

Let us apply this to the two species.

Structure (A): CH₃COOH (acetic acid) Draw its resonance. The carbonyl group ( $\ce{C=O}$ ) can push a lone pair from the adjacent hydroxyl oxygen toward the carbonyl carbon, forming a structure with a $\ce{C-O^-}$ single bond and a $\ce{C=OH^+}$ double bond.

\ce{CH3-C(=O)-OH \longleftrightarrow CH3-C(-O^-)=OH^+}

These two forms are not equivalent. The left (major) contributor has a neutral $\ce{C=O}$ and an $\ce{O-H}$ ; the right (minor) contributor has a positive charge on oxygen and a negative charge on the other oxygen — a high-energy separation of charge. The real molecule is mostly the left form, with only a tiny contribution from the right. Resonance stabilisation exists, but it is modest.

Structure (B): CH₃COO⁻ (acetate ion) Now draw its resonance. The negative charge on one oxygen can be delocalised onto the other oxygen through the $\pi$ system:

\ce{CH3-C(=O)-O^- \longleftrightarrow CH3-C(-O^-)=O}

These two structures are perfectly equivalent — same atoms, same bond orders, same charge distribution, just swapped. The real acetate ion is a hybrid with both $\ce{C-O}$ bonds identical (bond order 1.5) and the negative charge shared equally between the two oxygens. This is textbook maximum resonance stabilisation. …

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

AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{CH_3-CH(CH_3)-CH_2-CH_2^+}$
›Reveal solutionSolution
Carbocation stability follows the number of alkyl groups attached to the positively charged carbon (hyperconjugation + inductive effect): tertiary > secondary > primary.

Concept and Intuition

A carbocation is stabilised by electron density donated from neighbouring alkyl groups, both through hyperconjugation and the +I inductive effect. More alkyl substituents on the cationic carbon means more stabilisation, so tertiary cations are most stable, followed by secondary, then primary.

Step-by-Step Solution

I) $CH_3\text{-}CH(CH_3)\text{-}CH_2\text{-}CH_2^+$ : the positive carbon is a terminal $CH_2^+$ attached to only one other carbon — a primary carbocation.

II) $CH_3\text{-}C^+(CH_3)\text{-}CH_2\text{-}CH_3$ : the positive carbon is attached to three carbon groups — a tertiary carbocation. …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\xrightarrow[\mathrm{(ii)\ H_3O^+}]{\mathrm{(i)\ KMnO_4/OH^-}}$
›Reveal solutionSolution
I = benzoic acid, II = bromoacetic acid, III = phenylacetic acid; the $\alpha$ -halogen in II makes it the strongest acid, the extra $\mathrm{CH_2}$ insulating the phenyl ring in III makes it the weakest. Answer: (C).

Concept and Intuition

Carboxylic-acid strength is governed by how well the conjugate base's negative charge is stabilised. An electron-withdrawing group (like a halogen) sitting directly on or very near the carboxyl carbon pulls electron density away and stabilises the carboxylate strongly (inductive effect, which falls off fast with distance). A phenyl ring attached through an intervening $\mathrm{CH_2}$ helps much less than a phenyl ring attached directly to the carboxyl carbon, because the inductive effect must travel through one more bond and there's no direct conjugation either way.

Step-by-Step Solution

Reaction A: n-Propylbenzene, $\mathrm{C_6H_5CH_2CH_2CH_3}$ , treated with hot alkaline $\mathrm{KMnO_4}$ oxidises the entire alkyl side chain (regardless of its length) down to a single $-COOH$ directly attached to the ring; acidification (ii) liberates the free acid. So I $= \mathrm{C_6H_5COOH}$ (benzoic acid).

Reaction B: Acetic acid with $\mathrm{Br_2/red\ P}$ is the Hell–Volhard–Zelinsky reaction — it $\alpha$ -brominates the carboxylic acid; hydrolysis (ii) gives $\mathrm{BrCH_2COOH}$ . So II $=$ bromoacetic acid, with $\mathrm{Br}$ sitting right next to $-COOH$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$Q. Arrange the following free radicals in the correct order of their stability$
(i) $\mathrm{CH_2=CH^{\bullet}}$
(ii) $\mathrm{CH_3^{\bullet}}$
(iii) $\mathrm{CH_3-\overset{\bullet}{C}H-CH_3}$
(iv) $\mathrm{(CH_3)_3C^{\bullet}}$ (A) i > ii > iii > iv (B) iv > iii > ii > i (C) i < ii < iii < iv (D) iv > iii > i > ii
›Reveal solutionSolution
The stability of carbon free radicals increases with the number of alkyl groups attached to the radical center due to hyperconjugation and the inductive effect. The correct order is tertiary > secondary > primary > vinyl, so (B) iv > iii > ii > i is correct.

Concept & Intuition: Why alkyl groups stabilize free radicals

A free radical has an unpaired electron. Alkyl groups (like –CH₃) are electron-donating through both the inductive effect (pushing electron density) and hyperconjugation (delocalizing the unpaired electron into adjacent C–H σ bonds). More alkyl groups mean more ways to spread out the unpaired electron, lowering the energy and increasing stability. A vinyl radical (CH₂=CH•) is less stable than a methyl radical because the unpaired electron is in an sp² orbital near a double bond, which actually withdraws electron density (the carbon is more electronegative due to higher s-character), making the radical more reactive.

Step-by-step reasoning

Identify the radical types

(i) CH₂=CH• → vinyl radical (sp² hybridized, adjacent to a π bond)

(ii) CH₃• → methyl radical (primary, no alkyl substituents)

(iii) CH₃–•CH–CH₃ → isopropyl radical (secondary, two alkyl groups)

(iv) (CH₃)₃C• → tert-butyl radical (tertiary, three alkyl groups)

Apply the stability rule for alkyl radicals

The general order is:

tertiary > secondary > primary > methyl > vinyl

This is because each alkyl group donates electron density and provides hyperconjugative structures. For example, the tert-butyl radical has 9 C–H bonds that can hyperconjugate, the isopropyl has 6, the methyl has 3, and the vinyl has essentially none (and actually suffers from inductive withdrawal).

Compare the given radicals

(iv) is tertiary → most stable

(iii) is secondary → next

(ii) is primary (methyl) → less stable

(i) is vinyl → least stable

Match with the options

The correct decreasing order is: iv > iii > ii > i. …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$\mathrm{C_2H_2}$
›Reveal solutionSolution
Only terminal alkynes (with an H directly on the sp carbon) are meaningfully acidic; 2-butyne (fully internal) is least acidic, propyne is intermediate, and acetylene is the most acidic — giving II < III < I.

Concept and Intuition

The acidity of alkyne C–H bonds arises because the conjugate base (the acetylide anion) sits on an sp-hybridized carbon, which has 50% s-character — s-orbitals hold electron density closer to the nucleus, so the extra electron pair in the anion is stabilized more effectively than in sp² or sp³ carbanions. This makes terminal alkynes (with H directly bonded to an sp carbon) noticeably acidic, while alkynes substituted with alkyl groups on both triple-bond carbons have no such acidic hydrogen at all. Among terminal alkynes, an electron-donating alkyl group elsewhere on the molecule slightly destabilizes the negative charge of the conjugate base (since alkyl groups push electron density in, and a carbanion wants to disperse negative charge, not concentrate more of it), making substituted terminal alkynes marginally less acidic than acetylene itself.

Step-by-Step Solution

I. $C_2H_2$ (acetylene, $HC\equiv CH$ ): both carbons bear an acidic H directly on sp carbons.

II. $(CH_3)_2C_2$ (2-butyne, $CH_3-C\equiv C-CH_3$ ): both triple-bond carbons are substituted with methyl groups — there is no H on either sp carbon at all, so it has essentially no acidic C–H comparable to a terminal alkyne; its acidity here is negligible (methyl C–H bonds are ordinary sp³ C–H, far less acidic).

III. $(CH_3)C_2H$ (propyne, $CH_3-C\equiv C-H$ ): one sp carbon bears an acidic H, the other bears a methyl group. …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. Identify the correct trend of acidic strength for the given alcohols:$
(i) $\mathrm{CH_3CH_2CH_2OH}$ (propan-1-ol, a straight-chain primary alcohol)
(ii) a secondary alcohol drawn as a zig-zag chain with $\mathrm{OH}$ on an internal carbon, $\mathrm{CH_3CH(OH)CH_2CH_3}$ (butan-2-ol)
(iii) $\mathrm{(CH_3)_3COH}$ (tert-butanol, a tertiary alcohol) (A)
(i) >
(iii) >
(ii) (B)
(i) >
(ii) >
(iii) (C)
(iii) >
(i) >
(ii) (D)
(iii) >
(ii) > (i)
›Reveal solutionSolution
Alcohol acidity falls as alkyl branching increases (more $+I$ -donating groups destabilise the alkoxide and hinder its solvation), so the order is primary > secondary > tertiary: (i) > (ii) > (iii).

Concept and Intuition

An alcohol's acidity is governed by how well its conjugate base, the alkoxide ion $\mathrm{RO^-}$ , is stabilised. Alkyl groups are electron-donating ( $+I$ effect), so more alkyl substitution around the oxygen-bearing carbon pushes more electron density onto the already-negative oxygen, making the alkoxide less stable (higher energy) and hence the parent alcohol less willing to give up its proton. Bulkier alkoxides are also less well solvated in solution, further reducing their stability. Both effects work in the same direction: acidity falls as we go from primary to secondary to tertiary alcohols.

Step-by-Step Solution

(i) Propan-1-ol: primary alcohol, one alkyl group ( $\mathrm{CH_3CH_2-}$ ) attached to the C-OH carbon — least $+I$ donation, most stable/most accessible alkoxide, most acidic.

(ii) Butan-2-ol: secondary alcohol, two alkyl groups attached — more $+I$ donation than (i), less acidic than (i). …
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AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$Q. Assertion (A): Tertiary carbocations are more reactive than secondary and primary carbocations. Reason (R): Hyper conjugation, as well as inductive effect due to additional alkyl groups stabilize tertiary carbocations. (A) Both A and R are true and R is a correct explanation for A (B) Both A and R are true but R is not a correct explanation for A (C) A is true, R is false (D) A is false, R is true$
›Reveal solutionSolution
Hyperconjugation and inductive donation really do stabilise tertiary carbocations (Reason is true), but a more stable carbocation is less reactive, not more — so the Assertion (which claims tertiary carbocations are "more reactive") is false.

Concept and Intuition

Stability and reactivity of a reactive intermediate like a carbocation are inversely related: the more stabilised (lower energy) a carbocation is, the less driving force it has to react further, i.e., it is less reactive as an electrophile. Tertiary carbocations are the most stable of the simple alkyl carbocations because:

Hyperconjugation: more adjacent C–H (and C–C) sigma bonds can donate electron density into the empty p-orbital on the cationic carbon, spreading out the positive charge. A tertiary carbocation has more such adjacent bonds than secondary or primary.

Inductive effect: the extra alkyl groups are electron-donating (+I) relative to hydrogen, further offsetting the positive charge.

Both effects genuinely make tertiary carbocations more stable — but "more stable" is the opposite conclusion from "more reactive." A tertiary carbocation forms faster (because the transition state leading to it is lower in energy, e.g., in SN1/E1), but once formed it is comparatively unreactive (it survives longer, e.g., can even be observed under super-acid conditions) compared to a less-stabilised primary carbocation, which is so unstable it reacts almost instantly with whatever is available.

Step-by-Step Solution

Evaluate the Assertion literally: "Tertiary carbocations are more reactive than secondary and primary carbocations." Reactivity of a carbocation (as an electrophilic intermediate) scales inversely with its stability — the more stable species is less reactive. Since tertiary carbocations are the most stable, they are actually the least reactive of the three, not the most. So the Assertion is false. …
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Q.Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain. (A) CH3COOH and (B) CH3COO^-

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Q.Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.
(A) CH3COOH and (B) CH3COO^-

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