Identify the most stable species in the following set of ions giving reasons: (i) CH3^+, CH2Br^+, CHBr2^+, CBr

NCERT Exemplar · Q43

Q. Identify the most stable species in the following set of ions giving reasons:

(i) CH3^+, CH2Br^+, CHBr2^+, CBr3^+

(ii) CH3^-, CH2Cl^-, CHCl2^-, CCl3^-

Sikkim CbseShort· 3mImportance★★★★★

82% · 107/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

--- …

Start your 14-day free trial to unlock the full solution →

The stability of carbocations and carbanions is governed by the inductive effect of substituents. For carbocations, electron-donating groups (like alkyl) stabilise, while electron-withdrawing groups (like halogens) destabilise. For carbanions, the reverse is true. The most stable species are: (i) CH₃⁺ and (ii) CCl₃⁻.

The Concept: Inductive Effect and Charge Stability

The key to this problem lies in understanding how substituents influence the stability of a charged carbon atom through the inductive effect. This is the permanent polarisation of a sigma bond due to electronegativity differences. A substituent can either push electron density toward the charged carbon (electron-donating) or pull it away (electron-withdrawing).

For a carbocation (positively charged carbon), stability increases when electron density is donated to it. This neutralises the positive charge, making the ion less reactive and more stable. Electron-donating groups (EDGs) like alkyl groups ( $-CH_3$ ) are stabilising. Electron-withdrawing groups (EWGs) like halogens ( $-Br$ , $-Cl$ ) are destabilising because they pull electron density away, intensifying the positive charge.

For a carbanion (negatively charged carbon), the opposite is true. Stability increases when electron density is withdrawn from it. This disperses the negative charge, making the ion less reactive. Electron-withdrawing groups (EWGs) are stabilising, while electron-donating groups (EDGs) are destabilising because they concentrate the negative charge.

Watch out

A common mistake is to apply the same logic to both carbocations and carbanions. Remember: a positive charge wants electrons, a negative charge wants to get rid of them. The effect of a substituent is exactly opposite in the two cases.

Now, let's apply this to each series.

(i) Carbocations: $\ce{CH3+}$ , $\ce{CH2Br+}$ , $\ce{CHBr2+}$ , $\ce{CBr3+}$

Identify the substituent effect. Bromine is more electronegative than carbon. Therefore, each $-Br$ group acts as an electron-withdrawing group via the inductive effect. It pulls electron density away from the positively charged carbon.
Analyze the series. As we replace hydrogen atoms (which have a negligible inductive effect) with bromine atoms, we are adding more electron-withdrawing groups to the carbocation centre.
- $\ce{CH3+}$ : No Br atoms. The positive charge is only stabilised by the three H atoms (weak +I effect).
- $\ce{CH2Br+}$ : One Br atom pulls electron density away, making the positive charge more intense and less stable.
- $\ce{CHBr2+}$ : Two Br atoms pull even more electron density away, further destabilising the ion.
- $\ce{CBr3+}$ : Three Br atoms exert the strongest electron-withdrawing pull, making this the most unstable carbocation in the series.
Conclusion. The stability decreases as the number of bromine atoms increases. The most stable species is the one with the fewest electron-withdrawing groups. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ42 — Draw the resonance structures of the following compounds: (i) CH2=CH—C…→Next questionQ44 — Give three points of differences between inductive effect and resonanc…

Q.Identify the most stable species in the following set of ions giving reasons:

The Concept: Inductive Effect and Charge Stability

(i) Carbocations: CHX3X+\ce{CH3+}CHX3​X+, CHX2BrX+\ce{CH2Br+}CHX2​BrX+, CHBrX2X+\ce{CHBr2+}CHBrX2​X+, CBrX3X+\ce{CBr3+}CBrX3​X+

🎓Unlock everything free for 14 days

(i) Carbocations: $\ce{CH3+}$ , $\ce{CH2Br+}$ , $\ce{CHBr2+}$ , $\ce{CBr3+}$

Unlock everything free for 14 days