Arrange the following carbanions in order of their decreasing stability. (A) H3C-C≡C^- (B) H-C≡C^- (C) H3C-CH2

NCERT Exemplar · Q8

Q. Arrange the following carbanions in order of their decreasing stability. (A) H3C-C\equivC^- (B) H-C\equivC^- (C) H3C-CH2^- (A) A > B > C (B) B > A > C (C) C > B > A (D) C > A > B

Arunachal CbseMCQ· 1mImportance★★★★★

52% · 47/90 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 carbanions is primarily determined by the s-character of the carbon atom bearing the negative charge and the inductive effect of attached groups. Higher s-character and electron-withdrawing groups stabilise carbanions. The order of decreasing stability is H-C≡C $^-$ > H $_3$ C-C≡C $^-$ > H $_3$ C-CH $_2^-$ .

Let's understand the factors that influence the stability of carbanions. A carbanion is a species where a carbon atom carries a negative charge. For a carbanion to be stable, this negative charge must be effectively dispersed or delocalised.

The two main factors governing carbanion stability in this context are:

Hybridisation (s-character): The more s-character a hybrid orbital has, the closer the electrons in that orbital are to the nucleus. This means the carbon atom is more electronegative. A more electronegative carbon atom can better accommodate a negative charge, leading to greater stability.
- sp hybridised carbon has 50% s-character.
- sp $^2$ hybridised carbon has 33.3% s-character.
- sp $^3$ hybridised carbon has 25% s-character. Therefore, the order of electronegativity is sp > sp $^2$ > sp $^3$ . This implies that a negative charge on an sp carbon is more stable than on an sp $^2$ carbon, which is more stable than on an sp $^3$ carbon.
Inductive Effect: Alkyl groups (like -CH $_3$ , -CH $_2$ CH $_3$ ) are electron-donating groups (+I effect). They push electron density towards the negatively charged carbon. This intensifies the negative charge, making the carbanion less stable. Conversely, electron-withdrawing groups (-I effect) would stabilise a carbanion by pulling electron density away from the negatively charged carbon.

Now let's apply these concepts to the given carbanions.

Analyze Carbanion (B): H-C≡C $^-$
- The carbon atom bearing the negative charge is part of a triple bond, meaning it is sp hybridised.
- sp hybridisation provides 50% s-character, making this carbon highly electronegative and very capable of stabilising the negative charge.
- There is no significant inductive effect from the hydrogen atom.
- This carbanion is an acetylide ion.
Analyze Carbanion (A): H $_3$ C-C≡C $^-$
- The carbon atom bearing the negative charge is also part of a triple bond, so it is sp hybridised, just like in (B). This contributes significantly to its stability.
- However, there is a methyl group (H $_3$ C-) attached to the sp-hybridised carbon. Methyl groups are electron-donating (+I effect).
- This electron-donating effect pushes electron density towards the already negatively charged carbon, destabilising it compared to (B).
Analyze Carbanion (C): H $_3$ C-CH $_2^-$
- The carbon atom bearing the negative charge is part of a single bond, meaning it is sp $^3$ hybridised.
- sp $^3$ hybridisation provides only 25% s-character, making this carbon much less electronegative than the sp-hybridised carbons in (A) and (B). It is less capable of accommodating a negative charge. …

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

←Previous questionQ7 — Arrange the following hydrogen halides in order of their decreasing re…→Next questionQ9 — Arrange the following alkyl halides in decreasing order of the rate of…

Q.Arrange the following carbanions in order of their decreasing stability. (A) H3C-C≡C^- (B) H-C≡C^- (C) H3C-CH2^- (A) A > B > C (B) B > A > C (C) C > B > A (D) C > A > B

🎓Unlock everything free for 14 days

Q.Arrange the following carbanions in order of their decreasing stability.
(A) H3C-C≡C^-
(B) H-C≡C^-
(C) H3C-CH2^-
(A) A > B > C
(B) B > A > C
(C) C > B > A
(D) C > A > B

Unlock everything free for 14 days