Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?

Exercises · 8.9

Q. Which of the two: O2NCH2CH2O- or CH3CH2O- is expected to be more stable and why?

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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 key idea is that the nitro group ( $-NO_2$ ) exerts a strong electron-withdrawing inductive effect ( $-I$ ), which stabilises the negative charge on the oxygen in $O_2NCH_2CH_2O^-$ by pulling electron density away. In contrast, the ethyl group ( $-CH_2CH_3$ ) in $CH_3CH_2O^-$ is electron-donating ( $+I$ ), which destabilises the negative charge. Therefore, $O_2NCH_2CH_2O^-$ is more stable.

Why this approach works: The Inductive Effect

When you have a negatively charged oxygen (an alkoxide ion), the stability of that ion depends on how well the negative charge is dispersed or delocalised. A charge that is concentrated on one atom is less stable than one that is spread out.

The inductive effect is the transmission of charge through a chain of atoms via sigma bonds. An electron-withdrawing group ( $-I$ effect) pulls electron density towards itself, away from the negative charge. This reduces the charge density on the oxygen, making the ion more stable. An electron-donating group ( $+I$ effect) pushes electron density towards the negative charge, making it even more concentrated and less stable.

Here, we compare two alkoxide ions. The only difference is the group attached to the two-carbon chain: a nitro group ( $-NO_2$ ) versus an ethyl group ( $-CH_2CH_3$ ).

Step-by-step reasoning

Identify the groups attached to the negative charge.

In $O_2NCH_2CH_2O^-$ , the group attached to the $-CH_2CH_2O^-$ part is $-NO_2$ .

In $CH_3CH_2O^-$ , the group attached to the $-O^-$ is $-CH_2CH_3$ .
Determine the inductive effect of each group.
- The nitro group ( $-NO_2$ ) is a classic strong electron-withdrawing group ( $-I$ effect). The highly electronegative nitrogen and oxygen atoms pull electron density towards themselves through the sigma bonds.
- The ethyl group ( $-CH_2CH_3$ ) is an electron-donating group ( $+I$ effect). Alkyl groups are generally electron-donating compared to hydrogen, pushing electron density away from themselves.
Trace the effect on the negative charge.
- For $O_2NCH_2CH_2O^-$ : The $-I$ effect of the $-NO_2$ group pulls electron density away from the negatively charged oxygen. This reduces the localised negative charge on the oxygen, making the ion more stable.
- For $CH_3CH_2O^-$ : The $+I$ effect of the $-CH_2CH_3$ group pushes electron density towards the negatively charged oxygen. This increases the localised negative charge, making the ion less stable.
Consider the distance of the effect. …

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

CBSE 2026Set 56/1/11 markMCQ
$pK_b$
›Reveal solutionSolution
The key idea is that $pK_b$ is inversely related to basicity; electron-donating groups increase basicity (lower $pK_b$ ), while electron-withdrawing groups decrease it (higher $pK_b$ ). Among the given aniline derivatives, N,N-dimethylaniline is the most basic due to the +I effect of two methyl groups, so it has the lowest $pK_b$ . The correct option is (A).

Why $pK_b$ and Basicity Are Inversely Related

The $pK_b$ of an amine is the negative logarithm of its base dissociation constant $K_b$ :
$pK_b = -\log K_b$
A lower $pK_b$ means a larger $K_b$ , which means the amine is a stronger base — it more readily accepts a proton. So the question "which has the lowest $pK_b$ ?" is really asking: "which is the strongest base among these?"

The Inductive Effect: The Engine of Basicity in Anilines

In aniline ( $C_6H_5NH_2$ ), the lone pair on nitrogen is partially delocalised into the benzene ring through resonance. This makes aniline a weaker base than aliphatic amines. Now, when we attach substituents to the ring or to the nitrogen, we alter this basicity through the inductive effect (and also resonance, but here the key difference is inductive).
$+I$
Electron-donating groups (like $CH_3$ ) push electron density toward the nitrogen, making the lone pair more available for protonation — increasing basicity, hence decreasing $pK_b$ .

Electron-withdrawing groups (like $NO_2$ ) pull electron density away, decreasing basicity, hence increasing $pK_b$ .

Step-by-Step Comparison

Identify the parent structure — All four compounds are derivatives of aniline ( $C_6H_5NH_2$ ). The nitrogen's lone pair is the basic site.

Compare the substituents on nitrogen — Options (A), (B), and (C) differ only in how many methyl groups are attached to the nitrogen:

(C) $C_6H_5NH_2$ — no methyl groups (just H)

(B) $C_6H_5NH(CH_3)$ — one methyl group

(A) $C_6H_5N(CH_3)_2$ — two methyl groups

Each methyl group has a +I (inductive) effect: it pushes electron density toward the nitrogen. More methyl groups → more electron density on N → stronger base → lower $pK_b$ .

So the order of basicity among these three is:

$(A) > (B) > (C)$
And therefore the order of $pK_b$ is:

$(A) < (B) < (C)$ …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. -I effect is shown by :$
(a) -Cl
(b) -Br
(c) both
(a) and
(b)
(d) -CH3
›Reveal solutionSolution

-Cl and -Br are both more electronegative than carbon, so both withdraw electron density through the sigma-bond framework and show a -I (negative inductive) effect; -CH3 is electron-donating (+I), not -I.

The inductive effect is the polarisation of the sigma-bond electron cloud along a chain, caused by the electronegativity difference between an atom/group and carbon. Groups more electronegative than carbon (such as the halogens -F, -Cl, -Br, -I, and groups like -NO2, -CN, -OH) pull electron density towards themselves through the sigma bonds -- this is called the -I effect (electron-withdrawing).

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The correct relative order of +I effect of alkyl groups is:$
(a) -C(CH3)3 > -CH(CH3)2 > -CH2CH3 > -CH3
(b) -CH3 > -CH2CH3 > -CH(CH3)2 > -C(CH3)3
(c) -CH2CH3 > -CH3 > -C(CH3)3 > -CH(CH3)2
(d) -CH(CH3)2 > -C(CH3)3 > -CH2CH3 > -CH3
›Reveal solutionSolution
+I effect (electron release through sigma bonds) grows with the number of alkyl substituents on the carbon attached to the chain, so tert-butyl > isopropyl > ethyl > methyl.

The +I effect measures how strongly a group pushes electron density towards the rest of the molecule through sigma bonds. For alkyl groups, this arises because each additional alkyl substituent on the attached carbon supplies more hyperconjugative/inductive electron density than a hydrogen would.

Comparing the four groups by how many carbon substituents sit on the first carbon:

-CH3 (methyl): 0 additional carbons on that carbon — weakest +I effect.

-CH2CH3 (ethyl): 1 additional carbon. …
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Q.Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?

Why this approach works: The Inductive Effect

Step-by-step reasoning

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