Arrange the following compounds in increasing order of their property as indicated: (i) Acetaldehyde, Acetone,

Exercises · 8.12

Q. Arrange the following compounds in increasing order of their property as indicated:

(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii)

CH_3CH_2CH(Br)COOH

CH_3CH(Br)CH_2COOH

(CH_3)_2CHCOOH

CH_3CH_2CH_2COOH

(acid strength)

(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

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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 reactivity toward HCN depends on steric hindrance around the carbonyl carbon, while acid strength depends on the inductive effect of substituents. For (i) the order is Di‑tert‑butyl ketone < Methyl tert‑butyl ketone < Acetone < Acetaldehyde; for (ii) the order is $(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$ ; for (iii) the order is 4‑Methoxybenzoic acid < Benzoic acid < 4‑Nitrobenzoic acid < 3,4‑Dinitrobenzoic acid.

The Concept: Inductive Effect and Steric Hindrance

Before we touch a single compound, understand the two forces at play.

Inductive effect is the permanent shift of sigma electrons along a carbon chain due to a difference in electronegativity. An electron‑withdrawing group (EWG) like –NO₂ or –Br pulls electron density toward itself, creating a partial positive charge on the carbon to which it is attached. This makes a nearby carboxylic acid group more willing to lose its proton — stronger acid. An electron‑releasing group (ERG) like –CH₃ or –OCH₃ pushes electron density away, making the acid weaker.

Steric hindrance is the physical blocking of a reaction site by bulky groups. For nucleophilic addition to a carbonyl (like HCN attack), the nucleophile must approach the carbonyl carbon. If that carbon is surrounded by large groups, the approach is hindered and reactivity drops.

Now let’s apply these to each part.

(i) Reactivity towards HCN

HCN adds to the carbonyl group via nucleophilic addition. The rate depends on how easily the nucleophile (CN⁻) can reach the carbonyl carbon. Two factors matter: the electrophilicity of the carbonyl carbon (more positive = faster) and steric hindrance around it.

Acetaldehyde has one small –CH₃ group and one –H. The carbonyl carbon is quite exposed and also fairly electrophilic because the –H doesn’t donate electrons. Acetone has two –CH₃ groups — more steric hindrance and also more electron donation (methyl is weakly electron‑releasing), so it reacts slower than acetaldehyde.

Methyl tert‑butyl ketone has one –CH₃ and one –C(CH₃)₃. The tert‑butyl group is huge — it blocks one side of the carbonyl. Di‑tert‑butyl ketone has two such giants. The carbonyl carbon is almost buried. Reactivity plummets.

Watch out

A common mistake is to think that more alkyl groups always increase reactivity because they donate electrons. But here, steric hindrance dominates. Even though alkyl groups are weak electron donors (which would slightly reduce electrophilicity), the physical blocking is far more important.

So the order from least reactive to most reactive is:

Di‑tert‑butyl ketone (most hindered)
Methyl tert‑butyl ketone
Acetone
Acetaldehyde (least hindered, most reactive)

(ii) Acid strength of substituted butanoic acids

Acid strength is measured by the stability of the conjugate base (the carboxylate anion). An electron‑withdrawing group stabilises the negative charge by pulling electron density toward itself. The closer the EWG is to the –COOH group, the stronger its effect.

Let’s list the compounds:

$CH_3CH_2CH_2COOH$ — butanoic acid, no substituent. This is our reference.
$(CH_3)_2CHCOOH$ — isobutyric acid. Two methyl groups at the α‑carbon. Methyl is weakly electron‑releasing, so this acid is weaker than butanoic acid.
$CH_3CH(Br)CH_2COOH$ — bromine is at the β‑carbon (one carbon away from –COOH). The –I effect of Br is felt, but attenuated by distance.
$CH_3CH_2CH(Br)COOH$ — bromine is at the α‑carbon (directly attached to the carbon bearing –COOH). This gives the strongest electron‑withdrawing effect.

Tip

The inductive effect drops off rapidly with distance. An α‑substituent has about twice the effect of a β‑substituent. So α‑bromo acid is significantly stronger than β‑bromo acid. …

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Method: Inductive Effect Analysis

This method uses the inductive effect (electron-withdrawing or electron-donating through sigma bonds) to predict relative reactivity or acidity.

(i) Reactivity towards HCN — Nucleophilic Addition Reactivity

Method: Compare steric hindrance and electron density at the carbonyl carbon.

Steps:

Identify the reaction: HCN adds to the carbonyl group ( $\ce{C=O}$ ) via nucleophilic attack. Reactivity depends on:
- Electrophilicity of carbonyl carbon (more positive = more reactive)
- Steric hindrance around carbonyl (bulkier groups slow attack)
Analyze each compound:
- Acetaldehyde ( $\ce{CH3CHO}$ ): Least steric hindrance; one small alkyl group.
- Acetone ( $\ce{CH3COCH3}$ ): Two methyl groups — moderate hindrance.
- Methyl tert-butyl ketone ( $\ce{CH3COC(CH3)3}$ ): One very bulky group — high hindrance.
- Di-tert-butyl ketone ( $\ce{(CH3)3CCOC(CH3)3}$ ): Two extremely bulky groups — maximum hindrance.
Order by increasing reactivity:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

Key concept: Steric hindrance dominates over inductive effects here — bulkier groups block nucleophilic attack.

(ii) Acid Strength — Inductive Effect of Halogen

Method: Compare the distance of electron-withdrawing Br from the carboxyl group.

Steps:

Recall: Acid strength increases with electron-withdrawing groups (EWG) near the $-\ce{COOH}$ group. The closer the EWG, the stronger the acid.
Analyze each compound:
- $\ce{CH3CH2CH(Br)COOH}$ : Br on $\alpha$ -carbon (directly attached to carboxyl carbon) — strongest inductive withdrawal.
- $\ce{CH3CH(Br)CH2COOH}$ : Br on $\beta$ -carbon — weaker effect (distance reduces inductive pull).
- $\ce{CH3CH2CH2COOH}$ : No Br — weakest acid (only alkyl groups, which are electron-donating).
- $(\ce{CH3})2\ce{CHCOOH}$ : Two methyl groups (electron-donating) — makes it weaker than straight-chain butanoic acid.
Order by increasing acid strength:

$(\ce{CH3})2\ce{CHCOOH} < \ce{CH3CH2CH2COOH} < \ce{CH3CH(Br)CH2COOH} < \ce{CH3CH2CH(Br)COOH}$

Key concept: Inductive effect decreases rapidly with distance — $\alpha$ -halo acids are much stronger than $\beta$ -halo acids.

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(i) Reactivity towards HCN: Acetaldehyde, Acetone, Di‑tert‑butyl ketone, Methyl tert‑butyl ketone

✗ Mistake 1: Ignoring steric hindrance and only considering +I effect of alkyl groups

Why it’s wrong: Students often think more alkyl groups = more +I effect = less electrophilic carbonyl = lower reactivity. That’s partially correct, but steric bulk around the carbonyl carbon matters even more for nucleophilic addition (like HCN).
How to avoid: Remember: HCN addition requires the nucleophile to approach the carbonyl carbon. Bulky groups block this approach. So steric hindrance dominates over electronic effects here.

✗ Mistake 2: Forgetting that aldehydes are more reactive than ketones

Why it’s wrong: Acetaldehyde (an aldehyde) has one alkyl group; ketones have two. Aldehydes are always more reactive towards nucleophiles than ketones (less steric hindrance, more electrophilic carbonyl).
How to avoid: Always rank aldehydes before ketones in nucleophilic addition reactivity.

✓ Correct order (increasing reactivity):

Di‑tert‑butyl ketone < Methyl tert‑butyl ketone < Acetone < Acetaldehyde

Reason:

Di‑tert‑butyl ketone: two huge tert‑butyl groups → extreme steric hindrance → least reactive.
Methyl tert‑butyl ketone: one tert‑butyl + one methyl → still bulky.
Acetone: two methyl groups → moderate hindrance.
Acetaldehyde: one small H + one methyl → least hindered, most reactive.

(ii) Acid strength: $CH_3CH_2CH(Br)COOH$ , $CH_3CH(Br)CH_2COOH$ , $(CH_3)_2CHCOOH$ , $CH_3CH_2CH_2COOH$

✗ Mistake 1: Thinking that more Br atoms = stronger acid, regardless of position

Why it’s wrong: The inductive effect of Br is distance‑dependent. The closer Br is to the –COOH group, the stronger the acid. A Br on the $\alpha$ ‑carbon (adjacent to COOH) stabilises the conjugate base much more than a Br on the $\beta$ ‑carbon.
How to avoid: Always check the distance of the electron‑withdrawing group from the carboxyl group. Use the rule: closer = stronger acid.

✗ Mistake 2: Forgetting that alkyl groups are electron‑donating (+I)

Why it’s wrong: $(CH_3)_2CHCOOH$ has two methyl groups donating electrons, making it weaker than the straight‑chain $CH_3CH_2CH_2COOH$ .
How to avoid: Compare the number and position of alkyl groups. More branching near COOH = weaker acid (due to +I effect).

✓ Correct order (increasing acid strength):

$(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$

Reason:

$(CH_3)_2CHCOOH$ : two +I methyl groups → weakest.
$CH_3CH_2CH_2COOH$ : no Br, one alkyl chain → slightly stronger.
$CH_3CH(Br)CH_2COOH$ : Br on $\beta$ ‑carbon → weak inductive effect.
$CH_3CH_2CH(Br)COOH$ : Br on $\alpha$ ‑carbon → strong inductive effect → strongest.

(iii) Acid strength: Benzoic acid, 4‑Nitrobenzoic acid, 3,4‑Dinitrobenzoic acid, 4‑Methoxybenzoic acid

✗ Mistake 1: Confusing electron‑withdrawing and electron‑donating groups

Why it’s wrong: –NO $_2$ is strongly electron‑withdrawing (‑I and ‑M), so it increases acid strength. –OCH $_3$ is electron‑donating (+M, weak ‑I), so it decreases acid strength. Students often reverse these.
How to avoid: Memorise common groups:
- Withdrawing (increase acidity): –NO $_2$ , –CN, –COOH, –F, –Cl, –Br, –I …

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Q.Arrange the following compounds in increasing order of their property as indicated:

The Concept: Inductive Effect and Steric Hindrance

(i) Reactivity towards HCN

(ii) Acid strength of substituted butanoic acids

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