Arrange the following compounds in the increasing order of their property indicated : (Any two) (i) Acetaldehy

Question

Q. Arrange the following compounds in the increasing order of their property indicated : (Any two)

(i) Acetaldehyde, Benzaldehyde, Acetophenone, Acetone (Reactivity towards HCN)

(ii)

(CH_3)_2CHCOOH

CH_3CH_2CH(Br)COOH

CH_3CH(Br)CH_2COOH

(Acidic strength)

(iii)

CH_3CH_2OH

CH_3CHO

CH_3COOH

(Boiling point)

CBSECBSE Class XII Board 2022Subjective· 2mImportance★★★★★

✓ Free question

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.

Watch out

Do not confuse inductive effect with resonance effect. Inductive effect operates through sigma bonds and is distance-dependent. Resonance effect operates through pi bonds and can act over long distances. For example, $-\text{NO}_2$ is both strongly electron-withdrawing inductively (through sigma bonds) and by resonance (through pi bonds). But $-\text{Cl}$ is electron-withdrawing inductively but electron-donating by resonance — the net effect on acidity depends on which dominates.

The Key Takeaway

Inductive effect on acidity: Electron-withdrawing groups (EWGs) increase acidity by stabilising the conjugate base through sigma-bond polarisation. Electron-donating groups (EDGs) decrease acidity. The effect is strongest when the group is closest to the acidic site and diminishes with distance.

\text{Acidity} \propto \text{Number and strength of EWGs near acidic site}

The inductive effect on acidity is a recurring theme in the CBSE Class 11 Chemistry syllabus (organic chemistry fundamentals) and is tested repeatedly in board exams, JEE Main and NEET through questions ranking carboxylic acids or comparing substituted acetic acids by pKa. Searches for "inductive effect definition class 11 chemistry" or "inductive effect vs resonance effect" usually land on exactly this distance-dependent, sigma-bond explanation.

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.

5. The "Why" Behind Common Exam Patterns

Example: Acidity of carboxylic acids

Question: Why is $\text{Cl-CH}_2\text{COOH}$ more acidic than $\text{CH}_3\text{COOH}$ ?

Reasoning:

$-\text{Cl}$ is electronegative → withdraws electron density through $\sigma$ bonds.
This stabilizes the conjugate base ( $\text{Cl-CH}_2\text{COO}^-$ ) by dispersing the negative charge.
Stabilized conjugate base → lower energy → equilibrium shifts toward dissociation.
Hence, higher $K_a$ (stronger acid).

Key formula:

$K_a(\text{Cl-CH}_2\text{COOH}) > K_a(\text{CH}_3\text{COOH})$

Example: Basicity of amines

Question: Why is $\text{NH}_3$ more basic than $\text{CF}_3\text{NH}_2$ ?

Reasoning:

$-\text{CF}_3$ is strongly electron-withdrawing (inductive).
It pulls electron density away from the nitrogen lone pair.
The lone pair is less available for protonation.
Hence, lower basicity.

Summary: The Core Logic

Step	What Happens	Why
1	Electronegativity difference	Different nuclear charges attract electrons unequally
2	Bond polarization	$\sigma$ electrons shift toward more electronegative atom
3	Transmission along chain	Partial charges polarize adjacent bonds
4	Attenuation with distance	Coulombic force decreases as $1/r^2$
5	Observable effect	Changes in acidity, basicity, reactivity

Final takeaway: The inductive effect is electrostatic in origin — it's about how charge imbalances propagate through the sigma framework of a molecule. The formulas and constants are just tools to quantify this fundamental physical phenomenon.

(i) Reactivity towards HCN

Concept: Nucleophilic addition to carbonyl compounds depends on electrophilicity of the carbonyl carbon and steric hindrance around it.

Reasoning:

Electron-donating groups (alkyl, aryl) decrease carbonyl electrophilicity by +I effect; electron-withdrawing groups increase it.
Bulkier substituents hinder nucleophilic attack.
Acetaldehyde ( $CH_3CHO$ ): one alkyl group, least hindered → most reactive.
Acetone ( $(CH_3)_2CO$ ): two alkyl groups, more +I effect and steric hindrance than acetaldehyde.
Benzaldehyde ( $C_6H_5CHO$ ): aryl group donates electrons by resonance but is bulkier than methyl.
Acetophenone ( $C_6H_5COCH_3$ ): both aryl and alkyl groups, maximum +I/resonance effect and steric crowding → least reactive.

✓Final answer

The increasing order of reactivity is: Acetophenone < Acetone < Benzaldehyde < Acetaldehyde .

(ii) Acidic Strength

Concept: Acidity of carboxylic acids increases when electron-withdrawing groups stabilize the carboxylate anion; the effect weakens with distance.

Reasoning:

Bromine is electron-withdrawing (−I effect), stabilizing $COO^-$ and increasing acidity.
$CH_3CH_2CH(Br)COOH$ : Br is on the α-carbon (closest to $COOH$ ) → strongest −I effect → most acidic.
$CH_3CH(Br)CH_2COOH$ : Br is on the β-carbon → weaker −I effect → intermediate acidity.
$(CH_3)_2CHCOOH$ : no Br, only alkyl groups with +I effect → least acidic.

✓Final answer

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

(iii) Boiling Point

Concept: Boiling point depends on the strength of intermolecular forces—hydrogen bonding dominates over dipole-dipole and van der Waals forces.

Reasoning:

$CH_3COOH$ forms strong dimeric hydrogen bonds (two H-bonds per dimer) → highest boiling point.
$CH_3CH_2OH$ forms hydrogen bonds but weaker than carboxylic acid dimers.
$CH_3CHO$ has dipole-dipole interactions only (no H-bonding from carbonyl) → lowest boiling point.

✓Final answer

CH_3CHO < CH_3CH_2OH < CH_3COOH

(i) Nucleophilic addition to carbonyls is governed by steric hindrance and electronic effects; electron-withdrawing groups and smaller substituents accelerate attack. Acetophenone < Acetone < Benzaldehyde < Acetaldehyde. (ii) Acidic strength increases when electron-withdrawing groups stabilize the carboxylate anion; proximity matters. $(CH_3)_2CHCOOH$ < $CH_3CH(Br)CH_2COOH$ < $CH_3CH_2CH(Br)COOH$ .

(i) Reactivity towards HCN

The addition of HCN to a carbonyl is a nucleophilic attack by cyanide ion ( $CN^-$ ) on the electrophilic carbonyl carbon. Two factors control how fast this happens: steric hindrance around the carbonyl and the electronic nature of the substituents.

A carbonyl carbon bearing bulky groups is harder to approach. Electron-donating groups (alkyl, aryl) push electron density onto the carbonyl carbon, making it less electrophilic and less reactive. Electron-withdrawing groups do the opposite.

1. Acetaldehyde ( $CH_3CHO$ )

One small methyl group and one hydrogen on the carbonyl. Minimal steric crowding, and the methyl is only weakly electron-donating. This is the most reactive.

2. Benzaldehyde ( $C_6H_5CHO$ )

One phenyl group and one hydrogen. The phenyl ring is bulkier than a methyl, but it also withdraws electrons by resonance (the carbonyl oxygen can conjugate with the ring, pulling electron density away from the carbonyl carbon). The net effect is moderate reactivity, less than acetaldehyde but more than ketones.

3. Acetone ( $(CH_3)_2CO$ )

Two methyl groups. Both donate electrons inductively, reducing the electrophilicity of the carbonyl. Steric hindrance is moderate. Reactivity drops.

4. Acetophenone ( $C_6H_5COCH_3$ )

One phenyl and one methyl. The phenyl provides both steric bulk and resonance stabilization of the carbonyl (the lone pair on oxygen delocalizes into the aromatic ring, making the carbonyl less electrophilic). This is the least reactive.

Tip

Aldehydes are almost always more reactive than ketones toward nucleophiles: they have one less alkyl group, so less steric hindrance and less electron donation.

Increasing order of reactivity:

\text{Acetophenone} < \text{Acetone} < \text{Benzaldehyde} < \text{Acetaldehyde}

(ii) Acidic strength

Acidity of a carboxylic acid depends on how stable the conjugate base (the carboxylate anion, $RCOO^-$ ) is. Electron-withdrawing groups stabilize the negative charge by pulling electron density away; the closer they are to the carboxyl group, the stronger the effect.

1. $(CH_3)_2CHCOOH$ (2-methylpropanoic acid)

The isopropyl group is electron-donating (three alkyl branches). It destabilizes the carboxylate anion by pushing electron density toward the negative charge. This is the weakest acid.

2. $CH_3CH(Br)CH_2COOH$ (3-bromobutanoic acid)

Bromine is electron-withdrawing (high electronegativity), but it sits on the $\beta$ -carbon, two bonds away from the carboxyl group. The inductive effect weakens with distance. Acidity increases relative to the first compound, but not dramatically.

3. $CH_3CH_2CH(Br)COOH$ (2-bromobutanoic acid)

Bromine is now on the $\alpha$ -carbon, directly adjacent to the carboxyl group. The inductive withdrawal is much stronger, stabilizing the carboxylate anion significantly. This is the strongest acid.

Watch out

A common mistake is to ignore the position of the substituent. An electron-withdrawing group on the $\gamma$ - or $\delta$ -carbon has almost no effect on acidity; proximity is everything.

Increasing order of acidic strength:

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

(iii) Boiling point

Boiling point is determined by the strength of intermolecular forces. For small organic molecules, hydrogen bonding dominates when it is present. The more extensive the hydrogen bonding network, the higher the boiling point.

1. $CH_3CHO$ (Acetaldehyde)

An aldehyde has a polar carbonyl group, so dipole–dipole interactions exist. But there is no O–H or N–H bond, so no hydrogen bonding. Boiling point is the lowest of the three.

2. $CH_3CH_2OH$ (Ethanol)

Alcohols form hydrogen bonds: the O–H can donate a hydrogen bond, and the oxygen can accept one. Each molecule can participate in multiple H-bonds, creating a network. Boiling point is significantly higher than acetaldehyde.

3. $CH_3COOH$ (Acetic acid)

Carboxylic acids form dimers in the liquid phase: two molecules associate via a pair of hydrogen bonds (each carbonyl oxygen accepts a bond from the other molecule's O–H). This dimer is exceptionally stable and effectively doubles the molecular weight being vaporized. Acetic acid has the highest boiling point.

Compound	Intermolecular force	Boiling point (°C)
$CH_3CHO$	Dipole–dipole	20
$CH_3CH_2OH$	Hydrogen bonding	78
$CH_3COOH$	Dimer (double H-bond)	118

Increasing order of boiling point:

CH_3CHO < CH_3CH_2OH < CH_3COOH

✓Final answer

(CH_3)_2CHCOOH

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:
$(A) < (B) < (C)$

Now consider option (D) — 4-nitroaniline has a nitro group ( $NO_2$ ) attached para to the $NH_2$ group. The nitro group is a powerful electron-withdrawing group (both by -I and -M effects). It pulls electron density away from the nitrogen through the ring, making the lone pair much less available for protonation. This makes 4-nitroaniline a much weaker base than aniline itself.

Watch out

A common mistake is to think that any substituent on the ring increases basicity. The nitro group is strongly electron-withdrawing — it decreases basicity dramatically. 4-nitroaniline has a $pK_b$ around 13–14, while aniline itself is around 9.4.

Rank all four — Combining the inductive effect on nitrogen and the withdrawing effect on the ring:

Strongest base (lowest $pK_b$ ): N,N-dimethylaniline (A) — two +I methyl groups

Next: N-methylaniline (B) — one +I methyl group

Next: aniline (C) — no alkyl groups

Weakest base (highest $pK_b$ ): 4-nitroaniline (D) — strong -I/-M nitro group

Tip

You don't need to memorise exact $pK_b$ values. Just remember: alkyl groups on nitrogen increase basicity (lower $pK_b$ ), and electron-withdrawing groups on the ring decrease basicity (raise $pK_b$ ). The compound with the most alkyl groups on N and no withdrawing groups on the ring will have the lowest $pK_b$ .

✓Final answer
$pK_b$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Arrange the following in the increasing order of their boiling points : i. Pentane ii. 2-methylbutane iii. 2, 2-dimethylpropane.$
(a) A) iii < ii < i
(b) B) i < ii < iii
(c) C) ii < iii < i
(d) D) ii < i < iii
›Reveal solutionSolution
[!TLDR]

A) iii < ii < i

Why

Boiling point decreases with increasing branching (less surface area, weaker van der Waals forces); 2,2-dimethylpropane (most branched) has the lowest boiling point and pentane (unbranched) the highest, so increasing order is iii < ii < i.

[!ANSWER]

A) iii < ii < i
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).

Both chlorine and bromine are significantly more electronegative than carbon, so both -Cl and -Br withdraw electron density from the carbon chain and show a -I effect.

In contrast, alkyl groups such as -CH3 are relatively electron-releasing (due to hyperconjugation/inductive donation) compared with the carbon they are attached to, so they show a +I (positive inductive, electron-donating) effect, not -I.

✓Final answer
$The correct option is (c) both (a) and (b).$
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.

-CH(CH3)2 (isopropyl): 2 additional carbons.

-C(CH3)3 (tert-butyl): 3 additional carbons — strongest +I effect.

So the +I effect strengthens as: CH3 < CH2CH3 < CH(CH3)2 < C(CH3)3, i.e., in decreasing order: C(CH3)3 > CH(CH3)2 > CH2CH3 > CH3.

✓Final answer
$The answer is (a): -C(CH3)3 > -CH(CH3)2 > -CH2CH3 > -CH3.$

Q.Arrange the following compounds in the increasing order of their property indicated : (Any two)

(i) Reactivity towards HCN

(ii) Acidic strength

(iii) Boiling point

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