Write reasons for the following statements : (i) Benzoic acid does not undergo Friedel-Crafts reaction. (ii) O

Question

Q. Write reasons for the following statements :

(i) Benzoic acid does not undergo Friedel-Crafts reaction.

(ii) Oxidation of aldehydes is easier than that of ketones.

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.

The key idea is that the carboxyl group in benzoic acid is strongly deactivating (meta-directing, electron-withdrawing), which prevents Friedel-Crafts alkylation/acylation; and aldehydes have a less hindered, more reactive carbonyl carbon with a hydrogen atom that can be easily abstracted, making oxidation much easier than for ketones.

Why Benzoic Acid Does Not Undergo Friedel-Crafts Reaction

The Concept: Inductive Effect and Ring Deactivation

Friedel-Crafts reactions (both alkylation and acylation) require the aromatic ring to be electron-rich enough to attack the electrophile (a carbocation or an acylium ion). The catalyst, typically anhydrous AlCl₃, generates this strong electrophile.

Now look at benzoic acid: the carboxyl group ( $-COOH$ ) is a strong electron-withdrawing group via the inductive effect (and also resonance, but the inductive effect dominates here). The oxygen atoms are highly electronegative, pulling electron density away from the ring through the sigma bonds. This makes the benzene ring electron-deficient — it becomes a poor nucleophile.

The ring is too deactivated.

The electron density on the ring is so low that it cannot attack the electrophile generated by AlCl₃. Even if a tiny amount of product formed, the reaction would be extremely slow and impractical.
The carboxyl group coordinates with AlCl₃.

This is the real killer. The Lewis acid AlCl₃ has a strong affinity for the lone pairs on the oxygen atoms of the carboxyl group. It forms a stable complex with the $-COOH$ group. This does two things:
- It consumes the catalyst — AlCl₃ is no longer free to generate the electrophile.
- It makes the carboxyl group even more electron-withdrawing (now it's a positively charged complex), further deactivating the ring.
The complex is bulky and blocks the ortho positions.

The AlCl₃–carboxyl complex is large and sterically hinders approach of the electrophile to the ortho positions. The meta positions are also deactivated.

Watch out

A common mistake is to think that because benzene itself undergoes Friedel-Crafts, any substituted benzene will too. But strongly deactivating groups (like $-NO_2$ , $-CN$ , $-COOH$ , $-SO_3H$ ) prevent Friedel-Crafts reactions entirely. The reaction simply does not proceed under normal conditions.

Tip

If you ever need to introduce an alkyl or acyl group onto a benzoic acid derivative, you must first protect the carboxyl group (e.g., convert it to an ester or acid chloride), then do the Friedel-Crafts, and finally deprotect. But even then, the ester group is still deactivating, so yields are poor.

Why Oxidation of Aldehydes is Easier Than That of Ketones

The Concept: The Carbonyl Carbon's Reactivity and the Hydrogen Factor

Both aldehydes and ketones contain the carbonyl group ( $C=O$ ), but the key difference is that aldehydes have at least one hydrogen atom directly attached to the carbonyl carbon, while ketones have two carbon groups attached.

Steric hindrance is lower in aldehydes.

In an aldehyde, the carbonyl carbon is bonded to one small hydrogen atom and one R group. In a ketone, it is bonded to two R groups (which are bulkier). An oxidizing agent (like $KMnO_4$ , $K_2Cr_2O_7$ , or Tollens' reagent) must approach and attack this carbon. The less hindered aldehyde carbon is much more accessible.
The aldehyde hydrogen is easily abstracted.

This is the most important reason. In the oxidation mechanism (e.g., with chromic acid or Tollens' reagent), the first step is often the removal of the aldehyde hydrogen as a hydride ion ( $H^-$ ) or as a proton, depending on the reagent. This hydrogen is unique to aldehydes — ketones don't have it. Once that hydrogen is gone, the carbonyl carbon becomes a carbocation-like species that readily forms a carboxylic acid.
The intermediate is more stable for aldehydes.

For aldehydes, oxidation leads to a hydrate (gem-diol) intermediate in many mechanisms (e.g., with $CrO_3$ ). This hydrate is relatively stable and easily formed. For ketones, the hydrate is much less stable (due to steric and electronic reasons) and does not form readily under mild conditions.
Ketones require harsh conditions.

Ketones can be oxidized, but only by strong oxidizing agents (like hot, concentrated $KMnO_4$ or $HNO_3$ ) that break carbon-carbon bonds. This is a destructive oxidation — it cleaves the molecule into smaller fragments (e.g., a mixture of carboxylic acids). This is not a clean, selective reaction like aldehyde oxidation.

RCHO \xrightarrow{[O]} RCOOH

Watch out

A common error is to think that because both have a carbonyl group, they oxidize similarly. They do not. Aldehydes are among the most easily oxidized organic compounds; ketones are among the most resistant to oxidation. This is a key distinguishing test in the lab (Tollens' test, Fehling's test).

✓Final answer

Benzoic acid does not undergo Friedel-Crafts reaction because the carboxyl group strongly deactivates the ring and forms an inactive complex with the AlCl₃ catalyst. Oxidation of aldehydes is easier than that of ketones because aldehydes have a less hindered carbonyl carbon with a readily abstractable hydrogen, allowing mild oxidizing agents to work, whereas ketones require harsh conditions that break C-C bonds.

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.Write reasons for the following statements :

Why Benzoic Acid Does Not Undergo Friedel-Crafts Reaction

The Concept: Inductive Effect and Ring Deactivation

Why Oxidation of Aldehydes is Easier Than That of Ketones

The Concept: The Carbonyl Carbon's Reactivity and the Hydrogen Factor

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