(a) Give equations of the following reactions: (i) Phenol is treated with conc. $HNO_3$. (ii) Propene is treated with $B_2H_6$ followed by $H_2O_2/OH^-$. (iii) Sodium t-butoxide is treated with $CH_3Cl$. (b) How will you distinguish between butan-1-ol and butan-2-ol? (c) Arrange the following in increasing order of acidity: Phenol, ethanol, water OR (a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride? (b) Write the structure of the major product obtained from dinitration of 3-methylphenol. (c) Write the reaction involved in Kolbe's reaction.

(a) Give equations of the following reactions: (i) Phenol is treated with conc. HNO₃. (ii) Propene is treated

Question

Q. (a) Give equations of the following reactions:

(i) Phenol is treated with conc.

HNO_3

(ii) Propene is treated with

B_2H_6

followed by

H_2O_2/OH^-

(iii) Sodium t-butoxide is treated with

CH_3Cl

(b) How will you distinguish between butan-1-ol and butan-2-ol?

(OR)

(a) How can you obtain Phenol from

(i) Cumene,

(ii) Benzene sulphonic acid,

(iii) Benzene diazonium chloride?

(b) Write the structure of the major product obtained from dinitration of 3-methylphenol.

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Part (a)Concept understanding — Acidity Of Phenol

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?

Acidity of Phenol: Why It's More Acidic Than Alcohols

Let's build this from first principles — understanding why phenol is acidic is the key to mastering organic chemistry.

1. The Core Observation

Phenol ( $\ce{C6H5OH}$ ) has a pKa ≈ 10, while ethanol ( $\ce{CH3CH2OH}$ ) has a pKa ≈ 16.

This means phenol is about 1 million times more acidic than a typical alcohol.

The question: Why does the O–H bond in phenol break so much more easily?

2. The Key: Stability of the Conjugate Base

Acidity is determined by the stability of the conjugate base after losing $\ce{H+}$ .

Alcohol conjugate base: $\ce{CH3CH2O-}$ (alkoxide ion) — negative charge is localized on oxygen.
Phenol conjugate base: $\ce{C6H5O-}$ (phenoxide ion) — negative charge is delocalized into the benzene ring.

The Resonance Explanation

The phenoxide ion has multiple resonance structures:

\ce{C6H5O-} \leftrightarrow \text{(several resonance forms where negative charge moves to ortho/para carbons)}

Draw the structures mentally:

One structure has the negative charge on oxygen.
Other structures show the negative charge on carbon atoms at the ortho and para positions of the ring.

This delocalization spreads the negative charge over more atoms, making the ion more stable.

Key principle: The more stable the conjugate base, the stronger the acid.

3. Why Alcohols Can't Do This

In an alkoxide ion ( $\ce{RO-}$ ), the negative charge is stuck on oxygen.

There are no empty p-orbitals or conjugated $\pi$ systems nearby to accept the charge.

Result: The alkoxide is less stable, so the alcohol is less acidic.

4. The Inductive Effect Also Helps (But Resonance Dominates)

The benzene ring is slightly electron-withdrawing (due to its $sp^2$ carbons being more electronegative than $sp^3$ ).

This inductive effect pulls electron density away from the O–H bond, making the proton slightly more positive and easier to remove.

However, resonance stabilization of the conjugate base is the dominant factor — inductive effects alone cannot explain the million-fold difference.

5. The Quantitative Picture (pKa Values)

Compound	pKa	Conjugate base stability
Ethanol	~16	Localized charge on O
Phenol	~10	Delocalized charge via resonance
Acetic acid	~4.76	Even more resonance (two O atoms)

Part (b)Concept understanding — Kolbe Reaction

The Kolbe Reaction: From Intuition to Mechanism

Imagine you have a benzene ring with an –OH group attached — that's phenol. Now, phenol is weakly acidic, so when you treat it with a strong base like NaOH, you get sodium phenoxide. That oxygen now carries a full negative charge. That negative charge is the key.

The oxygen's lone pairs are strongly electron-donating. They push electron density into the ring, making the ortho and para positions much more reactive toward electrophiles. The ortho position, being right next to the oxygen, gets the most activation. This is the intuition: the negatively charged oxygen "wakes up" the ring, especially the ortho carbon, making it hungry for a positive or electron-deficient species.

Now, enter carbon dioxide. CO₂ is a weak electrophile — the carbon is partially positive because of the two oxygens pulling electrons away. Under normal conditions, CO₂ is too weak to attack phenol directly. But under high pressure (about 4–7 atm) and moderate heat (125–150°C), the activated ortho position of sodium phenoxide can attack the carbon of CO₂.

Important

The Kolbe reaction is the carboxylation of sodium phenoxide using CO₂ under pressure, followed by acidification, to give salicylic acid (2-hydroxybenzoic acid). The carboxyl group (–COOH) attaches exclusively at the ortho position relative to the –OH group.

The precise steps

Formation of sodium phenoxide

\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O}

Carboxylation under pressure The ortho carbon of the phenoxide ion attacks CO₂. A tetrahedral intermediate forms, which then rearranges to give sodium salicylate.

\text{C}_6\text{H}_5\text{ONa} + \text{CO}_2 \xrightarrow{125-150^\circ\text{C}, \text{pressure}} \text{o-HO-C}_6\text{H}_4\text{-COONa}

Acidification Treating with dilute HCl gives the free carboxylic acid.

\text{o-HO-C}_6\text{H}_4\text{-COONa} + \text{HCl} \rightarrow \text{o-HO-C}_6\text{H}_4\text{-COOH} + \text{NaCl}

Note

Why only ortho? The para position is also activated, but steric hindrance from the bulky –ONa group and the incoming CO₂ molecule makes para attack unfavourable. The ortho position is both electronically favoured and sterically accessible.

Why this reaction matters

Salicylic acid is the precursor to aspirin (acetylsalicylic acid). The Kolbe reaction is the industrial route to salicylic acid — cheap, simple, and high-yielding. Without it, aspirin would be far more expensive. …

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Part (a): phenol + conc. $\text{HNO}_3$ $\to$ picric acid; propene hydroboration $\to$ propan-1-ol; Na t-butoxide + $\text{CH}_3\text{Cl}$ $\to$ methyl t-butyl ether; Lucas test separates butan-1-ol/butan-2-ol; acidity ethanol < water < phenol. Part (b): phenol from cumene / NaOH fusion / diazonium hydrolysis; dinitration of 3-methylphenol $\to$ 4,6-dinitro-3-methylphenol; Kolbe gives salicylic acid.

Part (a)

Equations

(i) Phenol + conc. $\text{HNO}_3$ : the strongly activating $-\text{OH}$ allows tri-nitration:

\text{C}_6\text{H}_5\text{OH} + 3\text{HNO}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{2,4,6-trinitrophenol (picric acid)} + 3\text{H}_2\text{O}

(ii) Propene + $\text{B}_2\text{H}_6$ , then $\text{H}_2\text{O}_2/\text{OH}^-$ : hydroboration–oxidation adds water anti-Markovnikov:

\text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow[(ii)\ \text{H}_2\text{O}_2/\text{OH}^-]{(i)\ \text{B}_2\text{H}_6} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH (propan-1-ol)}

(iii) Sodium t-butoxide + $\text{CH}_3\text{Cl}$ (Williamson): the bulky alkoxide does SN2 on the primary $\text{CH}_3\text{Cl}$ :

(\text{CH}_3)_3\text{CO}^-\text{Na}^+ + \text{CH}_3\text{Cl} \to (\text{CH}_3)_3\text{C-O-CH}_3 + \text{NaCl}

Distinguishing butan-1-ol and butan-2-ol

Use the Lucas test (conc. HCl + anhydrous $\text{ZnCl}_2$ ): butan-2-ol ( $2^\circ$ ) forms a stabler carbocation and gives turbidity (cloudiness) within ~5 minutes, whereas butan-1-ol ( $1^\circ$ ) shows no turbidity at room temperature.

Increasing order of acidity

Acidity tracks conjugate-base stability. Ethoxide has a localised charge (destabilised by the +I ethyl group); hydroxide is more stable; phenoxide is resonance-stabilised over the ring:

$\text{ethanol} < \text{water} < \text{phenol}$ …

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

Showing the 12 most recent of 31 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Given below are two statements : Statement I : Alcohols are acidic in nature; The acidic character of alcohol is due to the polar nature of the O-H bond in it. Statement II : Alcohols are weaker acids than water. In the light of the above statements, choose the most appropriate answer from the options given below :$
(a) Statement I is incorrect but Statement II is correct
(b) Both Statement I and Statement II are correct
(c) Both Statement I and Statement II are incorrect
(d) Statement I is correct but Statement II is incorrect
›Reveal solutionSolution

Alcohols are weakly acidic because of the polar O–H bond, and they are weaker acids than water — both statements are correct, so option (b).

Statement I (correct): Alcohols show weak acidic character because the O–H bond is polar; the more electronegative oxygen pulls electron density from hydrogen, so it can be released as a proton:

$\text{R–O–H} \rightleftharpoons \text{R–O}^- + \text{H}^+$ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. p-nitrophenol is stronger acid than phenol because nitro group is:$
(a) Electron donating
(b) Electron withdrawing
(c) Acidic
(d) Basic
›Reveal solutionSolution

p-Nitrophenol is more acidic than phenol because the -NO2 group is electron-withdrawing, stabilising the conjugate-base phenoxide ion.

The acidity of phenols depends on how readily they lose the phenolic proton and how stable the resulting phenoxide (conjugate base) anion is. Substituents that WITHDRAW electron density (by −I inductive effect and/or −M resonance effect) delocalise and stabilise the negative charge on the phenoxide ion, increasing acidity. Substituents that DONATE electron density destabilise the negative charge (concentrate it), decreasing acidity.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Phenol is more acidic than ethanol because$
(a) ethoxide ion is more stable than phenoxide ion
(b) phenoxide ion is more stable than ethoxide ion
(c) phenol undergoes electrophilic substitution reaction
(d) phenol undergoes protonation easily
›Reveal solutionSolution

Acid strength tracks conjugate-base stability; the phenoxide ion is resonance-stabilised by the aromatic ring while the ethoxide ion is not, so phenol is the stronger acid.

Phenoxide ion, $C_6H_5O^-$ : the negative charge on oxygen can delocalise into the ring through resonance, placing partial negative charge at the ortho and para carbons as well — several resonance structures share and spread out the charge, lowering the ion's energy (stabilising it).

Ethoxide ion, $CH_3CH_2O^-$ : there is no adjacent $\pi$ system to delocalise into, so the negative charge stays fully localised on the single oxygen atom; the electron-donating (+I) ethyl group actually intensifies (destabilises) this concentrated negative charge further.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. When Phenol is distilled with zinc powder, it gives-$
(a)
(i) Benzene
(b)
(ii) Toluene
(c)
(iii) Benzaldehyde
(d)
(iv) Benzoic acid
›Reveal solutionSolution

On distillation with zinc dust, phenol is reduced (the –OH group is removed) to give benzene. Correct option: (i).

Concept. Zinc dust is a reducing agent that removes the hydroxyl group of phenol as zinc oxide, replacing the C–OH bond with a C–H bond.

Reaction.

$\text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO}$

…

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CBSE 2026Set SEM31 markMCQ
$Q. The order of acidic strength for the following compounds is: (I) phenol (C6H5OH), (II) 4-nitrophenol, (III) 4-methylphenol, (IV) 2-nitrophenol$
(a) II > I > III > IV
(b) II > IV > I > III
(c) IV > II > I > III
(d) IV > II > III > I
›Reveal solutionSolution
Electron-withdrawing -NO2 raises phenol acidity, electron-donating -CH3 lowers it: 4-nitrophenol > 2-nitrophenol > phenol > 4-methylphenol. Correct option (b).

Acidity of phenols depends on stabilisation of the phenoxide ion:

(II) 4-nitrophenol: -NO2 at para withdraws electrons by -I and -R, strongly stabilising the phenoxide -> most acidic (pKa ~ 7.1).

(IV) 2-nitrophenol: -NO2 at ortho is also strongly acidifying (pKa ~ 7.2); it is slightly less acidic than the para isomer largely due to intramolecular hydrogen bonding effects, but still far more acidic than phenol.

(I) phenol: reference (pKa ~ 10.0). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The correct order of increasing acidic strength is:$
(a) Phenol < Ethanol < Chloroacetic acid < Acetic acid.
(b) Ethanol < Phenol < Chloroacetic acid < Acetic acid.
(c) Ethanol < Phenol < Acetic acid < Chloroacetic acid.
(d) Chloroacetic acid < Acetic acid < Phenol < Ethanol.
›Reveal solutionSolution
Acidic strength increases in the order Ethanol (weakest) < Phenol < Acetic acid < Chloroacetic acid (strongest), based on how well each conjugate base stabilizes the negative charge.

Ethanol ( $pK_a \approx 16$ ): the ethoxide ion ( $C_2H_5O^-$ ) has no way to delocalize its negative charge — it is the weakest acid of the four.

Phenol ( $pK_a \approx 10$ ): the phenoxide ion is resonance-stabilized by delocalization of the negative charge into the benzene ring, making phenol far more acidic than ethanol, though still weaker than carboxylic acids.

Acetic acid ( $pK_a \approx 4.76$ ): the acetate ion is stabilized by resonance between the two equivalent C–O bonds (much stronger stabilization than phenoxide's ring delocalization), making carboxylic acids much stronger acids than phenols. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Phenol is more acidic than ethanol. Reason (R): The phenoxide ion formed after loss of proton from phenol is stabilized by resonance, whereas the ethoxide ion is not.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
›Reveal solutionSolution

Phenol is indeed more acidic than ethanol, and the reason given — resonance stabilization of the phenoxide ion versus no such stabilization for the ethoxide ion — is exactly the correct explanation.

Assertion: Phenol ( $pK_a \approx 10$ ) is more acidic than ethanol ( $pK_a \approx 16$ ). This is a well-established fact — TRUE.

Reason: When phenol loses its $-OH$ proton, the resulting phenoxide ion ( $C_6H_5O^-$ ) has its negative charge delocalized into the aromatic ring through resonance (the O⁻ lone pair conjugates with the ring's π system), spreading the charge over several ring carbons and stabilizing the ion. In contrast, the ethoxide ion ( $C_2H_5O^-$ ) formed from ethanol has no adjacent π system to delocalize into — the negative charge stays local …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one of the following is the Strongest acid ?$
(a) 4-nitrophenol
(b) 2-nitrophenol
(c) 3-nitrophenol
(d) 4-chlorophenol
›Reveal solutionSolution

A $-NO_2$ group at the para position withdraws electron density from the phenolic $-OH$ by both resonance and induction, and (unlike the ortho isomer) without any competing intramolecular hydrogen bonding to the un-ionised $-OH$ — so 4-nitrophenol stabilises its conjugate-base phenoxide ion the most and is the strongest acid of the four options.

Why nitrophenols are more acidic than phenol: the electron-withdrawing $-NO_2$ group pulls electron density away from the $-OH$ oxygen (destabilising the neutral acid, weakening the $O-H$ bond) and, once the proton is lost, delocalises the resulting negative charge of the phenoxide ion onto its own oxygen atoms by resonance — strongly stabilising the conjugate base and shifting the ionisation equilibrium further towards dissociation. Both effects make nitrophenols far more acidic than phenol itself.

…

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CBSE 2024Set B1 mark
$Q. Write True or False: Phenols is also called carbolic acids.$
›Reveal solutionSolution

Phenol (C6H5OH) is indeed also called carbolic acid, historically used as an antiseptic.

Phenol was one of the earliest antiseptics used in surgery (by Joseph Lister), and its common/trade name from that era, 'carbolic acid', is still widely used alongside the systematic name phenol. Its wea …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Mark the correct order of decreasing acid strength of the following compounds:$
(a) phenol (OH on benzene ring)
(b) p-nitrophenol (OH with NO2 group para to it on the ring)
(c) p-methoxyphenol (OH with OCH3 group para to it on the ring)
(a) b > a > c
(b) a > b > c
(c) c > a > b
(d) c > b > a
›Reveal solutionSolution
Acidity of a substituted phenol depends on how well the ring substituent stabilises the resulting phenoxide anion: electron-withdrawing groups (like -NO2) increase acid strength, electron-donating groups (like -OCH3) decrease it.

Phenol - the reference compound (moderate acidity, phenoxide stabilised only by ring delocalisation).

p-nitrophenol - the -NO2 group is strongly electron-withdrawing (by both resonance and induction) at the para position, which further delocalises and stabilises the negative charge on the phenoxide oxygen, making the O-H bond easier to break -> MORE acidic than phenol. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is most acidic ?$
(a) Benzyl alcohol
(b) Cyclohexanol
(c) Phenol
(d) m-Chlorophenol
›Reveal solutionSolution
Phenols are more acidic than simple alcohols because the phenoxide ion is resonance-stabilised; an electron-withdrawing substituent like –Cl further stabilises the phenoxide and increases acidity beyond phenol itself.

Acidity here depends on how well the conjugate base (the anion left after losing $H^+$ ) is stabilised:

Benzyl alcohol and cyclohexanol are simple alcohols; their alkoxide ions have no resonance stabilisation, so they are the weakest acids of the four (cyclohexanol is even weaker than benzyl alcohol since the benzylic system gives benzyl alcohol slight extra stabilisation, but both are far less acidic than any phenol).

Phenol is more acidic than alcohols because the phenoxide ion delocalises its negative charge into the aromatic ring via resonance. …
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CBSE 2024Set ANNUAL1 mark
$Q. Write the ascending order of acidic strength of the following compound. (I) Phenol (II) Meta nitro phenol (III) p-Nitro phenol$
›Reveal solutionSolution
The -NO2 group withdraws electron density and stabilises the phenoxide ion; this stabilisation is strongest when -NO2 is para (direct resonance) and weaker, but still present, when meta (only inductive).

Acidic strength of a substituted phenol depends on how well the conjugate base (phenoxide ion, C6H5O-) is stabilised.

Phenol (I): No substituent; the phenoxide ion is stabilised only by resonance delocalisation over the ring — weakest acid of the three (pKa ~ 10.0).

m-Nitrophenol (II): The -NO2 group at the meta position cannot conjugate directly with the O- (no resonance structure places the negative charge on the carbon bearing NO2), so it stabilises the phenoxide ion only through its electron-withdrawing inductive effect. This makes it more acidic than plain phenol, but less acidic than the para isomer (pKa ~ 8.3). …
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Q.(a) Give equations of the following reactions:

Part (a)

Equations

Distinguishing butan-1-ol and butan-2-ol

Increasing order of acidity

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