(a) Account for the following : (i) Phenol is a stronger acid than an alcohol. (ii) The boiling point of alcoh

Question

Q. (a) Account for the following :

(i) Phenol is a stronger acid than an alcohol.

(ii) The boiling point of alcohols decreases with increase in branching of alkyl chain.

(OR)

(b)

(i) Write the mechanism of the following reaction :

CH_3CH_2OH \xrightarrow[443\ K]{H^+} CH_2=CH_2 + H_2O

(ii) Write the equation involved in Reimer-Tiemann reaction.

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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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 — Reimer-Tiemann Reaction

The Intuition: Why Would Chloroform Add an Aldehyde?

Imagine you have a phenol molecule — a benzene ring with an –OH group. That –OH is not just sitting there; it's a powerful electron-donating group. It pushes electron density into the ring, especially onto the ortho and para positions. This makes those positions unusually reactive toward electrophiles (species that love electrons).

Now, chloroform ( $\text{CHCl}_3$ ) in the presence of a strong base like aqueous $\text{NaOH}$ does something dramatic. The base pulls off a proton from chloroform, generating a highly reactive species called dichlorocarbene ( $:\text{CCl}_2$ ). This carbene is a fierce electrophile — it has a sextet of electrons and desperately wants two more.

The phenol's ortho position, rich in electrons, is the perfect target. The carbene attacks there, and a cascade of reactions follows, ultimately converting that –CHCl₂ group into an aldehyde (–CHO). The product is salicylaldehyde (2-hydroxybenzaldehyde).

Note

The reaction is ortho-selective because the –OH group directs the incoming electrophile to the ortho position. Para substitution is possible but much less common under these conditions.

The Precise Statement

Reimer–Tiemann Reaction:

When phenol is treated with chloroform ( $\text{CHCl}_3$ ) and aqueous sodium hydroxide ( $\text{NaOH}$ ) at about 60–70 °C, followed by acidification, the formyl group (–CHO) is introduced at the ortho position relative to the –OH group. The major product is salicylaldehyde.

\ce{C6H5OH + CHCl3 + 3 NaOH ->[\Delta] \underset{\text{salicylaldehyde}}{o-HOC6H4CHO} + 3 NaCl + 2 H2O}

Step-by-Step Mechanism (Why It Works)

Generation of dichlorocarbene $\text{NaOH}$ deprotonates chloroform:

\ce{CHCl3 + OH- -> CCl3- + H2O}

The trichloromethyl anion loses a chloride ion to form the electrophilic carbene:

\ce{CCl3- -> :CCl2 + Cl-}

Attack on phenoxide ion

Phenol first reacts with $\text{NaOH}$ to form the more nucleophilic phenoxide ion ( $\ce{C6H5O-}$ ). The carbene attacks the ortho carbon of the phenoxide ring.
Rearrangement and hydrolysis

The intermediate undergoes ring-opening to a dichloromethyl phenol derivative, which then hydrolyzes under basic conditions to give the aldehyde.
Acidification

After the reaction, adding dilute acid converts the sodium salt of salicylaldehyde back to the free aldehyde.

Watch out

A common mistake: thinking the –CHO group comes directly from chloroform. It does not — the carbon of the aldehyde is the carbon from chloroform, but it arrives via the carbene intermediate, not as a pre-formed formyl group.

Key Points for Exams

Reagents: Phenol + $\text{CHCl}_3$ + aqueous $\text{NaOH}$ (not alcoholic $\text{NaOH}$ — that would give a different reaction).
Temperature: ~60–70 °C (reflux). Too low, the carbene doesn't form; too high, side reactions dominate.
Product: Salicylaldehyde (ortho-hydroxybenzaldehyde). A small amount of para-hydroxybenzaldehyde may also form, but ortho is the major product. …

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Phenol is more acidic than an alcohol because its conjugate base (phenoxide) is resonance-stabilised while an alkoxide is not; branching lowers alcohol boiling points by weakening van der Waals forces.
Ethanol dehydrates (protonation $\rightarrow$ carbocation $\rightarrow$ loss of $\beta$ -H) to ethene; the Reimer–Tiemann reaction turns phenol into salicylaldehyde using $CHCl_3$ and NaOH.

Part (a)

(i) Why phenol is a stronger acid than an alcohol

Acid strength is decided by the stability of the conjugate base left after losing $H^+$ .

Phenol $\rightarrow$ phenoxide: the negative charge on O is delocalised into the aromatic ring (resonance onto the o- and p-carbons), spreading and stabilising it.
Alcohol $\rightarrow$ alkoxide: the charge stays localised on oxygen, with no resonance — a higher-energy, less stable anion.

A more stable conjugate base means the acid ionises more readily, so phenol ( $pK_a \approx 10$ ) is far more acidic than a typical alcohol ( $pK_a \approx 16$ ).

Watch out

The dominant reason is resonance stabilisation of the phenoxide ion, not merely the ring's inductive effect.

(ii) Why branching lowers the boiling point of alcohols …

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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 37 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 mark
$Q. Assertion (A): Reimer-Tiemann reaction of phenol with chloroform in presence of NaOH at 340K gives salicylaldehyde as the major product. Reason (R): The reaction occurs through intermediate formation of dichlorocarbene.$
›Reveal solutionSolution

Both statements are true, and dichlorocarbene formation is exactly the mechanistic reason salicylaldehyde forms.

Assertion: In the Reimer–Tiemann reaction, phenol is treated with $CHCl_3$ and $NaOH$ at about 340 K; the product (after hydrolysis) is predominantly the ortho-hydroxybenzaldehyde, i.e. salicylaldehyde. True.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The reaction: Phenol + CHCl3 + KOH --(heat, delta)--> Salicylaldehyde, is$
(a) Gattermann-Koch reaction
(b) Sandmeyer reaction
(c) Reimer-Tiemann reaction
(d) Fittig reaction
›Reveal solutionSolution

This is the classic Reimer-Tiemann reaction, in which phenol reacts with chloroform and KOH to introduce a -CHO group ortho to the -OH, giving salicylaldehyde.

$\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{KOH} \xrightarrow{\Delta} \text{o-HOC}_6\text{H}_4\text{CHO (salicylaldehyde)} + \text{KCl} + \text{H}_2\text{O}$

…

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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
$CHCl_3$
(a) benzaldehyde
(b) salicylaldehyde
(c) salicylic acid
(d) benzoic acid
›Reveal solutionSolution
This is the Reimer–Tiemann reaction: phenol reacts with chloroform under strongly basic conditions to install a formyl ( $-CHO$ ) group at the ortho position, giving salicylaldehyde after hydrolysis.

Mechanism outline

NaOH deprotonates $CHCl_3$ to form the trichloromethyl carbanion $:CCl_3^-$ , which rapidly loses $Cl^-$ to generate the highly reactive electrophile dichlorocarbene ( $:CCl_2$ ).

The phenoxide ion (from phenol + NaOH) attacks $:CCl_2$ preferentially at the ortho position (directed by the strongly activating $-O^-$ group), giving a dichloromethyl-substituted intermediate.

Hydrolysis of the $-CHCl_2$ group under the alkaline reaction conditions converts it to $-CHO$ .

Overall:

$C_6H_5OH \xrightarrow[\text{NaOH}]{CHCl_3} \text{o-hydroxybenzaldehyde (salicylaldehyde)}$

…
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Salicyaldehyde can be prepared from phenol by$
(a) Schotten-Baumann reaction
(b) Kolbe's reaction
(c) Reimer-Tiemann reaction
(d) Cannizzaro reaction
›Reveal solutionSolution

Dichlorocarbene, generated from CHCl3+NaOH, formylates phenol's activated ortho ring position, and hydrolysis of the resulting dichloromethyl intermediate reveals the aldehyde.

In the Reimer–Tiemann reaction, phenol is treated with chloroform and concentrated NaOH. Base generates dichlorocarbene (:CCl₂) from CHCl₃ (by α-elimination), which is attacked by the electron-rich phenoxide ring (ortho position preferred), forming a ortho-substituted dichloromethyl-phenol intermediate; hydrolysis of this gem-dihalide under the basic conditions (then acidification) re …

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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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Q.(a) Account for the following :

Part (a)

(i) Why phenol is a stronger acid than an alcohol

(ii) Why branching lowers the boiling point of alcohols …

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