What is the role of pyridine in the acylation reaction of amines?

NCERT Exemplar · Q37

Q. What is the role of pyridine in the acylation reaction of amines?

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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)

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Pyridine acts as a base to neutralise the HCl by-product and as a catalyst to activate the acylating agent, ensuring the acylation of amines proceeds smoothly without protonating the amine.

Why pyridine is needed in amine acylation

Acylation of an amine means replacing one of the amine’s hydrogen atoms with an acyl group ( $\ce{RCO-}$ ). The common reagent is an acid chloride (e.g., acetyl chloride, $\ce{CH3COCl}$ ). The reaction looks simple:

\ce{RNH2 + R'COCl -> R'CONHR + HCl}

But there’s a problem. The HCl produced is a strong acid. If you’re working with a primary or secondary amine, that amine is basic — it will immediately react with the HCl to form the corresponding ammonium salt:

\ce{RNH2 + HCl -> RNH3+Cl-}

Once the amine is protonated, it is no longer nucleophilic. The acylation stops. You lose yield and get a messy mixture.

Pyridine solves this in two ways.

Watch out

A common mistake is to think pyridine only speeds up the reaction. Its primary role is to prevent the amine from being destroyed by the acid by-product. Without it, the reaction often fails or gives poor yields.

Step-by-step reasoning

Pyridine is a weak base. It has a lone pair on nitrogen, but it is less basic than aliphatic amines (pKa of pyridinium ion ≈ 5.2 vs. ammonium ion ≈ 9–10). This means pyridine can accept a proton from HCl without being so basic that it deprotonates the amine or competes as a nucleophile.
It neutralises the HCl as it forms. The moment HCl is produced in the acylation, pyridine grabs the proton:

\ce{C5H5N + HCl -> C5H5NH+Cl-}

This keeps the reaction mixture neutral or slightly basic. The amine remains free and nucleophilic, ready to attack the next molecule of acid chloride.

It activates the acylating agent (catalytic role). Pyridine can also form a reactive intermediate with the acid chloride. The lone pair on pyridine attacks the carbonyl carbon of $\ce{RCOCl}$ , forming an acylpyridinium ion:

\ce{C5H5N + RCOCl -> [C5H5N+ - C(O)R] Cl-}

This intermediate is even more electrophilic than the original acid chloride because the positive charge on nitrogen pulls electron density away from the carbonyl. The amine then attacks this activated species more rapidly.

Tip

This activation is especially useful when the acid chloride is less reactive (e.g., benzoyl chloride) or when the amine is sterically hindered. Pyridine effectively “lends” its lone pair to make the carbonyl a better target. …

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Role of Pyridine in the Acylation of Amines

The Core Concept

In the acylation of amines (e.g., reaction with acetyl chloride or acetic anhydride), pyridine acts as a base and a catalyst. Its role is to:

Neutralize the HCl (or HX) acid byproduct — preventing the amine from being protonated and rendered unreactive.
Activate the acylating agent (e.g., acetyl chloride) by forming a more reactive acylpyridinium intermediate.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking pyridine is a solvent only

Why it's wrong: Pyridine is not inert — it actively participates in the reaction. It is a nucleophilic catalyst, not just a solvent.

How to avoid: Remember that pyridine's lone pair on nitrogen attacks the carbonyl carbon of the acyl halide, forming a positively charged intermediate that is more electrophilic than the original acyl halide. This accelerates the reaction.

✗ Mistake 2: Confusing pyridine with a stronger base like NaOH

Why it's wrong: Pyridine is a weak base ( $pK_b \approx 8.8$ ). It does not deprotonate the amine (which is more basic). Instead, it only neutralizes the acid byproduct.

How to avoid: Compare $pK_b$ values:

Pyridine: $pK_b \approx 8.8$
Aliphatic amines: $pK_b \approx 3\text{–}4$

Pyridine is too weak to deprotonate the amine — it only mops up the HCl produced.

✗ Mistake 3: Forgetting that pyridine prevents amine protonation

Why it's wrong: If the byproduct HCl is not removed, it will protonate the unreacted amine, forming an ammonium salt that cannot act as a nucleophile. This stops the reaction.

How to avoid: Always write the full mechanism:

Amine attacks acyl halide → forms amide + HCl.
Pyridine reacts with HCl → forms pyridinium chloride.
This keeps the amine free (unprotonated) for further reaction.

✗ Mistake 4: Saying pyridine is a catalyst in the same sense as an enzyme

Why it's wrong: Pyridine is consumed in the first step (forming the acylpyridinium intermediate) but regenerated when it picks up a proton from the amine's tetrahedral intermediate. So it is a true catalyst — but only if you track the full cycle.

How to avoid: Draw the catalytic cycle:

Pyridine + acyl halide → acylpyridinium (activated)
Amine attacks acylpyridinium → tetrahedral intermediate
Intermediate collapses, releasing pyridine (regenerated) + amide + H⁺

✗ Mistake 5: Using pyridine when the amine is already very reactive …

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

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{C}_4 \mathrm{H}_7 \mathrm{OH}$
›Reveal solutionSolution
The question asks for the incorrect statement among four organic chemistry claims. The key is to recall that dehydration ease follows Tertiary > Secondary > Primary, opposite to option (C), making (C) the false statement.

Concept & Intuition

This problem tests your grasp of alcohol reactivity, isomerism, and acidity trends. The trick is to spot the reversal of a well-known order: dehydration of alcohols with acid favors more substituted carbocations (tertiary > secondary > primary) because the reaction proceeds via a carbocation intermediate. Option (C) states the opposite, so it’s the clear outlier. The other options require checking structural isomer counts, the electron-withdrawing effect of nitro vs. methyl groups, and the role of acid in esterification.

Step-by-Step Reasoning

Evaluate option (A): Cyclic $\mathrm{C}_4 \mathrm{H}_7 \mathrm{OH}$ — a cyclobutanol derivative. The formula suggests a four-carbon ring with one OH group. Structural isomers include:

Cyclobutanol (OH on a 4-membered ring)

1-Methylcyclopropanol (OH on a 3-membered ring with a methyl)

2-Methylcyclopropanol (OH on a different carbon of the cyclopropane)

Cyclopropylmethanol (OH on a side chain) These are four distinct structural isomers, so (A) is correct.

Evaluate option (B): p-Nitrophenol vs. p-cresol. The nitro group ( $-\mathrm{NO}_2$ ) is strongly electron-withdrawing, stabilizing the phenoxide ion by resonance, making p-nitrophenol more acidic. The methyl group ( $-\mathrm{CH}_3$ ) in p-cresol is electron-donating, destabilizing the phenoxide, so p-cresol is less acidic. Thus (B) is correct. …
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KCET 2026Set D31 markMCQ
$Q. Carboxylic acids are more acidic than phenols because (A) Formation of dimers (B) Intermolecular hydrogen bonding (C) More covalent nature (D) More resonance stabilisation of their conjugate base$
›Reveal solutionSolution

Acid strength tracks how well the conjugate base's negative charge is stabilised; the carboxylate ion is stabilised more effectively by resonance than the phenoxide ion is.

Step 1 — Conjugate base of a carboxylic acid

Deprotonating $R-COOH$ gives the carboxylate ion $R-COO^-$ . Its two C–O bonds are equivalent by resonance: the negative charge and the double-bond character are shared equally between both oxygens, so each oxygen carries only half a unit of negative charge. This symmetric, low-energy delocalisation is very effective at stabilising the anion.

Step 2 — Conjugate base of phenol

Deprotonating phenol gives the phenoxide ion, $C_6H_5-O^-$ . Its negative charge can delocalise into the benzene ring through resonance, but this places some negative charge on ring carbons (less electronegative than oxygen) and, more importantly, disrupts the ring's aromatic stability in some resonance contributors — a less favourable and less effective stabilisation than the symmetric carboxylate case.

Step 3 — Why the other options are wrong …
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KCET 2025Set D-41 markMCQ
$Q. Phenol can be distinguished from propanol by using the reagent (A) Bromine water (B) Iron metal (C) Iodine in alcohol (D) Sodium metal$
›Reveal solutionSolution
Phenol's ring is so activated by the –OH that bromine water alone brominates it three times, precipitating white 2,4,6-tribromophenol; propanol does nothing.

Step 1 — What a distinguishing test must do.

It must give an observable change with one compound and no change (or a clearly different change) with the other. Both phenol and propanol contain an –OH group, so any test that keys on the –OH alone will fail. The reagent must exploit what makes them different: phenol's –OH is attached to a benzene ring; propanol's is on an alkyl chain.

Step 2 — Why phenol reacts with bromine water.

In phenol, the oxygen's lone pair is delocalised into the ring ( $+M$ effect):
$\mathrm{C_6H_5\ddot{O}H \longleftrightarrow \text{resonance structures with } -\text{ve charge at } o\text{- and } p\text{-positions}}$
This floods the ortho and para carbons with electron density, making the ring enormously more nucleophilic than benzene. It is so activated that it attacks even the weakly polarised Br $_2$ present in bromine water — no Lewis-acid catalyst (FeBr $_3$ /AlBr $_3$ ) is needed, and the reaction does not stop at one substitution:
$\mathrm{C_6H_5OH + 3Br_2 \xrightarrow{\ \text{aq.}\ } 2,4,6\text{-tribromophenol} \downarrow + 3HBr}$
2,4,6-Tribromophenol is insoluble in water, so it appears immediately as a white precipitate — and the brown colour of the bromine water is discharged. Both changes are unmistakable.

Step 3 — Why propanol does not.

$\mathrm{CH_3CH_2CH_2OH}$ is a saturated alcohol: no aromatic ring, no C=C, nothing for an electrophile to attack. Bromine water is not decolourised and no precipitate forms.

The contrast is therefore complete: white ppt ⇒ phenol; no change ⇒ propanol.

Step 4 — Why the other reagents fail.

(D) Sodium metal — the classic trap. Both phenol and propanol have an acidic O–H and both liberate hydrogen gas: …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. Two statements, One Assertion and the other Reason, are given. Which one of the following is the correct option? Assertion: The acid strength of 4 compounds in the descending order is p- Nitrophenol > p-Methoxyphenol > Phenol > p-chlorophenol. Reason: Electron withdrawing groups increase the acid strength while Electron donating groups decrease the acid strength of Phenol and its derivatives. (A) Assertion is correct but Reason is wrong. (B) Both Assertion and Reason are wrong. (C) Assertion is wrong but Reason is correct. (D) Both Assertion and Reason are correct.$
›Reveal solutionSolution
The key idea is that electron-withdrawing groups (EWGs) increase phenol acidity, while electron-donating groups (EDGs) decrease it. The given order is wrong because p-chlorophenol is more acidic than phenol, and p-methoxyphenol is less acidic than phenol. So the Assertion is false, but the Reason is true. The correct option is (C).

The problem asks us to judge both an Assertion (a specific ordering of acid strengths) and a Reason (a general principle about substituent effects). We need to check each separately.

Concept & Intuition

Phenol’s acidity comes from the stability of the phenoxide ion (the conjugate base). The negative charge on oxygen can be delocalized into the benzene ring. Any substituent that pulls electron density away from the oxygen (electron-withdrawing group, EWG) stabilizes the phenoxide ion, making the phenol more acidic. Any substituent that pushes electron density toward the oxygen (electron-donating group, EDG) destabilizes the phenoxide ion, making the phenol less acidic. This is the core principle behind the Reason.

Now, let’s test the Assertion’s order:

p-Nitrophenol > p-Methoxyphenol > Phenol > p-Chlorophenol.

p-Nitrophenol – The nitro group (−NO₂) is a strong EWG (by both inductive and resonance effects). It strongly stabilizes the phenoxide ion. So p-nitrophenol is indeed the most acidic among these four. ✓

p-Methoxyphenol – The methoxy group (−OCH₃) is a strong EDG (by resonance, it donates electrons into the ring). This destabilizes the phenoxide ion, making p-methoxyphenol less acidic than phenol. So placing it second (more acidic than phenol) is already wrong. ✗

Phenol – This is the reference compound. Its acidity is moderate.

p-Chlorophenol – Chlorine (−Cl) is a weak EWG (inductive withdrawal outweighs its weak resonance donation). It slightly stabilizes the phenoxide ion, so p-chlorophenol is more acidic than phenol. Placing it last (least acidic) is also wrong. ✗

So the correct descending order should be:

p-Nitrophenol > p-Chlorophenol > Phenol > p-Methoxyphenol. …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{O}-\mathrm{H}$
›Reveal solutionSolution
The question asks for the incorrect statement. By analyzing each option, we find that (C) contains a factual error in the mechanism of acid-catalyzed hydration of ethene, making it the wrong statement.

Concept & Intuition:

This problem tests your understanding of organic reaction mechanisms, acidity trends, and the dual role of alcohols as nucleophiles and electrophiles. The key is to recall the exact sequence of steps in well-known reactions and to recognize when a statement misrepresents the order of bond-making and bond-breaking. A common pitfall is confusing the role of the oxonium ion (protonated alcohol/water) with the carbocation that forms after it loses water.

Step-by-step analysis:

Option (A): Cresols are methylphenols. The methyl group is an electron-donating group (EDG) via hyperconjugation and induction. EDGs increase electron density on the benzene ring, which destabilizes the phenoxide ion (negative charge is less well dispersed). This makes cresols less acidic than phenol. The reasoning given is correct. So (A) is a true statement.

Option (B): Ethanol has an O–H bond. When the O–H bond breaks heterolytically, the oxygen retains the electron pair, forming an ethoxide ion — this is a nucleophile (donates electrons). When ethanol gets protonated (oxygen accepts a proton), it becomes $\text{CH}_3\text{CH}_2\text{OH}_2^+$ , which is electron-deficient and can act as an electrophile (accepts electrons). The statement is accurate. So (B) is true.

Option (C): In acid-catalyzed hydration of ethene:

Step 1: Ethene reacts with $\text{H}_3\text{O}^+$ (the acid). The proton adds to the double bond, forming a carbocation (not an oxonium ion). …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{pK}_{\mathrm{a}}$
›Reveal solutionSolution
Electron-withdrawing groups lower a phenol's $\mathrm{p}K_a$ and electron-donating groups raise it. Ordering the four para-substituted phenols by decreasing $\mathrm{p}K_a$ gives $A>C>D>B$ , so the correct option is (A).

Concept

The $\mathrm{p}K_a$ of a phenol measures how readily it loses its $-\mathrm{OH}$ proton: the more stable the resulting phenoxide ion, the stronger the acid and the lower the $\mathrm{p}K_a$ . A substituent on the ring shifts this stability:

Electron-withdrawing groups (EWG) such as $-\mathrm{NO_2}$ and $-\mathrm{Cl}$ pull electron density away, spread the negative charge of the phenoxide and stabilise it $\Rightarrow$ stronger acid $\Rightarrow$ lower $\mathrm{p}K_a$ .

Electron-donating groups (EDG) such as $-\mathrm{NH_2}$ and $-\mathrm{OCH_3}$ push electron density in, concentrate the negative charge and destabilise the phenoxide $\Rightarrow$ weaker acid $\Rightarrow$ higher $\mathrm{p}K_a$ .

Because the question asks for decreasing $\mathrm{p}K_a$ , we start with the least acidic (highest $\mathrm{p}K_a$ ) and end with the most acidic (lowest $\mathrm{p}K_a$ ).

Solution

Assign each labelled compound its para substituent and its approximate $\mathrm{p}K_a$ :

A ( $-\mathrm{NH_2}$ ): strong EDG (nitrogen lone pair donates strongly into the ring), most destabilised phenoxide $\Rightarrow$ highest $\mathrm{p}K_a\approx10.3$ .

C ( $-\mathrm{OCH_3}$ ): EDG by resonance, but slightly weaker net donor than $-\mathrm{NH_2}$ here $\Rightarrow$ $\mathrm{p}K_a\approx10.2$ , just below A. …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. Arrange the following compounds in the decreasing order of their acidic strength. (I) m-cresol (II) Phenol (III) m-aminophenol (IV) m-methoxyphenol (A) I > III > IV > II (B) I > II > III > IV (C) IV > III > II > I (D) III > I > II > IV$
›Reveal solutionSolution
At meta only inductive effects operate: –OCH $_3$ and –NH $_2$ are $-I$ (acid-strengthening), –CH $_3$ is $+I$ (acid-weakening), giving IV > III > II > I.

Phenol acidity depends on how well the substituent stabilizes the phenoxide anion. All substituents here are meta to –OH, so their resonance interaction with the oxygen-bearing carbon is not effective; the dominant effect is induction.

(IV) m-methoxyphenol: –OCH $_3$ has a strong $-I$ (electron-withdrawing) inductive effect at meta (oxygen is highly electronegative) → stabilizes the anion most → most acidic (pK $_a\approx 9.65$ ). …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{pK}_{\mathrm{a}}$
›Reveal solutionSolution
The key idea is to evaluate each statement about phenol acidity, substituent effects, steam distillation, and oxidation; the incorrect statement is C, because p‑nitrophenol is less steam volatile than o‑nitrophenol due to stronger intermolecular hydrogen bonding.

Concept & Intuition

This question tests four distinct concepts in organic chemistry:

Acidity of phenols depends on the stability of the phenoxide ion. Electron-withdrawing groups (EWG) increase acidity (lower pKₐ), while electron-donating groups (EDG) decrease acidity (higher pKₐ).

Ortho-substituent effects can involve both electronic and steric factors, plus intramolecular hydrogen bonding.

Steam distillation separates compounds based on volatility; stronger intermolecular forces (like hydrogen bonding) reduce volatility.

Oxidation of phenols with strong oxidants like dichromate can yield quinones (conjugated diketones).

Let’s examine each statement carefully.

Statement A: “Phenol has lower pKₐ than p‑cresol”

p‑Cresol has a methyl group at the para position. Methyl is an electron-donating group (+I effect).

This destabilizes the phenoxide ion (increases negative charge density), making p‑cresol less acidic than phenol.

Hence phenol (pKₐ ≈ 10) is more acidic than p‑cresol (pKₐ ≈ 10.2). So A is correct.

Statement B: “2‑Chlorophenol is more acidic than phenol”

Chlorine is an electron-withdrawing group (−I effect), which stabilizes the phenoxide ion by delocalizing the negative charge.

Additionally, ortho‑chlorophenol can form an intramolecular hydrogen bond between the –OH and –Cl, which further stabilizes the conjugate base.

Therefore, 2‑chlorophenol (pKₐ ≈ 8.5) is indeed more acidic than phenol (pKₐ ≈ 10). So B is correct.

Statement C: “Ortho and para nitrophenols can be separated by steam distillation since p‑Nitrophenol is more steam volatile than o‑Nitrophenol”

o‑Nitrophenol has intramolecular hydrogen bonding (between –OH and –NO₂), reducing its ability to form intermolecular hydrogen bonds with water. This makes it more steam volatile. …
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COMEDK 2024Set 2024-M1 markMCQ
$Q. From the following compounds, identify the one which is most acidic. (A) B (B) D (C) A (D) C$
›Reveal solutionSolution
p-Nitrophenol [C] is most acidic — the para nitro group stabilises the phenoxide anion by resonance/–M, far more than plain phenol or the alcohols.

Compare the four –OH compounds:

[B] dicyclohexyl-carbinol and [D] cyclohexanol are simple alcohols; their alkoxides get no resonance stabilisation, so they are the weakest acids.

[A] phenol: the phenoxide is resonance-stabilised over the aromatic ring, so phenol is markedly more acidic than alcohols. …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{m}$
›Reveal solutionSolution

(A) m-Cresol (3-methylphenol) - the electron-releasing -CH3 group destabilises the phenoxide ion, so m-cresol IS a weaker acid than phenol. CORRECT statement. (B) The acidity of phenol is attributed to the electron-withdrawing (-I) effect of the sp2/attached oxygen plus resonance stabilisation of the phenoxide ion. As stated in NCERT terms this is taken as a correct statement. (C) Acetylation of salicylic acid (with acetic anhydride) gives 2-acetoxybenzoic acid = aspirin. CORRECT statement. (D) Phenol heated with conc. H2SO4 gives PHENOL-4-SULPHONIC ACID (p-hydroxybenzenesulphonic acid), NOT s

Concept: check each statement about phenol.

(A) m-Cresol (3-methylphenol) - the electron-releasing -CH3 group destabilises the phenoxide ion, so m-cresol IS a weaker acid than phenol. CORRECT statement.

(B) The acidity of phenol is attributed to the electron-withdrawing (-I) effect of the sp2/attached oxygen plus resonance stabilisation of the phenoxide ion. As stated in NCERT terms this is taken as a correct statement. …
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Q.What is the role of pyridine in the acylation reaction of amines?

Concept understanding — Acidity Of Phenol

Inductive Effect on Acidity – From Intuition to Precision

The Core Intuition

The Precise Statement

Why Distance Matters

Why this formula?

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

1. The Core Observation

2. The Key: Stability of the Conjugate Base

The Resonance Explanation

3. Why Alcohols Can't Do This

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

5. The Quantitative Picture (pKa Values)

Why pyridine is needed in amine acylation

Step-by-step reasoning

Concept: Acylation of Amines & the Role of a Base Scavenger

Method Name: Base Scavenger / Acid Neutraliser Method

Steps of the Role:

Role of Pyridine in the Acylation of Amines

The Core Concept

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking pyridine is a solvent only

✗ Mistake 2: Confusing pyridine with a stronger base like NaOH

✗ Mistake 3: Forgetting that pyridine prevents amine protonation

✗ Mistake 4: Saying pyridine is a catalyst in the same sense as an enzyme

✗ Mistake 5: Using pyridine when the amine is already very reactive …

Step 1 — Conjugate base of a carboxylic acid

Step 2 — Conjugate base of phenol

Step 3 — Why the other options are wrong …

Concept

Solution

Q.What is the role of pyridine in the acylation reaction of amines?

Why pyridine is needed in amine acylation

Step-by-step reasoning

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