Q.Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

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

1. 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.

2. 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. …

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 …

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):

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:

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: …

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.

1. 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. ✓

2. 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. ✗

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

4. 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. …

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:

1. 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.

2. 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.

3. 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). …

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 $\mathrm{p}K_a$:

1. A ($-\mathrm{NH_2}$): strong EDG (nitrogen lone pair donates strongly into the ring), most destabilised phenoxide $\Rightarrow$ highest $\mathrm{p}K_a\approx10.3$.
2. 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. …

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$_a\approx 9.65$). …

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.

1. 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.

2. 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.

3. 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. …

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. …

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. …