Skip to content
NCERT Exemplar · Q11

Q.Which of the following compounds will react with sodium hydroxide solution in water?

(i) C6H5OHC_6H_5OH
(ii) C6H5CH2OHC_6H_5CH_2OH
(iii) (CH3)3COH(CH_3)_3COH
(iv) C2H5OHC_2H_5OH
CBSEMCQ· 1mImportance★★★★★
47% · 63/135 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The key idea is that only phenols are acidic enough to react with aqueous NaOH, forming a salt. Among the given options, only phenol (C6H5OHC_6H_5OH) does this — the others are alcohols that are too weakly acidic. The correct option is (i).

Why This Question Tests Acidity, Not Just "Reaction with Base"

At first glance, the question seems simple: which compound reacts with NaOH in water? But the trap is that all the compounds have an –OH group. The difference lies in how acidic that –OH hydrogen is. Aqueous NaOH is a strong base, but it’s in water — so the reaction is essentially an acid–base neutralisation. Only compounds with a pKapK_a low enough (roughly below 14) will donate a proton to hydroxide.

Alcohols like ethanol, tert-butanol, and benzyl alcohol have pKapK_a values around 15–18. They are weaker acids than water (pKa≈15.7pK_a \approx 15.7). In water, they do not react with NaOH because the equilibrium lies heavily towards the alcohol and hydroxide, not the alkoxide. Phenol, however, has a pKapK_a of about 10 — it is a stronger acid than water, so it reacts readily.

Watch out

A common mistake is to think that any –OH group will react with NaOH. But alcohols are not acidic enough in water. The reaction ROH+OH−⇌RO−+H2OROH + OH^- \rightleftharpoons RO^- + H_2O has an equilibrium constant Keq=Ka(ROH)/KwK_{eq} = K_a(ROH)/K_w. For ethanol, Keq≈10−16/10−14=10−2K_{eq} \approx 10^{-16}/10^{-14} = 10^{-2} — negligible. For phenol, Keq≈10−10/10−14=104K_{eq} \approx 10^{-10}/10^{-14} = 10^4 — essentially complete.

Step-by-Step Reasoning

1. Identify the nature of each –OH group.

  • (i) C6H5OHC_6H_5OH: Phenol — the –OH is directly attached to an aromatic ring.
  • (ii) C6H5CH2OHC_6H_5CH_2OH: Benzyl alcohol — the –OH is on a carbon next to the ring, not on it.
  • (iii) (CH3)3COH(CH_3)_3COH: tert-Butanol — a tertiary aliphatic alcohol.
  • (iv) C2H5OHC_2H_5OH: Ethanol — a primary aliphatic alcohol.

2. Recall the acidity trend.

Phenols are more acidic than alcohols because the phenoxide ion (C6H5O−C_6H_5O^-) is stabilised by resonance delocalisation of the negative charge into the aromatic ring. Alcohols have no such stabilisation — their conjugate bases are localised on oxygen, making them much less stable.

pKa values (approximate):C6H5OH≈10,  C2H5OH≈16,  (CH3)3COH≈18,  C6H5CH2OH≈15pK_a \text{ values (approximate)}: \quad C_6H_5OH \approx 10,\; C_2H_5OH \approx 16,\; (CH_3)_3COH \approx 18,\; C_6H_5CH_2OH \approx 15

3. Apply the condition for reaction with aqueous NaOH. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.