Q.Explain why is the OH group in phenols more strongly held as compared to OH group in alcohols.

Concept and Intuition

Acidity here is governed by how well the conjugate base (the corresponding phenoxide/alkoxide anion) is stabilised:

• p-Nitrophenol: the para $-$NO$_2$ group is strongly electron-withdrawing by resonance, delocalising and stabilising the negative charge of the phenoxide ion extensively $\Rightarrow$ most acidic of the four (lowest $pK_a$).
• Phenol: the phenoxide anion is stabilised by resonance delocalisation into the ring, but there is no additional electron-withdrawing substituent $\Rightarrow$ moderately acidic, $pK_a \approx 10.0$.
• p-Cresol: the para $-$CH$_3$ group is electron-donating (+I/hyperconjugation), which destabilises the phenoxide anion slightly relative to phenol $\Rightarrow$ marginally less acidic than phenol, $pK_a$ slightly higher ($\approx 10.2$).
• Ethanol: the ethoxide anion has no aromatic ring to delocalise charge into at all, so it is far less stabilised $\Rightarrow$ ethanol is a much weaker acid, with a much higher $pK_a$ ($\approx 15.9$).

Step-by-Step Solution

1. Rank acidity qualitatively: p-Nitrophenol (most acidic, EWG) > Phenol > p-Cresol (EDG lowers acidity slightly vs phenol) > Ethanol (no ring stabilisation, weakest acid, highest $pK_a$). …

Concept and Intuition

Acid strength of an -OH-bearing compound correlates with how well its conjugate base (the anion formed after losing $\mathrm{H^+}$) is stabilized. Stabilization by resonance is most effective when the contributing resonance structures are equivalent (same energy) — equivalent structures each contribute equally and substantially to the true (hybrid) structure, spreading the negative charge symmetrically and lowering the energy a great deal.

• In a carboxylate ion ($\mathrm{RCOO^-}$), the negative charge is delocalized over the two oxygen atoms via two resonance structures that are mirror images of each other — completely equivalent, both localizing charge on the (highly electronegative) oxygen. This gives maximal stabilization.
• In a phenoxide ion, resonance delocalizes the negative charge onto oxygen (one structure) and onto ring carbons (at ortho and para positions, three additional structures). These carbon-centred structures are not equivalent to the oxygen-centred one — they put negative charge on a less electronegative atom (carbon) and are higher in energy, contributing less to overall stabilization.

Step-by-Step Solution

1. Confirm Assertion: carboxylic acids (pKₐ ≈ 3–5) are indeed markedly more acidic than phenols (pKₐ ≈ 10) — true. …

Concept and Intuition

The acidity of a substituted phenol is governed by how well the ring substituent stabilises

the phenoxide (conjugate base) anion. Electron-withdrawing groups (like $\mathrm{-NO_2}$)

stabilise the negative charge and increase acidity; this effect is strongest when the group is

at the ortho/para position, where it can directly delocalise the negative charge through

resonance. At the meta position, $\mathrm{-NO_2}$ can only act inductively (no direct

resonance path to the phenoxide oxygen), so its acid-strengthening effect is smaller than at

para. Electron-donating groups (like $\mathrm{-CH_3}$) destabilise the phenoxide anion (push

electron density in, the opposite of what's needed) and so decrease acidity relative to plain

phenol.

Step-by-Step Solution

1. p-Nitrophenol (IV): $\mathrm{-NO_2}$ at para position — full resonance stabilisation of the phenoxide ion — strongest acid of the four.
2. m-Nitrophenol (III): $\mathrm{-NO_2}$ at meta — only inductive stabilisation, weaker than para but still more acidic than unsubstituted phenol. …

1. Understand the claim in Statement I

   Statement I says the C–O bond length in methanol is 136 pm and in phenol it is 142 pm — meaning phenol has a longer C–O bond. Intuitively, a longer bond is weaker. But we know from organic chemistry that phenol’s C–O bond has partial double-bond character due to resonance between the oxygen lone pairs and the aromatic ring. A double bond is shorter than a single bond. Therefore, phenol’s C–O bond should be shorter than methanol’s, not longer.

   • In methanol (CH₃OH), the C–O bond is a pure single bond (sp³ carbon, sp³ oxygen). Typical C–O single bond length is ~143 pm.
   • In phenol (C₆H₅OH), the oxygen’s lone pairs delocalize into the ring, giving the C–O bond about 30–40% double-bond character. This shortens it to roughly 136–137 pm. So Statement I is incorrect — it has the lengths reversed.

2. Examine Statement II about bond angles

   Statement II claims the C–O–H bond angle in methanol and phenol is “almost the same.” Let’s check:

   • In methanol, the oxygen is sp³ hybridized (two lone pairs, two sigma bonds), so the ideal bond angle is near the tetrahedral angle (~109.5°). The actual C–O–H angle in methanol is about 108.9°.
   • In phenol, the oxygen is also sp³ hybridized in the local sense, but one of its lone pairs is partially delocalized into the π system of the ring. This delocalization reduces the electron density on oxygen, which slightly changes the balance of repulsions. More importantly, the oxygen’s hybridization shifts toward sp² character (since the lone pair involved in resonance is in a p orbital). With sp² hybridization, the C–O–H angle would be closer to 120°. Experimentally, the C–O–H angle in phenol is about 109°? Actually, careful measurements show it is around 109° as well — but wait, that seems contradictory. Let’s check data:
     • Methanol: C–O–H ≈ 108.9°
     • Phenol: C–O–H ≈ 109.0° (some sources say 109.5°) So they are indeed very close. However, the reason they are close is not because nothing changes — it’s because the oxygen’s hybridization doesn’t fully become sp²; the resonance is partial. But the statement itself is about the fact that they are almost the same, which is true. But — there is a subtlety: The C–O–H angle in phenol is actually slightly larger than in methanol? Some data: methanol 108.9°, phenol 109.2°. That is “almost the same.” So Statement II is correct in its claim.

3. Re-evaluate with reliable data

   Let’s confirm bond lengths: …

Concept and Intuition

Phenol is acidic because the phenoxide ion ($\mathrm{C_6H_5O^-}$) is stabilized by resonance delocalization of the negative charge into the ring. Any substituent that makes the ring more electron-rich (an electron-donating group, EDG) works against this delocalization/stabilization of the negative charge, and so decreases acidity. Conversely, an electron-withdrawing group (EWG) stabilizes the anion further and increases acidity.

Step-by-Step Solution

1. Phenol (I): the reference compound, moderately acidic ($pK_a \approx 10$).
2. p-Cresol (II): the para $\mathrm{-CH_3}$ group is a weak electron donor (hyperconjugation/+I), pushing electron density into the ring and slightly destabilizing the phenoxide -- makes it less acidic than phenol.
3. 4-Methoxyphenol (III): the para $\mathrm{-OCH_3}$ group is a strong $+M$ (resonance) electron donor -- its lone pair conjugates directly with the ring, pushing even more electron density onto the ring (and especially onto the position para to it, i.e. right where the phenolic oxygen sits), destabilizing the phenoxide anion more than $\mathrm{-CH_3}$ does. …

Concept and Intuition

A phenol's acidity depends on how stable its conjugate base (the phenoxide ion) is. Electron-withdrawing groups (like $-NO_2$) at ortho/para positions delocalize the negative charge of the phenoxide oxygen into the substituent via resonance, spreading out (stabilizing) the charge and making the O–H bond easier to ionize. Electron-donating groups (like $-CH_3$) do the opposite — they push electron density onto the ring/oxygen, destabilizing the phenoxide and making the phenol less acidic than plain phenol.

Step-by-Step Solution

1. Rank the substituent effects: $-NO_2$ (strong EWG, both resonance and inductive) $\gg$ $-H$ (no effect, plain phenol) $>$ $-CH_3$ (weak EDG, destabilizes phenoxide).
2. Compare the number/position of $-NO_2$ groups: 4-nitrophenol has one $-NO_2$ at the para position (good resonance overlap with the phenoxide oxygen) — more acidic than phenol. 2,4,6-trinitrophenol has THREE $-NO_2$ groups, at both ortho positions AND the para position, all able to resonance-stabilize the phenoxide charge simultaneously. …