NCERT Solutions for Class 12 Chemistry Chapter 7 – Alcohols, Phenols and Ethers (English Medium)

👉 Propanol contains an –OH (hydroxyl) group.

✅ Oxygen is highly electronegative, so the O–H bond is polar.

✅ This allows propanol molecules to attract each other strongly through hydrogen bonding:

O–H···O (between two propanol molecules)

✅ Stronger intermolecular forces mean more heat energy is needed to separate molecules, so the boiling point increases.

👉 Butane is a non-polar hydrocarbon (only C–C and C–H bonds).

✅ It does not have an electronegative atom like O, N, or F.

✅ So, it shows only weak London dispersion (van der Waals) forces.

✅ Weak forces require less energy to break, so butane boils at a lower temperature.

👉 Alcohols contain the –OH (hydroxyl) group.

✅ The O–H bond is polar (oxygen attracts electrons strongly).
✅ So alcohol molecules can interact strongly with water .

Water is also polar and forms hydrogen bonds.

✅ Alcohols form hydrogen bonds such as:

R–O–H ··· O–H₂ (between alcohol and water)

✅ These strong attractions help alcohol molecules mix well with water, so solubility is higher.

👉 Hydrocarbons have only C–C and C–H bonds , so they are non-polar .

✅ They cannot form hydrogen bonds with water.
✅ Only weak interactions exist, so hydrocarbons do not mix well with water and remain mostly insoluble.

Hydroboration:
👉 CH₃–CH=CH₂ + BH₃ → (after 3 additions) (CH₃CH₂CH₂)₃B

Oxidation:
👉 (CH₃CH₂CH₂)₃B + H₂O₂/OH⁻ → 3 CH₃CH₂CH₂OH + B(OH)₃

✅ Product formed: propan-1-ol (anti-Markovnikov alcohol)

Phenol loses H⁺ to form phenoxide ion (ArO⁻).
✅ More stable phenoxide ion → more acidic phenol.

---

The nitro group is strongly electron-withdrawing:
• −I effect: pulls electron density through sigma bonds
• −M effect: withdraws electron density by resonance

✅ This reduces the negative charge density on oxygen and stabilizes the phenoxide ion by resonance (charge delocalization).
👉 इसलिए H⁺ निकलना आसान हो जाता है → acidity increases.

---

The methoxy group shows:
• +M effect (lone pair donation): donates electron density into the ring

✅ This increases electron density in the ring and makes the phenoxide ion less stable (negative charge is not relieved).
👉 इसलिए H⁺ निकलना comparatively मुश्किल → acidity decreases.

---

In ortho-nitrophenol, the nitro group is close enough to give extra stabilization (often discussed as “ortho effect”), so acidity is further enhanced compared to methoxy.

Ethanol (CH₃CH₂OH) contains an –OH group.

✅ The O–H bond is polar, so ethanol molecules attract each other strongly by hydrogen bonding:

👉 CH₃CH₂O–H ··· O–CH₂CH₃

✅ Strong attractions mean more heat energy is needed to separate molecules → higher boiling point.

---

Methoxymethane (CH₃OCH₃) has oxygen, but it has no O–H bond.

✅ So it cannot donate hydrogen for hydrogen bonding between its own molecules.
👉 It mainly has dipole–dipole and dispersion forces, which are weaker than hydrogen bonding.

✅ Therefore, it boils at a lower temperature than ethanol.

In Ar–O–R, oxygen has lone pairs.
✅ One lone pair is donated into the benzene ring through resonance:

👉 Ar–O–R ⇌ resonance forms where oxygen forms a π-bond with the ring

This resonance donation pushes electron density into the ring.
✅ More electron density = the ring can attack an electrophile (E⁺) more easily.
👉 इसलिए reaction rate increases, ring becomes activated.

---

When oxygen donates by resonance, the resonance structures place extra electron density (negative character) mainly at:

✅ ortho positions
✅ para position

So, during electrophilic substitution, the intermediate (sigma complex) formed by attack at ortho/para is more stabilized by resonance than attack at meta.

👉 Ortho/para attack gives a more resonance-stabilized carbocation intermediate, hence preferred.

---

Oxygen is electronegative, so –OR shows a −I effect (pulls electrons by sigma bond).
But in aryl ethers:
✅ +M (resonance donation) is stronger than −I, so overall the ring is activated and ortho/para directing.