Page No 344: - Chapter 11 - Alcohols Phenols & Ethers - Exercise Solutions - Ncert Solutions class 12 - Chemistry

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Question 11.1:

Write IUPAC names of the following compounds:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x) 

(xi) 

(xii)

Answer:

(i) 2, 2, 4-Trimethylpentan-3-ol

(ii) 5-Ethylheptane-2, 4-diol

(iii) Butane-2, 3-diol

(iv) Propane-1, 2, 3-triol

(v) 2-Methylphenol

(vi) 4-Methylphenol

(vii) 2, 5-Dimethylphenol

(viii) 2, 6-Dimethylphenol

(ix) 1-Methoxy-2-methylpropane

(x) Ethoxybenzene

(xi) 1-Phenoxyheptane

(xii) 2-Ethoxybutane

Question 11.2:

Write structures of the compounds whose IUPAC names are as follows:

(i) 2-Methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

(iii) 3,5-Dimethylhexane −1, 3, 5-triol

(iv) 2,3 − Diethylphenol

(v) 1 − Ethoxypropane

(vi) 2-Ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

(x) 3-Chloromethylpentan-1-ol.

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Question 11.3:

(i) Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names.

(ii) Classify the isomers of alcohols in Question 11.3 (i) as primary, secondary and tertiary alcohols.

Answer:

(i) The structures of all isomeric alcohols of molecular formula, C₅H₁₂O are shown below:

(a)

Pentan-1-ol (1°)

(b)

2-Methylbutan-1-ol (1°)

(c)

3-Methylbutan-1-ol (1°)

(d)

2, 2-Dimethylpropan-1-ol (1°)

(e)

Pentan-2-ol (2°)

(f)

3-Methylbutan-2-ol (2°)

(g)

Pentan-3-ol (2°)

(h)

2-Methylbutan-2-ol (3°)

(ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol;

3-Methylbutan-1-ol; 2, 2−Dimethylpropan-1-ol

Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol;

Pentan-3-ol

Tertiary alcohol: 2-methylbutan-2-ol

Question 11.4:

Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Answer:

Propanol undergoes intermolecular H-bonding because of the presence of −OH group. On the other hand, butane does not

Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

Question 11.5:

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Answer:

Alcohols form H-bonds with water due to the presence of −OH group. However, hydrocarbons cannot form H-bonds with water.

As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

Question 11.6:

What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Answer:

The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH₃)₂ to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Question 11.7:

Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.

Answer:

Question 11.8:

While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Answer:

Intramolecular H-bonding is present in o-nitrophenol. In p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, o-nitrophenol is steam volatile.

Question 11.9:

Give the equations of reactions for the preparation of phenol from cumene.

Answer:

To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide.

Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products.

Question 11.10:

Write chemical reaction for the preparation of phenol from chlorobenzene.

Answer:

Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

Question 11.11:

Write the mechanism of hydration of ethene to yield ethanol.

Answer:

The mechanism of hydration of ethene to form ethanol involves three steps.

Step 1:

Protonation of ethene to form carbocation by electrophilic attack of H₃O⁺:

Step 2:

Nucleophilic attack of water on carbocation:

Step 3:

Deprotonation to form ethanol:

Question 11.12:

You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.

Answer: