Chapter 11 Alcohols Phenols and Ethers Solutions
Question - 11 : - Which of the following is an appropriate set of reactantsfor the preparation of 1-methoxy-4-nitrobenzene and why?
Answer - 11 : -
(i)
(ii)
Answer
Set (ii) is an appropriate set of reactants for thepreparation of 1-methoxy-4-nitrobenzene.
In set (i), sodium methoxide (CH3ONa)is a strong nucleophile as well as a strong base. Hence, an eliminationreaction predominates over a substitution reaction.
Question - 12 : - Predict the products of the following reactions:
Answer - 12 : - 
Answer
(iv)
Write IUPAC names of the following compounds:
Answer - 13 : -
(I)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
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 - 14 : - Write structures of the compounds whose IUPAC names are asfollows:
Answer - 14 : -
(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)
(iv)
(v)
(vi)
Question - 15 : - (i) Drawthe structures of all isomeric alcohols of molecular formula C5H12Oand give their IUPAC names.
Answer - 15 : - (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.
Answer
(i) Thestructures of all isomeric alcohols of molecular formula, C5H12Oare 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) Primaryalcohol: 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
Explain why propanol has higher boiling point than that ofthe hydrocarbon, butane?
Answer - 16 : -
Propanol undergoes intermolecular H-bonding because of thepresence of −OH group. On the other hand, butane does not
Therefore, extra energy is required to break hydrogenbonds. For this reason, propanol has a higher boiling point than hydrocarbonbutane.
Question - 17 : - Alcohols are comparatively more soluble in water thanhydrocarbons of comparable molecular masses. Explain this fact.
Answer - 17 : -
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 inwater than hydrocarbons of comparable molecular masses.
Question - 18 : - What is meant by hydroboration-oxidation reaction?Illustrate it with an example.
Answer - 18 : -
The addition of borane followed by oxidationis known as the hydroboration-oxidation reaction. For example, propan-1-ol isproduced by the hydroboration-oxidation reaction of propene. In this reaction,propene reacts with diborane (BH3)2 toform trialkyl borane as an addition product. This addition product is oxidizedto alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
Question - 19 : - Give the structures and IUPAC names ofmonohydric phenols of molecular formula, C7H8O.
Answer - 19 : - 
Question - 20 : - While separating a mixture of ortho and para nitrophenolsby steam distillation, name the isomer which will be steam volatile. Givereason.
Answer - 20 : -
Intramolecular H-bonding is present in o-nitrophenol.In p-nitrophenol, the molecules are strongly associated due to thepresence of intermolecular bonding. Hence, o-nitrophenol is steamvolatile.
