Chapter 7 Alcohols Phenols and Ethers NCERT Solutions Class 12 Chemistry- PDF Download
Exercises
7.1. Write IUPAC name of the following compound:
Solution
(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) 2,5-DimethylphenoI
(vii) 4-Methylphenol
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane
7.2. 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) 4-chloro-3-ethylbutan-1-ol
Solution
(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) 4-chloro-3-ethylbutan-1-ol
7.3. (i) Draw the structures of all the isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question (a) primary, secondary and tertiary alcohols.'
Solution
(i) The molecular formula C5H12O represents eight isomeric alkanols. These are:
(a) Pentan-1-ol (1°)
CH3-CH2-CH2-CH2-CH2-OH
(b) 3-Methylbutan-1-ol (1°)
(c) 2-Methylbutan-1-ol (1°)
(d) 2, 2-Dimethylpropan-1-ol (1°)
(e) Pentan-2-ol (2°)
(f) 3-Methylbutan-2-ol (2°)
(g) 2-Methylbutan-2-ol (3°)
(h) Pentan-3-ol (2°)
(ii) Primary: (a), (b), (c) and (d); Secondary: (e), (f) and (h); Tertiary: (g)
7.4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Solution
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.
7.5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain the fact.
Solution
Water and alcohols both are polar in nature. When an alcohol is dissolved in water, it forms H-bonds with water molecule by breaking the H-bond already existing between water molecules.
Hydrocarbons are non-polar in nature and do not forms H-bonds with w ater molecules. Therefore, alcohols are readily soluble in water whereas hydrocarbons are not.
Hydrogen bonding among alcohol and water.
7.6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Solution
The addition of diborane to alkenes to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation.
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.
7.7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Solution
The three isomers are:
Solution
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.
Solution
To prepare phenol, cumene is first oxidised in the presence of air to cumene hydroperoxide.
The cumene hydroperoxide thus formed, is treated with dilute acid to prepare phenol and acetone as by-products.
7.10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Solution
Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.
7.11. Write the mechanism of hydration of ethene to yield ethanol.
Solution
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+:
H2O + H+ → H3O+
Step 2: Nucleophilic attack of water on carbocation formed:
Step 3: Deprotonation to form corresponding alcohols:
7.12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Solution
7.13. Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-1-ol using a suitable alkyl halide?
Solution
(i) 1-phenylethanol can be prepared from ethenylbenzene by addition of H2O in the presence of dil. H2SO4.
(ii) Hydrolysis of cyclohexylmethyl bromide by aqueous NaOH gives cyclohexylmethanol.
(iii) When 1-chloropentane is treated with NaOH, pentan-1-ol is produced.
CH3CH2CH2CH2CH2Cl + NaOH → CH3CH2CH2CH2CH2OH + NaCl
7.14. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Solution
The acidic nature of phenol can be represented by the following two reactions:
(a) Reaction with sodium: Phenol reacts with sodium to give sodium phenoxide, liberating H2 gas.
(b) Reaction with NaOH: Phenol reacts with NaOH to form sodium phenoxide and water as a by- product.
Phenol is more acidic than ethanol. This is due to the reason that after the loss of a proton phenol form phenoxide ion which is stabilized by resonance, while ethoxide ion formed after the loss of a proton from ethanol, does not.
The resonating structures of phenoxide ion are shown as below:
The lone pair of electrons on oxygen delocalizes into the benzene (mesomeric effect) which reduces the electron density in the O—H bond. The O—H bonds are weaker and therefore breaks easily whereas in ethanol the electron releasing inductive effect of the alkyl group increases the electron density on the O—H bond. This strengthens the bond so the bond does break not break easily. Therefore making ethanol less acidic than phenol.
7.15. Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol?
Solution
The nitro group is an electron withdrawing group. The presence of the group in the ortho-position decreases the electron density in the O–H bond. As a result, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed of the loss of proton is stabilised by resonance. Hence, ortho, nitrophenol is a stronger acid.
On the other hand, methoxy group is an electron releasing group. Thus, it increases the electron density in the O–H bond and hence, the proton can not be given out easily.
For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.
7.16. Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Solution
The −OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.
As a result, the benzene ring is activated towards electrophilic substitution.
7.17. Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 acid with phenol
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Solution
