NCERT Solutions for Class 12 Chemistry Chapter 11 - Alcohols, Phenols and Ethers

Question 1:

Give IUPAC names of the following compounds:

Answer:
  1. 4-Chloro-2, 3-dimethylpentan-1-ol
  2. ‘2-Ethoxypropane
  3. 2, 6-Dimethylphenol
  4. 1-Ethoxy-2-nitrocyclohexane
Question 2:

Give the structures and IUPAC names of products expected from the following reactions :

  1. Catalytic reduction of butanal
  2. Hydration of propene in the presence of dilute sulphuric acid.
  3. Reaction of propanone with methyl magnesium bromide followed by hydrolysis.
Answer:

Question 3:

Arrange the following sets of compounds in order of their increasing boiling points:

  1. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
  2. Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Answer:

Increasing order of boiling points :

  1. Methanol < ethanol < propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol
  2. n-Butane < ethoxyethane < pentanal < pentan-1-ol
Question 4:

Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2,4,6-trinitrophenol, nitrophenol, 3,5-dinitrophenol, phenol, 4-methyl phenol.

Answer:

Increasing order of acid strength is :
Propan-1-ol, 4-methyl phenol, phenol, 3-nitro phenol, 3,5-dinitrophenol, 2,4,6-trinitrophenol.

Question 5:

Write the structures of the major products expected from the following reactions:

  1. Mononitration of 3-methyl phenol
  2. Dinitration of 3-methyl phenol
  3. Mononitration of phenyl ethanoate
Answer:

Both OH and CH3 groups are o- and p- directing. Therefore, position 2,4 and 6 are activated. But due to steric hindrance, substitution does not occur at position 2, i.e., in between two groups.

Question 6:

The following in not an appropriate reaction for the preparation of t-butyl ether:

  1. What would be the major product of the reaction ?
  2. Write a suitable reaction for the preparation of tert-butyl ethyl ether.
Answer:
  1. The major product of the given reaction is 2-methylprop-1-ene. This is because sodium ethoxide is a strong nucleophile as well as a strong base. Therefore, elimination reaction predominates over substitution.

  2. To prepare tert-butyl ethyl ether, the alkyl halide should be 1° (chloroethane) and the nucleophile should be sodium tert-butoxide.

Question 7:

Give the major products that are formed by heating each of the following ethers with HI.

Answer:

Question 8:

Classify the following as primary, secondary and tertiary alcohols :

Answer:

(i) Primary alcohol (ii) Primary alcohol (iii) Primary alcohol (iv) Secondary alcohol (v) Secondary alcohol (vi) Tertiary alcohol.

Question 9:

Identify allylic alcohols in the above examples.

Answer:

(ii) and (vi) are allylic alcohols.

Question 10:

Write structures of the products of the following reactions :

Answer:

Question 11:

Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl—ZnCl2, (b) HBr and (c) SOCl2.
(i) Butan-1-ol (ii) Methylbutan-2-ol

Answer:

(a) HCl—ZnCl2. It is Lucas reagent. Butan-1-ol does not react with HCl­—ZnCl2 at room temperature. However, cloudiness appears only upon heating.

Question 12:

Predict the major product of acid catalysed dehydration of (i) 1-methyl cyclohexanol and (ii) butan-1-ol

Answer:

(i) Acid catalysed dehydration of 1-methyl cyclohexanol gives two products I and II.But product I is highly substituted and hence, according to Saytzeff rule, is major product.

(ii) The acid-catalysed dehydration of butan-1-ol produces but-2-ene as the major product and but-1-ene as the minor product. This is because dehydration of alcohols occurs through the formation of carbocation intermediates. During the reaction, it first forms 1°-carbocation (I) which undergoes 1, 2-hydride shift to form 2°-carbocation (II). This then loses a proton to form more stable but-2-ene instead of but-1-ene. This is because but-2-ene is more stable and, according to Saytzeff rule, is preferably formed.

Question 13:

Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Answer:

The resonating structures of phenoxide ion and ortho and para nitrophenoxide ions are :

It is clear from the above structures that due to –I effect of –NO2 group, o- and p-nitrophenoxide ions are more stable because they have additional resonance structures (IV′) and (V′′) than phenoxide ion. Hence, o- and p-nitrophenols are more acidic than phenol.

Question 14:

Write the equations involved in the following reactions :

  1. Reimer-Tiemann reaction
  2. Kolbe’s reaction
Answer:
  1. Reimer-Tiemann reaction. When phenol is refluxed with chloroform in the presence of aqueous caustic alkali at 340 K, an aldehydic group (CHO) gets introduced in the ring at a position ortho to the phenolic group. Ortho hydroxy benzaldehyde or salicylaldehyde is formed as the product of the reaction.

    This reaction is called Reimer-Tiemann reaction. In addition to o-salicylaldehyde, small amount of p-salicylaldehyde is also formed but the major product is ortho.
  2. Kolbe’s reaction. When phenol is treated wth carbon dioxide at about 400 K under pressure, salicylic acid is obtained as the product.

    In set (B), nucleophilic attack of 4-nitrophenoxide ion on methyl bromide gives the desired product.
    A small amount of para isomer is also obtained and if the temperature be allowed to rise above 410 K, the para isomer dominates.
Question 15:

Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Answer:

Question 16:

Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why ?

Answer:

In set (B), nucleophilic attack of 4-nitrophenoxide ion on methyl bromide gives the desired product.

Question 17:

Predict the products of the following reactions :

Answer:

Both the alkyl groups attached to O atom are primary and therefore, attack of Br– ion occurs on the smaller methyl group.

Question 18:

Write IUPAC names of the following compounds :

Answer:
  1. 2,2,4-Trimethylpentan-3-ol
  2. 5-Ethylheptane-2,4-diol
  3. Butane-2,3-diol
  4. Propane-1,2,3-triol
  5. 2-Methylphenol
  6. 4-Methylphenol
  7. 2,5-Dimethylphenol
  8. 2,6-Dimethylphenol
  9. 1-Methoxy-2-methylpropane
  10. Ethoxybenzene
  11. 1-Phenoxyheptane
  12. 2-Ethoxybutane
Question 19:

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

  1. 2-Methylbutan-2-ol
  2. 1-Phenylpropan-2-ol
  3. 3, 5-Dimethylhexane-1, 3, 5-triol
  4. 2, 3-Diethylphenol
  5. 1-Ethoxypropane
  6. 2-Ethoxy-3-methylpentane
  7. Cyclohexylmethanol
  8. 3-Cyclohexylpentan-3-ol
  9. Cyclopent-3-en-1-ol
  10. 3-Chloromethylpentan-1-ol
Answer:

Question 20:

(a) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(b) Classify the isomers of alcohols in question 3(a) as primary, secondary and tertiary alcohols.
Ans. The isomeric alcohols of molecular formula C5H12O are :

Answer:

The isomeric alcohols of molecular formula C5H12O are :

Question 21:

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

Answer:

The molecules of propanol are held together by intermolecular hydrogen bonding while butane molecules have only weak van der Waals forces of attraction. Since hydrogen bonds are stronger than van der Waals forces, therefore, propanol has higher boiling point than butane.

Question 22:

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

Answer:

Alcohols can form hydrogen bonds with water and break the hydrogen bonds existing between water molecules. Hence, they are soluble in water.

On the other hand, hydrocarbons cannot form hydrogen bonds with water molecules and hence are insoluble in water.

Question 23:

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

Answer:

The addition of diborane to alkenes to form trialkylboranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation reaction.

Question 24:

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

Answer:

hree isomers are possible

Question 25:

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

Answer:

o-Nitrophenol is steam volatile due to chelation because of intramolecular hydrogen bonding. On the other hand, p-nitrophenol is not steam volatile because of intermolecular hydrogen bonding. Hence, o-nitrophenol can be separated from p-nitrophenol by steam distillation.

Question 26:

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

Answer:

Question 27:

Write chemical reaction for the preparation of phenol from chlorobenzene.

Answer:

Question 28:

Write the mechanism of hydration of ethene to yield ethanol.

Answer:

Question 29:

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

Answer:

Question 30:

Show how will you synthesize

  1. 1-phenylethanol from a suitable alkene,
  2. cyclohexylmethanol using an alkyl halide by an SN2 reaction,
  3. pentan-1-ol using a suitable alkyl halide ?
Answer:

Question 31:

Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Answer:

The reactions showing acidic character of phenol are :

  1. Reaction with sodium. Phenol reacts with sodium to give H2 gas

  2. Reaction with NaOH. Phenol dissolves in NaOH to give sodium phenoxide and water

Difference in chemical behaviours of phenols and alcohols. Both alcohol and phenol contain hydroxyl (OH) group attached to an alkyl or aryl group respectively. Their structures are

Oxygen is more electronegative than both carbon and hydrogen and therefore, the distribution of electrons of C—O and O—H bonds result in slightly more electron density near the oxygen atom. As a result, an alcohol molecule is dipolar. For example, the dipole moment of methanol has been found to 1.71D. Phenol has smaller dipole moment (1.54 D) than methanol. This is due to he electron attracting benzene ring in phenol. Phenols are distinctly acidic in nature. They turn blue litmus red and react with alkalies to form salts.

But the acidic character of alcohols is so weak that they are considered practically to be neutral. They donot turn blue litmus red and not react with alkalies to form salts.

Question 32:

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Answer:

Due to strong —R and —I effect of —NO2 group, electron density in O—H bond decreases and hence the loss of a proton becomes easy.

The two negative charges repel each other and therefore, destabilize the o-methoxyphenoxide ion. Thus, o-nitrophenol is more acidic than o-methoxyphenol.

Question 33:

Explain how does the —OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Answer:

Phenol is a resonance hybrid of the following structures :

As a result of +R effect of the OH group, the electron density in the benzene ring increases and thereby facilitating the attack of the electrophile. Hence, presence of OH group activates the benzene ring towards electrophilic substitution reaction. Thus, because the —OH group is an activating group, these reactions occur at a faster rate than reactions of benzene itself. The —OH group is ortho-para directing and therefore, incoming group comes at ortho or para position. This is due to the reason that because of electronic effects caused by —OH group, the ortho and para positions become electron rich as shown.

Question 34:

Give equations of the following reactions :

  1. Oxidation of propan-1-ol with alkaline KMnO4 solution.
  2. Bromine in CS2 with phenol.
  3. Dilute HNO3 with phenol.
  4. Treating phenol with chloroform in presence of aqueous NaOH.
Answer:

Question 35:

Write mechanism of acid dehydration of ethanol to yield ethene.

Answer:

The mechanism is shown below :

  1. Alcohol combines with a proton to form a protonated alcohol.

  2. The protonated alcohol loses a water molecule to form a carbocation.

  3. The carbocation then eliminates a proton and undergoes rearrangement of elements to form the alkene.

Question 36:

How are the following conversions carried out ?

  1. Propene Propan-2-ol
  2. Benzyl chloride Benzyl alcohol
  3. Ethyl magnesium chloride Propan-1-ol
  4. Methyl magnesium bromide 2-Methylpropan-2-ol
Answer:

Question 37:

Name the reagents used in the following reactions :

  1. Oxidation of a primary alcohol to carboxylic acid
  2. Oxidation of a primary alcohol to aldehyde
  3. Bromination of phenol to 2,4,6-tribromophenol
  4. Benzyl alcohol to benzoic acid
  5. Dehydration of propan-2-ol to propene
  6. Butan-2-one to butan-2-ol
Answer:

Question 38:

Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Answer:

The boiling point of ethanol is higher than methoxymethane because of the presence of strong intermolecular hydrogen bonding between ethanol molecules. Because of hydrogen bonding, energy has to be supplied to overcome the forces of attraction between molecules and therefore, boiling point is high.

However, no such hydrogen bonding exists in methoxymethane.

Question 39:

Give IUPAC names of the following ethers :

Answer:

(i) 1-Methoxy-2-methylpropane (ii) 2-Chloro-1-methoxyethane (iii) 4-Nitroanisole (iv) 1-Methoxypropane (v) 4-Ethoxy-1, 1-dimethylcyclohexane (vi) Ethoxybenzene

Question 40:

Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis :

  1. 1-Propoxypropane
  2. Ethoxybenzene
  3. 2-Methoxy-2-methylpropane
  4. 1-Methoxyethane
Answer:

Question 41:

Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.

Answer:

The main limitation of Williamson’s synthesis is that for preparing unsymmetrical ethers, the halide used should preferably be primary because secondary and tertiary alkyl halides may form alkenes as major products due to elimination process. For 2° and 3° alkyl halide, elimination competes over substitution. If a 3° alkyl halide is used, an alkene is the only product and no ether is formed. For example, if we want to prepare ethyl tert-butyl ether, then we should take ethyl bromide and sodium tert-butoxide.

Question 42:

How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.

Answer:

Williamson synthesis

Question 43:

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Answer:

Acid-catalysed dehydration of 1° alcohols to ethers takes place by SN2 reaction involving nucleophilic attack by the alcohol molecule on the protonated alcohol molecule as :

However, under these conditions, 2° and 3° alcohols give alkenes rather than ethers. This is because of the steric hindrance, nucleophilic attack by the alcohol on the protonated alcohol molecule does not take place. Instead of this, the protonated 2° and 3° alcohols lose a molecule of water to form stable 2° and 3° carbocations. These carbocations then prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecule to form ethers.

Question 44:

Write the equation of the reaction of hydrogen iodide with

  1. 1-propoxypropane,
  2. methoxybenzene and
  3. benzyl ethyl ether.
Answer:

Question 45:

Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para position in benzene rin

Answer:

The alkoxy group increases the electron density on the benzene ring and therefore, activates the aromatic ring towards electrophilic substitution reaction as given below :

As is clear, structures III, IV and V show high electron density at ortho and para positions and therefore, direct the incoming substituents to o- and p-position in the benzene ring.

Question 46:

Write mechanism of the reaction of HI with methoxymethane.

Answer:

(i) Ether molecule gets protonated

(ii) The protonated ether undergoes SN2 attack by I– ion

If HI is in excess, the methanol formed in step 2 is also converted into methyl iodide as:

Question 47:

Write equations of the following reactions.

  1. Friedel Crafts reaction – alkylation of anisole
  2. Nitration of anisole
  3. Bromination of anisole in ethanoic acid medium
  4. Friedel Craft’s acetylation of anisole
Answer:
  1. Friedel Craft reaction-alkylation of anisole. Anisole reacts with methyl chloride in the presence of AlCl3 (Lewis acid) as catalyst introducing alkyl group at o- and p-positions.

  2. Nitration. Anisole on treating with a mixture of conc. HNO3 and H2SO4 gives a mixture of o- and p-nitro product.

  3. Bromination. Bromination of anisole gives o- and p-bromo product.

  4. Friedel Craft’s acetylation. Anisole reacts with acetyl chloride in the presence of AlCl3 (Lewis acid) to give o- and p-methoxy acetophenone.

Question 48:

Show how would you synthesize the following alcohols from appropriate alkenes ?

Answer:

The alkenes for synthesising the alcohols can be predicted by first dehydrating the alcohol to give single alkene or a mixture of alkenes. If a mixture of alkenes is possible, then find out which alkene gives the desired alcohol. It must be remembered that acid catalysed addition of H2O to alkenes occurs in accordance with Markovnikov’s rule.

Question 49:

When 3-methylbutan-2-ol is treated with HBr following reaction takes place;

Give a mechanism for this reaction.

Answer:

(i) Alcohol gets protonated.

Helpline