2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers

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Karnataka 2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers

2nd PUC Chemistry Alcohols, Phenols and Ethers NCERT Textbook Questions

Question 1.
Write IUPAC names of the following compounds:

Answer:
(i) 2, 2, 4 – Trimethyl pentan – 3 – ol
(ii) 5 – Ethyl heptane – 2, 4 – diol
(iii) Butane – 2, 3 – diol
(iv) Propane – 1, 2, 3 – triol
(v) 2 – Methyl phenol
(vi) 4 – Methyl phenol
(vii) 2,5 – Dimethyl phenol
(viii) 2,6 – Dimethyl phenol
(ix) 1 – Methoxy – 2 – Methyl propane
(x) Ethoxy benzene
(xi) 1 – Phenoxyheptane
(xii) 2 – Ethoxy butane

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methyl butan-2-ol
(ii) phenyl propan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-trioi
(iv) 2,3 – Diethyl phenol
(v) 1 – Ethoxy propane
(vi) 2-Ethoxy-3-methyipentane
(vii) Cyclohexylmethanol
(viii) 3-CycIohexyipentan-3-oI
(ix) Cyclopent-3-en-l-ol
(x) 3-Chloromethylpentan-l-ol.
Answer:

Question 3.
1. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names.
2. Classify the. isomers of alcohols in question 3 (i) as primary, secondary and tertiary alcohols.
Answer:
1. The structures of all isomeric alcohols of molecular formula, C₅H₁₂O are shown below
(a)

2. Primary alcohol:
Pentan -1 -ol; 2-Methylbutan-1- ol;
3-Methylbutan- 1-ol; 2,2—Dimethylproparv4-ol;

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

Tertiary alcohol:
2 – Methylbutan -2 -ol

Question 4.
Explain why propanol has a higher boiling point than of hydrocarbon, butane?
Answer:
Propane undergoes intermolecular H – bonding because of the presence of the -OH group. On the other hand, butane does not.

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

Question 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 6.
What is meant by hydroboration- oxidation reaction? Illustrate it with an example.
Answer:
The addition of diborane to the alkene to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohol is called hydroboration-oxidation. For example:

The alcohols obtained by this process appear to have been formed by the direct addition of water to the alkene against the MarkownikofFs rule.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Orthonitrophenol is steam volatile while para-nitrophenol is not. This is on ac¬count of chelation (intramolecular H—bonding) in the molecule of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol, which is not steam volatile and its molecules are linked by intermolecular H—bonding.

Question 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 hydroperoxide.

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

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Chlorobenzene is fused with NaOH to produce sodium phenoxide, which gives phenol on acidification.

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
The mechanism of hydration of ethane to form ethanol involves three steps.
Step 1:
Protonation of ethane to form carbocation by the electrophilic attack of H₃O⁺
H₂O + H⁺ → H₃O⁺

Step 2:
Nucleophilic attack of water on carbocation:

Step 3:
Deprotonation to form ethanol

Question 12.
You are given benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Question 13.
Show how well you synthesise:
(i) 1-phenyl ethanol from a suitable alkene.
(ii) cyclohexyl methanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
Answer:

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
The acidic nature of phenol can be represented by the following two reactions
1. Phenol reacts with sodium to give sodium phenoxide, liberating H₂

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Question 15.
Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?
Answer:

The nitro-group is the art electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. Also, the O-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, O-nitrophenol is a stronger acid.

On the other hand, the methoxy group is an electron releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For this reason, O-nitrophenol is more acidic than O-methoxy phenol.

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution?
Answer:
The -OH group exerts a +R effect on the benzene ring under the effect of attacking electrophile. As a result, there is an increase in the electron density in the ring particularly at ortho and para positions, therefore electrophilic substitution occurs mainly at o-and p-positions. (For resonance hybrid structures of phenol refer to NCERT textbook.)

Question 17.
Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO₄ solution.
(ii) Bromine in CS₂ with phenol.
(iii) Dilute HNO₃ with phenol,
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:

Question 18.
Explain the following with an example.
1. Kolbe’s reaction.
2. Reimer-Tiemann reaction
3. Williamson ether synthesis.
4. unsymmetrical ether.
Answer:
1. Kolbe’s reaction:
When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon-di-oxide, followed by acidification, undergoes electrophilic substitution to give o-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.

2. Reimer-Tiemann reaction: When phenol is treated with chloroform (CHCl₃) in the presence of sodium hydroxide, a – CHo group is introduced at the ortho position of the benzene ring.

This reaction is known as the Reimer – Tiemann reaction.
The intermediate is hydrolyzed in the presence of alkalis to produce salicylic dehyde.

3. Williamson ether synthesis:
Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.
R-X + R- ONa → R-O-R + NaX
This reaction involves S_N2 attack of the alkoxides ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

Ifthealkyl halide is 2° or 3°, then elimination competes over substitution.

4. Unsymmetrical ether:
An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For e.g: ethyl methyl ether

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer:
The mechanisms of acid dehydration of ethanol to yield ethene involves the following three steps.
Steps 1:
Protonation of ethanol to form ethyl oxonium ion

Steps 2:
Formation of carbocation (rate-determining step)

Steps 3:
Elimination of proton to form ethene

The acid consumed in step 1 is released in step 3. After the formation of ethene, it is removed to shift the equilibrium in the forwarding direction.

Question 20.
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:
1. If propene is allowed to react with water in the presence of an acid as a catalyst, then propan – 2-ol is obtained.

2. If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

3. When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propan-l-ol on hydrolysis.

4. When methyl magnesium bromide is treated with propane, an adduct is the product which give 2- methyl propane – 2-ol hydrolysis.

Question 21.
Name the reagents used in the following reactions:

1. Oxidation of primary alcohol to a carboxylic acid.
2. Oxidation of 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:

1. Acidified KMnO₄
2. Pyridinium chlorochromate (PCC)
3. Bromine water
4. Acidified KMnO₄
5. 85% phosphoric acid
6. NaBH₄ or Li AiH₄

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxy -methane.
Answer:
The boiling point of ethanol is higher than methoxymethane because of the presence of strong intermolecular hydrogen bonding between ethanol molecules. As a result, ethanol exists as associates molecules. However, no such H-bonding is present in methoxy methane.

Question 23.
Give IUPAC names of the following ethers:
1.

2. CH₃OCH₂CH₂CL
3. O₂N-C₆H₄-OCH₃(p)
4. CH₃CH₂CH₂OCH₃
5.

6.

Answer:

1. 1- Ethoxy – 2 – methyl propane
2. 2 – chloro -1 – methoxy ethane
3. 4 – Nitroanisole
4. 1 – Methoxypropane
5. 1 – Ethoxy – 4,4 – dimethyl cyclohexane
6. Ethoxy benzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxy propane
(ii) Ethoxy benzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer:
The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide

But if secondary or tertiary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence they react with alkyl halides, which results in an elimination reaction.

Question 26.
How is 1-propoxypropane synthesised from propan-l-ol? Write the mechanism of this reaction.
Answer:
1-Propoxy propane can be synthesised from propan-l-ol by dehydration.

The mechanism of this rxn involves the following three steps.
Steps 1: Protonation

Step 2: Nucleophilic attack

Step 3: Deprotonation

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
The formation of ethers by dehydration of alcohol is a bimolecular reaction (S_N2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohol, the alkyl group is hindered. As a result, elimination dominates over substitution. Hence, in place of ethers, alkenes are formed.

Question 28.
Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane
(ii) methoxy benzene and
(iii) benzyl ethyl ether.
Answer:

Question 29.
Explain the fact that in aryl alkyl ethers
1. the alkoxy group activates the benzene ring towards electrophilic substitution and
2. it directs the incoming substituents to ortho and para positions in the benzene ring.
Answer:
1.

In aryl alkyl ethers, due to +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

2. It can be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 30.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
The mechanism of the reaction of HI with methoxymethane involves the following steps.
Step 1: Protonation of methoxymethane

Step2: Nucleophilic attack of I

Step 3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide.

Question 31.
Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.
Answer:

Question 32.
Show how would you synthesise the following alcohols from appropriate alkenes?

Answer:

Acid hydration of Pent-2-ene produces Penton – 3-ol along with penton-2-ol. Hence first rxn using Pent-l-ene is preferred.

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from the 3rd carbon atom.
Answer:
The mechanism of the given reaction involves the following steps:
Step 1: Protonation

Step 2: Formation of 2° carbocation by the elimination of water molecule

Step 3: Rearrangement by the hydride-ion shift

Step 4: Nucleophilic attack