Gujarat Board GSEB Textbook Solutions Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers Textbook Questions and Answers, Additional Important Questions, Notes Pdf.

Gujarat Board Textbook Solutions Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers

GSEB Class 12 Chemistry Alcohols, Phenols and Ehers InText Questions and Answers

Question 1.
Classify the following as primary, secondary and tertiary alcohols:

Answer:
Primary alcohols : i, ii, iii
Secondary alcohols : iv and v
Tertiary alcohols : vi

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols : ii and vi

Question 3.
Name the following compounds according to IUPAC system.

Answer:
i. 3-Chloromethyl-2-isopropylpentan-1 -ol
ii. 2,5-Dimethylhexane-l,3-diol
iii. 3-Bromocyclohexanol
iv. Hex-l-en-3-ol
v. 2-Bromo-3 -methylbut-2-en-1 -ol

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Answer:

Question 5.
Write structures of the products of the following reactions:

Answer:

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with
(a) HCl – ZnCl₂
(b) HBr
(c) SOCl₂
(i) Butan-1-ol
(ii) 2-Methylbutan-2-ol
Answer:
(i) Butan-l-ol

(ii) 2-Methylbutan-2-ol

Question 7.
Predict the major product of acid catalysed dehydration of
(i) l-methyl cyclohexanol
(ii) butan-1-ol
Answer:
(i) 1-Methylcyclohexane
(ii) But-1-ene

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(i) Orthonitro phenol:

(ii) Paranitro Phenol:

Question 9.
Write the equations involved in the following reactions:
i. Reimer – Tiemann reaction
ii. Kolbe’s reaction
Answer:

Question 10.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methyl pentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 11.
Which of the following is an appropriate set of reactants for the preparation of 1 -methoxy-4- nitrobenzene and why?

Answer:

Aryl halides have very low reactivity towards nucleophilic substitution reaction.

Question 12.
Predict the products of the following reactions:
i. CH₃ – CH₂ – CH₂ – O – CH₃ + HBr →

Answer:
i. CH₃ – CH₂ – CH₂OH + CH₃ + Br

iv. (CH₃)₃ – I + C₂H₅OH

GSEB Class 12 Chemistry Alcohols, Phenols and Ethers Text Book Questions and Answers

Question 1.
Write IUPAC names of the following compounds:

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 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:

Question 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 3(i) as primary, secondary and tertiary alcohols.
Answer:
(i)

(ii) (a) primary, (b) primary, (c) primary, (d) secondary, (e) secondary, (f) secondary, (g) tertiary.

Question 4.
Explain why propanol has a higher boiling point than that of hydrocarbon, butane?
Answer:
Hydrogen bonding is present in propanol.

Question 5.
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 H—bond exists 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 6.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
Hydroboration-oxidation reaction:
Diborane, B₂H₆ or (BH₃)₂. being electron deficient acts as an electrophile and reacts with alkenes to yield trialkvlboranes, R₃B. These are oxidised to alcohols on reaction with hydrogen peroxide in presence of alkali.

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:
o-Nitrophenol is steam volatile because of intramolecular hydrogen bonding.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:

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

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of alkeiies : Alkenes may be hydrated by absorption in concentrated sulphuric acid, followed by hydrolysis of alkyl sulphate.
Example:

CH₃CH₂HSO₄ + H₂O → CH₃CH₂OH + H₂SO₄
Alkenes required for hydration is obtained by cracking of petroleum. In case of uns mmetrical alkenes, the addition reaction takes place in accordance with Markownikov’s rule.

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 will you synthesise:
(i) 1 -phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an S_N2 reaction.
Answer:

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:

Phenol is stronger than alcohol because the phenoxide ion formed from phenol is resonance stabilised while the alkoxide ion from alcohol is not resonance stabilised.

Question 15.
Explain why is ortho n itrophenol more acidic than ortho methoxyphenol.
Answer:
Due to electron withdrawing effect of nitro group and electron releasing effect of methoxy group.

Question 16.
Explain how does the – OH group attached to a carbon of benzene ring activate it towards electrophilic substitution.
Answer:
A non-bonded electron pair of oxygen atom of the phenolic OH interacts with the n system of the benzene ring and gets delocalised. The increased electron density in the benzene ring activates it towards electrophilic substitution reactions.

Question 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 with phenol.
(iv) Treating phenol wih chloroform in presence of aqueous NaOH.
Answer:

Question 18.
Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
Answer:
(i) Kolbe’s reaction : When sodium phenoxide is heated 10413 K with carbon dioxide under a pressure of about 4-7 atmospheres, the sodium salt of salicylic acid is formed. The salt on acidification yields salicylic acid (2-hydroxy benzoic acid).

Salicylic acid is used for the manufacture of aspirin (2-acetoxy benzoic acid)

Aspirin possesses analgesic, anti-inflammatory and antipyretic properties.

(ii) Reimer-Tiemann reaction : When phenol is treated with chloroform and alkali at 340 K. a phenolic aldehyde known as salicylaldehyde(o-hydroxy benzaldehyde) is formed. A small amount of p-isomer is also obtained.

But when phenol is treated with carbon tetrachioride in the presence of aqueous alkali yields o-hydroxybenzoic acid (as the major component).

(iii) Williamson ether synthesis : When sodium alkoxide is heated with alkyl halide, an ether is formed. This reaction is called Williamson’s synthesis. This method is most useful in preparation of mixed ether.
Example:

(iv) Unsymmetrical ether : Ethers in which the alkyl groups attached to oxygen are different, then these are named as mixed ethers. Example: C₂H₅OCH₃ Ethylmethyl ether.

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer:
(i) Formation of protonated alcohols:

(ii) Formation of carbocation

This step is slow and hence is the rate determining step of the reaction.

(iii) Loss of proton by the carbocation to form the alkene molecule.

Alkene is removed from the reaction mixture to drive the equilibrium to the right side.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium bromide → Propan-l-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:
(i) Hydration of propene.

(ii) By nucleophilic substitution of – Cl in benzyl chloride using dilute NaOH.

Question 21.
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:

1. Acidified K₂Cr₂O₇ or neutral acidic or alkaline KMnO₄.
2. Pyridinium chlorochromate (pec) in CH₂Cl₂ or Cu at 573 K.
3. Bromine water (Br₂/H₂O)
4. Acidified or alkaline KMnO₄
5. Cone. H₂SO₄ at 443K or 85% phosphoric acid at 443K.
6. Ni/H₂ or NaBH₄ or LiAlH₄.

Question 22.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol undergoes intermolecular hydrogen bonding while methoxymethane (ether) does not.

Question 23.
Give IUPAC names of the following ethers:

Answer:
(i) 1-Ethoxy-2-methylpropane.
(ii) 2-Chloro-l-dimethoxyethane.
(iii) 4-Nitroanisole.
(iv) 1-Methoxypropane.
(v) 1-Ethoxy-4,4-dimethyl cyclohexane.
(vi) Ethoxybenzene.

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(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:
Williamson ether synthesis: When sodium alkoxide is heated with alkyl halide, an ether is formed. This reaction is called Williamson’s synthesis. This method is most useful in preparation of mixed ether.
Example:

Question 26.
How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Answer:

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
During the acid dehydration of secondary or tertiary alcohols alkenes are formed.

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

Question 29.
Explain the fact that in ary! alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
(i) The alkoxy group is electron donating and hence increases the electron density on the benzene ring and activates it towards electrophilic substitution.

(ii) In the resonance structures, the electron density is more at ortho and para positions.

Question 30.
Show how would you synthesise the following alcohols from appropriate alkenes:

Answer:

Question 31.
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 tertiaiy carbocation by a hydride ion shift from 3rd carbon atom.)
Answer:

GSEB Class 12 Chemistry Alcohols, Phenols and Ehers Additional Important Questions and Answers

Question 1.
Methanol is known as ‘wood alcohol’ because it is produced by the destructive distillation of wood.
(i) How is methanol manufactured?
(ii) Methanol is usually used for denaturing ethanol. Give reason.
(iii) Suggest a method to convert methanol to ethanol.
Answer:
(i) Methanol is prepared industrially from water gas. Water gas when mixed with half its volume of hydrogen and passed over oxides of Zn, Cr and Cu catalyst under 200 atm at 623 – 670 K gives methanol.

(ii) Methanol is a poisonous substance and hence it can be used to make alcohols unfit for drink¬ing. Hence the reason.

(iii) Methanol can be converted into ethanol using the following steps.

Question 2.
What is absolute alcohol? How is it prepared?
Answer:
100% ethanol is known as Absolute alcohol. Benzene is mixed in absolute spirit and fractionally distilled. Azeotropic mixture of 74% water, 18 5% alcohol and 74% ben¬zene is distilled on 64-8% alcohol. After the removal of water at 68-2°C a binary mixture of alcohol (32-4%) and benzene (67-6%) is distilled. When the entire benzene is removed then at 78-l°C impure alcohol is distilled. It contains 100% alcohol.

Question 3.
A compound (A) reacts with thionyl chloride to give compound (B). (B) reacts with magnesium to form a Grignard reagent which is treated with acetone and the product on hydrolysis gives 2- methyl-2-butanol. Identify (A) and (B).
Answer:

Question 4.
Complete the following reactions.

Answer:

Question 5.
Phenol is acidic in nature. Give reason.
Answer:
Phenol ionises as
The phenoxide ion formed is more resonance stabilised than phenol. Thus it has a strong tendency to undergo ionisation and hence acts as an acid.

Question 6.
The teacher asked the students to conduct nitration of phenol. Ramu used dil.HNO₃ while John used a mixture of conc.HNO₃ and conc.H₂SO₄.
(i) What are the products obtained in each case?
(ii) Write chemical equations.
Answer:
(i) When phenol is heated with dil.HNO₃ gives o and p-nitrophenol while nitrating mixture gives picric acid.

(ii)

Question 7.
(i) Identify the functional isomer of ethanol from the following, a. Ethanol, b. Propanol, c. Dimethyl ether.
(ii) Suggest a method to prepare the functional isomer of ethanol.
(iii) How will you distinguish the isomers?
Answer:
(i) Dimethyl ether.
(ii)
(iii) Ethanol gives iodoform test while dimethyl ether doesn’t form yellow precipitate of iodoform when treated with f and NaOH

Question 8.
Suggest a method to distinguish the following set of compounds.
(i) Ethanol and methanol.
(ii) Ethanol and phenol.
(iii) 1-propanol and 2-propanol
Also, explain the chemistry behind.
Answer:
(i) Methanol and ethanol are distinguished by the iodoform test.
Ethanol when treated with iodine and sodium carbonate gives yellow crystals of iodoform.
CH₃H₂OH + 4l₂ + 3Na₂CO₃ → CHI₃ + HCOONa + 5 Nal + 2H₂O + 3CO₂.
Methanol does not answer iodoform test.

(ii) Phenol reacts with bromine water to give a yellow precipitate of 2,4.6-tribromophenol. While ethanol does not answer this test.

(iii) 1-propanol does not answer iodoform reaction while 2-propanol answers iodoform reaction.

Question 9.
Write the reaction of phenol with ferric chloride.
Answer:
Phenol reacts with FeCl₃ to form a water-soluble coloured complex. The colour varies from violet to red including green and blue.
6C₆H₅OH + FeCl₃ → (C₆H₅O)₆ Fe + 3HCl + 3H⁺