CBSE Class 12 Chemistry Alcohols Phenols Ethers Questions Answers

CBSE Class 12 Chemistry Alcohols Phenols Ethers Questions Answers. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Class 12 Chemistry: Alcohols, Phenols and Ethers Questions and Answers

ALCOHOLS, PHENOLS AND ETHERS

1 Mark Questions

Give the IUPAC name of the following.
CH₃–C(CH₃)=CH–CH₂OH

Answer: 3-Methylbut-2-en-1-ol

Phenols are much more acidic than alcohols. Why?

Answer: The phenoxide ion is more resonance stabilised than the alkoxide ion.

Give the IUPAC name of the following compound:
CH₃–C(CH₃)=C(Br)–CH₂OH

Answer: 2-Bromo-3-methylbut-2-en-1-ol

What happens when phenol is treated with excess of bromine water?

Answer: It gives 2,4,6-tribromophenol (a white precipitate).

Write the chemical equation for Williamson synthesis.

Answer: R–X + R’–ONa → R–O–R’ + NaX

Mention one use of methanol.

Answer: As a denaturant for ethanol.

The boiling point of ethanol is higher than that of methoxymethane. Why?

Answer: Ethanol has intermolecular hydrogen bonding; methoxymethane does not have H-bonding.

Name a substance that can be used as an antiseptic as well as a disinfectant.

Answer: Phenol. A 0.1% solution of phenol is used as an antiseptic and a 1% solution of phenol is used as a disinfectant.

Write the IUPAC name of (CH₃)₃C–OH.

Answer: 2-Methylpropan-2-ol

What are nucleophiles?

Answer: Nucleophiles are species with a high electron density (electron-rich species) which can donate an electron pair to an electron-deficient centre.

Which catalyst is used in the Friedel–Crafts reaction?

Answer: Anhydrous AlCl₃.

Write a test to distinguish between primary, secondary and tertiary alcohols.

Answer: Lucas test.

Write the IUPAC name of CH₃OH.

Answer: Methanol.

Write the IUPAC name of CH₃–O–CH₃.

Answer: Methoxymethane.

Write the IUPAC name of CH₃CH₂–O–CH₃.

Answer: Methoxyethane.

Short Answer Questions (2 Marks)

Phenol has a smaller dipole moment than methanol. Why?

Answer: Due to the electron-withdrawing nature of the phenyl group, the C–O bond in phenol is less polar. In methanol, the methyl group is electron releasing, so the C–O bond is more polar.

Explain why phenol does not undergo substitution of the –OH group like alcohol.

Answer: The C–O bond in phenol has some double bond character due to resonance and hence cannot be easily replaced by a nucleophile. In contrast, the C–O bond in alcohol is a pure single bond and hence can be easily replaced by a nucleophile.

Give a test to distinguish between phenol and benzyl alcohol.

Answer: Phenol gives a violet colour with neutral ferric chloride, while benzyl alcohol does not give this colour.

Give a test to distinguish between ethanol and phenol.

Answer: Phenol gives a violet colour with neutral ferric chloride solution (it is also acidic enough to dissolve in aqueous NaOH), but ethanol gives no such colour.

Write the Williamson synthesis reaction.

Answer: R–X + R’–ONa → R–O–R’ + NaX

3 Marks Questions

Write the reaction of phenol with zinc.

Answer: C₆H₅OH + Zn → C₆H₆ + ZnO

(i) Explain the mechanism of addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(ii) Explain the mechanism of acid catalysed dehydration of an alcohol forming an alkene.
(iii) Explain the mechanism of acid catalysed hydration of an alkene forming an alcohol.

Answer: (i) Grignard’s reagent is an alkyl magnesium halide. The alkyl group has a partial negative charge, whereas the magnesium group has a partial positive charge. The alkyl group attacks the carbon of the carbonyl group to form an addition compound.

H₂C=O  +  CH₃MgBr  →  CH₃–CH₂–OMgBr (adduct)

(methanal)   (in the carbonyl group C is δ+ and O is δ−; in CH₃MgBr the carbon is δ− and Mg is δ+)

Grignard’s reagent acts as a nucleophilic agent and attacks electrophilic carbon atoms to yield a carbon–carbon bond. The addition of the nucleophile is an irreversible process due to the high pK_a value of the alkyl group.

CH₃CH₂OMgBr  +  H₂O/H⁺  →  CH₃CH₂OH (ethanol)  +  Mg(OH)Br (magnesium hydroxy bromide)

(ii) When heated with concentrated sulphuric acid, phosphoric acid or boric acid, alcohols undergo dehydration to form alkenes. The mechanism of this reaction involves the protonation of alcohol, followed by loss of a water molecule and a proton.

1. CH₃–CH₂–OH + H⁺ ⇄ CH₃–CH₂–O⁺H₂ (oxonium ion)   [protonation]
2. CH₃–CH₂–O⁺H₂ ⇄ CH₃–C⁺H₂ (carbocation) + H₂O   [loss of water]
3. CH₃–C⁺H₂ ⇄ H₂C=CH₂ (alkene) + H⁺   [loss of proton]

During the dehydration of alcohol, the intermediate carbocation may undergo re-arrangement, resulting in the formation of a more stable carbocation.

(iii) Some reactive alkenes like 2-methylpropene undergo direct hydration in the presence of mineral acids which act as catalysts. The addition of water to the double bond takes place in accordance with Markovnikov’s rule.

1. Step I (slow): R₁R₂C=CH₂ + H–O⁺H₂ ⇄ R₁R₂C⁺–CH₃ + H₂O (the proton adds to the double bond to give a carbocation)
2. Step II (fast): R₁R₂C⁺–CH₃ + H₂O ⇄ R₁R₂C(CH₃)–O⁺H₂ (water attacks the carbocation)
3. Step III: R₁R₂C(CH₃)–O⁺H₂ + H₂O ⇄ R₁R₂C(CH₃)–OH + H₃O⁺ (loss of a proton gives the alcohol) [this step was not shown in the source figure]

Explain the following:
(i) The boiling point of ethanol is higher than that of methoxymethane.
(ii) Phenol is more acidic than ethanol.
(iii) o- and p-Nitrophenols are more acidic than phenol.

Answer: (i) Ethanol has a hydrogen attached to an oxygen atom, so it forms intermolecular hydrogen bonds and exists as associated molecules. Hence it has a higher boiling point than methoxymethane, which does not form hydrogen bonds.
(ii) The phenoxide ion left after the release of a proton is stabilised by resonance, whereas the ethoxide ion is not. Moreover, the ethoxide ion is destabilised by the +I effect of the ethyl group.
(iii) Due to the −I and −R effects of the NO₂ group, the resulting nitrophenoxide ion is more stabilised (the negative charge is spread over the ring and the nitro group), so the nitrophenols lose H⁺ more easily.

How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol
(ii) Methyl magnesium bromide to 2-methylpropan-2-ol

Answer: (i) C₆H₅CH₂Cl + aq. KOH → C₆H₅CH₂OH + KCl
(ii) CH₃MgBr + (CH₃)₂C=O → (CH₃)₃C–OMgBr; then H₂O/H⁺ → (CH₃)₃C–OH (2-methylpropan-2-ol)

5 Marks Questions

(a) An organic compound ‘A’ with molecular formula C₈H₈O gives positive DNP and iodoform tests. It does not reduce Tollens’ or Fehling’s reagent and does not decolourise Br₂/H₂O also. On oxidation with chromic acid it gives a carboxylic acid (B) with molecular formula C₇H₆O₂. Determine the structures of ‘A’ and ‘B’.
(b) Complete the following reactions by identifying A, B and C:

1. A + H₂(g) → (Pd/BaSO₄) (CH₃)₂CHCHO
2. (CH₃)₃C–CO–CH₃ + NaOI → B + C

Answer: (a) A = Acetophenone (C₆H₅COCH₃), B = Benzoic acid (C₆H₅COOH).
(b) (i) A = (CH₃)₂CHCOCl (Rosenmund reduction).
(ii) B = (CH₃)₃C–COONa and C = CHI₃ (iodoform).

An organic compound ‘A’ with molecular formula C₃H₆ on treatment with aq. H₂SO₄ gives ‘B’, which on treatment with HCl/ZnCl₂ gives ‘C’. The compound ‘C’ on treatment with ethanolic KOH gives back the compound ‘A’. Identify the compounds A, B and C.

Answer: A = Propene, B = Propan-2-ol, C = 2-Chloropropane.

CH₃–CH=CH₂  ⟶ aq. H₂SO₄ (hydration)   CH₃–CH(OH)–CH₃  ⟶ HCl/ZnCl₂   CH₃–CHCl–CH₃  ⟶ ethanolic KOH   CH₃–CH=CH₂

ALCOHOLS AND PHENOLS: 1 Mark Questions

What is absolute alcohol?

Answer: 100% C₂H₅OH is known as absolute alcohol.

What is rectified spirit?

Answer: 95% C₂H₅OH is known as rectified spirit.

Lower alcohols are soluble in water but higher alcohols are not soluble. Why?

Answer: Due to the large hydrocarbon part, which is hydrophobic.

Why is ether insoluble in water?

Answer: Due to the absence of hydrogen bonding with water (ethers cannot form hydrogen bonds among themselves or donate hydrogen for bonding with water).

How is alcohol made unfit for drinking purposes?

Answer: By adding CH₃OH (methanol).

Phenol is an acid but does not react with NaHCO₃. Why?

Answer: Phenol is a very weak acid (weaker than carbonic acid).

Diethyl ether does not react with Na. Why?

Answer: Due to the absence of acidic (active) hydrogen.

Propanol has a higher boiling point than that of n-butane. Why?

Answer: In propanol there is hydrogen bonding, but in n-butane there is no hydrogen bonding.

A mixture of o- and p-nitrophenol is separated by steam distillation. Why?

Answer: Due to intramolecular hydrogen bonding in o-nitrophenol, it is steam volatile, whereas p-nitrophenol (intermolecular H-bonding) is not.

Out of o- and p-nitrophenol, which one is more volatile?

Answer: o-Nitrophenol, due to intramolecular hydrogen bonding.

What is the IUPAC name of dimethyl ether?

Answer: Methoxymethane

What is the IUPAC name of isobutyl alcohol?

Answer: 2-Methylpropan-1-ol

What is anisole?

Answer: Methyl phenyl ether (C₆H₅–O–CH₃)

Give an example of a 3° alcohol.

Answer: 2-Methylpropan-2-ol

What is phenetole?

Answer: Ethyl phenyl ether (C₆H₅–O–C₂H₅)

Dimethyl ether is more volatile than ethanol. Why?

Answer: Due to the absence of hydrogen bonding in dimethyl ether.

Which product is obtained by dehydration of ethanol?

Answer: Ethene.

Out of but-2-en-1-ol and 2-butanol, which one shows cis–trans isomerism?

Answer: But-2-en-1-ol shows cis–trans isomerism.

Solution :

2 − Bromo-3-methyl-but-2-enol.

Q.5 Write chemical equation Williamson synthesis.

Ans:- R-X+R- O- Na --------- R-O-R + NaCl

Q.6 Mention one uses of methanol.

Q.7 The boiling point of ethanol is higher than that methoxy methane.

Ans:-Ethanol has inter molecular hydrogen bonding, methoxymehane does not have H-bonding.

Q.10 What is Nucleophiles.

Ans:- The species which has high electrons density.

Q.11 Which catalyst are used in Friedel craft reaction?

Ans:- Anh.AlCl3.

Ans:- Methoxy ethane. 

Short answers questions (2marks)

Q.2 Explain why Phenol do not undergo substitution of OH group like alcohol.

Ans:- phenol turns blue litmus red,but Ethanol donot have effect on litmus paper.

Q.5Write theWilliamson synthesis reaction