Alcohols and Phenols 4 - Class 12 Chemistry MCQ

This set of Class 12 Chemistry Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Alcohols and Phenols – 4”.

1. The oxidation of alcohol involves the cleavage of _____ bonds.
a) O-H
b) C-H
c) O-H and C-H
d) Hydrogen
View Answer

Answer: c
Explanation: Oxidation of alcohols involves the cleavage of both O-H and C-H bonds so as to form a carbon-oxygen double bond. This results in the release of dihydrogen and as a result is also known as dehydrogenation reactions.

2. Which of the following oxidising agents is used to obtain carboxylic acids directly from alcohols?
a) Acidified KMnO₄
b) Aqueous KMnO₄
c) Alkaline KMnO₄
d) Anhydrous CrO₃
View Answer

Answer: a
Explanation: Strong oxidising agents like acidified potassium permanganate or acidified potassium dichromate convert alcohol directly to carboxylic acid. However, only aldehyde can be obtained by using CrO₃ as the oxidising agent in an anhydrous medium.

3. Which of the following compounds gives a ketone when its vapours are passes over heated copper at 573K?
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
View Answer

Answer: b
Explanation: Secondary alcohols undergo dehydrogenation when its vapours are passed over Cu at 573K to give ketone. Primary and tertiary alcohols under the same conditions give aldehydes and alkenes respectively.

4. Ortho and para isomers of nitrophenol can be separated by ________ distillation.
a) fractional
b) steam
c) vacuum
d) zone
View Answer

Answer: b
Explanation: o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding between O of nitro group and H of hydroxyl group. On the other hand, p-Nitrophenol is present as associated molecules due to intermolecular hydrogen bonding, thus making it less volatile. Hence, a mixture of both can be separated by steam distillation.

5. Identify the product of the following reaction.

a) o- and p-Nitrophenol
b) 2- and 4-Phenolsulphonic acid
c) p-Nitrosophenol
d) 2,4,6-Trinitrophenol
View Answer

Answer: d
Explanation: Picric acid is commonly formed by treating phenol first with concentrated sulphuric acid to form phenol-2,4-disulphinc acid, and then with concentrated nitric acid. It can also be formed treating phenol with only concentrated nitric acid, but the yield of the product in this case in very poor.

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6. What is formed when phenol is reacted with bromine water?
a) White precipitate
b) Colourless gas
c) Brown liquid
d) No reaction
View Answer

Answer: a
Explanation: Phenol on treatment with bromine water gives a white precipitate which is actually 2,4,6-Tribromophenol.

7. Which of the following is incorrect regarding Kolbe’s reaction?
a) The final product is Salicylic acid
b) CO₂ acts as an electrophile
c) Acidification of phenol leads to formation of main product
d) Phenoxide ion undergoes electrophilic attack
View Answer

Answer: c
Explanation: The first step of Kolbe’s reaction is the formation of sodium phenoxide, which is formed by treating phenol with NaOH. This phenoxide undergoes substitution by a weak electrophile, CO₂, to form the main product, i.e., 2-Hydroxybenzoic acid or salicylic acid.

8. What is the final major product of Reimer-Tiemann reaction?
a) Salicylic acid
b) o-Salicylaldehyde
c) m-Salicylaldehyde
d) p-Salicylaldehyde
View Answer

Answer: b
Explanation: Phenol is first treated with chloroform and NaOH to form an intermediate substituted benzal chloride (with-CHO group substituted at ortho position of aromatic ring). This is then hydrolysed in the presence of an alkali to form 2-Hydroxy benzaldehyde or o-Salicylaldehyde.

9. Which compound can covert phenol back to benzene?
a) CrO₃
b) Zinc dust
c) Pyridine
d) Strong acid
View Answer

Answer: b
Explanation: When heated with zinc dust, the C-O bond in phenol is broken and it is reduced to benzene and zinc oxide.

10. In the presence of air, phenols slowly start to form dark coloured mixtures.
a) True
b) False
View Answer

Answer: a
Explanation: Phenols are slowly oxidised in the presence of air to form dark coloured mixtures containing quinones.

11. Lucas test was conducted on three compounds A, B and C. Compounds A and B showed turbidity at room temperature, but compound C became turbid only after heating. Which of the compounds has a tertiary structure?
a) A
b) A and B
c) C
d) Cannot be determined
View Answer

Answer: d
Explanation: It is said that A and B show turbidity at room temperature, but it is not mentioned whether the appearance of turbidity is immediate or after some time. So compounds A and B may be tertiary or secondary depending on whether turbidity appears immediately or after 5 minutes respectively. Compound C is may be primary.

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