Skip to main content
Chapter 8 of 15
Practice Quiz

Hydroxy Compounds and Ethers

Tamil Nadu Board · Class 12 · Chemistry

Practice quiz for Hydroxy Compounds and Ethers — Tamil Nadu Board Class 12 Chemistry. MCQs and questions with answers to test your preparation.

45 questions37 flashcards5 concepts

Interactive on Super Tutor

Studying Hydroxy Compounds and Ethers? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for practice quiz and more.

1,000+ Class 12 students started this chapter today

A step-by-step mechanism showing the E1 dehydration of a secondary alcohol (e.g., propan-2-ol) to an alkene, including protonation, loss of water to form a carbocation, and deprotonation.
Super Tutor

This is just one of 5+ visuals inside Super Tutor's Hydroxy Compounds and Ethers chapter

Explore the full set

Quick Quiz: Hydroxy Compounds and Ethers

0/4

Tap an answer to check it instantly. No sign-up needed for these 4.

1

An alcohol X gives blue colour in Victor Meyer's test and 3.7 g of X when treated with metallic sodium liberates 560 mL of H₂ at STP. What is the correct structure of X?

2

Which of the following reactions is NOT a correct application of Williamson's ether synthesis?

3

Arrange the following compounds in DECREASING order of acidity: (I) Phenol (II) p-nitrophenol (III) Ethanol (IV) p-cresol

4

In the hydroboration-oxidation of propene, the product and the type of addition are respectively:

45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

The dehydration of 3,3-dimethylbutan-2-ol with H₂SO₄ gives predominantly 2,3-dimethylbut-2-ene. Which intermediate explains this?

Show answer

A secondary carbocation that undergoes 1,2-hydride shift to form a more stable tertiary carbocation

Step 1 – 3,3-Dimethylbutan-2-ol: CH₃CH(OH)C(CH₃)₃. Protonation of –OH by H₂SO₄ gives oxonium ion, which loses water to form a secondary carbocation at C-2. Step 2 – The secondary carbocation (less stable) undergoes a 1,2-hydride shift: a hydrogen with its bonding electrons migrates from C-3 to C-2. This converts the secondary carbocation to a MORE stable tertiary carbocation at C-3. Step 3 – The tertiary carbocation at C-3 then loses a proton (Saytzeff's rule) to give the most substituted alkene: 2,3-dimethylbut-2-ene (4 alkyl substituents, 64% yield). Step 4 – Option C is wrong: a tertiary

2multiple choice
1 marks

Which reagent selectively oxidises a primary alcohol to an aldehyde WITHOUT further oxidation to carboxylic acid?

Show answer

Pyridinium chlorochromate (PCC)

Step 1 – The oxidation of primary alcohol normally proceeds: R-CH₂OH → R-CHO → R-COOH. The challenge is to STOP at aldehyde stage. Step 2 – PCC (Pyridinium chlorochromate, C₅H₅NH⁺CrO₃Cl⁻) is a mild oxidising agent that selectively oxidises primary alcohols to aldehydes and secondary alcohols to ketones. It does NOT further oxidise aldehydes to carboxylic acids because the reaction conditions (anhydrous, mild) do not support the hydration of aldehyde needed for further oxidation. Step 3 – Acidified K₂Cr₂O₇ (Option A) and KMnO₄/H₂SO₄ (Option C) are STRONG oxidising agents. They oxidise primary

3multiple choice
1 marks

Phenol reacts with CHCl₃ in the presence of NaOH to give salicylaldehyde (2-hydroxybenzaldehyde). This reaction is known as:

Show answer

Riemer-Tiemann reaction

Step 1 – The Riemer-Tiemann reaction involves treating phenol with chloroform (CHCl₃) in the presence of NaOH. A –CHO group is introduced at the ORTHO position of the phenol ring. Step 2 – Mechanism: CHCl₃ + NaOH → :CCl₂ (dichlorocarbene, a highly reactive electrophile). This electrophile attacks the activated benzene ring at the ortho position (–OH is ortho/para director). Step 3 – The intermediate substituted benzal dichloride is hydrolysed by NaOH to give the ortho-hydroxy benzaldehyde (salicylaldehyde). Step 4 – Do not confuse with: Kolbe-Schmitt (uses CO₂/NaOH, gives salicylic acid), S

4multiple choice
1 marks

In the preparation of phenol from cumene, the intermediate that is hydrolysed with dilute acid to give phenol and acetone is:

Show answer

Cumene hydroperoxide

Step 1 – Cumene (isopropylbenzene) is prepared by Friedel-Crafts alkylation of benzene with propene in presence of H₃PO₄ at 523K. Step 2 – When air is passed through a mixture of cumene and 5% aqueous Na₂CO₃, the benzylic C-H bond is oxidised by O₂. The secondary benzylic carbon readily forms a radical, which reacts with O₂ to form cumene hydroperoxide: C₆H₅C(CH₃)₂–O–O–H. Step 3 – Cumene hydroperoxide on treatment with dilute H₂SO₄ undergoes cleavage (Hock rearrangement) to give phenol (C₆H₅OH) and acetone (CH₃COCH₃). Step 4 – This is the industrial (Dow's cumene) process for phenol product

+41 more questions available

Practice All

Frequently Asked Questions

What are the important topics in Hydroxy Compounds and Ethers for Tamil Nadu Board Class 12 Chemistry?
Key topics in Hydroxy Compounds and Ethers include Hydroxy Compounds and Ethers - Comprehensive Overview, Classification and Properties of Alcohols and Phenols, Mind map showing classification of alcohols based on hydroxyl group attachment and substituent pattern.. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Hydroxy Compounds and Ethers — Tamil Nadu Board Class 12 Chemistry?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

For serious students

Get the full Hydroxy Compounds and Ethers chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Tamil Nadu Board Class 12 Chemistry.