ChemistryCBSE Class 12 · Sample paper, session 2025-26 · 2 marks · Very short answer

CBSE Class 12 Chemistry sample paper 2025-26: question 21

From the official CBSE sample question paper for 2025-26, section B, question 21, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q21 · 2 marks · Very short answerAlcohols, Phenols and Ethers

The mechanism of formation of alcohols from alkenes is given below. Rectify the errors in the mechanism and rewrite the corrected steps

Figure in the paper: Three-step mechanism drawn with curved arrows. STEP 1: C=C double bond attacks the H of $H_{3} O^{+}$ (H-O+(H)-H) in equilibrium giving a carbocation (-CH-C+<) and $H_{2} O$ . STEP 2: carbocation (H-C-C+) + $H_{2} O$ in equilibrium giving protonated alcohol (-C-C(H)-O+(H)-H). STEP 3: protonated alcohol + $H_{2} O$ giving alcohol (-CH-C(OH)-) + $H_{3} O^{+}$ . Arrows and charges drawn as printed. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Recall the NCERT mechanism of acid-catalysed hydration: protonation, nucleophilic attack, deprotonation.

Hint 2

Check each step: does the arrow start from electrons, where should the + charge sit, and is the step reversible?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

**Step 1 – Protonation of the alkene (electrophilic attack by H $_{3}$ O $^{+}$ ):** $> C=C < + H_{3} O^{+} ⇌ > CH - C + < + H_{2} O$ . The curved arrow starts at the π bond and points to H of H $_{3}$ O $^{+}$ . A second arrow goes from the H–O bond to O $^{+}$ , which becomes neutral H $_{2}$ O. The carbocation is formed on the carbon that gives the more stable carbocation (Markovnikov).
½ mark
Step 2 – Nucleophilic attack of water on the carbocation: $> CH - C + < + H_{2} \ddot{O} ⇌ > CH - C < (O + H_{2})$ . The arrow goes from a lone pair on the O of water to the positive carbon, giving the protonated alcohol (oxonium ion).
½ mark
Step 3 – Deprotonation: $> CH - C < (O + H_{2}) + H_{2} O ⇌ > CH - C < (OH) + H_{3} O^{+}$ . The arrow goes from a lone pair of a water molecule to an H on O $^{+}$ , and the O–H bond electrons move back onto oxygen. This step is also reversible.
½ mark
Corrections: every step is an equilibrium; each curved arrow starts from an electron source (π bond or lone pair) and ends at the electron-deficient atom. The positive charge sits on carbon in step 1 and on oxygen in the protonated alcohol, and H $_{3}$ O $^{+}$ is regenerated, so it acts only as a catalyst.
½ mark

Answer: Corrected mechanism: (1) alkene + H₃O⁺ ⇌ carbocation + H₂O; (2) carbocation + H₂O ⇌ protonated alcohol; (3) protonated alcohol + H₂O ⇌ alcohol + H₃O⁺, with arrows from electron-rich to electron-poor sites and all steps reversible.

Where the marks are

Correct step 1: protonation of alkene to carbocation with proper arrows (½ mark)
Correct step 2: nucleophilic attack of water to give protonated alcohol (½ mark)
Correct step 3: deprotonation by water regenerating H₃O⁺ (½ mark)
Correct charges, arrow directions and reversibility shown (½ mark)

Revise the chapter

Chapter: Alcohols, Phenols and Ethers (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Alcohols, Phenols and Ethers, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Chemistry sample paper 2025-26: question 21

Hints

Step-by-step solution

Revise the chapter

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