Answer the following questions : (i) Draw structure of the methyl hemiacetal of methanal. (ii) There are two -NH_2 groups in semicarbazide. However only one is involved in the formation of semicarbazones. Give reason. (iii) How will you convert ethanol to 3-hydroxybutanal ? (iv) Complete the following equation : cyclohexanol CrO_3 ? (v) Write the final product formed when phthalic acid is treated with NH_3 followed by strong heating.

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2025: question 32(b)

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerAldehydes, Ketones and Carboxylic Acids

Answer the following questions :

(i)
Draw structure of the methyl hemiacetal of methanal.
1 mark
(ii)
There are two $- N H_{2}$ groups in semicarbazide. However only one is involved in the formation of semicarbazones. Give reason.
1 mark
(iii)
How will you convert ethanol to 3-hydroxybutanal ?
1 mark
(iv)
Complete the following equation : cyclohexanol $C r O_{3}$ ?
1 mark
(v)
Write the final product formed when phthalic acid is treated with $N H_{3}$ followed by strong heating.
1 mark

Figure in the paper: In (iv), cyclohexanol: a cyclohexane ring with an OH group, reacting with CrO3. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

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

Hint 1

A hemiacetal forms when one alcohol molecule adds across C=O. Ethanol must first become an aldehyde that has α-hydrogens.

Hint 2

Think about which nitrogen's lone pair can be delocalised onto the carbonyl oxygen in semicarbazide.

Hint 3

$Cr O_{3}$ oxidises secondary alcohols to ketones. Ammonium salts of dicarboxylic acids lose water and then ammonia on strong heating.

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.

(i) Methanal adds one molecule of methanol across its C=O bond in the presence of dry HCl to form a hemiacetal:
$HCHO + C H_{3} OH dry HCl H_{2} C (OH) (OC H_{3})$
Structure: a central $C H_{2}$ carbon bonded to one $- OH$ group and one $- OC H_{3}$ group, written $HO - C H_{2} - O - C H_{3}$ (methoxymethanol).
1 mark
(ii) In semicarbazide, $H_{2} N - CO - NH - N H_{2}$ , the $- N H_{2}$ group attached directly to the carbonyl carbon has its lone pair delocalised into the C=O group by resonance ( $H_{2} N - C (= O) - \leftrightarrow H_{2} N^{+} = C (- O^{-}) -$ ). That nitrogen is therefore not nucleophilic. The terminal $- N H_{2}$ (on the hydrazine nitrogen) takes no part in resonance, so its lone pair stays free and only this group attacks the carbonyl compound to form the semicarbazone.
1 mark
(iii) Oxidise ethanol to ethanal with PCC (or Cu at 573 K). Then carry out aldol condensation of ethanal with dilute NaOH:
$C H_{3} C H_{2} OH PCC C H_{3} CHO$
$2C H_{3} CHO dil. NaOH C H_{3} - CH (OH) - C H_{2} - CHO$
The product is 3-hydroxybutanal.
1 mark
(iv) $Cr O_{3}$ oxidises the secondary alcohol to a ketone:
$cyclohexanol (C_{6} H_{11} OH) C r O_{3} cyclohexanone (C_{6} H_{10} O)$
The product is a six-membered ring with a C=O group in place of the CH–OH group.
1 mark
(v) Phthalic acid first forms ammonium phthalate with $N H_{3}$ . On heating this gives phthalamide, and on strong heating a molecule of $N H_{3}$ is lost to give the cyclic imide:
$C_{6} H_{4} (COOH)_{2} N H_{3} C_{6} H_{4} (COON H_{4})_{2} Δ, - 2 H_{2} O C_{6} H_{4} (CON H_{2})_{2} strong Δ, - N H_{3} C_{6} H_{4} (CO)_{2} NH$
The final product is phthalimide.
1 mark

Answer: (i) $HO - C H_{2} - OC H_{3}$ ; (ii) the $- N H_{2}$ next to C=O is tied up in resonance, so only the terminal $- N H_{2}$ is nucleophilic; (iii) ethanol → ethanal (PCC) → aldol with dil. NaOH → 3-hydroxybutanal; (iv) cyclohexanone; (v) phthalimide

Where the marks are

Correct structure of methyl hemiacetal, $HOC H_{2} OC H_{3}$ (1 mark)
Resonance of the $- N H_{2}$ attached to C=O makes it non-nucleophilic, so only the other one reacts (1 mark)
Oxidation of ethanol to ethanal followed by aldol condensation with dil. NaOH (1 mark)
Cyclohexanone as the product (1 mark)
Phthalimide as the final product (1 mark)

Revise the chapter

Chapter: Aldehydes, Ketones and Carboxylic Acids (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Chemistry 2025: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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