CBSE Important Questions Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids

Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids is part of the CBSE Class 12 Chemistry syllabus. All three families contain the carbonyl group. Students learn their nomenclature and structure, methods of preparation, physical properties, nucleophilic addition reactions, reactions due to the alpha hydrogen, oxidation and reduction, the acidity of carboxylic acids, and important named reactions.

These CBSE important questions for Class 12 Chemistry Chapter 8 cover why benzophenone does not give the iodoform test while acetophenone does, the esterification reaction and the role of dry HCl gas, the aldol condensation, the Cannizzaro reaction, distinguishing tests for aldehydes and ketones, and the effect of substituents on the acidity of carboxylic acids. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Explain why benzophenone does not give a yellow precipitate with I₂ in alkaline solution, while acetophenone does.

Answer: This is the iodoform test, which gives a yellow precipitate of iodoform (CHI₃).
The test is positive only for compounds that have a CH₃CO− group (methyl ketone) attached to a carbon or hydrogen, or for alcohols that can be oxidised to such a group.
• Acetophenone is C₆H₅COCH₃. It has the CH₃CO− group, so the three hydrogens of the methyl group are replaced by iodine and the compound then splits to give the yellow precipitate of CHI₃.
• Benzophenone is C₆H₅COC₆H₅. Both groups attached to the carbonyl are phenyl groups, and there is no methyl group next to the carbonyl. So the reaction cannot take place and no yellow precipitate forms.

Q2. Alcohols react with carboxylic acids in the presence of a few drops of concentrated H₂SO₄ or dry HCl gas to form esters. Name this reaction, write the general equation, and explain the role of dry HCl gas.

Answer: Name: This is the Fischer esterification reaction (or simply esterification).
General equation:
R-COOH + R′-OH ⇌ R-COO-R′ + H₂O (in the presence of H⁺)
Role of dry HCl gas:
1. It acts as a catalyst by supplying H⁺ ions. The proton attaches to the carbonyl oxygen of the acid, making the carbonyl carbon more electrophilic and so easier for the alcohol to attack.
2. Being dry, it does not add water to the mixture. Since esterification is a reversible reaction, keeping water out pushes the equilibrium towards the ester and increases the yield.

Q3. What is the aldol condensation? Write the reaction for acetaldehyde.

Answer: Aldol condensation is a reaction in which two molecules of an aldehyde or ketone that have at least one α-hydrogen combine in the presence of a dilute base to give a β-hydroxy aldehyde or ketone, which on heating loses water to give an α,β-unsaturated compound.
For acetaldehyde:
2CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal, the "aldol")
On heating: CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O (but-2-enal)
The product is called an "aldol" because it contains both an aldehyde and an alcohol group.
Compounds without an α-hydrogen, such as benzaldehyde and formaldehyde, cannot undergo this reaction.

Q4. What is the Cannizzaro reaction? Which aldehydes undergo it?

Answer: The Cannizzaro reaction is a disproportionation reaction. When an aldehyde without an α-hydrogen is treated with concentrated alkali, one molecule is oxidised to a carboxylic acid salt and another is reduced to an alcohol.
Example with formaldehyde:
2HCHO + NaOH → CH₃OH + HCOONa
Which aldehydes: Only those with no α-hydrogen, such as formaldehyde (HCHO), benzaldehyde (C₆H₅CHO) and trimethylacetaldehyde.
Aldehydes that do have an α-hydrogen undergo aldol condensation instead.

Q5. How will you distinguish between an aldehyde and a ketone? Name two tests.

Answer: Aldehydes are easily oxidised while ketones are not, and this difference is the basis of both tests.
1. Tollen's test (silver mirror test): The compound is warmed with Tollen's reagent (ammoniacal silver nitrate).
• Aldehyde: gives a bright silver mirror on the inside of the tube.
• Ketone: no reaction.
2. Fehling's test: The compound is warmed with Fehling's solution (an alkaline copper complex).
• Aldehyde: gives a reddish-brown precipitate of Cu₂O.
• Ketone: no reaction.
(Note that aromatic aldehydes such as benzaldehyde give the Tollen's test but not the Fehling's test.)

Q6. Why is the boiling point of a carboxylic acid higher than that of an alcohol of comparable molecular mass?

Answer: Carboxylic acids form stronger and more extensive hydrogen bonds than alcohols.
In the liquid state, two molecules of a carboxylic acid join through two hydrogen bonds to form a stable cyclic dimer. In effect, the molecules move about in pairs, so the particle moving is twice as heavy.
An alcohol molecule can form hydrogen bonds too, but only a single chain, not a double-bonded dimer.
Breaking the two hydrogen bonds of the dimer needs more energy, so the carboxylic acid boils at a higher temperature. For example, acetic acid boils at 391 K while propan-1-ol, of similar mass, boils at 370 K.

Q7. Arrange the following in increasing order of acidic strength: acetic acid, chloroacetic acid, trichloroacetic acid, formic acid.

Answer: Acetic acid < formic acid < chloroacetic acid < trichloroacetic acid
The order follows the stability of the carboxylate ion left after losing the proton.
• Trichloroacetic acid (CCl₃COOH) is strongest. Three chlorine atoms pull electron density away strongly (−I effect) and spread out the negative charge, making the ion very stable.
• Chloroacetic acid has only one chlorine, so the effect is weaker.
• Formic acid (HCOOH) has only a hydrogen, with no electron-releasing group at all.
• Acetic acid (CH₃COOH) is weakest, because the methyl group releases electrons (+I effect) and destabilises the carboxylate ion.

Q8. Why is the carbonyl carbon of an aldehyde or ketone open to nucleophilic attack?

Answer: The C=O bond is strongly polar, because oxygen is far more electronegative than carbon. The shared electrons are pulled towards oxygen, giving the carbon a partial positive charge (δ+) and the oxygen a partial negative charge (δ−).
A carbon with a partial positive charge naturally attracts nucleophiles, which are electron-rich species.
Also, the carbonyl carbon is sp² hybridised and planar, so a nucleophile can approach it easily from above or below the plane without much hindrance.
This is why the typical reaction of aldehydes and ketones is nucleophilic addition, while alkenes, whose C=C is non-polar, undergo electrophilic addition instead.

Q9. Why are aldehydes more reactive than ketones towards nucleophilic addition?

Answer: Two reasons, both pointing the same way.
1. Electronic (inductive) effect: A ketone has two alkyl groups attached to the carbonyl carbon, while an aldehyde has only one (plus a hydrogen). Alkyl groups release electrons (+I effect), which reduces the positive charge on the carbonyl carbon. Less positive charge means less attraction for the nucleophile.
2. Steric effect: The two bulky alkyl groups of a ketone block the approach of the nucleophile. An aldehyde has a small hydrogen in that place, so the carbon is much more exposed.
So the order of reactivity is HCHO > CH₃CHO > CH₃COCH₃, with formaldehyde the most reactive of all.

Q10. What happens when acetaldehyde is treated with (a) dilute NaOH and (b) Tollen's reagent?

Answer: (a) With dilute NaOH: Acetaldehyde has α-hydrogens, so it undergoes aldol condensation.
2CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal)
On warming this loses water to give but-2-enal, CH₃CH=CHCHO.
(b) With Tollen's reagent: Acetaldehyde is oxidised to acetic acid (as the acetate ion), and the silver ions in the reagent are reduced to metallic silver, which deposits as a bright silver mirror.
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 4NH₃ + 2H₂O

Q11. What is the HVZ reaction? Write its equation.

Answer: HVZ stands for Hell-Volhard-Zelinsky. In this reaction, a carboxylic acid that has an α-hydrogen is treated with chlorine or bromine in the presence of a small amount of red phosphorus. The α-hydrogen is replaced by the halogen.
Equation:
CH₃COOH + Cl₂ → ClCH₂COOH + HCl (in the presence of red P)
By using more halogen, two or three α-hydrogens can be replaced, giving dichloroacetic and trichloroacetic acid.
Importance: The α-halo acids produced are very useful starting materials, since the halogen can be replaced by other groups to make α-hydroxy acids and amino acids.

Q12. How can you prepare benzaldehyde from toluene? Name two methods.

Answer: Method 1: Etard reaction
Toluene is treated with chromyl chloride (CrO₂Cl₂) in carbon disulphide or carbon tetrachloride. A brown chromium complex forms, which on hydrolysis gives benzaldehyde.
C₆H₅CH₃ + CrO₂Cl₂ → complex → (on hydrolysis) C₆H₅CHO
Method 2: Side chain chlorination followed by hydrolysis
Toluene is chlorinated in the presence of light to give benzal chloride, which is then hydrolysed.
C₆H₅CH₃ + 2Cl₂ → C₆H₅CHCl₂ → (hydrolysis) C₆H₅CHO
In both methods the control is important: the oxidation must stop at the aldehyde stage and not continue on to benzoic acid.

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