CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 15

Answer: (b) 2-Methylbutanal

Explanation:
1. Number the chain from the CHO carbon, which is always C-1.
2. The longest chain has four carbons (butanal) and the CH₃ group is on C-2.
3. So the name is 2-methylbutanal.

Teacher's Note:
1. The aldehyde carbon is always C-1, so no locant is needed for it.
2. Choose the longest chain that contains CHO.

Answer: (c) Pentan-2-one

Explanation:
1. The chain has five carbons, so the parent is pentane.
2. Number from the end that gives C=O the lower locant: the carbonyl is on C-2.
3. Name: pentan-2-one.

Teacher's Note:
1. Ketones take the suffix –one and the C=O gets the lowest locant.

Answer: (a) But-2-enal

Explanation:
1. The aldehyde carbon is C-1 and the chain has four carbons.
2. The double bond lies between C-2 and C-3, so the name is but-2-enal.

Teacher's Note:
1. Number from the CHO carbon; the double bond locant follows from that.

Answer: (d) Propanone

Explanation:
1. Propanal is CH₃CH₂CHO, molecular formula C₃H₆O.
2. Propanone, CH₃COCH₃, also has the formula C₃H₆O but a different functional group (ketone).

Teacher's Note:
1. Aldehydes and ketones with the same number of carbons are functional isomers.
2. Check that the molecular formula is the same.

Answer: (a) sp², 120°

Explanation:
1. The carbonyl carbon forms three σ bonds and one π bond.
2. Three σ bonds in one plane mean sp² hybridisation with 120° angles.

Teacher's Note:
1. Count the σ bonds plus lone pairs to get the hybridisation.

Answer: (b) Pyridinium chlorochromate (PCC)

Explanation:
1. PCC is a mild oxidising agent in a non-aqueous medium.
2. It stops at RCHO.
3. The other reagents oxidise the aldehyde further to RCOOH.

Teacher's Note:
1. Mild oxidants (PCC, CrO₃ in pyridine) give aldehydes.
2. Strong oxidants give carboxylic acids.

Answer: (d) Propanone

Explanation:
1. Addition of water follows Markovnikov's rule, giving the enol CH₃C(OH)=CH₂.
2. The enol tautomerises to the ketone CH₃COCH₃.

Teacher's Note:
1. Only ethyne gives an aldehyde on hydration.
2. All other alkynes give ketones.

Answer: (c) Prevent further reduction of the aldehyde to the alcohol

Explanation:
1. BaSO₄ acts as a partial poison for the Pd catalyst.
2. This lowers the activity so that RCOCl is reduced to RCHO but RCHO is not reduced to RCH₂OH.

Teacher's Note:
1. Think of BaSO₄ as a catalyst poison that gives selectivity.

Answer: (a) Methanal

Explanation:
1. The method needs the acyl chloride RCOCl.
2. For methanal the acyl chloride would be HCOCl (formyl chloride), which is unstable and decomposes to CO and HCl.

Teacher's Note:
1. Rosenmund reduction starts from an acid chloride.

Answer: (d) Ethanal

Explanation:
1. SnCl₂/HCl reduces CH₃CN to the imine CH₃CH=NH (Stephen reaction).
2. Hydrolysis of the imine gives CH₃CHO and NH₃.

Teacher's Note:
1. Nitrile → imine → aldehyde is the Stephen reduction.

Answer: (b) Benzaldehyde

Explanation:
1. This is the Etard reaction.
2. CrO₂Cl₂ oxidises the CH₃ group to a chromium complex, which is hydrolysed to –CHO.
3. The reaction stops at the aldehyde stage.

Teacher's Note:
1. Remember: Etard = toluene → benzaldehyde with CrO₂Cl₂.

Answer: (c) Benzaldehyde

Explanation:
1. This is the Gattermann–Koch reaction.
2. The electrophile (formed from CO and HCl) substitutes a ring hydrogen by –CHO.

Teacher's Note:
1. Formylation of benzene uses CO and HCl.

Answer: (a) CH₃COCl in the presence of anhydrous AlCl₃

Explanation:
1. This is Friedel–Crafts acylation.
2. The acylium ion CH₃CO⁺ attacks benzene to give C₆H₅COCH₃.

Teacher's Note:
1. Acyl chloride with AlCl₃ gives a ketone; alkyl chloride gives an alkylbenzene.

Answer: (d) Propanone

Explanation:
1. 2CH₃COCl + (CH₃)₂Cd → 2CH₃COCH₃ + CdCl₂.
2. Dialkylcadmium does not attack the ketone formed, so the reaction stops at the ketone.

Teacher's Note:
1. Dialkylcadmium is less reactive than a Grignard reagent.

Answer: (c) Propan-1-ol

Explanation:
1. Propan-1-ol forms intermolecular hydrogen bonds.
2. Propanal and propanone have only dipole–dipole attraction, and butane is non-polar.
3. Boiling points: 370 K, 322 K, 329 K and 273 K respectively.

Teacher's Note:
1. Hydrogen bonding raises the boiling point most.

Answer: (b) Form hydrogen bonds with water

Explanation:
1. The carbonyl oxygen accepts hydrogen bonds from water molecules.
2. This makes the lower members freely soluble.

Teacher's Note:
1. Solubility falls as the alkyl part gets bigger.

Answer: (a) Butan-2-one

Explanation:
1. CH₃CN + C₂H₅MgBr gives CH₃(C₂H₅)C=NMgBr.
2. Hydrolysis gives CH₃COC₂H₅, which is butan-2-one.

Teacher's Note:
1. Nitrile + Grignard reagent → ketone (after hydrolysis).

Answer: (d) A 40% aqueous solution of methanal

Explanation:
1. Methanal is a gas at room temperature.
2. Its 40% aqueous solution is called formalin and is used to preserve biological specimens.

Teacher's Note:
1. Remember formalin = 40% HCHO in water.

Answer: (b) Oxalic acid

Explanation:
1. Ethanedioic acid has two carbons joined and two COOH groups: oxalic acid.
2. Malonic is propanedioic, succinic is butanedioic and adipic is hexanedioic acid.

Teacher's Note:
1. Remember the order: oxalic, malonic, succinic, glutaric, adipic.

Answer: (c) Ethanoic acid

Explanation:
1. CH₃CN + 2H₂O → CH₃COOH + NH₃ (the amide forms first).
2. In acid the ammonia is obtained as the ammonium salt.

Teacher's Note:
1. Hydrolysis of RCN gives RCOOH.