JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 09

Answer: (a) \(\ce{H2}\), \(\ce{Pd}\)/\(\ce{BaSO4}\)

Explanation:
1. Hot alkaline or acidic \(\ce{KMnO4}\) oxidises the whole side chain of cumene to benzoic acid (A), and \(\ce{PCl5}\) converts it to benzoyl chloride (B).
2. Benzoyl chloride is reduced to benzaldehyde by hydrogen over poisoned palladium (\(\ce{Pd}\)/\(\ce{BaSO4}\)), the Rosenmund reduction.

Teacher's Note:
1. The poison (\(\ce{BaSO4}\)) stops the aldehyde from being reduced further to the alcohol.
2. Gattermann-Koch uses \(\ce{CO}\)/\(\ce{HCl}\) directly on benzene, so it cannot start from an acyl chloride.

Answer: (d) all of them

Explanation:
1. (a) A non-enolisable aldehyde such as benzaldehyde disproportionates in base to the carboxylate and the alcohol.
2. (b) Propanal self-condenses to \(\ce{CH3CH2CH=C(CH3)CHO}\), and (c) HVZ brominates or chlorinates the alpha carbon of an acid.
3. (d) The Hofmann reaction removes the carbonyl carbon and gives the amine with one carbon less, so all four matches are right.

Teacher's Note:
1. Each reaction name should be linked to its signature feature: Cannizzaro needs no alpha hydrogen, aldol needs alpha hydrogen, HVZ needs red P and halogen, Hofmann loses one carbon.
2. Such "all correct" questions are solved fastest by checking the odd-looking item first.

Answer: (c) Butanal

Explanation:
1. Reductive ozonolysis of but-2-ene gives two molecules of ethanal (P).
2. Aldol condensation on warming with base gives but-2-enal (Q), \(\ce{CH3CH=CHCHO}\).
3. One equivalent of \(\ce{H2}\) over \(\ce{Pd}\)/\(\ce{C}\) saturates the \(\ce{C=C}\) bond and leaves the aldehyde group, so R is butanal.

Teacher's Note:
1. The carbon-carbon double bond is hydrogenated far more easily than the carbonyl group over palladium.
2. Remember the sequence ozonolysis, aldol condensation, selective hydrogenation as a common multi-step pattern.

Answer: (a) a and b only

Explanation:
1. (a) Benzene has three double bonds in the Kekule picture, and ozonolysis cleaves them to give three molecules of glyoxal.
2. (b) \(\ce{SeO2}\) oxidises the \(\ce{CH3}\) group next to the carbonyl to give \(\ce{OHC-CHO}\) (Riley oxidation).
3. (c) Hydration of ethyne stops at ethanal, so it does not give glyoxal. Only (a) and (b) are correct.

Teacher's Note:
1. Selenium dioxide is the standard reagent for oxidising an active methylene or methyl group next to a carbonyl to a carbonyl.
2. Ethyne with mercury(II) and acid gives ethanal, the key aldehyde from an alkyne.

Answer: (d) a, b, c, d, e

Explanation:
1. Phthalaldehyde has no alpha hydrogen and undergoes an intramolecular Cannizzaro reaction in conc. base.
2. 2-Methylcyclohexanone has alpha hydrogens and condenses in base on heating.
3. The Etard, Gattermann-Koch and Gattermann aldehyde syntheses are all correctly described, so all five pairs are right.

Teacher's Note:
1. Learn the reagent fingerprints: \(\ce{CrO2Cl2}\) is Etard, \(\ce{CO}\)/\(\ce{HCl}\) is Gattermann-Koch, \(\ce{HCN}\)/\(\ce{HCl}\) is Gattermann.
2. Gattermann-Koch fails for phenols and for nitrobenzene-type deactivated rings.

Answer: (d) The polarity of the group is \(\ce{C^{\delta-}=O^{\delta+}}\).

Explanation:
1. Oxygen is more electronegative and pulls the pi electrons toward itself, so oxygen carries the partial negative charge.
2. The correct polarity is \(\ce{C^{\delta+}=O^{\delta-}}\), so statement (d) has the charges reversed.

Teacher's Note:
1. The partial positive carbon is why nucleophiles attack the carbonyl carbon.
2. Aldehydes and ketones have higher dipole moments than ethers of similar mass, which is why their boiling points are also higher.

Answer: (d) Urea (\(\ce{H2N-CO-NH2}\)) gives a semicarbazone

Explanation:
1. Ammonia derivatives \(\ce{NH2-Z}\) add to the carbonyl group and lose water to give \(\ce{C=N-Z}\) compounds.
2. The semicarbazone is formed from semicarbazide, \(\ce{H2N-NH-CO-NH2}\), and not from urea.

Teacher's Note:
1. Semicarbazide has an extra \(\ce{-NH-}\) before the amide group, which is what makes the derivative a hydrazone type.
2. Urea has no free hydrazine nitrogen and does not give a semicarbazone.

Answer: (c) 2-Acetylcyclohexane-1,3-dione

Explanation:
1. The hydrogen on carbon 2 of 2-acetylcyclohexane-1,3-dione lies between three carbonyl groups.
2. The enolate formed after its removal is stabilised by delocalisation over all three oxygens, so it is by far the most acidic of the four.

Teacher's Note:
1. The more carbonyl groups that flank a carbon-hydrogen bond, the greater the acidity.
2. In 1,4-dione each alpha carbon is next to only one carbonyl, so it is similar to a simple ketone.

Answer: (d) Phthalimide

Explanation:
1. Both methyl groups oxidise to carboxylate, giving phthalate (A).
2. Ammonia gives the ammonium salt or amide acid (B), and strong heating loses water to give the cyclic imide, phthalimide.

Teacher's Note:
1. Phthalimide is the starting compound of the Gabriel synthesis of primary amines.
2. Heating the diammonium salt or amide of a 1,2-dicarboxylic acid gives the five-membered cyclic imide.

Answer: (d) (i) \(\ce{BH3}\)/THF, (ii) \(\ce{H2O2}\)/\(\ce{OH-}\), (iii) PCC

Explanation:
1. Hydroboration-oxidation puts the \(\ce{OH}\) on the less substituted carbon, giving cyclohexylmethanol, a primary alcohol.
2. PCC then oxidises the primary alcohol to the aldehyde.
3. Ozonolysis would cut the double bond and give cyclohexanone, and acid hydration would give a tertiary alcohol.

Teacher's Note:
1. Hydroboration is anti-Markovnikov, so the product here is the primary alcohol.
2. PCC is the standard reagent for stopping at the aldehyde stage.

Answer: (b) a, b, c only

Explanation:
1. Aliphatic aldehydes and methanoic acid reduce Fehling's and Benedict's solutions, and aldehydes in general reduce Tollens' reagent.
2. Benzaldehyde (aromatic) does not reduce Fehling's solution, and ketones such as propanone do not reduce Tollens' reagent, so (d) and (e) are wrong.

Teacher's Note:
1. Fehling and Benedict work only for aliphatic aldehydes.
2. Tollens' reagent works for all aldehydes, and also for alpha-hydroxy ketones such as benzoin.

Answer: (d) All of these

Explanation:
1. The Schmidt reaction converts an acid to an amine with loss of \(\ce{CO2}\).
2. The Curtius rearrangement of an acyl azide gives an isocyanate and then an amine with one carbon less, and the Hofmann reaction similarly loses the carbonyl carbon.

Teacher's Note:
1. All three reactions pass through an isocyanate or similar nitrene-type rearrangement with migration of R from carbon to nitrogen.
2. The result is always an amine with one carbon fewer than the starting acid or amide.

Answer: (b) Ethyl 3-oxobutanoate

Explanation:
1. Propanone arises from decarboxylation of 3-oxobutanoic acid (acetoacetic acid), so B is \(\ce{CH3COCH2COOH}\).
2. Hydrolysis of an ester that releases ethanol must be the ethyl ester, so A is ethyl 3-oxobutanoate.

Teacher's Note:
1. Beta-keto acids lose \(\ce{CO2}\) readily on warming through a cyclic six-membered transition state.
2. The alcohol released tells you the alkyl group of the ester, here ethyl.

Answer: (b) \(\ce{CH3CH2CHO}\)

Explanation:
1. The iodoform reaction needs a \(\ce{CH3CO-}\) group or a \(\ce{CH3CH(OH)-}\) group, which ethanal, ethanol and propanone all contain.
2. Propanal has the group \(\ce{CH3CH2CO-}\) and no methyl ketone unit, so it gives no iodoform.

Teacher's Note:
1. Ethanal is the only aldehyde that gives the iodoform test, and ethanol is the only primary alcohol that does.
2. Check for the \(\ce{CH3-C=O}\) or \(\ce{CH3-CH(OH)-}\) unit.

Answer: (b) Butan-2-one

Explanation:
1. Hydrazone formation needs a carbonyl group, which rules out ethanol.
2. Iodoform needs a \(\ce{CH3CO-}\) unit, which rules out 1-cyclohexylpropan-1-one, and Tollens' test must be negative, which rules out the aldehyde 3-oxobutanal. Butan-2-one fits all three conditions.

Teacher's Note:
1. A methyl ketone gives iodoform and a hydrazone but not Tollens' test.
2. Any compound containing a free \(\ce{CHO}\) group would give a silver mirror.

Answer: (c) 3-Hydroxy-2-methylpentanal

Explanation:
1. The enolate of propanal attacks the carbonyl carbon of a second propanal.
2. The new bond is between the alpha carbon of one molecule and the carbonyl carbon of the other, giving \(\ce{CH3CH2CH(OH)CH(CH3)CHO}\), which is 3-hydroxy-2-methylpentanal.

Teacher's Note:
1. In an aldol product the new carbon-carbon bond lies between the alpha carbon and the former carbonyl carbon.
2. The methyl group at carbon 2 comes from the alpha carbon of propanal.

Answer: (b) \(\ce{CH3CH2NH2}\)

Explanation:
1. A is propanoyl chloride and B is propanamide.
2. Hofmann bromamide degradation removes the carbonyl carbon, so \(\ce{CH3CH2CONH2}\) gives \(\ce{CH3CH2NH2}\), ethylamine.

Teacher's Note:
1. The amine from a Hofmann reaction has one carbon fewer than the amide.
2. Count the carbons of the amide, subtract one and keep the \(\ce{NH2}\) on the last carbon.

Answer: (a) Tollens' reagent

Explanation:
1. Both ethanal and benzaldehyde reduce Tollens' reagent, so it gives no difference.
2. Ethanal reduces Fehling's and Benedict's solutions and gives the iodoform test, but benzaldehyde does neither, so those tests separate them.

Teacher's Note:
1. Tollens' reagent reacts with aromatic aldehydes too, while Fehling's does not.
2. This pair is a standard distinction question.

Answer: (a) The order of reactivity towards hydrolysis is ethanoyl chloride, ethanoic anhydride, ethyl ethanoate, ethanamide

Explanation:
1. (a) Acid chloride is more reactive than anhydride, ester and amide, in that order.
2. (b) is wrong because benzaldehyde is the least reactive aldehyde, and (c) is wrong because phenol is more reactive than anisole.
3. (d) is wrong because the tertiary alcohol reacts fastest with the Lucas reagent.

Teacher's Note:
1. Resonance with the ring lowers the electrophilicity of aromatic aldehydes.
2. Phenol is more activated than anisole because the phenoxide-like oxygen donates more strongly.

Answer: (d) Iodoform test

Explanation:
1. Ethanal has a \(\ce{CH3CO-}\) group and gives iodoform, whereas benzaldehyde does not.
2. Tollens' and Baeyer's reagents react with both, and Hinsberg's reagent is for amines.

Teacher's Note:
1. Iodoform is the cleanest test to separate ethanal from other aldehydes.
2. Hinsberg's reagent is benzenesulphonyl chloride and is used for amines only.

Answer: (a) Benzaldehyde with \(\ce{HCN}\)

Explanation:
1. HCN attacks the planar carbonyl group from either face with equal probability, giving equal amounts of both enantiomers of the cyanohydrin.
2. In (b) the stereocentre is not touched and in (c) an \(\ce{S_N2}\) reaction inverts a single centre, so neither gives a racemate.

Teacher's Note:
1. Addition to a planar carbonyl group creates a new stereocentre as a racemic mixture.
2. Reactions at a carbon next to, but not at, the stereocentre leave the configuration unchanged.

Answer: (d) I, III, II, IV

Explanation:
1. Chloral is most electrophilic because of the three chlorine atoms.
2. Aldehydes are more reactive than ketones, and the aryl group of acetophenone lowers reactivity further by resonance, so the order is I, III, II, IV.

Teacher's Note:
1. Reactivity = more positive carbonyl carbon and less steric hindrance.
2. Aryl ketones are the least reactive of the common carbonyl compounds.

Answer: (c) Sodium 2-hydroxyethanoate

Explanation:
1. Glyoxal has no alpha hydrogen, so it undergoes an intramolecular Cannizzaro reaction.
2. One carbonyl is oxidised to carboxylate and the other reduced to alcohol, giving \(\ce{HOCH2COONa}\).

Teacher's Note:
1. The intramolecular version of Cannizzaro occurs in dialdehydes without alpha hydrogens.
2. The product contains both an alcohol and a carboxylate group.