Chapter 8: Aldehydes, Ketones and Carboxylic Acids Case Study Questions | chemistry Class 12 CBSE

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Chapter 8: Aldehydes, Ketones and Carboxylic Acids Case Study Questions | chemistry Class 12 CBSE | ByteNotes.in

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Priya works in a food-processing laboratory. She receives an unknown organic compound and must identify whether it is an aldehyde or a ketone. She performs Tollens' test and Fehling's test. She then prepares its 2,4-dinitrophenylhydrazone derivative for characterisation. In a subsequent step she reduces the compound using NaBH4 and separately uses LiAlH4. The compound gives a yellow precipitate with iodoform reagent (I2/NaOH). On drastic oxidation with chromic acid, the compound yields a carboxylic acid with one fewer carbon atom. Based on these observations, the compound is identified as a methyl ketone. It is widely used as an industrial solvent.

Question 1: Which tests confirm that the compound is a ketone and not an aldehyde?

The compound does not give a silver mirror with Tollens' reagent and does not give a reddish-brown precipitate with Fehling's reagent, ruling out aldehyde.
Ketones do not reduce these mild oxidising agents, so both tests give negative results for the compound.

Question 2: Explain why the compound gives a positive iodoform test and identify it.

The compound contains a CH3CO– group (methyl ketone) which reacts with I2/NaOH (sodium hypoiodite) to give CHI3 (yellow precipitate of iodoform).
Since oxidation gives an acid with one fewer carbon, and considering industrial solvent use, the compound is propanone (acetone, CH3COCH3).

Question 3: Compare the products obtained when propanone is reduced by (i) NaBH4, (ii) LiAlH4, and (iii) zinc amalgam with concentrated HCl. Write the name of each product and the type of reduction involved.

NaBH4 reduces propanone to propan-2-ol (secondary alcohol) — nucleophilic addition of hydride; NaBH4 is a mild, selective reducing agent.
LiAlH4 also reduces propanone to propan-2-ol (secondary alcohol) — it is a stronger hydride donor but gives the same product for ketones.
Zinc amalgam with conc. HCl (Clemmensen reduction) reduces the carbonyl group completely to a –CH2– group, giving propane — this converts the carbonyl to a methylene group.

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A fragrance chemist is studying aromatic aldehydes used in perfumery. Benzaldehyde, salicylaldehyde, vanillin and cinnamaldehyde are some naturally occurring aromatic aldehydes. She notices that benzaldehyde can be prepared in the lab by the Gatterman-Koch reaction, Rosenmund reduction, and by the side-chain chlorination of toluene followed by hydrolysis. She also observes that benzaldehyde undergoes the Cannizzaro reaction with concentrated NaOH, and gives a positive 2,4-DNP test but a negative Fehling's test. It is used extensively in perfumery and dye industries.

Question 1: Why does benzaldehyde give a negative Fehling's test even though it is an aldehyde?

Fehling's test is given by aliphatic aldehydes. Aromatic aldehydes like benzaldehyde do not reduce Fehling's reagent due to the resonance stabilisation of the benzaldehyde molecule making the carbonyl carbon less reactive toward the Cu2+ complex.
Therefore benzaldehyde gives a negative Fehling's test.

Question 2: Write the balanced equation for the Cannizzaro reaction of benzaldehyde with concentrated NaOH.

2C6H5CHO + conc. NaOH → C6H5CH2OH + C6H5COONa
One molecule is oxidised to sodium benzoate (carboxylate salt) and the other is reduced to benzyl alcohol — this is a disproportionation reaction of aldehydes lacking alpha-hydrogen.

Question 3: Describe the preparation of benzaldehyde by (i) the Gatterman-Koch reaction and (ii) the Rosenmund reduction, writing the reagents and conditions for each.

Gatterman-Koch reaction: Benzene is treated with CO and HCl in the presence of anhydrous AlCl3 (or CuCl) as catalyst. The electrophilic formylation gives benzaldehyde directly.
Rosenmund reduction: Benzoyl chloride (C6H5COCl) is hydrogenated over palladium on barium sulphate (Pd-BaSO4) catalyst with H2 gas. The catalyst is partially poisoned to prevent over-reduction, giving benzaldehyde.
Rosenmund reduction is an example of selective reduction of acyl chloride to aldehyde.

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In an industrial plant, acetic acid (ethanoic acid) is one of the most important carboxylic acids manufactured. It is obtained by the oxidation of ethanol and acetaldehyde. Acetic acid is used as a solvent and as vinegar in the food industry. A chemist notices that acetic acid has a much higher boiling point (391 K) than propan-1-ol (370 K) despite both having similar molecular masses. He also finds that acetic acid exists as a dimer even in the vapour phase. The chemist further studies various substituted acetic acids and finds that electron-withdrawing groups increase the acidity.

Question 1: Explain why acetic acid has a higher boiling point than propan-1-ol despite similar molecular masses.

Acetic acid molecules form strong intermolecular hydrogen bonds and exist as dimers even in the vapour phase, due to the cyclic hydrogen-bonded structure between two –COOH groups.
This more extensive association requires more energy to overcome, resulting in a higher boiling point than propan-1-ol.

Question 2: Arrange the following acids in decreasing order of acid strength and justify: CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH.

Decreasing acid strength: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH.
More chlorine atoms attached to the alpha-carbon exert a stronger inductive electron-withdrawing effect, stabilising the carboxylate anion more, thus increasing acidity.

Question 3: Explain why carboxylic acids are stronger acids than phenols despite phenoxide ion having more resonance structures. Support your answer with a comparison of their conjugate bases.

In the carboxylate ion (conjugate base of RCOOH), the negative charge is delocalised equally over two electronegative oxygen atoms through two equivalent resonance structures, providing high stabilisation.
In the phenoxide ion (conjugate base of phenol), the negative charge is delocalised over one oxygen and several less-electronegative carbon atoms through non-equivalent resonance structures — this delocalisation is less effective.
Since the carboxylate ion is more stabilised than the phenoxide ion, carboxylic acids ionise more readily, making them stronger acids than phenols.

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A student is performing reactions of ethanal in the organic chemistry laboratory. Ethanal (CH3CHO) is a volatile liquid with a pungent odour. The student observes that ethanal forms an aldol product when treated with dilute NaOH, undergoes nucleophilic addition with HCN to give a cyanohydrin, reacts with hydroxylamine to give an oxime, and forms a silver mirror with Tollens' reagent. The student also notes that in the cross-aldol condensation with propanal, multiple products are obtained. The α-carbon of ethanal is more acidic than a typical C–H bond.

Question 1: Write the product of the aldol condensation of ethanal and name it.

Two molecules of ethanal react in dilute NaOH to give 3-hydroxybutanal (aldol), which on heating loses water to give but-2-enal (crotonaldehyde, CH3CH=CHCHO).
The final product is but-2-enal, the aldol condensation product.

Question 2: Explain the mechanism by which HCN adds to ethanal to give a cyanohydrin.

HCN alone adds very slowly; it is catalysed by base (OH–) which generates CN– ion, a stronger nucleophile.
CN– attacks the electrophilic carbonyl carbon of ethanal from a direction perpendicular to the sp2 plane, changing hybridisation to sp3. The resulting alkoxide intermediate is protonated by H+ from the medium to give 2-hydroxypropanenitrile (lactonitrile), the cyanohydrin.

Question 3: When ethanal and propanal are mixed and subjected to aldol condensation, four products form. Write the structural formulas and names of all four products.

Self-aldol of ethanal: CH3CH=CHCHO — but-2-enal (from two ethanal molecules).
Self-aldol of propanal: CH3CH2CH=C(CH3)CHO — 2-methylpent-2-enal (from two propanal molecules).
Cross product (ethanal as nucleophile, propanal as electrophile): CH3CH2CH=CHCHO — pent-2-enal.
Cross product (propanal as nucleophile, ethanal as electrophile): CH3CH2C(CH3)=CHCHO — 2-methylbut-2-enal.

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A pharmaceutical researcher needs to prepare various carboxylic acid derivatives starting from available raw materials. She plans to convert benzene to benzoic acid, prepare adipic acid (hexanedioic acid) for nylon-6,6 manufacture, and also obtain p-nitrobenzoic acid from toluene. She uses Grignard reagents, oxidation reactions, and nitrile hydrolysis. She also observes that carboxylic acids undergo esterification with alcohols in the presence of mineral acid catalyst via a nucleophilic acyl substitution mechanism.

Question 1: How can benzoic acid be prepared from bromobenzene using a Grignard reagent? Give the reagents and conditions.

Bromobenzene is treated with Mg in dry ether to form phenylmagnesium bromide (C6H5MgBr).
The Grignard reagent is reacted with dry CO2 (dry ice); acidification with H3O+ gives benzoic acid (C6H5COOH).

Question 2: Describe the mechanism of esterification of acetic acid with ethanol using H2SO4 as catalyst.

H+ protonates the carbonyl oxygen of acetic acid, activating the carbonyl carbon towards nucleophilic attack by ethanol, forming a tetrahedral intermediate.
Proton transfer converts –OH to –+OH2 (a better leaving group), which leaves as water, and the protonated ester loses H+ to give ethyl acetate (ethyl ethanoate).

Question 3: How would you convert toluene to p-nitrobenzoic acid in not more than three steps? Write the reagents and reactions at each step.

Step 1: Nitration of toluene — treat toluene with conc. HNO3/conc. H2SO4 at low temperature (below 300 K) to get p-nitrotoluene (major product due to ortho/para directing –CH3 group).
Step 2: Oxidation — treat p-nitrotoluene with KMnO4/KOH under heat (or acidic KMnO4) to oxidise the –CH3 group to –COOH, giving dipotassium p-nitrbenzoate; acidify with H3O+ to get p-nitrobenzoic acid.
The –NO2 group is electron-withdrawing and resistant to oxidation; the entire methyl side chain is converted to –COOH.

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Chapter 8: Aldehydes, Ketones and Carboxylic Acids | Case Study Questions

Passage

Question 1: Which tests confirm that the compound is a ketone and not an aldehyde?

Question 2: Explain why the compound gives a positive iodoform test and identify it.

Question 3: Compare the products obtained when propanone is reduced by (i) NaBH4, (ii) LiAlH4, and (iii) zinc amalgam with concentrated HCl. Write the name of each product and the type of reduction involved.

Passage

Question 1: Why does benzaldehyde give a negative Fehling's test even though it is an aldehyde?

Question 2: Write the balanced equation for the Cannizzaro reaction of benzaldehyde with concentrated NaOH.

Question 3: Describe the preparation of benzaldehyde by (i) the Gatterman-Koch reaction and (ii) the Rosenmund reduction, writing the reagents and conditions for each.

Passage

Question 1: Explain why acetic acid has a higher boiling point than propan-1-ol despite similar molecular masses.

Question 2: Arrange the following acids in decreasing order of acid strength and justify: CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH.

Question 3: Explain why carboxylic acids are stronger acids than phenols despite phenoxide ion having more resonance structures. Support your answer with a comparison of their conjugate bases.

Passage

Question 1: Write the product of the aldol condensation of ethanal and name it.

Question 2: Explain the mechanism by which HCN adds to ethanal to give a cyanohydrin.

Question 3: When ethanal and propanal are mixed and subjected to aldol condensation, four products form. Write the structural formulas and names of all four products.

Passage

Question 1: How can benzoic acid be prepared from bromobenzene using a Grignard reagent? Give the reagents and conditions.

Question 2: Describe the mechanism of esterification of acetic acid with ethanol using H2SO4 as catalyst.

Question 3: How would you convert toluene to p-nitrobenzoic acid in not more than three steps? Write the reagents and reactions at each step.

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