RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
The carbonyl group C=O is one of the most reactive functional groups in organic chemistry, and this chapter is really a study of what nucleophiles do to it. You will trace the mechanism of nucleophilic addition, learn how to tell an aldehyde from a ketone with simple test-tube reactions, and see how oxidising a carbonyl one step further gives the carboxylic acid group, whose acidity is governed by the stability of its conjugate base.
Which of the following gives a positive Tollens' test?
Answer
Ethanal. Tollens' reagent oxidises the C–H bond directly attached to the carbonyl carbon of an aldehyde, depositing metallic silver as a mirror. Acetone, benzophenone and butan-2-one are all ketones; the carbonyl carbon of a ketone bears two alkyl/aryl groups and no hydrogen, so there is no easily removable hydrogen for the oxidant to attack, and none of them reduces Tollens' reagent.
Arrange in decreasing order of acidity: acetic acid, chloroacetic acid, trichloroacetic acid, formic acid.
Answer
Trichloroacetic acid > chloroacetic acid > formic acid > acetic acid. Each electronegative chlorine on the α-carbon withdraws electron density inductively and stabilises the carboxylate anion further, so acidity rises with the number of chlorines: trichloro- > chloro- > unsubstituted acetic acid. Formic acid (HCOOH) has no alkyl group at all, so it lacks the mild electron-donating effect of the methyl group in acetic acid, making it more acidic than acetic acid but less than the chlorinated acids.
Benzaldehyde on treatment with concentrated NaOH gives:
Answer
Benzyl alcohol and sodium benzoate in equal amounts. Benzaldehyde has no α-hydrogen (the carbon next to CHO is part of the aromatic ring), so it cannot form an enolate for aldol condensation. Instead concentrated base drives the Cannizzaro reaction: one molecule is reduced to C₆H₅CH₂OH while a second is oxidised to C₆H₅COO⁻Na⁺, in a 1:1 ratio, via hydride transfer between two molecules of the aldehyde.
Propanoic acid is treated with Br₂ in the presence of a small amount of red phosphorus. The product is:
Answer
2-Bromopropanoic acid. This is the Hell–Volhard–Zelinsky reaction: red phosphorus converts a trace of the acid to the acid bromide, which enolises much more readily than the acid itself, and bromine adds at the resulting α-carbon (C-2, adjacent to –COOH). The final product, after the enol form re-tautomerises, is the α-brominated carboxylic acid.
Assertion (A): Aldehydes are generally more reactive than ketones towards nucleophilic addition. Reason (R): A ketone carbonyl carbon is flanked by two alkyl groups, which are bulkier and more electron-donating than the single alkyl group (or hydrogen) on an aldehyde carbon.
Answer
Both A and R are true and R is the correct explanation of A. Two alkyl groups on a ketone carbonyl carbon create more steric hindrance to an incoming nucleophile than the single alkyl group (or a plain hydrogen, as in formaldehyde) on an aldehyde. They also push electron density onto the carbonyl carbon more strongly by the +I effect, making it less electrophilic and less attractive to a nucleophile. Both the steric and the electronic factor cited in R work together to make ketones less reactive than aldehydes.
Write the aldol condensation of acetaldehyde with itself under dilute NaOH, showing both the aldol product and the final product on heating.
Answer
Step 1 (aldol addition): 2CH₃CHO --dil. NaOH--> CH₃CH(OH)CH₂CHO (3-hydroxybutanal). Hydroxide removes an α-hydrogen from one molecule to give a resonance-stabilised enolate, which attacks the carbonyl carbon of a second acetaldehyde molecule; protonation gives the β-hydroxy aldehyde. Step 2 (dehydration on heating): CH₃CH(OH)CH₂CHO --heat--> CH₃CH=CHCHO + H₂O, giving but-2-enal (crotonaldehyde), the α,β-unsaturated carbonyl product.
Distinguish between propanal and propan-2-one using (a) Tollens' reagent, (b) iodoform test, (c) 2,4-DNP reagent, stating what is observed with each and for which compound.
Answer
(a) Tollens' reagent: propanal, an aldehyde, gives a bright silver mirror on warming as it is oxidised to propanoate while Ag⁺ is reduced to Ag(0); propan-2-one, a ketone, gives no mirror because it has no hydrogen on the carbonyl carbon to be removed. (b) Iodoform test: propan-2-one, which has a CH₃–CO– group, gives a yellow precipitate of iodoform (CHI₃) with I₂/NaOH; propanal, CH₃CH₂CHO, lacks a methyl ketone/appropriate CH₃CH(OH)– pattern and gives a negative test. (c) 2,4-DNP reagent: both compounds, having a carbonyl group, give a yellow-orange precipitate of the hydrazone, so this test confirms the presence of C=O in either but does not distinguish the two.
Explain why formic acid (HCOOH) reduces Tollens' reagent and decolourises Br₂ water, unlike acetic acid.
Answer
Formic acid contains a –CHO-like hydrogen directly bonded to the carboxyl carbon; structurally it can be written as H–COOH, which behaves partly like an aldehyde because that hydrogen is oxidisable. It is therefore readily oxidised by Tollens' reagent to give a silver mirror (with formation of CO₂) and by bromine water, which it decolourises as it is oxidised to carbonic acid. Acetic acid, CH₃COOH, has only unreactive methyl hydrogens on the far side of the molecule and no hydrogen on the carboxyl carbon itself, so it shows neither reaction.
(i) Describe the mechanism of nucleophilic addition of HCN to acetaldehyde, and explain why the reaction needs a trace of base. (ii) Compare the relative rates of nucleophilic addition of HCN to acetaldehyde, acetone and benzaldehyde, giving reasons. (iii) How would you prepare acetaldehyde from ethyne in one step?
Answer
(i) Cyanide ion, the true nucleophile, attacks the electrophilic carbonyl carbon of CH₃CHO, and the C=O π electrons shift fully onto oxygen, giving a tetrahedral alkoxide intermediate CH₃CH(O⁻)(CN). This alkoxide is then protonated (by HCN or water) to give the product, 2-hydroxypropanenitrile, CH₃CH(OH)CN. HCN itself is a very weak acid and dissociates poorly, so the concentration of free CN⁻ generated is too low for a fast reaction; a trace of base such as NaOH or KCN removes a proton from HCN, boosting [CN⁻] and speeding up the nucleophilic attack. (ii) Rate: acetaldehyde > benzaldehyde > acetone (roughly). Acetaldehyde has only a small methyl group and a hydrogen at the carbonyl carbon, offering the least steric hindrance and the least electron donation, so it is the fastest of the three. Benzaldehyde's phenyl ring is bulkier and also donates electron density into the carbonyl by resonance, lowering the carbon's electrophilicity, so it reacts more slowly than acetaldehyde but is still an aldehyde, so it typically outpaces acetone. Acetone carries two electron-donating, sterically demanding methyl groups, making it the least reactive of the three towards nucleophilic addition. (iii) Ethyne is passed into warm dilute H₂SO₄ containing HgSO₄ as catalyst (acid-catalysed mercury-ion hydration): HC≡CH + H₂O --HgSO₄, H₂SO₄--> CH₃CHO. Water adds across the triple bond according to Markovnikov's rule to give the unstable enol CH₂=CHOH, which immediately tautomerises to the more stable keto form, acetaldehyde.
(i) Explain why carboxylic acids are much stronger acids than phenols, comparing the stability of their respective anions. (ii) Arrange the following in increasing order of acidic strength and justify: benzoic acid, 4-nitrobenzoic acid, 4-methoxybenzoic acid. (iii) Describe, with equation, how esters are prepared from carboxylic acids and name the catalyst and its role.
Answer
(i) In the carboxylate anion RCOO⁻ the negative charge is delocalised equally over two identical, equivalent oxygen atoms by resonance, confirmed experimentally by the two equal C–O bond lengths in the anion; this symmetric, extensive delocalisation gives strong stabilisation. In the phenoxide ion the negative charge is delocalised into the aromatic ring, placing partial negative charge on carbon atoms, which are far less electronegative and thus less able to stabilise negative charge than oxygen. Because the carboxylate's stabilisation (charge shared between two oxygens) is more effective than the phenoxide's (charge partly on carbon), carboxylic acids ionise far more readily and are much stronger acids than phenols. (ii) Increasing acidic strength: 4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid. The para-methoxy group is electron-donating by resonance, which destabilises the developing negative charge on the carboxylate and makes the parent acid the weakest of the three. The para-nitro group is strongly electron-withdrawing by both induction and resonance, stabilising the carboxylate anion strongly and making this acid the strongest. Unsubstituted benzoic acid falls in between. (iii) Fischer esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O, carried out by refluxing the acid and alcohol with a small amount of concentrated H₂SO₄ (or dry HCl gas). The acid catalyst protonates the carbonyl oxygen, making the carbonyl carbon much more electrophilic so the alcohol oxygen can add to it; after proton transfers, water is eliminated to give the ester and regenerate the catalyst. Because the reaction is an equilibrium, excess alcohol or removal of water (e.g. by distillation) is used to push the yield of ester higher.
Four test tubes contain, in random order, ethanal, propan-2-one, benzaldehyde and propanoic acid. A student runs three tests. Test 1 (Tollens' reagent): two tubes give a silver mirror. Test 2 (iodoform, I₂/NaOH): one of the two Tollens'-positive tubes also gives a yellow precipitate, and the Tollens'-negative tube that is not acidic also gives a yellow precipitate. Test 3 (litmus): exactly one tube turns blue litmus red. (a) Identify all four compounds with reasoning. (b) Write the iodoform equation for the ketone identified. (c) Which tube would decolourise both Tollens' reagent's silver-forming step and give no iodoform test — describe its full identification path. (d) Suggest a test to distinguish benzaldehyde from ethanal directly, since both are Tollens'-positive.
Answer
(a) Litmus turns red only for the acid, so propanoic acid is identified directly by test 3. Two tubes are Tollens'-positive, so those two must be the aldehydes: ethanal and benzaldehyde. Of the Tollens'-positive tubes, the one also giving a positive iodoform test has a CH₃CO– or CH₃CH(OH)– pattern, which matches ethanal (CH₃CHO); the other Tollens'-positive tube, giving no iodoform, is benzaldehyde. The Tollens'-negative, non-acidic tube that still gives iodoform must be the methyl ketone, propan-2-one. (b) CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O. (c) No single tube here is both Tollens'-negative and iodoform-negative among the four, since propanoic acid is identified by litmus alone, is Tollens'-negative and iodoform-negative — its full path is: does not turn Tollens' reagent, does not give iodoform (no CH₃CO or CH₃CH(OH) unit), but turns blue litmus red, confirming it as the carboxylic acid, propanoic acid. (d) Warm each with Fehling's solution: ethanal, an aliphatic aldehyde, gives the brick-red Cu₂O precipitate, while benzaldehyde, an aromatic aldehyde, gives no reaction with Fehling's solution, cleanly telling the two apart despite both being Tollens'-positive.
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