CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Case Study Questions Set 02

Case Study 1

Reductive alkylation is the term applied to the process of introducing alkyl groups into ammonia or a primary or secondary amine by means of an aldehyde or ketone in the presence of a reducing agent. The present discussion is limited to those reductive alkylations in which the reducing agent is hydrogen and a catalyst or “nascent” hydrogen, usually from a metal acid combination; most of these reductive alkylations have been carried out with hydrogen and a catalyst. The principal variation excluded is that in which the reducing agent is formic acid or one of its derivatives; this modification is known as the Leuckart reaction. The process of reductive alkylation of ammonia consists in the addition of ammonia to a carbonyl compound and reduction of the addition compound or its dehydration product. The reaction usually is carried out in ethanol solution when the reduction is to be effected catalytically:
RCHO + NH₃ ⇄ RCHOHNH₂ \( \xrightarrow{\text{2[H]}} \) RCH₂NH₂
RCHOHNH₂ ⇄ RCH=NH \( \xrightarrow{\text{2[H]}} \) RCH₂NH₂
Since the primary amine is formed in the presence of the aldehyde it may react in the same way as ammonia, yielding an addition compound, a Schiff’s base (RCH=NCH₂R) and finally, a secondary amine. Similarly, the primary amine may react with the imine, forming an addition product which also is reduced to a secondary amine. Finally, the secondary amine may react with either the aldehyde or the imine to give products which are reduced to tertiary amines.
RCH=NH + RCH₂NH₂ ⇄ RCH(NH₂)NHCH₂R \( \xrightarrow{\text{2[H]}} \) (RCH₂)₂NH + NH₃
(RCH₂)₂NH + RCHO ⇄ (RCH₂)₂NCH(OH)R \( \xrightarrow{\text{2[H]}} \) (RCH₂)₃N + H₂O
(RCH₂)₂NH + RCH=NH ⇄ (RCH₂)₂NCH(NH₂)R \( \xrightarrow{\text{2[H]}} \) (RCH₂)₃N + NH₃
Similar reactions may occur when the carbonyl compound employed is a ketone.

Case Study 2

Aldehydes and ketones are highly reactive compounds. Since both these classes of organic compounds have the same functional group, i.e., polarized carbonyl group, they show a number of common reactions. However, the presence of a H-atom on the carbonyl group of aldehydes make them much more reactive than ketones. At the same time, this H-atom is responsible for many reactions in which aldehydes differ from ketones. The carbonyl group (>C=O) undergoes nucleophilic addition reactions due to electronegativity difference between carbon and oxygen atoms. Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins. Further aldehydes and ketones having atleast one α-H atom in the presence of dilute alkali as a catalyst form β-hydroxy aldehyde (aldol) or β-hydroxy ketones (ketol). This reaction is known as aldol condensation. Aldehydes which do not have an α-hydrogen undergo disproportionation reaction in the presence of concentrated alkali giving a mixture of alcohol and salt of carboxylic acid. This reaction is called Cannizzaro reaction. Cannizzaro reaction involves a hydride ion shift from the carbonyl carbon that is attacked by the base to another carbonyl carbon. Since, there is no hydrogen attached to the carbonyl carbon in a ketone therefore it does not undergo Cannizzaro reaction.

Question 6. Propose the mechanism for the following reaction:
CH₃–CH₂–CHO + HCN \( \xrightarrow{H^+} \) CH₃–CH₂–CH(OH)–CN
Answer:
The reaction proceeds through the nucleophilic attack of CN⁻ ion as follows.
Step 1 (slow): CN⁻ attacks the carbonyl carbon (C^δ+) and the π-electrons shift to the oxygen atom, giving a tetrahedral alkoxide intermediate:
CH₃CH₂CHO + CN⁻ ⇄ CH₃CH₂CH(O⁻)CN
Step 2 (fast): The intermediate takes up a proton (H⁺) to give the cyanohydrin:
CH₃CH₂CH(O⁻)CN + H⁺ ⇄ CH₃CH₂CH(OH)CN

Teacher's Note:
1. In the mechanism, always show the nucleophile attacking carbon first (slow step) and the proton adding to oxygen afterwards (fast step).

Question 7. Propanone is less reactive than ethanal towards nucleophilic addition reactions. Why?
Answer:
The methyl group due to its +I effect reduces the magnitude of positive charge on carbonyl carbon atom. Moreover, it also hinders the approach of the nucleophile. Since in propanone, there are two methyl groups while in acetaldehyde there is one methyl group, therefore, propanone is less reactive than acetaldehyde towards nucleophilic addition reactions.

Teacher's Note:
1. Always quote both reasons – electronic (+I effect) and steric (crowding) – to get full marks.

Question 8. How will you convert ethanal to 2-hydroxy propanoic acid?
Answer:
CH₃CHO (Ethanal) \( \xrightarrow[H^+]{+HCN} \) CH₃–CH(OH)–CN \( \xrightarrow{H_2O/H^+} \) CH₃–CH(OH)–COOH (2-Hydroxy propanoic acid)

Teacher's Note:
1. Cyanohydrin formation followed by hydrolysis adds one carbon and gives an α-hydroxy acid – here lactic acid.

Question 9. Which of the following compounds will undergo Cannizzaro reaction?
(i) (CH₃)₃C–CHO    (ii) CH₃–CH₂–CH(CH₃)–CHO
Write the structures of expected product of the Cannizzaro reaction.
Answer:
(i) will undergo Cannizzaro reaction as it does not contain α-hydrogen.
2(CH₃)₃C–CHO (2,2-Dimethylpropanal) \( \xrightarrow{\text{conc. NaOH}} \) (CH₃)₃C–CH₂OH (2,2-Dimethylpropan-1-ol) + (CH₃)₃C–COO⁻Na⁺ (Sodium-2,2-dimethylpropanoate)

Teacher's Note:
1. Compound (ii) has one α-hydrogen, which is enough for it to undergo aldol condensation instead of Cannizzaro reaction.

Question 10. What happens when acetone undergoes aldol condensation? Write the reaction and name of the products.
Answer:
2CH₃COCH₃ (Acetone) \( \xrightarrow{Ba(OH)_2} \) (CH₃)₂C(OH)–CH₂–CO–CH₃ (4-Hydroxy-4-methylpentan-2-one, Ketol)
(CH₃)₂C(OH)–CH₂–CO–CH₃ \( \xrightarrow[-H_2O]{\Delta} \) (CH₃)₂C=CH–CO–CH₃ (4-Methylpent-3-en-2-one)

Teacher's Note:
1. The ketol formed is a β-hydroxy ketone; on heating it loses water easily to give an α,β-unsaturated ketone.

Case Study 3

Carboxylic acids are the most acidic amongst all the organic compounds studied so far. However, carboxylic acids are much weaker acids than the mineral acids.
Carboxylic acids are more acidic than alcohols and phenols due to the less electrophilic nature of carboxyl carbon which puts a partial positive charge on the hydroxyl O-atom. The value of K_a is a measure of the acidic strength of an acid. Greater the value of K_a, greater is the tendency of the acid to ionize and hence stronger is the acid. The acidic strength of saturated aliphatic carboxylic acids depends mainly upon the inductive effect of the substituent and its position with respect to –COOH group. Electron-donating substituents tend to decrease whereas electron-withdrawing substituents tend to increase the acidic strength. The acidic strength of aromatic carboxylic acids, on the other hand depends upon both the inductive and the resonance effect of the substituents.

Question 11. What is meant by ‘acidity constant’ K_a? How is it expressed?
Answer:
Acidity constant K_a is a measure of the strength of the acid. It is expressed as,
\( K_a = \frac{[RCOO^-][H_3O^+]}{[RCOOH]} \)

Teacher's Note:
1. Larger K_a means smaller pK_a and a stronger acid – questions often switch between the two.

Question 12. What makes ethanoic acid a stronger acid than ethanol?
Answer:
Ethanoate ion obtained by loss of a proton from ethanoic acid is stabilized by resonance but ethoxide ion obtained by loss of a proton from ethanol does not.

Teacher's Note:
1. Acid strength questions are best answered by comparing the stability of the conjugate bases.

Question 13. Why is pK_a of chloroacetic acid lower than pK_a of acetic acid?
Answer:
Due to –I effect of Cl atom and +I effect of CH₃ group the electron density in the O–H bond in chloroacetic acid is much weaker than in acetic acid and hence loses a proton more easily than acetic acid.

Teacher's Note:
1. Lower pK_a means stronger acid; the −I effect of Cl also stabilises the chloroacetate ion.

Question 14. Arrange the following in the decreasing order of their acidic character:
CH₃CH₂OH, CH₃COOH, ClCH₂COOH, FCH₂COOH, C₆H₅CH₂COOH
Answer:
FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH

Teacher's Note:
1. F is more electronegative than Cl, so its −I effect is stronger; the phenyl group is also weakly electron-withdrawing compared with H.

Question 15. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give two reasons.
Answer:
(i) Resonating structures of carboxylate ion are more stable than phenoxide ion.
(ii) Negative charge is dispersing on two electronegative oxygens in carboxylate ion whereas it is on one oxygen atom in phenoxide ion.

Teacher's Note:
1. The number of resonance structures matters less than where the negative charge sits – on oxygen it is stable, on carbon it is not.

Case Study 4

Aldol condensation is not confined to condensation between two molecules of the same aldehyde or ketone but can also take place between two different aldehydes or ketones or between one aldehyde and one ketone. Such condensations are called crossed aldol condensations. If both the aldehydes/ketones contain α-hydrogens, such crossed aldol condensations are not of any synthetic value since a mixture of four products is obtained which is difficult to separate. However, if one of the carbonyl compounds does not possess α-hydrogens then crossed aldol condensations are of great synthetic utility. Further, crossed aldol condensation between an aromatic aldehyde and an aliphatic aldehyde or a ketone or even an aliphatic esters are called Claisen-Schmidt condensations. Besides these, intermolecular condensations, intramolecular aldol condensation can also occur between compounds containing two aldehyde/ketone groups or one aldehyde and one ketone group.

Question 16. Presence of α-hydrogen in aldehydes and ketones is essential for aldol condensation. Give reason.
Answer:
Aldol condensation involves nucleophilic attack by carbanion of one molecule of an aldehyde/ketone on the carbonyl group of the other. For carbanion formation from an aldehyde/ketone, the α-hydrogen must be acidic so that it can be easily abstracted by a weak base. This is possible only if α-hydrogens are acidic and the carbanion formed is stabilized by resonance due to electron-withdrawing effect of the carbonyl group.
HO⁻ + H–CH(R)–CO–R′ → H₂O + ⁻CH(R)–CO–R′ ↔ CH(R)=C(O⁻)–R′ (Carbanion)

Teacher's Note:
1. Without an α-H, no carbanion (enolate) can form, so there is no nucleophile to start the condensation.

Question 17. Write the structure of the crossed aldol product in which butanal acts as an electrophile and propanal acts as a nucleophile.
Answer:
The carbanion of propanal (formed at its α-carbon) attacks the carbonyl carbon of butanal :
CH₃CH₂CH₂–CHO (Butanal, electrophile) + H–CH(CH₃)–CHO (Propanal, nucleophile) \( \xrightarrow{Dil.\ NaOH} \) ⁶CH₃⁵CH₂⁴CH₂–³CH(OH)–²CH(CH₃)–¹CHO (3-Hydroxy-2-methylhexanal)

Teacher's Note:
1. The nucleophile keeps its CHO group; the electrophile’s C=O becomes the CH–OH of the product.

Question 18. Explain Claisen-Schmidt condensation between one mole each of benzaldehyde and acetone.
Answer:
The carbanion of acetone (formed by removal of an α-H by dil. NaOH) attacks the carbonyl carbon of benzaldehyde, which has no α-H. The β-hydroxy ketone formed loses water to give benzalacetone :
C₆H₅CHO + H–CH₂COCH₃ \( \xrightarrow{Dil.\ NaOH} \) [C₆H₅CH(OH)CH₂COCH₃] \( \xrightarrow{-H_2O} \) C₆H₅CH=CH–CO–CH₃ (Benzalacetone)

Teacher's Note:
1. The dehydration is very easy here because the C=C formed is conjugated with both the ring and the C=O group.

Question 19. With one example in each case, differentiate between intermolecular and intramolecular aldol condensation reactions.
Answer:
(i) Intermolecular aldol condensation takes place between two separate molecules :
CH₃CHO + H–CH₂CHO \( \xrightarrow{Dil.\ NaOH} \) CH₃CH(OH)CH₂CHO (3-Hydroxybutanal)
(ii) Intramolecular aldol condensation takes place between two carbonyl groups of the same molecule, giving a ring :
OHC–CH₂CH₂CH₂CH₂–CHO (Hexane-1, 6-dial) \( \xrightarrow{Dil.\ NaOH} \) [cyclic aldol] \( \xrightarrow{-H_2O} \) Cyclopent-1-enecarbaldehyde (a five-membered ring with C=C at C-1 carrying the CHO group)

Teacher's Note:
1. Intramolecular aldol reactions are favoured when they form five- or six-membered rings.

Case Study 5

Except formic acid, both aliphatic and aromatic carboxylic acids can be prepared conveniently in excellent yields by using Grignard reagents. Formic acid and acetic acid show some different properties. Whereas formic acid acts as a reducing agent but acetic acid does not. Further acetic acid shows Hell-Vohlard Zelinsky reaction but formic acid does not.

Question 20. With a chemical equation, explain the formation of benzoic acid using a suitable Grignard reagent.
Answer:
C₆H₅–MgBr (Phenylmag bromide) + O=C=O \( \xrightarrow{Dry\ ether} \) [C₆H₅–CO–OMgBr] \( \xrightarrow[-Mg(OH)Br]{H^+/H_2O} \) C₆H₅–COOH (Benzoic acid)

Teacher's Note:
1. The nucleophilic carbon of the Grignard reagent attacks the carbon of CO₂.

Question 21. Which out of formic acid and acetic acid is a stronger acid and why ?
Answer:
Formic acid is a stronger acid than acetic acid because +I-effect of the CH₃ group destabilizes acetate ion relative to formate ion :
H–COO⁻ (Formate ion, more stable) ; CH₃→COO⁻ (Acetate ion, destabilized by +I-effect of CH₃ group)

Teacher's Note:
1. pK_a: HCOOH 3.75, CH₃COOH 4.76; the lower value belongs to the stronger acid.

Question 22. Give an example of Hell-Vohlard Zelinsky reactions. Why CH₃COOH gives this reaction but HCOOH does not ?
Answer:
Carboxylic acids containing α-hydrogens on treatment with Cl₂ or Br₂ in presence of red P undergo Hell-Vohlard-Zelinsky (HVZ) reaction to give α-haloacids. For example,
CH₃COOH + Br₂ \( \xrightarrow{Red\ P} \) BrCH₂COOH (α-Bromoacetic acid) + HBr ; HCOOH \( \xrightarrow{Br_2/red\ P} \) No reaction
CH₃COOH contains α-hydrogens and hence gives HVZ reaction but HCOOH does not contain an α-hydrogen and hence does not undergo HVZ reaction.

Teacher's Note:
1. HCOOH has no α-carbon at all: its only H on carbon is attached to the carboxyl carbon itself.

Question 23. Formic acid reduces Tollens’ reagent and Fehling’s solution but acetic acid does not. Explain, why so ?
Answer:
Formic acid contains a H-atom attached to >C=O group and hence can be regarded as an aldehyde. Therefore, it acts as a reducing agent and hence reduces Tollens’ reagent and Fehling’s solution but acetic acid does not.
H–COOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ \( \xrightarrow{Heat} \) 2Ag↓ (Silver mirror) + 4NH₃ + CO₂ + 2H₂O
H–COOH + 2Cu²⁺ + 4OH⁻ \( \xrightarrow{Heat} \) Cu₂O↓ (Red ppt.) + CO₂ + 3H₂O

Teacher's Note:
1. Formic acid is oxidised all the way to CO₂ and H₂O, which is why it is a good reducing agent.

Case Study 6

Two useful reactions of aldehydes are Cannizzaro reaction and Aldol condensation. Whereas Cannizzaro reaction is characteristic of aldehydes containing no α-hydrogens in presence of concentrated alkali, Aldol condensation is characteristic of aldehydes containing α-hydrogens in presence of dilute alkali. Besides aldehydes, ketones containing α-hydrogens also undergo Aldol condensation. These aldols on heating with dilute acid or alkali undergo dehydration to give the corresponding α, β-unsaturated aldehydes or ketones. Besides simple Cannizzaro reactions and Aldol condensations, Cross Cannizzaro reactions and Cross aldol condensations are also quite common reactions.

Question 24. Predict the products (A and B) of the following reaction.
3-Chlorobenzaldehyde (m-ClC₆H₄CHO) + KOH (50%) → A + B
Answer:
2 m-ClC₆H₄CHO + KOH (50%) \( \xrightarrow[reaction]{Cannizzaro} \) m-ClC₆H₄COOK + m-ClC₆H₄CH₂OH
A = Potassium 3-chlorobenzoate ; B = 3-Chlorobenzyl alcohol

Teacher's Note:
1. The Cl on the ring does not take part; only the CHO group is oxidised or reduced.

Question 25. An equimolar mixture of benzaldehyde and formaldehyde is treated with a concentrated solution of KOH. Write the products and explain which of the two is oxidised and which of the two is reduced ?
Answer:
C₆H₅CHO (Benzaldehyde, reduced) + HCHO (Formaldehyde, oxidised) \( \xrightarrow{50\%\ KOH} \) C₆H₅CH₂OH (Benzyl alcohol) + HCOOK (Pot. formate)
In this crossed Cannizzaro reaction, formaldehyde, being more reactive, is attacked by OH⁻ and transfers a hydride ion to benzaldehyde. Hence formaldehyde is oxidised to formate and benzaldehyde is reduced to benzyl alcohol.

Teacher's Note:
1. In any crossed Cannizzaro reaction with HCHO, it is HCHO that ends up as formate.

Question 26. Write the product of crossed aldol condensation between acetaldehyde and propronaldehyde when acetaldehyde acts as the electrophile and propronaldehyde acts as nucleophile ?
Answer:
CH₃CH=O (Acetaldehyde, electrophile) + CH₃–CH₂–CHO (Propionaldehyde, nucleophile) \( \xrightarrow{Dil\ NaOH} \) ⁴CH₃–³CH(OH)–²CH(CH₃)–¹CHO (3-Hydroxy-2-methylbutanal)

Teacher's Note:
1. The α-carbon of the nucleophile (here C-2 of propanal) forms the new C–C bond.

Question 27. How is mesityl oxide obtained from acetone : Indicate the conditions and the name of the reaction.
Answer:
By aldol condensation of acetone in presence of Ba(OH)₂, followed by heating with dilute acid :
(CH₃)₂C=O + H–CH₂–CO–CH₃ \( \xrightarrow[(Aldol\ condensation)]{Ba(OH)_2} \) (CH₃)₂C(OH)CH₂COCH₃ (Diacetone alcohol) \( \xrightarrow[\Delta]{H^+/H_2O} \) (CH₃)₂C=CH–CO–CH₃ (Mesityl oxide) + H₂O

Teacher's Note:
1. Mesityl oxide is 4-methylpent-3-en-2-one; its C=C is conjugated with the C=O group.

Case Study 7

Besides preparation of aldehydes from alcohols and carboxylic acids, other important reactions include Rosenmund reduction, Stephen reduction, reduction of nitriles and esters with DIBAL-H, oxidation of toluene and its derivatives with chromyl chloride (Etard reaction) or chromium trioxide in acetic anhydride and Gattermann-Koch formylation or simply Gattermann formylation.

Case Study 8

The acid strength of carboxylic acids is expressed either in terms of K_a or pK_a. The nature of substituents affect the acid strength of both aliphatic and aromatic acids. Whereas electron-withdrawing groups (EWG) increase while electron-donating groups (EDG) decrease the acid strength of substituted acids. Besides the nature, position of the substituent w.r.t. the –COOH group also affects the acid strength. In case of aliphatic acids, farther the substituent from the –COOH group, weaker is its effect. In case of aromatic acids, the effect is more pronounced at p- than at m-position. The o-substituted benzoic acids (ortho effect) are usually stronger acids than benzoic acid regardless of the nature of the substituent whether electron-donating or electron-withdrawing.