CUET UG Chemistry: Organic Chemistry - Aldehydes and Ketones, Nucleophilic Addition, Cannizzaro, Aldol

Must?Know

• Aldehydes are more reactive than ketones in nucleophilic addition reactions due to lower steric hindrance and greater polarity of the carbonyl group; e.g., acetaldehyde reacts faster with HCN than acetone.
• Nucleophilic addition of HCN to aldehydes and ketones forms cyanohydrins; e.g., CH?CHO + HCN-CH?CH(OH)CN (acetaldehyde cyanohydrin).
• The reaction of aldehydes with NaHSO? forms crystalline bisulfite adducts; useful for purification; e.g., formaldehyde forms HCHO·NaHSO?.
• Grignard reagents add to aldehydes to form secondary and tertiary alcohols; e.g., CH?CHO + CH?MgBr-CH?CH(OMgBr)CH?-CH?CH(OH)CH? after hydrolysis.
• Aldehydes reduce Tollen’s reagent (ammoniacal AgNO?) to silver mirror; ketones do not; e.g., CH?CHO gives silver mirror, CH?COCH? does not.
• Fehling’s solution is reduced by aliphatic aldehydes to red Cu?O precipitate; aromatic aldehydes and ketones do not react; e.g., CH?CHO gives red ppt, C?H?CHO does not.
• Benedict’s test is positive for aliphatic aldehydes; e.g., glucose (an aldose) gives red Cu?O.
• The Iodoform test is given by acetaldehyde and methyl ketones (CH?CO–); e.g., CH?COCH? + 3I? + 4NaOH-CHI? (yellow ppt) + CH?COONa + 3NaI + 3H?O.
• Formaldehyde does not undergo Cannizzaro reaction with concentrated NaOH; it undergoes cross-Cannizzaro with aldehydes lacking ?-hydrogen.
• Aldehydes without ?-hydrogen undergo Cannizzaro reaction: 2HCHO + NaOH-CH?OH + HCOONa (methanol and sodium formate).
• Benzaldehyde undergoes Cannizzaro reaction: 2C?H?CHO + NaOH-C?H?CH?OH + C?H?COONa.
• Aldol condensation occurs in aldehydes/ketones with ?-hydrogen; e.g., two molecules of acetaldehyde in dilute NaOH form aldol (CH?CH(OH)CH?CHO).
• Acetone undergoes aldol condensation: 2CH?COCH (CH?)?C(OH)CH?COCH? (diacetone alcohol).
• Aldol products from aldehydes are ?-hydroxy aldehydes; from ketones, ?-hydroxy ketones.
• Dehydration of aldol gives ?,?-unsaturated carbonyl compound; e.g., CH?CH(OH)CH?CHO-CH?CH=CHCHO (crotonaldehyde).
• Cross aldol between acetaldehyde and acetone gives four products due to two possible enolates and two carbonyls.
• Clemmensen reduction converts carbonyl to methylene using Zn(Hg)/conc. HCl; e.g., CH?COCH?-CH?CH?CH?.
• Wolff-Kishner reduction uses NH?NH?/KOH in ethylene glycol to convert carbonyl to CH?; e.g., acetophenone-ethylbenzene.
• Formaldehyde is a gas at room temperature; acetaldehyde boils at 294 K; acetone at 329 K (verify from NCERT).
• Aldehydes have higher boiling points than hydrocarbons but lower than alcohols of comparable mass due to dipole-dipole interactions.

Difficulty Level

Intermediate — requires understanding of reaction mechanisms, conditions, and ability to distinguish between similar reactions (e.g., aldol vs. Cannizzaro).

Common CUET Traps

• Trap: Assuming all aldehydes give iodoform test. Avoid: Only acetaldehyde (CH?CHO) and methyl ketones (R–CO–CH?) give iodoform test.
• Trap: Thinking benzaldehyde undergoes aldol condensation. Avoid: Benzaldehyde lacks ?-hydrogen, so it undergoes Cannizzaro, not aldol.
• Trap: Believing ketones reduce Tollen’s reagent. Avoid: No ketone reduces Tollen’s reagent; only aldehydes do.

Practice MCQs

1. Which compound gives a silver mirror with Tollen’s reagent?
   A. Acetone
   B. Acetaldehyde
   C. Benzophenone
   D. Diethyl ketone
   Answer: B
   Explanation: Acetaldehyde is an aldehyde and reduces Tollen’s reagent to metallic silver.
   Why others fail: Acetone and other ketones do not reduce Tollen’s reagent.

2. Which of the following undergoes Cannizzaro reaction?
   A. CH?CHO
   B. CH?COCH?
   C. HCHO
   D. CH?CH?CHO
   Answer: C
   Explanation: Formaldehyde (HCHO) lacks ?-hydrogen and undergoes Cannizzaro reaction.
   Why others fail: Acetaldehyde and propanal have ?-hydrogen, so they undergo aldol, not Cannizzaro.

3. The product formed when acetaldehyde reacts with HCN is:
   A. CH?CH?CN
   B. CH?CH(OH)CN
   C. CH?C(CN)?
   D. CH?CH=NOH
   Answer: B
   Explanation: Nucleophilic addition of HCN to acetaldehyde gives 2-hydroxypropanenitrile.
   Why others fail: Option A suggests reduction, not addition; C is gem-dinitrile, not formed here.

4. Which of the following gives iodoform test?
   A. Propanal
   B. Butanal
   C. 3-Pentanone
   D. Acetone
   Answer: D
   Explanation: Acetone (CH?COCH?) is a methyl ketone and gives yellow iodoform precipitate.
   Why others fail: 3-Pentanone is CH?CH?COCH?CH? — no CH?CO– group directly attached to methyl, so negative test.

5. In the presence of dilute NaOH, two molecules of acetaldehyde form:
   A. Acetic acid
   B. Ethanol
   C. Aldol (3-hydroxybutanal)
   D. Crotonaldehyde
   Answer: C
   Explanation: Aldol condensation of acetaldehyde gives 3-hydroxybutanal.
   Why others fail: Crotonaldehyde (D) is the dehydrated product, not the initial adduct.

Last?Minute Revision

• Aldehydes reduce Tollen’s reagent; ketones do not.
• Only aldehydes without ?-H undergo Cannizzaro (e.g., HCHO, C?H?CHO).
• Aldol requires ?-hydrogen; e.g., CH?CHO yes, C?H?CHO no.
• Iodoform test: positive for CH?CHO and R–CO–CH?.
• HCN addition gives cyanohydrin: RRC(OH)CN.
• NaHSO? adducts are crystalline; used to purify carbonyls.
• Grignard + formaldehyde-primary alcohol; other aldehydes-secondary.
• Grignard + ketone-tertiary alcohol.
• Clemmensen: Zn(Hg)/HCl reduces C=O to CH?.
• Wolff-Kishner: NH?NH?/KOH/heat reduces C=O to CH?.
• Acetaldehyde b.p. = 294 K; acetone = 329 K (verify from NCERT).
• Formaldehyde is gas; others are liquids.
• Cross aldol of two different carbonyls may give up to 4 products.
• Dehydration of aldol gives ?,?-unsaturated carbonyl.
• Benzaldehyde + acetaldehyde-cinnamaldehyde via cross-aldol followed by dehydration.
• Fehling’s test: only aliphatic aldehydes positive.
• Benedict’s test: used for urinary glucose detection.
• Mnemonic: "Aldol has ?-H, Cannizzaro doesn’t" — use to distinguish.