ORGANIC CHEMISTRY

Answer: d. (CH₃)₂C(OH)CO₂H

Step-by-step reasoning:

Mechanism:

1. Nucleophilic addition of HCN to ketone → cyanohydrin
2. Acid hydrolysis: -CN → -COOH
3. Product: 2-hydroxy-2-methylpropanoic acid

Answer: b. 2-bromobutane

Analysis:

Why 2-bromobutane?

• Step 1: Nucleophilic substitution → 2-butanol (secondary alcohol)
• Step 2: Secondary alcohol oxidizes → butanone (ketone, C₄H₈O)
• Ketone cannot oxidize further → “only C₄H₈O”

Why not others?

• a. 1-bromobutane: Would give butanal then butanoic acid (not “only”)
• c. 1-bromo-2-methylpropane: Primary alcohol → aldehyde → acid
• d. 2-bromo-2-methylpropane: Tertiary alcohol → no oxidation

Answer: a. BrCH₂CH₂CH₂CH₂Br

Synthetic pathway:

Key points:

• Need a dihalide to get dinitrile (two -CN groups)
• Reduction: Each -CN → -CH₂NH₂
• Product: Hexamethylenediamine (monomer for nylon 6,6)
• Chain length: (CH₂)₄ becomes (CH₂)₆ after reduction (adds 2 CH₂ per CN)

Answer: C. Q and R with 2,4-DNPH; R only with Fehling’s

Analysis of burnt sugar structure:

Test results:

Conclusion: 2,4-DNPH tests for ANY carbonyl → both Q and R. Fehling’s only for aldehydes → only Q.

Answer: d. CH₃CH(OH)CN

Synthetic route:

Why this works:

• Adds one carbon via HCN
• Cyanohydrin intermediate
• Hydrolysis gives hydroxy acid (lactic acid analog)
• 3-carbon product: CH₃CH(OH)COOH (2-hydroxypropanoic acid)

Answer: c. dehydrogenation

Step analysis:

Why dehydrogenation?

• Dehydrogenation: Removal of hydrogen
• Not dehydration: No water removed (that would give ethene)
• Not condensation: No small molecule elimination + bond formation
• Not hydrogenation: That would add hydrogen

Common method: Copper catalyst at 300°C: CH₃CH₂OH → CH₃CHO + H₂

Answer: a. propanal

Reasoning:

Correction: Actually, propanal oxidizes to propanoic acid. But aldehydes can undergo disproportionation (Cannizzaro reaction) with concentrated alkali to give alcohol + acid. However, the “sweet-smelling liquid” suggests ester formation.

Better explanation: Some oxidation of propanal gives propanoic acid. Unreacted propanal + propanoic acid (with acid catalyst) could theoretically give ester (sweet smell). But this is unlikely under normal conditions.

Most likely intended answer: Propanal oxidizes to propanoic acid. Excess propanal + acid catalyst could form ester (though not typical).

Answer: C. Step 1: KCN(aq/alcoholic), Step 2: H₂SO₄(aq)

Synthetic route:

Why C is correct:

• KCN substitutes Br with CN (nucleophilic substitution)
• Acid hydrolysis converts -CN to -COOH
• Product: Butanedioic acid (succinic acid)

Why others wrong:

• A: HCN gas doesn’t substitute well; HCl won’t hydrolyze nitrile to acid
• B: HCO₂Na gives esters, not nitriles
• D: NaOH gives diol, oxidation gives dialdehyde/diacid but messy

Answer: d. Na₂CO₃(aq)

Analysis:

Test differentiation:

Conclusion: Na₂CO₃ gives immediate effervescence with carboxylic acid (Y) but not with alcohol/phenol (Z).

Answer: B. Step 1: HCN, NaCN(aq/alcoholic); Step 2: H₂SO₄(aq)

Synthetic pathway:

Mechanism:

1. Each aldehyde adds HCN → cyanohydrin
2. Double cyanohydrin formation
3. Acid hydrolysis converts both -CN to -COOH
4. Product: Tartaric acid (2,3-dihydroxybutanedioic acid)

Answer: Based on functional groups present.

Systematic approach:

1. Step 1: Na₂CO₃ test → identifies carboxylic acid (A)
2. Step 2: Tollen’s/Fehling’s → identifies aldehyde (B)
3. Step 3: 2,4-DNPH → positive for both B and C
4. Step 4: K₂Cr₂O₇ → identifies oxidizable alcohol (D)
5. Step 5: Iodoform test if needed for specific alcohols

Answer:

Answer:

Key points:

• Stage I: Free radical substitution (like alkane chlorination)
• Stage II: CN⁻ is a good nucleophile for SN2
• Stage III: Bromination at α-position to cyanide (acidic H)
• Final product: α-bromo nitrile (lachrymator – tear gas)

Answer:

Alternative for step vi: Hot conc. KMnO₄ could also cleave diol to dialdehyde/ketone.

Answer:

Complete route:

1. CH₃CH₂OH → CH₃CHO (oxidation, distill)
2. CH₃CHO + HCN → CH₃CH(OH)CN (cyanohydrin formation)
3. CH₃CH(OH)CN + 2H₂O → CH₃CH₂COOH + NH₃ (acid hydrolysis)

Answer:

Note: This gives 2-hydroxy-2-methylpropanoic acid, not lactic acid. For actual lactic acid (2-hydroxypropanoic acid), a different route is needed.

Better route to lactic acid:

1. Propene + HOBr → CH₃CH(OH)CH₂Br (anti-Markovnikov)
2. Substitution with KCN → CH₃CH(OH)CH₂CN
3. Hydrolysis → CH₃CH(OH)CH₂COOH (3-hydroxybutanoic acid?)

Answer:

Simpler route: Glycolic acid (HOCH₂COOH) already has 2 carbons. To get ethanoic acid (CH₃COOH), need to replace -OH with -H (reduction).

Actual method:

1. Convert -OH to good leaving group (e.g., with PBr₃ → BrCH₂COOH)
2. Reduction: BrCH₂COOH + 2[H] → CH₃COOH + HBr
3. Reducing agent: Could use LiAlH₄ (but reduces -COOH too) or other methods

Answer:

Better route:

1. CH₃OH → CH₃Cl (with HCl/ZnCl₂ or PCl₅)
2. CH₃Cl + KCN → CH₃CN (nucleophilic substitution)
3. CH₃CN + 2H₂O → CH₃COOH + NH₃ (acid hydrolysis)

Most efficient:

This adds one carbon via nitrile, then hydrolyzes to acid.

Answer: Based on standard reactions.

Answer: W (But-1-ene)

Reason:

• W: But-1-ene (alkene) → can undergo addition polymerization
• X: 1,2-dibromobutane (dihalide) → not typically polymerized
• Y: Butan-1-ol (alcohol) → could form polyesters but not alone
• Z: Butanoic acid (acid) → could form polyamides/polyesters with others

Polymerization of W:

Polybutene (similar to polypropylene)

Answer:

Answer:

Note: 2-oxobutanoic acid = CH₃CH₂COCOOH (keto acid).

Alternative route via ozonolysis:

1. But-1-ene → Ozonolysis → Propanal + Formaldehyde
2. Propanal oxidation → Propanoic acid
3. α-bromination → 2-bromopropanoic acid
4. Substitution with CN → then hydrolysis to keto acid? Complex.

Answer:

Key similarities:

• Both aim to produce target molecule
• Both use knowledge of organic reactions
• Both consider functional group interconversions

Key differences:

• Synthesis is practical (lab-focused); retrosynthesis is planning (design-focused)
• Retrosynthesis often reveals multiple possible routes
• Synthesis deals with actual experimental conditions