Which compound, on reaction with hydrogen cyanide, produces a compound with a chiral centre?
A. \(CH_3CHO\)
B. CH₃CH₂COCH₂CH₃
C. CH₃CO₂CH₃
D. HCHO
▶️ Answer/Explanation
Ans: A
1. Reaction with HCN: The addition of hydrogen cyanide (HCN) to a carbonyl compound (aldehyde or ketone) forms a cyanohydrin.
2. Chiral Centre Formation: A chiral centre is created when the carbonyl carbon (C=O) becomes a carbon atom bonded to four different groups. – For A (CH₃CHO): \[ CH_3CHO + HCN \to CH_3CH(OH)CN \] The product has a chiral centre (carbon bonded to -OH, -CN, -CH₃, and -H). – Other options (B, C, D) either do not react with HCN or produce symmetrical products without chiral centres.
3. Conclusion: Only A (CH₃CHO) forms a product with a chiral centre, making it the correct answer.
Which statement about \(H_2C=C(CH_3)CH_2CO_2CH_3\) is correct?
▶️ Answer/Explanation
Ans: D
The compound \(H_2C=C(CH_3)CH_2CO_2CH_3\) is methyl 3-methylbut-2-enoate, containing both an alkene (\(C=C\)) and an ester (\(CO_2CH_3\)) functional group.
- Option A: Hydrolysis of the ester would yield a carboxylic acid (\(CH_3C(CH_3)=CHCO_2H\)) and methanol, not a secondary alcohol. Incorrect.
- Option B: This ester cannot be synthesized from ethanoic acid (which would give a 2-carbon ester). Incorrect.
Option C
- : The compound lacks a methyl ketone group (\(CH_3CO-\)), which is required for a positive iodoform test.
Incorrect.
- Option D: Oxidative cleavage of the alkene with hot acidified KMnO₄ produces CH₃COCH₂COOH (3-oxobutanoic acid) and CO₂. Correct.
Thus, D is the correct statement.
Which statement about PVC is correct?
A. Combustion products of PVC are very alkaline and harmful to breathe in.
B. The empirical formula of PVC is the same as the empirical formula of the monomer.
C. Molecules of PVC are unsaturated.
D. The repeat unit of PVC is (CH₂CCl₂).
▶️ Answer/Explanation
Ans: B
In addition polymerization, the empirical formula of the polymer (PVC) remains the same as its monomer (vinyl chloride, C₂H₃Cl). Option B is correct because no atoms are lost during polymerization. Option A is incorrect (combustion produces acidic HCl gas, not alkaline products). Option C is wrong (PVC chains are saturated). Option D is incorrect (the repeat unit is −[CH₂−CHCl]−, not CH₂CCl₂).
Compound Y:
● changes the colour of acidified K₂Cr₂O₇ from orange to green
● has no effect on Fehling’s reagent
● produces an orange precipitate with 2,4-dinitrophenylhydrazine reagent.
What is compound Y?
▶️ Answer/Explanation
Ans: C
Analysis of tests:
- K₂Cr₂O₇ (orange → green): Indicates oxidation of an alcohol or aldehyde. However, since Fehling’s reagent is unaffected, it rules out aldehydes (which reduce Fehling’s). Thus, Y must be a secondary alcohol (oxidizable to a ketone).
- No reaction with Fehling’s: Confirms Y is not an aldehyde (e.g., CH₃CHO or HCOOH).
- 2,4-DNPH (orange precipitate): Confirms the presence of a carbonyl group (C=O), consistent with a ketone (or aldehyde, but already ruled out).
Conclusion: Y is CH₃COCH₃ (a ketone), as it fits all observations:
- Secondary alcohols (like isopropanol) oxidize to ketones (e.g., acetone), explaining the K₂Cr₂O₇ result.
- Ketones do not react with Fehling’s reagent.
- Ketones form orange precipitates with 2,4-DNPH.
Thus, the correct answer is C (CH₃COCH₃).