The diagram shows a synthetic route to produce 1-methylcyclohexanol.
What is reagent Y?
▶️ Answer/Explanation
Ans: A
The synthetic route involves the conversion of 1-methylcyclohexene to 1-methylcyclohexanol. This is achieved through hydroboration-oxidation, where the reagent Y must be aqueous NaOH (in the presence of \( H_2O_2 \)) to facilitate the anti-Markovnikov addition of water across the double bond. Cold dilute \( KMnO_4 \) (B) would form a diol, while hot concentrated \( KMnO_4 \) (D) would cleave the double bond. Ethanolic NaOH (C) is irrelevant here. Thus, the correct answer is A.
C₂H₅COOCH₃ is reacted with aqueous acid. The products from this reaction are reacted with LiAlH₄ to form two molecules Y and Z. What are the identities of molecules Y and Z?
▶️ Answer/Explanation
Ans: D
1. Step 1: Acid Hydrolysis of Ester (C₂H₅COOCH₃)
The ester undergoes hydrolysis with aqueous acid to form a carboxylic acid and an alcohol: \[ \text{C}_2\text{H}_5\text{COOCH}_3 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_2\text{H}_5\text{COOH} + \text{CH}_3\text{OH} \]
2. Step 2: Reduction with LiAlH₄
The products (propanoic acid and methanol) are then reduced by LiAlH₄: \[ \text{C}_2\text{H}_5\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{C}_2\text{H}_5\text{CH}_2\text{OH} \] \[ \text{CH}_3\text{OH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{OH} \ (\text{no change}) \]
3. Final Products
– Methanol (CH₃OH) remains unchanged.
– Propanoic acid is reduced to propan-1-ol (C₂H₅CH₂OH).
4. Conclusion
The two molecules formed are CH₃OH and C₂H₅CH₂OH, corresponding to option D.
Which alcohol gives only one possible oxidation product when warmed with dilute acidified potassium dichromate(VI)?
▶️ Answer/Explanation
Ans: B
Key Concept: Secondary alcohols (like butan-2-ol) oxidize to ketones, which cannot be further oxidized under these conditions. Primary alcohols (like butan-1-ol and 2-methylpropan-1-ol) oxidize first to aldehydes and then to carboxylic acids (two products). Tertiary alcohols (like 2-methylpropan-2-ol) cannot be oxidized.
Analysis of Options:
- A (butan-1-ol): Primary alcohol → forms butanal (aldehyde) and then butanoic acid (carboxylic acid). Two products.
- B (butan-2-ol): Secondary alcohol → forms only butanone (ketone). One product.
- C (2-methylpropan-1-ol): Primary alcohol → forms 2-methylpropanal and then 2-methylpropanoic acid. Two products.
- D (2-methylpropan-2-ol): Tertiary alcohol → no oxidation occurs. No product.
Thus, only butan-2-ol (B) yields a single oxidation product (butanone).
The diagram shows three reactions of ethanal. In each case, an excess of ethanal is used.
Observations are made after each of the three reactions. What are the colours of solution 1 and solids 2 and 3?
▶️ Answer/Explanation
Ans: B
1. Solution 1 (Fehling’s solution + ethanal): Ethanal reduces Fehling’s solution, producing a red precipitate (Cu₂O) and leaving a colorless solution.
2. Solid 2 (Tollen’s reagent + ethanal): Ethanal reduces Tollen’s reagent, forming a silver mirror (Ag).
3. Solid 3 (2,4-DNPH + ethanal): Ethanal reacts with 2,4-DNPH to form a yellow-orange precipitate (2,4-dinitrophenylhydrazone).
Thus, the correct combination is B (colorless, silver, yellow).