The compound ‘leaf alcohol’ is partly responsible for the smell of new-mown grass.
leaf alcohol: \( \text{CH}_3\text{CH}_2\text{CH=CHCH}_2\text{CH}_2\text{OH} \)
What will be formed when ‘leaf alcohol’ is oxidised using an excess of hot acidified \( \text{K}_2\text{Cr}_2\text{O}_7(\text{aq}) \)?
▶️ Answer/Explanation
Ans: C
1. Oxidation of Primary Alcohol: The \(-\text{CH}_2\text{OH}\) group in leaf alcohol is a primary alcohol, which oxidises to a carboxylic acid (\(-\text{CO}_2\text{H}\)) under strong conditions (hot acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \)).
2. Effect on the Double Bond: The \( \text{C=C} \) bond remains unaffected because \( \text{K}_2\text{Cr}_2\text{O}_7 \) does not oxidise alkenes under these conditions.
3. Final Product: The product is \( \text{CH}_3\text{CH}_2\text{CH=CHCH}_2\text{CO}_2\text{H} \) (hex-3-enoic acid).
Thus, the correct answer is C.
Compound X:
● does not react with Tollens’ reagent
● forms a yellow precipitate with alkaline \(I_2\)(aq)
● does not react with sodium.
What could be the identity of X?
▶️ Answer/Explanation
Ans: B
Analysis of observations:
- No reaction with Tollens’ reagent: X is not an aldehyde (rules out A, \(CH_3CHO\)).
- Yellow precipitate with alkaline \(I_2\)(aq): Indicates a methyl ketone (iodoform test, positive for \(C_2H_5COCH_3\)).
- No reaction with sodium: X is not an alcohol or carboxylic acid (rules out D, \(CH_3CHOHCH_3\)).
Conclusion: The only compound fitting all criteria is B (\(C_2H_5COCH_3\)), a methyl ketone that gives a positive iodoform test but does not react with Tollens’ reagent or sodium.
A sample of propanoic acid of mass 3.70 g reacts with an excess of magnesium. A second sample of propanoic acid of mass 3.70 g reacts with an excess of sodium. Both reactions go to completion forming a gas. Which row is correct?
▶️ Answer/Explanation
Ans: A
Step 1: Write the balanced equations
1. With Magnesium (Mg):
\[ 2CH_3CH_2COOH + Mg \rightarrow (CH_3CH_2COO)_2Mg + H_2 \]
2. With Sodium (Na):
\[ 2CH_3CH_2COOH + 2Na \rightarrow 2CH_3CH_2COONa + H_2 \]
Step 2: Calculate moles of propanoic acid
Molar mass of \( CH_3CH_2COOH = 74 \, \text{g/mol} \).
Moles of propanoic acid = \( \frac{3.70 \, \text{g}}{74 \, \text{g/mol}} = 0.05 \, \text{mol} \).
Step 3: Determine moles of gas produced
1. Reaction with Mg: 2 moles of propanoic acid produce 1 mole of \( H_2 \).
Thus, 0.05 moles produce \( \frac{0.05}{2} = 0.025 \, \text{mol} \) of \( H_2 \).
2. Reaction with Na: 2 moles of propanoic acid produce 1 mole of \( H_2 \).
Thus, 0.05 moles produce \( \frac{0.05}{2} = 0.025 \, \text{mol} \) of \( H_2 \).
Step 4: Compare with the table
Both reactions produce the same amount of gas (0.025 mol), which matches Row A.
Compound Y reacts with alkaline \(I_2(aq)\). When the products of this reaction are acidified, a dicarboxylic acid is produced. The formula of the dicarboxylic acid is HOOC–R–COOH where R consists of one or more \(CH_2\) groups. Which compound is Y?
▶️ Answer/Explanation
Ans: D
The reaction described is the iodoform test for methyl ketones or secondary alcohols with a methyl group adjacent to the hydroxyl-bearing carbon. When compound Y reacts with alkaline \(I_2(aq)\), it undergoes oxidation and cleavage to form a dicarboxylic acid upon acidification.
Key Points:
- Pentan-2,4-diol (Option D) has hydroxyl groups at positions 2 and 4. Oxidation of both hydroxyl groups (to ketones) followed by cleavage with \(I_2/NaOH\) yields a dicarboxylic acid with the structure HOOC–CH2–COOH (malonic acid).
- The other options do not produce a dicarboxylic acid under these conditions:
- A (1,4-diol) would give a monocarboxylic acid.
- B (1,5-diol) would give adipic acid (HOOC–(CH2)4–COOH), but this requires a different oxidative pathway.
- C (2,3-diol) would not yield a dicarboxylic acid with the given conditions.
Thus, pentan-2,4-diol (Option D) is the correct answer.