Question

An organic compound \( \text{C}_5\text{H}_{10}\text{O} \) does not reduces Tollen's reagent but forms addition compound with sodium hydrogen sulphite and gives the Iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid.

Hint:

Use chemical tests as "clues" to puzzle out organic structures.
\( \text{NaHSO}_3 \) test = Carbonyl group.
Negative Tollen's test = Ketone.
Positive Iodoform test = Terminal methyl group attached to the carbonyl.
Popoff's Rule helps in predicting the cleavage products of unsymmetrical ketones.

Answer 1

Step 1: Understanding the Concept:
By identifying the functional groups corresponding to specific chemical tests, the exact structure of an unknown compound can be deduced.
Step 2: Key Formula or Approach:
The approach is to interpret each qualitative test: \( \text{NaHSO}_3 \) test for carbonyls, Tollen's test for distinguishing aldehydes from ketones, and the Iodoform test for methyl ketones, then use Popoff's rule for oxidation cleavage.
Step 3: Detailed Explanation:
The given molecular formula is \( \text{C}_5\text{H}_{10}\text{O} \), which has a degree of unsaturation of 1, pointing to either a double bond or a ring.
Since the compound forms an addition compound with sodium hydrogen sulphite (\( \text{NaHSO}_3 \)), it must contain a carbonyl group (making it an aldehyde or a ketone).
The compound does not reduce Tollen's reagent, which confirms it is a ketone and not an aldehyde.
Furthermore, it gives a positive Iodoform test, indicating the presence of a methyl ketone group (\( \text{CH}_3-\text{CO}- \)).
A 5-carbon ketone containing a methyl ketone group must be Pentan-2-one (\( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \)).
To verify, let's look at the oxidation products.
Upon vigorous oxidation of ketones, the carbon-carbon bonds break according to Popoff's rule, where the carbonyl group tends to stay with the smaller alkyl fragment.
Oxidative cleavage of Pentan-2-one between C-2 and C-3 gives a 2-carbon fragment and a 3-carbon fragment.
These fragments oxidize to ethanoic acid (\( \text{CH}_3\text{COOH} \)) and propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)).
This perfectly matches the given products.
Step 4: Final Answer:
The organic compound is Pentan-2-one.