Q.An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.

The key idea here is Chain Isomerism in alkenes, combined with ozonolysis to locate the double bond.

Reasoning:

1. Ozonolysis cleaves the C=C bond, converting each doubly bonded carbon into a carbonyl group (aldehyde or ketone). The products are ethanal ($\text{CH}_3\text{CHO}$) and pentan-3-one ($\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$).
2. Reversing the reaction: the two carbonyl carbons were originally the doubly bonded carbons. So, the alkene must have the structure: $\text{CH}_3\text{CH}=\text{C}(\text{CH}_2\text{CH}_3)_2$. …

The key idea here is retrosynthetic analysis of ozonolysis. Ozonolysis of an alkene breaks the carbon-carbon double bond and replaces it with two carbonyl groups (aldehydes or ketones). If you know the products, you can reconstruct the original alkene by joining the carbonyl carbons back together with a double bond.

Ethanal is $\text{CH}_3\text{CHO}$ — that's a two-carbon aldehyde. Pentan-3-one is $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ — a five-carbon ketone with the carbonyl at position 3. So the alkene must have had a total of $2 + 5 = 7$ carbons, and the double bond was between the carbon that became the aldehyde carbon and the carbon that became the ketone carbon.

Let's work through it step by step.

1. Identify the fragments from the products.

   Ethanal: $\text{CH}_3-\text{CHO}$. The carbonyl carbon (the $\text{CHO}$ carbon) was one end of the original double bond.

   Pentan-3-one: $\text{CH}_3\text{CH}_2-\text{CO}-\text{CH}_2\text{CH}_3$. The carbonyl carbon (the $\text{CO}$ carbon) was the other end of the double bond.

2. Reconnect the two carbonyl carbons with a double bond.

   Remove the oxygen from each carbonyl and join the two carbons. That gives:

The left fragment (from ethanal) gives $\text{CH}_3\text{CH}=$ — a two-carbon chain with the double bond at the end.

The right fragment (from pentan-3-one) gives $=\text{C}(\text{CH}_2\text{CH}_3)_2$ — a central carbon attached to two ethyl groups.

3. Determine the longest chain and number it. The alkene has 7 carbons total: 2 from the ethanal-derived end and 5 from the pentan-3-one-derived end (the central carbon plus two ethyl groups). The full structure is:

To name it, find the longest chain that includes the double bond: starting from the left $\text{CH}_3$, the chain runs through the double bond and continues into one of the two ethyl groups on the far carbon (the other ethyl group becomes a substituent). That gives a 5-carbon chain:

$\underset{C1}{\text{CH}_3}-\underset{C2}{\text{CH}}=\underset{C3}{\text{C}}(\text{CH}_2\text{CH}_3)-\underset{C4}{\text{CH}_2}-\underset{C5}{\text{CH}_3}$ …