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

Explanation

To solve the problem, we need to determine the structure and IUPAC name of the alkene ‘A’ that upon ozonolysis yields a mixture of ethanal and pentan-3-one.

### Step-by-Step Solution:

**Step 1: Understand Ozonolysis**
Ozonolysis is a reaction where ozone (O₃) reacts with alkenes or alkynes, leading to the formation of carbonyl compounds (aldehydes or ketones). In this case, we are given that the products are ethanal (an aldehyde) and pentan-3-one (a ketone).

**Step 2: Analyze the Products**
The products of ozonolysis are:
- Ethanal (CH₃CHO)
- Pentan-3-one (C₅H₁₀O)

This indicates that the alkene must have a structure that, when cleaved, can yield these two products.

**Step 3: Determine the Structure of Alkene ‘A’**
To produce ethanal and pentan-3-one, we can deduce that the alkene must have a structure that allows for the formation of a 5-carbon chain (for pentan-3-one) and a 2-carbon chain (for ethanal) upon cleavage.

Let’s assume the structure of alkene ‘A’ is:
- A 5-carbon chain with a double bond, specifically at the 2-position.

Thus, the structure can be represented as:
- 3-Ethylpent-2-ene

**Step 4: Draw the Structure**
The structure of 3-ethylpent-2-ene can be drawn as follows:

```
CH3
|
CH3-CH-CH=CH-CH2-CH3
```

This structure shows a double bond between the second and third carbon atoms, with an ethyl group on the third carbon.

**Step 5: Write the IUPAC Name**
The IUPAC name of the compound is derived from its structure:
- The longest carbon chain has 5 carbons (pentane).
- The double bond is located at the second carbon (2-ene).
- There is an ethyl group on the third carbon (3-ethyl).

Thus, the IUPAC name of alkene ‘A’ is **3-ethylpent-2-ene**.

### Final Answer:
- Structure of ‘A’: 3-Ethylpent-2-ene
- IUPAC Name: 3-ethylpent-2-ene