An alkene ‘A’ contains three C – C, eight C – H `(sigma)` bonds and one C – C `(pi)` bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Explanation

To solve the problem step by step, we need to analyze the information given about the alkene 'A' and its ozonolysis products.

### Step 1: Analyze the bonds in alkene 'A'
- The alkene 'A' has three C-C (sigma) bonds and one C-C (pi) bond.
- Since it is an alkene, it must have at least one double bond, which is represented by the C-C (pi) bond.

### Step 2: Determine the number of carbon atoms
- The presence of three C-C (sigma) bonds suggests that there are four carbon atoms in the alkene. This is because each C-C bond connects two carbon atoms, and with three such bonds, we can visualize a chain of four carbon atoms.

### Step 3: Count the hydrogen atoms
- The alkene has eight C-H (sigma) bonds. Each carbon atom in an alkene is typically bonded to two hydrogen atoms (for the two carbons involved in the double bond) and one hydrogen atom for each of the other carbons.
- For four carbon atoms in an alkene, the general formula is CnH2n, where n is the number of carbon atoms.
- For n = 4, the formula gives us C4H8, which matches the eight C-H bonds.

### Step 4: Ozonolysis products
- The ozonolysis of alkene 'A' produces two moles of an aldehyde with a molar mass of 44 u.
- The molecular formula of the aldehyde can be deduced: Aldehydes with a molar mass of 44 u typically correspond to the formula C2H4O (as C2H4O has a molar mass of 44 g/mol).

### Step 5: Identify the structure of alkene 'A'
- The ozonolysis of butyrene (C4H8) can yield two moles of acetaldehyde (C2H4O) as products.
- Butyrene can be represented as but-2-ene (CH3-CH=CH-CH3), which fits the criteria of having three C-C bonds and one C-C (pi) bond.

### Step 6: Write the IUPAC name
- The IUPAC name of the alkene 'A' is but-2-ene.

### Final Answer:
The IUPAC name of the alkene 'A' is **but-2-ene**.

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