1 mole of a symmetrical alkene on ozonolysis gives 2 moles of an aldehyde with molecular mass 44w. The alkene is

But-2-ene

The correct answer is Option (4) → But-2-ene.

Let us go through the ozonolysis of a symmetrical alkene step by step, understanding how it produces two moles of aldehyde with a molecular mass of 44 g/mol, and why the correct alkene is but-2-ene.

Ozonolysis is a reaction where an alkene reacts with ozone \((O_3)\) to cleave the double bond between carbon atoms.

After ozonolysis, a reductive workup (typically with zinc and water or dimethyl sulfide) converts the cleaved bond into carbonyl groups (either aldehydes or ketones).

Aldehyde with Molar Mass 44 g/mol

The reaction produces two moles of aldehyde from one mole of the alkene.

The molecular mass of the aldehyde is 44 g/mol.

We can deduce that this aldehyde is acetaldehyde (ethanal, \(CH_3CHO\)). The molar mass of acetaldehyde is:

\(\text{Molecular mass of }CH_3CHO = 12 (C) + 3 (H) + 12 (C) + 1 (H) + 16 (O) = 44 \, \text{g/mol}\)

Identifying the Symmetrical Alkene

The alkene must be symmetrical, meaning that it is structured in such a way that when the double bond breaks, it produces two identical aldehyde molecules.

Based on the reaction information, the alkene forms two moles of acetaldehyde \((CH_3CHO)\), a 2-carbon aldehyde.

Step-by-Step Analysis of Each Option

(1) Ethene \((C_2H_4)\):

Ethene is the simplest alkene: \(CH_2=CH_2\)

Ozonolysis of ethene would give:

\(CH_2=CH_2 + O_3 \longrightarrow 2 \, \text{formaldehyde} \, (\text{HCHO})\)

Formaldehyde has a molecular mass of 30 g/mol, not 44 g/mol.

Therefore, ethene is not the correct answer.

(2) Propene \((C_3H_6)\):

Propene is a 3-carbon alkene: \(CH_3-CH=CH_2\)

Ozonolysis of propene would give acetaldehyde \((CH_3CHO)\) and formaldehyde \((HCHO)\)

\(CH_3-CH=CH_2 + O_3 \longrightarrow CH_3CHO + HCHO\)

While acetaldehyde has the desired molecular mass of 44 g/mol, ozonolysis of propene gives one mole of acetaldehyde and one mole of formaldehyde. This does not match the information that two moles of aldehyde with a mass of 44 g/mol are produced.

Thus, propene is not the correct answer.

(3) But-1-ene:

But-1-ene is a 4-carbon alkene with the double bond at the end:

\(CH_2=CH-CH_2-CH_3\)

Ozonolysis of but-1-ene would produce acetaldehyde \((CH_3CHO)\) and propanal \((CH_3CH_2CHO)\):

\(CH_2=CH-CH_2-CH_3 + O_3 \longrightarrow CH_3CHO + CH_3CH_2CHO\)

Propanal has a molecular mass of 58 g/mol, while acetaldehyde has a mass of 44 g/mol. Since two different aldehydes are produced, this does not match the requirement of two moles of aldehyde with a mass of 44 g/mol.

Thus, but-1-ene is not the correct answer.

(4) But-2-ene:

But-2-ene is a symmetrical alkene with the double bond in the middle: \(CH_3-CH=CH-CH_3\)

Ozonolysis of but-2-ene would cleave the double bond symmetrically and produce two moles of acetaldehyde \((CH_3CHO)\):

\(CH_3-CH=CH-CH_3 + O_3 \longrightarrow 2 \, CH_3CHO\)

Both molecules produced are acetaldehyde with a molecular mass of 44 g/mol.

This matches the condition of producing two moles of an aldehyde with a molecular mass of 44 g/mol.

Conclusion

The correct alkene is But-2-ene because ozonolysis of but-2-ene gives 2 moles of acetaldehyde \(CH_3CHO)\), each with a molecular mass of 44 g/mol.

Therefore, the answer is But-2-ene.