Question:
One mol of 2-Methylbuta-1,3-diene on ozonolysis gives
One mol of 2-Methylbuta-1,3-diene on ozonolysis gives
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In ozonolysis, each \(C=C\) bond is cleaved and both double-bonded carbon atoms are converted into carbonyl groups. Terminal \(=CH_2\) groups give methanal.
Updated On: Jun 18, 2026
- 2 moles of methanal and 1 mole of propanone
- 2 moles of methanal and 1 mole of 2-ketopropanal
- 1 mole of methanal, 1 mole of ethanol, and 1 mole of propanone
- 2 moles of ethanal and 1 mole of 2-ketopropanal
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The Correct Option is B
Solution and Explanation
Step 1: Write the structure of 2-Methylbuta-1,3-diene.
2-Methylbuta-1,3-diene is also called isoprene.
Its structure is \[ CH_2=C(CH_3)-CH=CH_2 \]
Step 2: Understand ozonolysis of double bonds.
Ozonolysis breaks each carbon-carbon double bond and converts the double-bonded carbon atoms into carbonyl compounds.
The molecule contains two double bonds: \[ CH_2=C(CH_3) \] and \[ CH=CH_2 \]
Step 3: Cleavage of the first double bond.
On cleavage of the double bond \[ CH_2=C(CH_3)- \] the terminal carbon gives \[ HCHO \] which is methanal.
The substituted carbon remains connected to the middle carbon and becomes part of a keto-aldehyde fragment.
Step 4: Cleavage of the second double bond.
On cleavage of the double bond \[ -CH=CH_2 \] the terminal carbon again gives \[ HCHO \] which is methanal.
The remaining middle fragment forms \[ CH_3COCHO \] which is 2-ketopropanal.
Step 5: Write the overall products.
Thus, \[ CH_2=C(CH_3)-CH=CH_2 \] on ozonolysis gives \[ 2HCHO + CH_3COCHO \] That is, \[ 2\ \text{moles of methanal} + 1\ \text{mole of 2-ketopropanal} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\text{2 moles of methanal and 1 mole of 2-ketopropanal}} \] Hence, the correct option is (2).
2-Methylbuta-1,3-diene is also called isoprene.
Its structure is \[ CH_2=C(CH_3)-CH=CH_2 \]
Step 2: Understand ozonolysis of double bonds.
Ozonolysis breaks each carbon-carbon double bond and converts the double-bonded carbon atoms into carbonyl compounds.
The molecule contains two double bonds: \[ CH_2=C(CH_3) \] and \[ CH=CH_2 \]
Step 3: Cleavage of the first double bond.
On cleavage of the double bond \[ CH_2=C(CH_3)- \] the terminal carbon gives \[ HCHO \] which is methanal.
The substituted carbon remains connected to the middle carbon and becomes part of a keto-aldehyde fragment.
Step 4: Cleavage of the second double bond.
On cleavage of the double bond \[ -CH=CH_2 \] the terminal carbon again gives \[ HCHO \] which is methanal.
The remaining middle fragment forms \[ CH_3COCHO \] which is 2-ketopropanal.
Step 5: Write the overall products.
Thus, \[ CH_2=C(CH_3)-CH=CH_2 \] on ozonolysis gives \[ 2HCHO + CH_3COCHO \] That is, \[ 2\ \text{moles of methanal} + 1\ \text{mole of 2-ketopropanal} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\text{2 moles of methanal and 1 mole of 2-ketopropanal}} \] Hence, the correct option is (2).
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