An alkene X on ozonolysis gives a mixture of Propan-2-one and methanal. What is X?
Propene
2-Methylpropene
2-Methylbut-1-ene
2-Methylbut-2-ene
To solve the problem, we need to determine which alkene (X) gives a mixture of Propan-2-one (acetone) and Methanal (formaldehyde) upon ozonolysis.
1. Understanding Ozonolysis:
Ozonolysis cleaves the double bond in alkenes and forms carbonyl compounds (aldehydes or ketones) at the positions where the double bond was present.
2. Analyze the products:
- Propan-2-one: $CH_3COCH_3$ (a ketone with 3 carbon atoms)
- Methanal: $HCHO$ (an aldehyde with 1 carbon atom)
3. Work backwards to deduce the structure of X:
To get these two products, the double bond must be between a carbon with two methyl groups (to give acetone) and a terminal CH₂ group (to give methanal).
4. Structure of X:
The correct structure is that of 2-Methylpropene ($CH_2=C(CH_3)_2$)
Upon ozonolysis:
$CH_2=C(CH_3)_2 \rightarrow CH_3COCH_3$ (Propan-2-one) + $HCHO$ (Methanal)
Final Answer:
The correct alkene X is 2-Methylpropene.
The reactions which cannot be applied to prepare an alkene by elimination, are
Choose the correct answer from the options given below:
An alkene X (\( C_4H_8 \)) on reaction with HBr gave Y (\( C_4H_9Br \)). Reaction of Y with benzene in the presence of anhydrous \( AlCl_3 \) gave Z which is resistant to oxidation with \( KMnO_4 + KOH \). What are X, Y, Z respectively?
In organic chemistry, an alkene is a hydrocarbon containing a carbon-carbon double bond.[1]
Alkene is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds.
Read More: Ozonolysis
Read More: Unsaturated Hydrocarbon