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:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.
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