MCQ
Which energy diagram best represents the given reaction ?
- A
- B
- C
- ✓
✓
Answer
Correct option: D.
d
$(d)$ It is possible to interconvert cis and trans alkenes, but the $\pi$ -bond must be broken first. This requires a considerable amount of energy - around $260\, kJ\, mol^{-1}$ .One way to break the $\pi$ -bond would be to promote an electron from the $\pi$ -orbital to the $\pi^*$ -orbital. If this were to happen, there would be one electron in the bonding $\pi$ -orbital and one in the antibonding $\pi^*$ -orbital and hence no overall bonding. Electromagnetic radiation of the correct energy could promote the electron from $HOMO$ to $LUMO$. The correct energy actually corresponds to light in the ultraviolet $(UV)$ region of the spectrum. Thus, shining $UV$ light on an alkene would promote an electron from its bonding $\pi$ -molecular orbital to its antibonding $\pi^*$ -molecular orbital, thereby breaking the $\pi$ -bond (but not the $\sigma$ - bond) and allowing rotation to occur.
$(d)$ It is possible to interconvert cis and trans alkenes, but the $\pi$ -bond must be broken first. This requires a considerable amount of energy - around $260\, kJ\, mol^{-1}$ .One way to break the $\pi$ -bond would be to promote an electron from the $\pi$ -orbital to the $\pi^*$ -orbital. If this were to happen, there would be one electron in the bonding $\pi$ -orbital and one in the antibonding $\pi^*$ -orbital and hence no overall bonding. Electromagnetic radiation of the correct energy could promote the electron from $HOMO$ to $LUMO$. The correct energy actually corresponds to light in the ultraviolet $(UV)$ region of the spectrum. Thus, shining $UV$ light on an alkene would promote an electron from its bonding $\pi$ -molecular orbital to its antibonding $\pi^*$ -molecular orbital, thereby breaking the $\pi$ -bond (but not the $\sigma$ - bond) and allowing rotation to occur.
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