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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The noble gases can be separated by
Which element gives maximum balanced oxide
Correct major product of reaction is
Correct statement for $3d-$ series is
→For a liquid normal boiling point is $-173\,^oC$ then at $2\ atm$ pressure it's boiling point should be nearly ......... $^oC.$ $(\Delta H_{vap} = 200\ cal/mole,\ R = 2 \ cal/mol-K,\,\,\ln \,2 = 0.7)$
→Consider the following statements: According the Werner's theory.
$(a)$ Ligands are connected to the metal ions by covalent bonds.
$(b)$ Secondary valencies have directional properties
$(c)$ Secondary valencies are non-ionisable Of these statements:
→$(a)$ Ligands are connected to the metal ions by covalent bonds.
$(b)$ Secondary valencies have directional properties
$(c)$ Secondary valencies are non-ionisable Of these statements:
When propionic acid is treated with aqueous sodium bicarbonate $C{O_2}$ is liberated. The ‘$C$’ of $C{O_2}$ comes from
→Which give positive iodoform test
Consider the Galvanic cell $\,Z{n^\Theta }|ZnS{O_4}||CuS{O_4}|C{u^ \oplus }$ the reaction at cathode is
→The $\mathrm{Eu}^{2+}$ ion is a strong reducing agent in spite of its ground state electronic configuration
→(outermost) : [Atomic number of $\mathrm{Eu}=63$ ]