Chemistry - Halogen Derivative Question with Solution | TestHub

Concept:-
Reductions with diimide:
- Isolated carbon-carbon double bonds can be reduced with hydrazine $\left({\mathrm{NH}}_2{\mathrm{NH}}_2\right)$ in the presence of oxygen or an oxidizing agent.
- The actual reducing agent in these reactions is the highly active species diimide, $\mathrm{HN}=\mathrm{NH}$, formed in situ by oxidation of hydrazine.
- This compound is a highly selective reducing agent which in many cases offers a useful alternative to catalytic hydrogenation for the reduction of carbon-carbon multiple bonds.
- The reagent diimide is prepared by oxidation of hydrazine, by decomposition of tosylhydrazide, or from azodicarboxylic acid.
- It is a suitable reducing agent for symmetrical double bonds such as $\mathrm{C}=\mathrm{C}$, and $\mathrm{N}=\mathrm{N}$, but unsymmetrical, more polar bonds such as $\mathrm{C}=\mathrm{N},\mathrm{C}=\mathrm{N},\mathrm{N}=\mathrm{O}$, and $\mathrm{S}=\mathrm{O}$ are not reduced. It is particularly suitable for reduction in the presence of other reactive functional groups, or in cases that are unsuccessful with catalytic hydrogenation.

- On reaction with diimide, only the less-conjugated alkene in the substrate is reduced by diimide without affecting the other double bonds or the epoxide ring, whereas catalytic hydrogenation promotes epoxide opening to give the phenol.

- The reactions are highly stereoselective, taking place by cis-addition of hydrogen in all cases. Transfer of the hydrogen atoms is thought to take place simultaneously through a cyclic six-membered transition state. This mechanism explains the stereospecificity of the reaction, and couples the driving force of nitrogen formation with the addition reaction. Concerted cis-transfer of hydrogen is a symmetry allowed process.

Explanation:-
- The reaction pathway is shown below:

- In the first step of the reaction, 1 eq. of hydrazine $\left({\mathrm{NH}}_2{\mathrm{NH}}_2\right)$ undergoes in situ by oxidation in the presence of oxygen and $\mathrm{Cu}(\mathrm{II})$ which act as an oxidizing agent to give 1 eq. of diimide ( $\mathrm{NH}=\mathrm{NH}$).
- The diimide reduces only the less-conjugated alkene in a substrate.
- But in the above substrate both the $\mathbf{C}=\mathbf{C}$ double bond is disubstituted, one of the $\mathbf{C}=\mathrm{C}$ double bond is in $\mathbf{Z}$ conformation while the other one is in $\mathbf{E}$ conformation.
- 1 eq. of diimide $(\mathrm{NH}=\mathrm{NH})$ reagent will the reduce the $\mathbf{E}$-alkene stereospecifically to give the final product.
Conclusion:-
- Hence, the major product $M$ formed in the following reaction is