Given: The compound $\left(\mathrm{s}\right),$ which on reaction with ${\mathrm{HNO}}_3$ will give the product having degree of rotation, $\left[\mathrm{\alpha }\right]\mathrm{D}=-52.7°$ is (are)

• A

  

• B

  

• Correct answer
  C

  

• Correct answer
  D

  

Solution

Now the structure, that shows $[\propto]_{D}=-52.7^{\circ}$ is

Hence, the correct answer is options: 'C', 'D'.