Question

Identify A, B, C, D, and E in the following schemes and write their structures:

• A. A is a vicinal dihalide and B has 2 $\pi$ bonds
• B. C is an aldehyde and D is a semicarbazone
• C. C is an ketone and D is a semicarbazone
• D. E has 4 alpha deuterium atoms

Answer 1

Answer: (A), (C), (D)

The correct options are
A A is a vicinal dihalide and B has 2 $\pi$ bonds
C C is an ketone and D is a semicarbazone
D E has 4 alpha deuterium atoms

In the first step, electrophilic addition across the double bond ensues to give (1,2-dibromoethyl)cyclohexane, which is the compound A. the vicinal dibromide A, undergoes elimination (dehydrohalogenation) to yield the terminal alkyne B – (ethynyl)cyclohexane. Further, a reaction with $HgS{O}_4/H_{2}S{O}_4$ produces cyclohexyl methyl ketone.


Carbonyls react with semicarbazide to give a special type of imine called semicarbazone. Hydrolysis of the semicarbazone with $NaOD/D_{2}O$ gives back the ketone except that all previously held $\alpha$ hydrogen atoms are replaced with deuterium.