Answer:
Benzophenone
Step-by-step explanation:
Given that:
an unknown compound’s semicarbazone melts at 162° -165°
The compound does not give a silver mirror in the Tollens’ test.
On reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms.
The objective is to identify the unknown compound.
Since the unknown compound result to a semicarbazone, we can deduce that the unknown compound is a carbonyl compound. A carbonyl compound is either an aldehyde or ketone in nature. Also, the absence of silver mirror in the Tollens’ test carried out in the reaction confirms that the compound is a ketone because ketones will never give a silver mirror in Tollens’ test.
Similarly, on reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms, that is an iodoform test. That implies that a keto-methyl group is absent in the unknown compound.
Finally, since the unknown compound melting point is between 162° -165° and Benzophenone semicarbazone melting point is 164°.
We can conclude that the unknown compound is Benzophenone semicarbazone