An unknown compound’s semicarbazone melts at 162-165 degree Celsius. The compound does not give a silver mirror in the Tollens’ test. On reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms. Identify the unknown. Explain your reasoning.

Benzophenone

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  

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