Ketone Formation

1-
a. One aldehyde gives a positive iodoform test. Which one is it?
• Ethanal
b. Explain why ethanal gives a positive iodoform test.
• Ethanal is the only aldehyde to give a the positive result because the reaction requires a methyl group connected to a carbon atom with a keto or an OH- substituent.
2- In semicarbazone formation only one of the two –NH2groups in semicarbazide undergoes nucleophilic addition to the carbonyl group. Explain the difference in the reactivity of these two –NH2 groups, using resonance structures.
• Within the semicarbazone formation there are two –NH2 groups. The one closest to the ketone partakes in the electron resonance and therefore is the more stable out of the two. It doesn’t want to interrupt the p-orbital it’s sharing.

Referring to the table of derivatives, which one aldehyde or ketone is most likely her unknown? Explain your reasoning, if she is unsure what conclusion to draw, what experiment should she do next?
• Judging by those two melting points, acetophenone is most likely her unknown. The listed melting point for Oxime is 60°C and the listed melting point for Semicarbazone is 203°C. The melting points she derived are marginally lower, which is an indicated of slight impurity. A decrease in melting point occurs with an impure substance. If she is not convinced by those two melting points, she can create a solid derivative of 2,4-DNP to obtain a third melting point to compare. Furthermore, she could perform the Tollens’ test or the Iodoform test just to be sure she is coming to the correct conclusions.
4- A student finds that the 2,4-DNP derivative of his unknown melts at 155-156°C, and the oxime melts at 68-69°C. He has only 20 minutes left, not enough time to prepare and recrystallize the semicarbazone. Does he have any other recourse to decide between the remaining possible