C₆H₁₂O₆ →(Zymase) A → _Δ ^(NaOI) B + CHI₃
The number of carbon atoms present in the product B is _______________.

Solution

Chemistry ma’am · 4 steps

Ethanol has only two carbons, and the iodoform test takes one of them away as CHI₃.

1. Identify A. Zymase, an enzyme from yeast, ferments glucose into ethanol and carbon dioxide: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. So A is ethanol, CH₃CH₂OH.
2. Iodoform reaction. NaOI first oxidises ethanol to acetaldehyde, CH₃CHO. The CH₃ group next to C=O is then tri-iodinated and split off as yellow iodoform, CHI₃.

   CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃ + HCOONa + 5NaI + 5H₂O

3. Count the carbons. Ethanol has 2 carbons. One leaves as CHI₃, so the other stays behind in B, which is sodium formate, HCOONa.
4. So. B is HCOONa, which has 1 carbon atom.

The common slip: 2. Taking B as sodium acetate gives 2 carbons, but that would mean no carbon left as CHI₃.

The idea to keep. In the haloform reaction the CH₃ next to C=O leaves as CHX₃. The salt left behind always has one carbon fewer than the methyl ketone or aldehyde it came from.

Answer from the NTA final answer key. Solution by PrepWiser.