C6H12O6→ZymaseA→ΔNaOIB+CHI3\mathrm{C_6H_{12}O_6\xrightarrow{Zymase}A\xrightarrow[\Delta]{NaOI}B+CHI_3}C6H12O6ZymaseANaOIΔB+CHI3
The number of carbon atoms present in the product B is ______.
Official answer: 1
A is ethanol; the iodoform reaction gives CHI3\mathrm{CHI_3}CHI3 and sodium formate (B), which has one carbon.
Answer: 1