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← JEE Main 2022 · 29 Jun · Shift 1
ChemistryInteger+4 / -1Question 90 of 90

JEE Main 2022 · 29 Jun · Shift 1, Chemistry question 90

C6H12O6→ZymaseA→ΔNaOIB+CHI3\mathrm{C_6H_{12}O_6\xrightarrow{Zymase}A\xrightarrow[\Delta]{NaOI}B+CHI_3}

The number of carbon atoms present in the product B is ______.

Answer and worked solution

Official answer: 1

A is ethanol; the iodoform reaction gives CHI3\mathrm{CHI_3} and sodium formate (B), which has one carbon.

Answer: 1