C 6 H 12 O 6 ⟶ Zymase A → ∆ NaOI B + CHI 3 The number of carbon atoms present in the product B is

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$C_{6} H_{12} O_{6} \overset{Zymase}{⟶} A \to_{∆}^{NaOI} B + {CHI}_{3}$

The number of carbon atoms present in the product $B$ is

ChemistryBiomoleculesJEE MainJEE Main 2022 (29 Jun Shift 1)

Solution:

1118 Upvotes Verified Answer

The correct answer is: 1

no. of carbon atoms present in $B$ is 1

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