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JEE Main 202229 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Morning Shift
Subject: Chemistry
Chapter: Biomolecules

JEE Main 2022 · 29 June · Morning Shift · Chemistry · Biomolecules

C _ 6 H _ 12 O _ 6 Zymase ⟶ A → _ ∆ ^ NaOI B + CHI _ 3 The number...

${C}_{6}{H}_{12}{O}_{6}\overset{Zymase}{⟶}A\mathop{\to}_{∆}^{NaOI}B + {CHI}_{3}$

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

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

Correct answer

1

Step-by-step solution

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

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

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