JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 19)
C6H12O6 $$\buildrel \text{Zymase} \over \longrightarrow $$ A $$\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\Delta ^\text{NaOI}} $$ B + CHI3
The number of carbon atoms present in the product B is _______________.
Answer
1
Explanation
$$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \stackrel{\text { Zymase }}{\longrightarrow} \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \stackrel{\mathrm{NaOl}}{\longrightarrow} \underset{\text { (B) }}{\mathrm{HCOONa}}+\mathrm{CHI}_3
$$
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