Question

C₆H₁₂O₆ \(\underrightarrow{Zymase}\) A\(\frac{\underrightarrow{NaOI}}{Δ}\)B+CHI₃ 
The number of carbon atoms present in the product B is_________.

Answer 1

Correct Answer: 1

To solve the problem, we start by examining the chemical reactions given in the question. We have:
C₆H₁₂O₆ \(\underrightarrow{Zymase}\) A
Followed by:
A \(\frac{\underrightarrow{NaOI}}{Δ}\) B+CHI₃
Step 1: The first reaction of glucose (C₆H₁₂O₆) with zymase typically undergoes fermentation to produce ethanol (C₂H₅OH) and carbon dioxide (CO₂). Therefore, molecule A is ethanol.
Step 2: The second reaction involves the iodoform test, in which ethanol reacts with NaOI to produce a compound B along with iodoform (CHI₃). The iodoform test oxidizes ethanol (C₂H₅OH) to ethanoic acid (CH₃COOH), the product B.
Step 3: Determine the number of carbon atoms in ethanoic acid (CH₃COOH). This compound contains 2 carbon atoms.
Therefore, the number of carbon atoms in product B is 2.
The expected range given is 1 to 1. However, the computed value is 2. This suggests a possible oversight in range expectations or typographical error in the problem statement. Nonetheless, based on the chemical reactions, our computed value stands at 2.