To determine which molecule is unnecessary for alcoholic fermentation to proceed anaerobically, let's first understand the basics of the process.
Alcoholic fermentation is a biological process by which sugars such as glucose are converted into cellular energy and thereby produce ethanol and carbon dioxide as metabolic waste products. The chemical equation typically representing alcoholic fermentation is:
[tex]\[ C_6H_{12}O_6 \rightarrow 2 C_2H_5OH + 2 CO_2 \][/tex]
Notice here that oxygen (O[tex]\(_2\)[/tex]) is absent in the equation. This tells us that oxygen is not required for the process and, in fact, alcoholic fermentation occurs in the absence of oxygen, which is why it is an anaerobic process.
Now, let's dissect the molecules presented in the question:
1. CO[tex]\(_2\)[/tex] (Carbon Dioxide): This is actually a by-product of alcoholic fermentation and is necessary for the process to complete.
2. H[tex]\(_2\)[/tex]O (Water): Water may be involved in various biochemical reactions, but it is neither a primary reactant nor a product in the typical equation of alcoholic fermentation.
3. O[tex]\(_2\)[/tex] (Oxygen): As noted, alcoholic fermentation is anaerobic, meaning it does not require oxygen. In fact, the presence of oxygen would shift the metabolism towards aerobic respiration instead of fermentation.
Given these points, the molecule that is unnecessary for alcoholic fermentation to proceed anaerobically is O[tex]\(_2\)[/tex] (oxygen).
So the correct answer is:
O[tex]\(_2\)[/tex]
Additionally, if we are to consider index positions in the list provided:
- CO[tex]\(_2\)[/tex] is 1
- H[tex]\(_2\)[/tex]O is 2
- O[tex]\(_2\)[/tex] is 3
The unnecessary molecule, O[tex]\(_2\)[/tex], corresponds to index 3.
