Redox and ester

1)
1. Balance the number of oxygen first (2O at LHS, but 1O at RHS)
Therefore, O2 → 2OH-
2. Balance the number of hydrogen atom. As it is in alkaline medium, we use water to balance the number of hydrogen atom.
Therefore, O2 + H2O → 2OH-
3. As this makes the numbre of oxygen not the same for the both side, we need to add more water and also increase the number of hydroxide ion, so that number of H and O can be balanced.
Therefore, O2 + 2H2O → 4OH-
4. Finally, we balance the charge of the equation. For LHS, there is no charges, but at RHS, there are 4 -ve charges, so we add 4 electrons to the LHS.
Therefore, the equation becomes O2 + 2 H2O + 4 e- → 4 OH-


2)
⊙-COOH [ ⊙ stands for a phenyl group(C6H5-) ]
⊙-COO-CH3
Why the first one is not an ester? Can’t -R(alkyl group) be H(from C0H1)?
Actually, R just represents an alkyl group (i.e. CnH2n+1). If the position is a H atom, we can simply give H for this site.
Moreover, the structure -COOH is an acid group, while an ester has the structure of RCOOR'.
Therefore, the first one is an acid (benzoic acid), while the second one is an ester (methyl benzoate).