An ester `A(C_4 H_8 O_2)`, on treatement with excess of methyl magnesium bormide followed by acidification, gives an alcohol `B` as the sole organic product. Alcohol `B` on oxidation with `NaOCl` followed by acidification gives acetice acid. Deduce the structures of `A` and `B`. Show the reactions involved.

Explanation

If the ester is of type, a `3^@` alcohol would be obtained with `G.R` , if the ester is of type, a `2^@` alcohol would be obtained with `G.R`. In both the cases, two `R"` groups (i.e., `Me` group) have come from the `G.R`.

Here, `3^@` alcohol `(I)` does not undergo oxidation, because hypohalide reactions (e.g., iodoform type reactions) are shown by the alcohols containing . group. Such group is not present in `3^@` alcohol `(I)`. Moreover, `3^@` alcohol does not undergo hypohalide oxidation. So, the ester is of `(H-overset(O)overset(||)(C )-OR)` type.Structure of ester : Total `C` atoms in ester are four. So `R` group in the ester can be propyl or isoptopyl (three `C` atoms). The ester may be