When treated with an equivalent of methanol, compound A, with the molecular formula C₄H₆C_l2O, forms the compound whose ¹H NMR spectrum is shown here. Identify compound A.

It is the proton nuclear magnetic resonance that helps in the identification of a molecule that shows the peak of the hydrogen of the molecule where equivalent proton produced single peak and non-equivalent proton different peaks as per the environment.

As per the given data peak observed at 3.6-3.8 ppm shows the presence of the ester group and the triplet peak is shown at 3.6-3.5 shows the presence of the methylene group adjacent to each other the sharp peak around 4 is due to the presence of the chloride ion and the around 2 is due to presence of the methyl group attached to the oxygen of ether. So, the possible structure is:

As given the compound, b is produced on the reaction of compound a with methanol that the ether is produced on the reaction of methanol with acyl chloride with by-product HCl hence, the compound a is: