Question

The question is given below​

Answer 1

Explanation:

1. Ethene and but-2-ene.(A and C)
2. Methanol.(D)
3. But-2-ene.(C)
4. Butanoic acid(B)
5. CH3OH + CHOOH = CH3OOCH + H20

(Methanol) + (Methanoic acid) = (Ester)

Now, Reaction of this Easter with a base (NaOH) will give both compounds back.

NaOH + CH3OOCH = CH3OH + CHOOH + Na.

Hence, The organic compound is Easter (Easter of methane).(Generally, Easter of ethane is studied).

And,the reaction is a saponification reaction.(Converse of Esterification).