Question

Product A will be

(A)- $C{{H}_{3}}CH(OH)-C{{H}_{2}}(OC{{H}_{3}})$
(B)- $C{{H}_{3}}CH(OC{{H}_{3}})-C{{H}_{2}}(OH)$
(C)- $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OC{{H}_{3}}$
(D)- $C{{H}_{3}}CH(OC{{H}_{3}})C{{H}_{3}}$

Answer 1

Hint: The reaction is a base catalysed nucleophilic substitution reaction due to the presence of the highly reactive epoxide compound, having high ring strain present in it.

Complete step by step answer:
-In the given reaction, the reactant is propylene oxide having a three-membered ring, with the oxygen bonded to two-adjacent carbon atoms. This cyclic ether is called an epoxide, which has a lot of strain and hence highly reactive in nature.
-The reaction of the epoxide with the sodium methoxide base in presence of the methanol solvent causes ring-opening, through the nucleophilic substitution on the electrophilic carbon centers on the epoxide. The reaction mechanism is as follows:
- The base, with $(-OC{{H}_{3}})$ as the nucleophile will attack through ${{S}_{N}}2$, that is from the backside, preferring the least hindered carbon atom. So, the carbon atom with only hydrogen attached to it is favoured.
-This leads to the opening of the strained ring, and the formation of the alkoxide ion, with the negatively charged oxygen atom in it.

- The methanol present further leads to the protonation of the oxygen atom on the alkoxide ion.

Thus, we get product A to be option (A)- $C{{H}_{3}}CH(OH)-C{{H}_{2}}(OC{{H}_{3}})$, that is, 1-methoxy-2-propanol or the propylene glycol methyl ether.

Note: The ring opening of the epoxide is a regioselective reaction, depending on whether it is a basic or acidic alcohol, involved in the nucleophilic substitution. As we have seen the basic alcohol favours the ${{S}_{N}}2$ mechanism, but the acidic alcohol favours the ${{S}_{N}}1$ mechanism. Also, the methanol solvent is present in equilibrium with the sodium methoxide and does not affect its attack.