Question

In the following reaction,
Carbonyl compound + MeOH $\xrightarrow{HCl}\$ acetal
Rate of the reaction is the highest for:
(A) acetone as substrate and methanol in stoichiometric amount
(B) propanal as substrate and methanol in stoichiometric amount
(C) acetone as substrate and methanol in excess
(D) propanal as substrate and methanol in excess

Answer 1

In the above chemical reaction, identify the incoming molecule and the molecule being attacked upon. Since the reaction takes place through ${{S}_{N}}2$ mechanism we need to take steric hindrance into consideration. The quantity of attacking molecules must be in excess in order to increase the rate of reaction and the steric hindrance in substrate be minimum.

Complete step-by-step answer:
The ${{S}_{N}}2$ reaction is a type of reaction mechanism commonly applied in the field of organic chemistry. In this mechanism, one bond is broken and another bond is formed at the same time i.e. simultaneously. ${{S}_{N}}2$ is a kind of nucleophilic substitution reaction mechanism where two molecules are involved in the rate determining step hence the name, substitution nucleophilic bimolecular (${{S}_{N}}2$). The rate of reaction in ${{S}_{N}}2$ mechanism can be increased by the following measures:
- Excess of attacking nucleophiles.
- Less steric hindrance of the molecule being attacked upon or rather substrate.
In the above reaction we can identify that methanol is the incoming nucleophile and the carbonyl compound is substrate. Based on the above explanation we can conclude that in order to increase the rate of reaction, the substrate is propanal and the quantity of methanol is to be taken in excess.

Therefore, the correct answer is option (D).

Note: In the above reaction, we considered propanal to have less steric hindrance than acetone thus choosing propanal to increase rate of ${{S}_{N}}2$ reaction. This is because in acetone, there are two methyl groups on either side of the carbonyl group making the site hindered for substitution. On the contrary, propanal has a methyl group on one side and hydrogen on the other side. Thus, the attacking side in propanal is less hindered.