Video Transcript
Some students react propyl ethanoate with ammonia to make two different types of product molecules. What are the names of these product molecules? (A) Propanamide and ethanol, (B) ethanamide and propan-1-ol, (C) ethanamide and propan-2-ol, (D) ethanamide and propanal, or (E) propanamide and propanal.
Propyl ethanoate is an ester, and esters have a general formula of RCOOR prime, where the R group is a hydrogen, an alkyl group, or an aryl group. The R prime group can be any alkyl or aryl group. And an ester linkage connects the R group to the R prime group. Based on the name of the molecule propyl ethanoate, we know that we have a three-carbon chain that’s bonded to the oxygen of the ester linkage. And the term “ethanoate” suggests that we have a two-carbon chain, where one of the carbons is a part of the ester linkage. When an ester, such as propyl ethanoate, is reacted with ammonia, the process is known as ester ammonolysis. Ammonolysis is a chemical reaction where ammonia reacts, breaking one or more bonds.
In an ester ammonolysis reaction, we have an ester that reacts with ammonia, or NH3. This reaction breaks the carbon-oxygen single bond of the ester linkage. During this process, a new bond is formed between the nitrogen of ammonia and the carbon of the ester linkage, giving a primary amide and an alcohol. It’s important to notice that the R group that is bonded to the carbon of the ester linkage is a part of the primary amide product. And the R prime group that is bonded to the oxygen of the ester linkage is always a part of the alcohol product.
And now we can apply this knowledge to the reaction of propyl ethanoate with ammonia. In this reaction, we know that the carbon-oxygen single bond of the ester linkage will be broken. And a new bond will be formed between the nitrogen of the ammonia and the carbon of the ester linkage. The formation of this bond gives us the primary amide product ethanamide. We also know that when the ester linkage bond is broken, the three-carbon chain bonded to the ester oxygen will form an alcohol product, in this case propan-1-ol. With this information in hand, we can correctly choose answer choice (B), ethanamide and propan-1-ol, as our correct answer.
