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Question Video: Identifying the Reactant of an Ammonolysis Reaction from the Products Benzamide and Ethanol Chemistry

Consider the following reaction: X + NH₃ ⟶ benzamide + ethanol. Which of the following is the reactant X? [A] C₂H₅COOC₆H₅ [B] C₆H₅COC₂H₅ [C] C₆H₅COOC₂H₅ [D] C₂H₅COC₆H₅ [E] C₆H₅CONH₂C₂H₅

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Video Transcript

Consider the following reaction: X plus NH3 react to form benzamide and ethanol. Which of the following is the reactant X? (A) C2H5COOC6H5, (B) C6H5COC2H5, (C) C6H5COOC2H5, (D) C2H5COC6H5, or (E) C6H5CONH2C2H5.

In this question, we will be examining what molecule will react with NH3, or ammonia, to form the primary amide, benzamide, and the alcohol, ethanol. Let’s start by deducing the structures of the two products. We can begin drawing the structure of benzamide based on the ending -amide. An amide functional group has the structure of a carbon with one double bond to oxygen and one single bond to a nitrogen. In order to fill the octets of each atom, we know that the carbon of the amide functional group has to have one more bond, giving it a total of four bonds. And the nitrogen atom needs two more bonds and one lone pair to have a full octet and a formal charge of zero.

The standard naming convention for amide functional groups is that groups bonded to the nitrogen atom are named N with a dash and then the functional group. For example, the structure of the molecule N,N-dimethylformamide has two methyl groups or CH3 groups directly bonded to the nitrogen atom. However, you can see in benzamide, there’s no such prefix. And when there’s no specified functional group bonded to the nitrogen, we know that there will be two hydrogen atoms, making it a primary amide. We also know that the prefix benz- means that we’ll have a benzene ring directly attached to the carbon of our amide. Now in the molecule ethanol, we have the ending of O-L, which tells us that it’s an alcohol, and the prefix ethan-, which tells us that it will be a two-carbon alkane chain.

And now that we’ve deduced the structure of the two products, we now have to determine the type of reaction that generates a primary amide and an alcohol as a product. In an ammonolysis reaction, ammonia reacts with another molecule, breaking one or more of the bonds. And the type of ammonolysis reaction where ammonia reacts with another molecule to form a primary amide and an alcohol is an ester ammonolysis. In an ester ammonolysis reaction, you have an ester with the general formula of RCOOR prime reacting with ammonia, NH3, to give you primary amide and alcohol products.

It’s important to notice that the R group that’s bonded to the carbon of the ester starting material is the same R group that’s bonded to the carbon in the primary amide product. You can also see that the R prime group bonded to the oxygen of the ester is the same R prime group that’s bonded to the oxygen of the alcohol. And finally, the NH2 group from the primary amide and the hydrogen from the alcohol come from the ammonia starting material.

And now, with this information, we should be able to answer the question. “Consider the following reaction: X plus NH3 react to form benzamide and ethanol. Which of the following is the reactant X?” We know that the NH2 of the primary amide and the hydrogen of the alcohol products come from the ammonia starting material. And we also know that the benzene ring bonded to the C double bond O of the amide product is the RCO of the ester starting material. And finally, the OCH2CH3 of the alcohol must be the OR prime of the starting ester. And since this molecule has a molecular formula of C6H5COOC2H5, we can correctly identify answer choice (C), C6H5COOC2H5, as the correct answer.

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