Distinguishing Test for Organic Compounds

Given the following pairs of samples, how do I distinguish between them?


1. benzamide and ethanamide
2. ethanoyl chloride and ethanoic anhydride


My Approach:


1. We could easily convert the amides to amines by reducing them with $\ce{LiAlH4}$. Now, the question reduces to identifying the methyl group against the benzene group.
2. I am really lost here but my guess would be that we are looking for some characteristic reaction for anhydrides.




Since this sounds like an assignment, I am not going to give it all away. 


Does you have to use qualitative chemical tests? I don't even teach those anymore. They were supplanted by spectroscopic techniques in most professional labs *decades* ago.


IR spectroscopy would answer both questions quickly. 


1. Benzamide would have peaks characterizing an aromatic ring, for example: $\ce{C}\_{sp^2}\ce{-H}$ stretches around $3050-3100\ \mathrm{cm^{-1}}$. Ethanamide will not have these peaks, but will instead have $\ce{C}\_{sp^3}\ce{-H}$ peaks $2800-3000\ \mathrm{cm^{-1}}$. There are other differences as well.
2. There are differences in the position (and sometimes number!) of carbonyl peaks between an anhydride and an acid halide. What are they?


So would mass spectrometry. Benzamide and ethanamide have different molar masses and different fragmentation patterns. Ethanoyl chloride... chlorine has two major isotopes. How can that help? 


NMR spectroscopy would also be useful. However, IR is cheap, fast, and easy.


 4

For the first case, we need to compare benzamide and ethanamide, One of the ways I can think of is comparing aniline and methylamine.


But we need to get there first, so we first react both our substrates with $\ce{Br2/NaOH}$ (Hoffman Bromamide degradation). As taken from [Chem Libretexts/Hoffman Degradation](https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hofmann_Degradation):



> 
> Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. eg:
> 
> 
> [![enter image description here](https://i.stack.imgur.com/r8NzJ.png)](https://i.stack.imgur.com/r8NzJ.png)
> 
> 
> 


Therefore, we now have aniline and methylamine. A way to compare these two is via diazotization, wherein we treat the compound with $\ce{cold NaNO2/HCl}$. After which we add β-napthol. Now, this reaction is known as the azo-dye test, where we see aromatic amines give an orange dye which is insoluble in water. Methylamine doesn't give this test and hence, we have distinguished the two compounds.


Case two, we have ethanoyl chloride and ethanoic anhydride, this is a much easier test, you need to add water to both substrates, one produces white fumes of HCl (ethanoyl chloride) during hydrolysis and the other substrate (ethanoic anhydride) shows no reaction but the water becomes acidic due to the formation of the carboxylic acid.


 3

Hydrolysis of the amides produces in the first case acetic acid which is a liquid miscible with water. In the second case it makes benzoic acid which is solid and insoluble in water.


Let's consider now the second group of compounds. Acetic anhydride does not react quickly with water, and it produces only acetic acid. Acetyl chloride reacts quickly with water, producing acetic acid and hydrochloric acid.


 2