Question

In step 5 of reaction 3, the reaction mixture containing the products shown below in ethyl acetate is extracted twice with HCl. Camino di HN. 0-methylaniline m-methylaniline O p-methylacetanilide

a) Which products (shown above) from reaction 3 can react with HCl in an acid-base reaction? Show the mechanism for this acid-base reaction and give the structures that would result from the protonation of these products. Which phase (organic or aqueous) would these protonated products be soluble in?

b) Which phase (organic or aqueous) would the p-methylacetanilide be found in?

Answer 1

The three isomers you have taken for the acetylation purpose are o-methyl aniline, m-methyl aniline, and p-,methyl aniline. It is known that p-methyl aniline undergoes easily than the remaining two isomers. This is because, the methyl group present in the para position can easily donate its electron density by hyperconjugation than those present in ortho or meta positions. Hence p-methylaniline undergoes acetylation easily than o- and m-methyl anilines(NOTE: If you add excess of acetic anhydride, then all will undergo acetylation!! Here only one equivalent is used. So only p-methyl aniline undergoes acetylation leaving the other two intact). Now, after the reaction, the reaction mixture contains o-methyl aniline, m-methyl aniline and p-methylacetanilide. It is important to note that o-methyl aniline and m-methyl anilines are amines and are basic in nature. Hence they can readily react with acids. Say HCl to give salts. But p-methyl aniline reacted with acetic anhydride(acetylation) and converted to p-methyl acetanilide. Now, sacetanilide is not basic in nature and will not react with HCl. Hence after the addition of HCl in step 5, the solution will contain o-methyl anilinium hydrochloride, m-methyl anilinium hydrochloride and p-methyl acetanilide. o-methyl anilinium hydrochloride and m-methyl anilinium hydrochloride are salts. Hence they are water soluble and ethyl acetate (organic solvent) insoluble. That means these two will be present in the aqueous phase. p-methyl acetanilide is not a salt and is an organic compound. It is water insoluble and ethylacetate soluble. Hence it will be present in the organic phase.

elemek + WK acetic anhydrides ri o methyl aniline m-methylaniline p-methyl aniline tha HN methyl aniline nomethyl aniline p-methyl autanilide lad. omethylanilinium hydrochloride methyl anilinium hydrochloride mary aukcilide. salls innlater soluble. Hence organic will be present in an aqueous us phase compound. anate, insoluble and organie solied soluble Hence will be present in organic phare.

The amine contains a lone pair. Hence the protonation takes place to this easily available lone pairs. The mechanism will be as shown below.

NHA ระ” 8,34 o-methyl aniline. “น วาระ fuo m-methyl acriline

Now there will be a question. Why not p-methyl acetanilide didn't undergo protonation eventhough it carries a nitrogen and a lone pair. This is because, the nitrogen atom in p-methyl acetanilide is attached to a strongly electron withdrawing carbonyl group and the lone pair is involved in resonance with the carbonyl group. Hence this lone pair is not easily available for protonation as in the case of o- and m-methyl anilines.

an HAN nitrogen tear Lone pair available due to is not resonance.

I believe the explanation is CLEAR!!!!