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Stuck on a Question

A sample of ethanedioic acid was treated with an excess of an unknown alcohol in the presence of a strong acid catalyst. The products of the reaction were separated and analysed in a time of flight (TOF) mass spectrometer. Two peaks were observed at m / z = 104 and 118.

(a) Identify the species responsible for the two peaks.

The answer's are [CH3OCOCOOH]+ and [CH3OCOCOOCH3] + but I'm not sure why. Cheers in advance.
Patterned it now. It’s a deceptively difficult 2 marker, and you wouldn’t be expected to know it for AS. Not sure how to close a thread, so if anyone in the future’s stuck.

Acid’s + Alcohol’s -> Esters + Water
The question’s talking about the Mr of the products of this reaction. The Mr of water’s 18, so we can definitely rule that out of the question.

The Mr of ethanedioic acid is 90 and we need to go from 90 to 104, and 118. Since the product in question is DEFINITELY an ester, you want to trial an error different alcohols, and see if your new ester will have an Mr of 104 firstly.

The ester formed from Ethanedioic Acid + Methanol gives you an Mr of 104, as shown in the image below. Since the difference between 118 and 104 is the same as the jump from 104 to 118, and also there is an extra carboxylate group on the other side of the acid, which can react with another alcohol, that gives your second product.

Since they are molecular ion peaks, don’t forget to add the charges. When writing out ester’s you work from alcohol group end to the carboxylic acid end. With diester’s it’s the same, but you add COOCH3 at the end.

A lot of work for 2 marks, so with these it’s always best to assess if you can instantly work it out, and if not, skipping, rather than wasting 4-5 minutes in your exam.

Hope this helps!💪🏾

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