Chemistry Mass Spectrogram PMT questions

hello can anyone pls explain how to do q 1,4,5 and 9 or any one of those would be a great help thanks.
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/4-Core-Organic-Chemistry/Set-G/MCQ/4.2.4%20Analytical%20Techniques%20MCQ.pdf
the questions are here

Reply 1

TypicalNerd

Original post

by koketenshi

(1. You need to calculate the Mr values for the full molecular ion and possible fragment ions.

A: The molecular formula is C5H10O2.
The Mr of the molecule is 5(12) + 10(1) + 2(16) = 102, which means the m/z ratio of the molecular ion is 102. But there are no sections of the molecule (CH3, CH3CH, (CH3)2CH, CHCH2 etc) that have an Mr of 29 and so this is not the correct option.

B: The molecular ion is C4H8O2. The Mr of the molecule is 4(12) + 8(1) + 2(16) = 88, so you cannot get any ions - molecular or fragment with an m/z of 102. That is to say we can safely eliminate B as an option.

C: This is an isomer of A, so we can be sure it forms a molecular ion with an m/z of 102. It also forms a fragment ion with an m/z of 29, since there is an ethyl group (CH3CH2 => CH3CH2^+ can form). This is therefore the correct answer.

D: This is an isomer of B, which was previously eliminated.

(4. Similar trick to last time.

Notice how C has a 2-propyl group (e.g (CH3)2CH-). This has an Mr of 43, e.g 2(12 + 3(1)) + 12 + 1 = 43.

(5. 31 is 2 more than 29, so find something similar in structure to an ethyl group (which leads to the formation of an ion with an m/z of 29) that maybe contains another atom.

Take for example a -CH2OH group, which is similar in structure to an ethyl group. Looking at the Mr, we have 12 + 2(1) + 16 + 1 = 31, so anything with such a group will do. The only option from the list where this is possible is D.

(9. We saw in (4. how something with a 2-propyl group forms an ion with an m/z of 43. It stands to reason that anything with a 1-propyl group would do likewise and so you need to spot a molecule without either group in it.

A is the only molecule without a propyl group and so should be the correct answer.

I would memorise the following:

Methyl fragment => m/z = 15
Ethyl fragment => m/z = 29
1-propyl or 2-propyl fragment => m/z = 43
Butyl fragments (e.g any fragments with a formula of C4H9-) => m/z = 57

Reply 2

koketenshi

hey man thanks for the help i kinda understand it better now but for q 4 i think u read it wrong as it asks for what is NOT likely to have a fragment ion at m/z 43. However, apart from that I understand the rest fairly better now so thanks.

Reply 3

TypicalNerd

Original post

by koketenshi

Yeah, I did misread Q4 (gotta love functioning on two hours sleep lol).

Ethanoic acid has one fragment with an m/z of 43 (the CH3CO - part), as does propan-1-ol (the CH3CH2CH2- part). This leaves A.

Reply 4

koketenshi

Original post

by TypicalNerd

Yeah, I did misread Q4 (gotta love functioning on two hours sleep lol).
Ethanoic acid has one fragment with an m/z of 43 (the CH3CO - part), as does propan-1-ol (the CH3CH2CH2- part). This leaves A.

thanks for the help!

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