Danyal124
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Help!!! Very confused could someone explain the answers I got the wrong answer but I worked logically in solving it
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Drogonmeister
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There is probably other ways to solve this but I used the C-O shift.
a) It has to be ester 1 when looking at the shifts of the proton groups.
4ppm correlates to HC-O and there is only 1 peak at 4ppm meaning there is probably only 1 proton group next to the C-O group. In ester 1 there is only the CH3-C-O group we are looking for and ester 2 has CH3CH2C-O which would produce too many peaks.
The other peaks at 2ppm and 1ppm correspond to the CH2CH3 groups on ester 1 due to the fact that they correlate to the HC-C=O and HC-R on the data
sheet.

b) The quartet group is in the CH2 which is bonded to the CH3 group of ester 2.
We can correlate this gain to the data sheet with HC-C=O which puts you in the 2-3ppm range. Then you use the other groups of ester 2 to determine that it is shifted more towards the lower end of 2ppm.

Hope this helps.
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charco
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(Original post by Danyal124)
Help!!! Very confused could someone explain the answers I got the wrong answer but I worked logically in solving it
When a CHx group is directly attached to an electronegative atom (such as oxygen) it's electron density is withdrawn and the protons are said to be deshielded.

This means that these protons experience a stronger external magnetic field, so the energy difference between the two spin states is greater and more energy is needed to cause transitions between spin states. Hence, the signal appears at higher ppm values (delta).

In these two esters, the first of them has a CH3CH2-O arrangement.

This means that the expected splitting pattern of a triplet (due to the CH3-) and a quartet (due to the -CH2-) will have the quartet shifted downfield to about 4.0 ppm.

In the second ester, there is a CH3 group directly attached to an oxygen, CH3-OCO-R.

This would have a singlet (due to the non split CH3) shifted up to about 4.0ppm.
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