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Study Notes: Structural Elucidation - Examples

The determination of the structures of organic compounds is carried out by the joint analysis of the MS, IR and NMR spectra.

The examples and problems that follow refer to organic compounds containing combinations of C, H, O and N. Each compound may contain one or two functional groups from ROH, ROR', RCOR', RCHO, RCONR'R'', RCOOR', RCOOH and RNH2. There will not be any aromatics.

Procedure:

  1. Obtain the Mr and use it to look up the molecular formula.

  2. Rule out functional groups that cannot belong to the molecular formula.

  3. Examine the IR spectrum to determine the presence of suspected functional groups. This is done by examining the peaks in the functional group region (4000 - 1100cm-1) for peaks in positions that are characteristic of functional groups that can fit with the molecular formula.

  4. List the (main) possible isomeric structures.

  5. Examine the NMR spectrum and note the number of peak groups and their integrals. This may indicate the number of methyl (CH3), methylene (CH2) and methine (CH) groups.

  6. Examine the NMR spectrum chemical shifts and attempt to determine the proximities of nearby protons to the functional group. If a functional group proton is suspected, it may be possible to locate it by consideration of the integral and chemical shift data.

  7. Examine the NMR spectrum for peak multiplicity in order to determine the number of protons on neighbouring groups. This will give the proton sequence that determines which isomer is the correct structure.

  8. Confirm the structure by checking the MS fragmentation pattern.


Example No. 1:      Mr = 60.0977

Mass spectrum example number 1

IR spectrum example number 1

NMR spectrum example number 1

Working out:

Mr = 60.0977 means that the molecular formula is C3H8O.

The molecular formula rules out the possiblity of a carboxylic acid or amine.
(Also - must not contain a carbonyl group or multiple C-C bonds, to accommodate all eight Hs in the structure)

IR      Strong and fairly broad absorption from 2800 to 3300 cm-1
               Hence, probably an alcohol (H-bonded in solution).

         Sharp peak at about 1100 cm-1 would be the C-O absorption of the alcohol.

Alcohol isomers CH3CH2CH2OH    CH3CH(OH)CH3    (CH3)2CHOH

NMR Shows three peaks:
 
delta4.0
multiplet (5 or 6 peaks)
integrates as 1.
 
delta3.4
broad single, peak
integrates as 1.
 
delta1.2
sharp, doublet
integrates as 6.
 

There are six equivalent protons at delta1.2 suggesting two CH3 groups arranged symmetrically (same chemical environment). The chemical shift agrees with CH3-C-OH.
Since the absorption is a doublet, there must be -CH- next to each of the two methyl groups. i.e. it must be between them.
 
     The structure so far would be:
CH3-CH-CH3
 

The multiplet at delta4.0 would belong to the -CH- since it integrates as 1 and would be extensively split by the two adjacent methyl groups.

 

The delta3.4 absorption would belong to -OH (integrates as 1) which is in the correct range (0.5 - 5.5) for a hydroxyl proton. The hydroxyl group would be attached at C2.
 
     The final structure is CH3CH(OH)CH3
2-propanol
 

The mass spectrum has the characteristic M - H2O (60 - 18 = 42) peak for an alcohol.
The M - CH3 peak (at 45) appears as would be expected for this structure.


Example No. 2:      Mr = 102.1360

Mass spectrum example number 2

IR spectrum example number 2

NMR spectrum example number 2

Working out:

Mr = 102.1360 means that the molecular formula is C5H10O2

The molecular formula rules out the possibility of an amine.

IR      Strong absorption 1750 cm-1

               Hence, carbonyl group of an ester.

         Broad absorption below 1300 cm-1 for the C-O absorption of the ester.

Ester isomers:   CH3CH2CH2COOCH3 (CH3)2CHCOOCH3
    CH3CH2COOCH2CH3  
    CH3COOCH2CH2CH3 CH3COOCH(CH3)2

 

NMR Four absorptions shown:
 
delta3.8
singlet
integrates as 3.
 
delta2.2
triplet
integrates as 2.
 
delta1.4
sextuplet
integrates as 2.
 
delta0.9
triplet
integrates as 3.
 

There are three protons at delta3.8 so it is probably CH3. The singlet shows that the protons are isolated: could be methyl ester. The chemical shift of delta3.8 is in good agreement with R-CO-O-CH3 (located at 3.7).

There are two protons at delta2.2 suggesting a CH2 group.
The chemical shift agrees with -C-CH2-CO-O-R'.

The triplet splitting further suggests that there is an additional CH2 neighbour (see absorption at delta1.4).


 
     The structure so far would be:
-CH2-CH2-CO-O-CH3
 

The sextuplet at delta1.4, integrating as two, would belong to -CH2 - and would be split by CH2 and CH3 (see delta0.9) groups on either side.
The chemical shift agrees well with -CH2-C-CO-O-R'.

The delta0.9 absorption would belong to CH3 (integrates as 3). It is split into a triplet by a neighbouring CH2.

       The final structure is CH3CH2CH2COOCH3 methyl butyrate
 

The mass spectrum shows: 71 arrow right M - 31 for removal of OCH3 from RCOOCH3
                                         59 arrow right appearance of CO+OCH3 from RCOOCH3

 

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