Part 8: SPECTROSCOPY

The following data is available from the question.

Note : Remember to cross reference things to confirm.... eg. IR mnay show C=C, use NMR to confrim that.

MS: M+ seen at 114 g/mol (even MW, no Cl or Br isotope patterns).

EA: Elemental analysis data for C and H (i.e. standard analysis) showed 63.14% C and 8.83% H  (therefore there is 28.03% something else, most likely oxygen : remember oxygen can't be detected in a combustion analysis).   Use the MW from the mass spectra (above) to get the partial molecular formula = C6H10 ?  (adds up to 72 + 10 = 82 so that leaves 32). The missing piece is consistent with 2 oxygen atoms (because 2 x 16 = 32) .... hence  C6H10O2
From here we get the IHD = 2

IR: Confirms the presence of oxygen. There is a C=O at about 1700 cm-1 and a typical broad OH band between 3400-2300cm-1.  C-O can be seen around 1300cm-1.  There is also a band to the right of the C=O at about 1650cm-1 to consider, a possible C=C.

13C nmr: The proton decoupled spectrum shows a total of 6 peaks indicating 6 types of C. By analysis of the chemical shifts, we have types in the region for 1 x C=O above 160 ppm, two types of C=C at 100-160ppm,  and 3 peaks between 0-40 ppm that are most likely from a hydrocarbon portion.

1H nmr: First of all we have 6 sets of peaks so we have 6 types of H showing up. After this, it's a good idea to tabulate the information to make sure you get it all correctly matched up:

 d/ppm multiplicity integration Inference 12 singlet 1 exchangeable => OH probable carboxylic acid RCO2H 7.2 multiplet 1 alkene C=C complex coupling (doublet of triplets) to 2H and 1H 5.8 doublet 1 alkene C=C coupled to 1H 2.2 quartet 2 CH2 coupled to 3H, slightly deshielded 1.7 hextet 2 CH2 coupled to 5H 1.0 triplet 3 CH3 coupled to 2H

The most significant structural information from this is :
• peaks at 5.8 and 7.2 ppm indicated a disubstituted alkene = C2H2-  (can't be aromatic since there are too few H and C)
• the alkene is most likely 1,2-disubstituted i.e. H-C=C-H rather than 1,1-disubstituted i.e. C=CH2 system.
• it's the coupling patterns and the chemical shifts that indicate this...if 1,1-disubstituted,
• the two vinyl H would have chemical shifts that were similar, and
• they would have similar types of coupling patterns.
• peaks between 1.0 and 2.2 ppm indicate 3 pieces that have to be joined as a CH3CH2CH2 unit
• peak at 12 ppm CO2H
Summary....
The MS indicated MW =  114 g/mol
The IR showed the presence of C=O, OH and C=C
13C and H NMR gives  H-C=C-H, CH3CH2CH2 and CO2H
Use this to confirm the molecular formula as C6H10O2

Altogether...

 With the pieces we have :   H-C=C-H, CH3CH2CH2 and CO2H We actually have the structure as we only have 1 middle piece, the H-C=C-H and two then end pieces. 2-hexenoic acid

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit.  You should be asking yourself : "Does my answer give me what the H-nmr shows ?"

Notes :

• Can't easily distinguish the cis- and trans- isomers except by the size of the coupling constant between the H atoms on C2 and C3.
• Moving the C=C to different positions in the chain would give significantly different coupling patterns and chemical shifts.