The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

determining the structure of a compound using spectra part 2

Molecular formula ive worked out from (C % 80.5, H % 10.1, N % 9.4)
C10H15N

mass spec shows molecular ion at 149
also has a peak at 134 of 100% and a few others (graphs attached)

NMR has 4 clusters (a b c d)

a: triplet at around 1.1-1.1ppm
b: quartet at around 3.2-3.4ppm
c: im not sure if its a quartet, the cluster looks a bit funny 6.6-6.7ppm
d: looks like a triplet but again not sure 7.1-7.3ppm

IR spec shows lots of absorbances, not sure which ones to consider, although i think it may be an aromatic

ive attached a data table in PDF form and the graphs are too big too attach, so ive got a link them

http://www.filefactory.com/file/af61faf/n/28_IR_pdf

http://www.filefactory.com/file/af61fd0/n/28_NMR_pdf

help please guys!
SPEC TABLES.pdf18.9KB
Ok I'm going to lunch now but the structure is C6H5N(CH2CH3)2, see if you can see why - I or someone else will explain later :tongue:
Reply 2
Avatar for MrCynical
MrCynical
OK, this compound obviously contains a C6H5 group (fragment of mass 77 in mass spec, and NMR shifts in the 6 to 8 region are usually aromatic protons).

The peaks in the 6.6-7.3ppm region aren't as simple as a triplet and quartet. What you've got here is a triplet from the proton directly on the opposite side of the ring from the nitrogen (the para proton) coupling to the two meta protons on each side of it. The two meta protons give a doublet of doublets from coupling to one para and one ortho proton each. The ortho protons give a doublet from coupling to the adjacent meta proton. The spectrum is a little confusing as some of these are overlapped to form the "quartet". You can also see traces of finer structure in the triplet from longer range coupling.

The triplet and quartet at 1.1, 3.3 again suggests an ethyl group. 3.3ppm is a bit higher shift than usual for an ethyl, suggesting it's attached to something electron withdrawing (i.e. the nitrogen). Based on the intensities of the peaks, given the whole aromatic system contains 5 protons, there must be two equivalent ethyl groups. There's only one way to assemble that structure given that molecular formula: C6H5N(CH2CH3)2

Related discussions

Latest

Trending

Trending

Articles for you