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#1 2009-09-06 23:57:07


NMR of 4-t-butyl-1-methylbenzene, unknown of C11H16

Yahoo!Answers in chemistry wrote:

The H NMR spectrum of a compound of formula C11H16 is shown here.
Draw its structure.
For a compound with MF = CxHyO, ignore the oxygen.
IHD = [(2*x + 2)-y]/2
IHD = [(2*11 + 2) - 16]/2
IHD = 24 - 16 = 8/2 = 4
Therefore there are 4 rings or double bonds.
The peaks around 7 ppm indicate a benzene ring, therefore it represents one ring and three double bonds.
From the integral, we can see there are 4 aromatic hydrogens, therefore the other peaks are 3Hs and 9Hs by measuring their heights (about 0.5 block per hydrogen).
Since hydrogens must come in 1H, 2H, and 3H units, the peak at 2.3 ppm is a CH3 group and the peak at 1.3 must be 3xCH3 or a t-butyl group.
Since there are four aromatic hydrogens, the benzene ring is di-substituted. The choices are 1,2-, 1,3-, or 1,4-substituted. 1,2- and 1,3-substitution would result in 4 aromatic signals, as no symmetry would be present. 1,4-substitution would result in two signals with agrees with this spectrum.
Putting this all together will result in 1-(tert-butyl)-4-methylbenzene.



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