1hgoac2h2o 2nabh4ho h o hgoac h hgoac hgoac hoh oh oh

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Unformatted text preview: oup. 3) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C6H12O2. First 2(6)+2-12=2/2=1 And based on the carbonyl peak in the IR we know this is our degree of unsaturation. Also we know that there must also be an ether since there is no OH peak in the IR. And based on the proton NMR we have two types of protons. One has to be connected to the ether. And for the rest to be all the same, there must be an isobutyl group. O O (4) Determine the structure of the following compound based on its mass, IR, and 1H NMR spectra. 114/12=9 carbons 114-108=6 hydrogens C9H6 2(9)+2-6= 7 degrees of unsaturation Based on the IR we know there is a carbonyl So –CH4 add O C8H2O 2(8)+2-2=16/2=8 Lets take o...
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