Unknown #12: C9H12 1) Give the molecular formula of your unknown and the degree

Question

Unknown #12: C9H12 1) Give the molecular formula of your unknown and the degree of unsaturation (show calculation). 2) Give the major IR absorbances (cm-1) of interest and assign each to a functional group. You may add additional comments as necessary. The table may have more lines than you need. (For example:absorp:1720, assignment: c=0 , comment:count for 1 unsaturation..) 3)Give NMR data as required by the table. The table may have more lines than you need. chemical shift? multiplicity? # of hydrogens ? assignment? comment? 4. Give the structure of your unknown and assign the hydrogens to their appropriate NMR signals (1, 2, etc as per first column of NMR table).

Explanation / Answer

First, determine the molecular weight using the analysis and the MS - C, 68.2; H, 13.6. Note that these don't add up to 100% - so there's something else in there. The molecular weight is 88, so number of C's = (88*0.682)/12 = 5; number of H's = (88*0.136)/1 = 12. C5H12 = 72. Something weighing 15 is missing => oxygen. So the molecular formula is C5H12O. The IR spectrum is suggestive of an alcohol (the big broad peak just below 3200 cm-1). From the 13C NMR sepctrum you can see there's 2 unique CH2 groups (appear as 2 triplets), 1 unique CH group (appears as 1 doublet) and one unique CH3 (appears as 1 quartet). The chemical shift of on of these doublets, up at ~55 ppm, suggests that this is also bound to an O. So far we've got HO-CH2-. In the proton NMR, the CH2 at 3.5 ppm (which is the one bound to O), appears as a triplet, suggesting it is bound to another CH2. So: HO-CH2-CH2- The large doublet at 1 ppm integrates to 6 protons. The only way that this is possible, is if there are two identical methyl groups coupling to a single CH ie -CH(CH3)2. If we clip these two portions together you get HO-CH2-CH2-CH(CH3)2. Also known as 3-methyl-1-butanol or iso-amyl alcohol.

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