Question

Organic Chemistry - Using the following IR anaylsis identify the major peaks and possible functional group identity. Using the two NMR spectra, discuss the integration of peaks, chemical shifts, and splitting. Using the mass spectra, identify molecular weight and important fragments to confirm the structure. Draw the fragments that correspond to each peak in the mass spectra. Draw the structure of the unknown that corresponds to all available data.

additional information: Unknown ‘J’ is a alkane.

H-NMR multiplet, 1OH triplet, 6H PPM C-NMR 15 PPM IR L00 T DOO 1000 sdo Mass Spec. 50O 4000 Mass Spec. 100 80- 60- M+ 20- 0 50 60 70 80 90 100 2 10 20 30 40 m/ z

Answer 1

n-heptane The alkane is n-heptane. HNMR 0.9 ppm , triplet, 6H for the two end -CH3 gronps 1.4 ppm, multiplet. 10H for all the CH2 groups 13C NMR 14 ppn the two CH3 carbons 22 ppm the two CH2 C attached to CH3 groups 29 ppm, the middle CH2 (4th carbon) 34 ppin 3and 5 C112 oups CH3- CH2-CH2-CH2- CH2- CH2- CH2- CH3 14 22 34 29 IR the sp3C-H sttechings at 2900-3000 the C-H def are at 1500 and 1350-1400 No other functional group pealks observed mass M+ ion at 1small peak peaks at 1 (M 29) CSH1+ dueto los of ethyl. 57M+- 43) C4H9+ due to loss of propyl group 43(MH -57) C3H+ due to loss of butyl group In alkanes , the propyl and butyl fraginents will be of highest intensity due to their 29 due to C2H5+ st ability. 2H6 n- heptame