Question

1. Given the 'H NMR spectrum below and a molecular formula of CH20, provide a structure. Note you must assign all of the signals in the spectrum below to receive full credit. IR: 1600-1650 cm' (multiple peaks) (5 pts.) 3H, S 3H. t 2H, d 2H, d 2H, PPM 2. Given the 'H NMR spectrum below and a molecular formula of CH.0, provide a structure. Note you must assign all of the signals in the spectrum below to receive full credit. IR: 1710 cm (5 pts.) 6H, al 1H, s 2H, d 1H, m PPM

Answer 1

To deduce the structure of any given unknown compound with molecular formula, first step is to determine the Degree of Unsaturation For the calculation of Degree of Unsaturation (DU) following formula can be used DU = Number of carbons +1 - (Number of Hydrogens + Number of monovalent atoms (like halogens)- Number of trivalent atoms (like N and P))/2 Oxygens can be ignored while calculating DU Since in the given questions only carbons, hydrogens and oxygens are present the formula to calculate DU can be further simplified as DU = Number of carbons +1 - (Number of Hydrogens)/2 For CH 120 the DU is = 9+1-(12)/2 = 10-6 = 4 Hence the compound should have 4 double bonds or 4 rings or a benzene ring in its structure In the IR spectrum, peaks around 1400-1600 cm-' is due tayaromatikoaz bestretching. Chemical Formula: CgH120 From the given spectra, the first unknown compound is methyl 4-ethylbenzoate In the NMR spectrum, by seeing 2 doublet peaks at 6-8 ppm is due to para disubstituted benzene ring triplet and quartet splitting peaks at around 2-3 ppm are due to presence of ethyl group 1H NMR values 6.86 7.13 3.81 1.18 6.86 2.72 7.13 PPM

For C5H100 the DU is = 5+1-(10)/2 = 6-5 = 1 Hence the compound should have 4 double bonds or 4 rings or a benzene ring in its structure. is due to carbonyl In the IR spectrum, peaks around 1710 cm C=O stretching. 3-methylbutanal Chemical Formula: C5H100 From the given spectra, the unknown compound is 3-methylbutanal In the NMR spectrum, by seeing a doublet peak and a multiplet peak with 1 proton integral in aliphatic region (1-2 ppm) one can identify the presence of isopropyl group. By seeing a peak at around 10 ppm one can identify the presence of aldehyde group. 1H NMR values 0.94 1.93.32 49.72 0.94 4 PPM