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Unformatted text preview: l stretch. So to add an O, subtract a methyl group C4H8O. 2(4)+2-8=2/2=1 Lastly there is a peak at 2740 which tells us that the carbonyl is due to an aldehyde. O (1) The following 1H NMR spectrum is of a compound of molecular formula C3H8O. Propose a structure for this compound. First you have a septet that integrates to 1 H and a doublet that integrates to 6 Hs. This is typical of an isopropyl group. Then the peak at 2.5ppm. is a singlet and represents a H on an OH group. OH 2) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C5H12O. 2(5)+2-12=0 so no double bonds or rings. Also there is no –OH or C=O peaks in the IR, so it has to be an ether. Looking downfield you have a triplet and a singlet. For there to be a singlet there must be only a methyl on one side of the ether. Thus giving us the following structure. O Looking at this structure, it explains the presence of the multiplets for the middle two CH2s. And finally the triplet upfield is for the terminal methyl gr...
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This note was uploaded on 02/07/2014 for the course CHEM 2311 taught by Professor Stenberg during the Spring '11 term at Northeastern.

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