Solutions_Manual_for_Organic_Chemistry_6th_Ed 287

Solutions_Manual_for_Organic_Chemistry_6th_Ed 287 - 1 3- 1...

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Unformatted text preview: 1 3- 1 8 continued (c ) b H d H H c Br (d) H a \ / C=C / Cl H b This compound has six types of protons. H and H are diastereotopic. a,b,c = & 7 .2; e f d & 5 .0; e,f = & 3 .6 == H c b This compound has three types of protons. Ha and Hb are diastereotopic. a,b = & 5 -6 ; c = & 7 -8 H \ 13-19 (a) CH3CH2 - O-H • • .. � . + H-B "" --- CH3CH2 - �-H + .B + H I') ". - --- CH3CH2 - �: H 1 . H 1 + H-8 ( b) CH 3CH2- O-H . •• ') " +:B - --- CH3CH2-R: .. � + H-B "" --- CH3CH2- O: . + :B 1 3 -20 The protons from the O H i n ethanol exchange with the deuteriums in D 20. Thus, the O H in ethanol is replaced with O D w hich does not absorb in the NMR. What happens to the H? It becomes H OD w hich can usually be seen as a broad singlet around & 5 . 25. (If the sol vent is C DCI3, t he i mmiscible H OD w ill float on top of the solvent, out of the spectrometer beam , and its signal will be missi ng.) R OH CH3CH2 OH + + D 20 - R OD + H OD D 20 2H 3H H OD Peaks from OH are usually broad because of rapid proton transfer. This is especial l y true in water, or HOD. I I I J I TMS I, I 10 9 8 7 6 5 & ( ppm) 4 I II 3 I I I 2 o I I 280 ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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