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Unformatted text preview: The other carbon of the C — O — C unit has a hydrogen whose signal is split into a triplet. This hydrogen must therefore be attached to a carbon that bears a methylene group. These data permit us to complete the structure by adding an additional carbon and the requisite number of hydrogens in such a way that the signals of the protons attached to the carbons of the ether linkage are not split further. The correct structure is ethyl propyl ether. 16.37 A good way to address this problem is to consider the dibromide derived by treatment of compound A with hydrogen bromide. The presence of an NMR signal equivalent to four protons in the aro- matic region at d 7.3 ppm indicates that this dibromide contains a disubstituted aromatic ring. The four remaining protons appear as a sharp singlet at d 4.7 ppm and are most reasonably contained in two equivalent methylene groups of the type ArCH 2 Br. Because the dibromide contains all the carbons and hydrogens of the starting material and is derived from it by treatment with hydrogen...
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This note was uploaded on 09/28/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.
- Fall '01
- Organic chemistry