Unformatted text preview: 18-73 (a) First, deduce what functional groups are present in A and B. The IR of A shows no alkene and no carbonyl: the strongest peak is at 1060 cm-1, possible a C-O bond. After acid hydrolysis of A, the IR of B shows a carbonyl at 17 15 cm-1: a ketone. (If it were an aldehyde, it would have aldehyde C-H around 2700-2800 cm-1, absent in the spectrum of B.) What functional group has C-O bonds and is hydrolyzed to a ketone? An acetal (ketal)! ° R'O OR '
'\.. R /, C
A / R-C-R R
=> II B mol. wt. 1 16 C6H 1 ZOZ
There is only one ketone of formula
° C4HgO: butan-2-one. H3C - C - CHzCH3
B II . ..t-.. A must have the same alkyl groups as B. A has one element of unsaturation and is missing only CZH4 from the partial structure above. The most likely structure is the ethylene ketal. Is this consistent with the NMR? �
I 03.9 (singlet,
\ O 4H) o 1.3 (singlet, 3 H) i H3C X �
0 1.6 A - CH3 } 00.9 (triplet, 3H) (quartet, 2H) What about the peaks in the ion at 1 16.
. M S at rnJz 87 and WI? The 87 peak is the loss of 29 from the molecular 'W
H3C rnJz116 CHz-CH3 plus two resonance forms with positive charge on the oxygen atoms .. . .._---I� .. The peak at rnJz 1 0 1 comes from loss of CH3 from the molecular ion at plus two resonance forms with positive charge on the oxygen atoms 1 16. rnJz 101
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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.
- Spring '10