Unformatted text preview: nd butanoic acid..
OH OH Only 1-butanol works, because there are no peaks corresponding to C=C and C=O. 2) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound.
phenylacetone, benzoic acid, acetophenone, benzyl alcohol, benzaldehyde First you can eliminate benzoic acid and benzyl alcohol because there is no –OH peak. Second you can eliminate benzaldehyde because there is no peak at 2740 cm-1 (aldehyde peak). That leaves phenylacetone and acetophenone.
O Acetophenone is the answer because the carbonyl peak is at 1700 cm-1 and a simple ketone like that on phenylacetone would absorb at a higher energy. 3) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound.
2-ethynylcyclohexanone, 2-methyl-2-cyclohexenone, acetophenone, cyclohexylmethyl alcohol, 4-ethylcyclohexanone. First, 2-ethynylcyclohexanone can be eliminated because there is no peak for a carbon carbon triple bond.
Second, cyclohexylmethyl alcohol is eliminated...
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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.
- Spring '11