A flavor chemist attempted to isolate and identify the major flavor component of a recently discovered wild berry. A sample of berries was pureed in a blender , and the resulting slurry was dissolved in ethyl acetate. The solution was first extracted with 0.1 M HCI, followed by extractions with aqueous sodium bicarbonate. The ethyl acetate portion was concentrated and thin layer chromatography of the resulting oil indicated two compounds. Compound 1 with Rf of 0.2, and Compound 2 with an Rf of 0.6. The two compounds were separated by column chromatography, and the top spot (Compound 2) was identified as the essential oil. Compound 2 had the following physical characteristics: Formula: C11H1403 IR: 3100, 2950, 1735, 800 cm-1 1H NMR: 7.5 ppm, multiplet, 4H 4.5 ppm, septet, 1H 2.1 ppm, singlet, 3H 1.4 ppm, doublet, 6H The nmr signal at 1.4 ppm is a doublet and integrates to six hydrogens. These six hydrogens are bonded to carbon atoms adjacent to carbon atom(s) how many other hydrogens? O A. 3 O B. 2 O C. 1 O D.5 O E. 6

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A flavor chemist attempted to isolate and identify the major flavor component of a recently discovered wild berry. A
sample of berries was pureed in a blender , and the resulting slurry was dissolved in ethyl acetate. The solution was first
extracted with 0.1 M HCI, followed by extractions with aqueous sodium bicarbonate.
The ethyl acetate portion was concentrated and thin layer chromatography of the resulting oil indicated two compounds.
Compound 1 with Rf of 0.2, and Compound 2 with an Rf of 0.6. The two compounds were separated by column
chromatography, and the top spot (Compound 2) was identified as the essential oil. Compound 2 had the following
physical characteristics:
Formula: C11H1403
IR: 3100, 2950, 1735, 800 cm-1
1H NMR: 7.5 ppm, multiplet, 4H
4.5 ppm, septet, 1H
2.1 ppm, singlet, 3H
1.4 ppm, doublet, 6H
The nmr signal at 1.4 ppm is a doublet and integrates to six hydrogens. These six hydrogens are bonded to carbon
atoms adjacent to carbon atom(s) how many other hydrogens?
А. З
В. 2
С. 1
D. 5
O E. 6
Transcribed Image Text:A flavor chemist attempted to isolate and identify the major flavor component of a recently discovered wild berry. A sample of berries was pureed in a blender , and the resulting slurry was dissolved in ethyl acetate. The solution was first extracted with 0.1 M HCI, followed by extractions with aqueous sodium bicarbonate. The ethyl acetate portion was concentrated and thin layer chromatography of the resulting oil indicated two compounds. Compound 1 with Rf of 0.2, and Compound 2 with an Rf of 0.6. The two compounds were separated by column chromatography, and the top spot (Compound 2) was identified as the essential oil. Compound 2 had the following physical characteristics: Formula: C11H1403 IR: 3100, 2950, 1735, 800 cm-1 1H NMR: 7.5 ppm, multiplet, 4H 4.5 ppm, septet, 1H 2.1 ppm, singlet, 3H 1.4 ppm, doublet, 6H The nmr signal at 1.4 ppm is a doublet and integrates to six hydrogens. These six hydrogens are bonded to carbon atoms adjacent to carbon atom(s) how many other hydrogens? А. З В. 2 С. 1 D. 5 O E. 6
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