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< Question 35 of 47 - 35. 6.38 points You did not receive full credit for this question in a previous attempt Check my work Click the "draw structure button to launch the drawing utility. Report problem Use the 'H NMR and IR data to determine the structure of the following compound Hint Compound B Molecular formula CH,0 IR absorption at 1718 cm-" IH NMR data: 1.10 (doublet, 6 H), 2.14 (singlet, 3H), and 2.58 (septet, 1 H) ppm Mall dil cript edit structure. com HEML

Answer 1

The molecular formula of the compound = C5H100 Double bond equivalence is DB = Η 2 C- Χ - 2 . +1= 5- Ν +- 2 10 — 2 +1=1 Only one unsaturation is present in the molecule. And IR absorption at 1718 cm-1. So a carbonyl group may be present in the molecule. If the group is aldehyde, -CHO proton should appear at around 9 ppm. But given that no such peak present. Hence keto group is present in the molecule. In H-NMR, a doublet (6H) at 1.10 ppm suggests that this group is coupled with -CH group. A septet at 2.58 ppm suggests that it is coupled with two -CH3 groups and 2.58 ppm suggests that this group is bonded to carbonyl group. a singlet (3H) at 2.14 ppm suggests that presence of isolated -CH3 groups that is bonded to carbonyl group Hence the structure of the molecule is 1.10, d HH2CH/ 2.14, s 2.58, septet CH3 1.10, d 3-methylbutan-2-one