Molecular Formula: C₃H₆O₂

Mass SpectrumData: UV-VIS Data:

Molecular Ion (M⁺) 74m/z No significant UV absorption > 220 nm.

Fragment 43 m/z (base peak)

Answer the following questions.

11. What is the degree of unsaturation for this compound?

a. 0

b. 1

c. 2

d. 3

12. The peak at 1746 cm^-1 on the IR spectrumtells us the compound contains a(n)

a. hydroxyl group

b. primary amine

c. carbonyl group

d. aromatic ring

13. The sharp intense peak at 1246 cm^-1 is usuallyindicative of _____________ stretch.

a. carbon-carbon single bond

b. carbon-nitrogen single bond

c. carbon-hydrogen single bond

d. carbon-oxygen single bond

14. Based on the information from the IR spectrum, themajor functional group in this compound is a(n) _____________.

a. alcohol

b. carboxylic acid

c. ester

d. aldehyde

15. Based on the ¹H NMR how many sets ofnon-equivalent protons are in this molecule?

a. 1

b. 2

c. 3

d. 4

16. The most deshielded peak in the ¹³C NMR ischaracteristic of the ______________ group.

a. vinyl carbon

b. sp-hybridized carbon atoms

c. carbonyl carbon atom

d. aromatic carbon atoms

17. The base peak at m/z 43 in the mass spectrum is dueto the loss of ___________ group.

a. OCH₃

b. C=O group

c. CH₃ group

d. O=CCH₃

18. The peak at 3.4 ppm in the ¹H NMR is moredownfield than the peak at 2.0 ppm because it is attached to_______________.

a. another carbon atom

b. an oxygen atom

c. more hydrogen atoms

d. more carbon atoms

19. The molecular ion peak at m/z 74 tells us the______________ of the compound.

a. molecular formula

b. melting point

c. molecular weight

d. empirical formula

20. The name and structure of the compound that best fitsthis data is

Transcribed Image Text:

Absorbance / % 0 50- 100- 4000 180 (3H) 160 3000 -M 140 2500 120 2000 nam Warenumbers/cm 100 PPM 1500 (3H) 2 PPM 18 1000 T 500 Absorbance / % 0 50- 100- 4000 180 (3H) 160 3000 -M 140 2500 120 2000 nam Warenumbers/cm 100 PPM 1500 (3H) 2 PPM 18 1000 T 500