Question

Introduction One of the most important tasks that chemists are faced with is the separation and identification of compounds from mixtures. Gratifyingly, chemists have multiple tools in the quest to identify unknown organic compounds. Infrared spectroscopy is useful for determining functional groups that may exist in a molecule. NMR spectroscopy tells information about the connectivity of C-H bonds, as well as the numbers of and chemical environments of both carbon and hydrogen (as well as many other NMR active nuclei, such as fluorine and phosphorus). Mass spectrometry is also a very powerful tool. Weakly ionizing methods such as MALDI (matrix-assisted laser desorptive ionization) and ESI (electrospray ionization) give molecular ions (gives you the molecular weight of the molecule), while strongly ionizing methods such as El (electron impact) and CI (chemical impact) generate significant fragmentation, which can be used to determine pieces that can fall off of the molecule in the mass spectrometer. Background After years of study, you have recently finished your graduate degree, and have just accepted a position at a small liberal arts college. You show up the first day and look into your new lab, and see that there are six unlabeled containers. The previous occupant has retired at the beginning of the summer, and is apparently somewhere in the middle of the Atlantic Ocean now enjoying retirement (and coincidentally cannot be reached). You're unsure whether or not the compounds are useful, but thankfully the department you're in has a GC-MS, NMR and IR spectrometer, and a number of Meltemps. Between these instruments and online resources, you're confident that you can identify these unlabeled compounds. Although there are no labels, you know based on the previous professor's research that the compounds are likely to be from one of the following classes of compounds: Alcohols and phenols, aldehydes and ketones, amines, or carboxylic acids, amides, and esters. You'll be provided with MS, IR, melting point, boiling point and NMR data on Blackboard. Directions Identify and assign a structure for each of the following unknowns. Make sure to analyze all relevant spectra (label all peaks as appropriate). Explain in detail how you have used your data to determine the identity of the unknown compounds. You will be graded for the success of your interpretation of your results and the logical conclusions you make based on that interpretation. NOTE: The numbers under 'H NMR spectra represent integration values.

Answer 1

The compound 1 is Pentan-1-ol

the compound 2 is pentan-3-one

compound 3 is ethyl acetate

compound 4 is diethylamine

compound 5 is 4-methylbenzoic acid

For the analysis of the structure see the attachment

2 f HO OH H. mole culast weight = 882 b.p=138° Mip. =- 78°C # H 1H - N/M2 % 13C-NMR aus 46 ppm, broad Singlet, 14.0-H a. 63 ppm b: 3.6 ppm, trüplet, QH, CH, bå 32 ppm do 2.1ppm, triplet, 2H, CH2 C. 28 ppm ed: 16-1.9 ppm, multiplet, 4H, 2X CH2 22 ppm to o'q ppm, treipiet, 3 H, GHz eå 14 pm IR 3100-3500 em: 0-H stretching, 2900 cm! Sp3c-h Stret - 26 Sunday 2 he H. b by CH3 moleculare weight = 86.1 b. p. = 102°C; m. p. =-39°C H TH-NMRO 13 C-NMR a : lippm, triplet, 3H, CH₂ tela Roma: 9 ppm bi 2.4 ppm, quartet, 2H, CH2 6:38 ppm li 212 Pom IRE 3000 cñli Sp3. C-H Streetching 1720 cml; Cao Stretching

3 H molecular weight = 881 b. p. = 77.1 m. p. =-83.6°6 HC eltz 'H-NMR: 13 C-NMR a) 21 ppm b) 170 PPM C) 61 ppm 4 ppm a: 2.1 ppm, Singlet, 3 H, CH3 C : 4,1 ppm, quartet, 2H, CH2 d: 1:3 ppm, trúplet, 3H, CH3 d) 16 ppm IR : 2950 cm i SPB CH Stretching, 1750 cm C =0 Street letching 30 Thursday b. ...... 4 a "О molecutart weight eart = 737 b. p=55.5oc, m... =-50°C H-NMRU .1.2 ppm, triplet, 6H, 2X3H, CH3 bi 2.6 ppm, quartiet, 41, 2X2H, CH2 Col5 ppm, Singlet ( broad), 1H, N-H 130 - NMR a: 16 PPM bi 44 PPM IRO ********************* 0 o 3200cm 2900 Color N-H stretching SP3 c-H stretching

5 botha Te moleculare Weights 1361 M.P. = 180-180°C, b. p. = 274-275°C 19 CH₃ it TH-NMRU a: 12.3 ppm, Singlet, IH, COO-H be do 8'1 ppm, aromotie e-H,1HX2 , doublet 7.2 ppm, aromatic C-H, THX2, doublet f: 2:3 ppm, Singlet, 3 H, CH₂ Sunday 13C-NMRO 6% 170 ppm, C % 124 ppm, d: 131 ppm, e: 129 ppm f. 21 ppm, g: : 143 ppm IRO 2800-3000 cmt camborylic acid 0-4 stretching Czo Stretching C=C Streetchi ching 1690 cm 1600 cm