PS5_soln - Problem Set 5 132 points April 30, 2009 17-5 +10...

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Problem Set 5 132 points April 30, 2009 17-5 +10 pts The sharp band at 2250 cm –1 is characteristic of a nitrile or alkyne group. No evidence is found in the 1600-1450 cm –1 range for an aromatic structure. Thus, the bands at about 3000 cm –1 are probably due to aliphatic hydrogens. The pair of absorptions between 1425 and 1475 cm –1 is compatible with one or more alkane groups. No evidence for amine or amide groups is seen. It therefore seems likely that the compound is an alkyl nitrile. Propanenitrile (CH 3 CH 2 C ± N) boils at 97°C and seems to be a probable candidate. 17-12a,d a . 1 2 22 2s i n () c p s c d n n ± ² = ³´ µ¶ ·¸ ¹º 2500 c c nm n == 1 2 2500 1.03 i n 4 5 ( ) 2.00 p d = ²³ ´ 821.17 p dn m = 1 2 2500 1.03 i n 6 0 2.00 p d = ´ 571.4 p m = d. 1 2 2500 1.003 i n 4 5 ( ) 2.8 p d = ´ = 652.64nm at 3000cm -1 , 870.19 nm at 1000cm -1 , 2610.56 nm +10 20-13b Table 20-3 reveals that for every 100 35 Cl atoms there are 32.5 37 Cl atoms. Thus, (M + 2) + /M + = (1 ² 98/100) + (1 ² 32.5/100) = 1.30 (M + 4) + /M + = (0 ² 98/100) ² (1 ² 32.5/100) = 0.32
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20-14 (a) Because all conditions except accelerating voltage are constant, Equation 20-9 can be abbreviated to ( m /z) s = K / V s and ( m /z) u = K / V u where the subscripts s and u designate standard and unknown respectively. Dividing one of these equations by the other gives the desired relationship. (/) / / ss u uu s mz KV V == (b) 69.00 0.965035 ( / ) 71.50 u u = = (c) The approximately half-integral m / z value suggests that the ion being studied in part (b) was doubly charged. This conclusion is in agreement with the fact that the molecular mass of the unknown is 143. The second conclusion is that the unknown must contain an odd number of nitrogen atoms. 20-18 +12 pts (4 ea) m / z = 84 due to 35 Cl 2 C + m / z = 85 due to 35 Cl 2 13 CH 2 + m / z = 86 due to 37 Cl 35 Cl 12 CH 2 + m / z = 87 due to 37 Cl 35 Cl 13 CH 2 + m / z = 88 due to 37 Cl 2 12 CH 2 + 20-19a-c +20pts +5 a. chemical ionization because the molecular ion is present and you can determine the molecular weight of the parent molecule +5 b.
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This note was uploaded on 04/12/2010 for the course ECE 380 taught by Professor Prof.a during the Spring '10 term at University of Illinois at Urbana–Champaign.

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PS5_soln - Problem Set 5 132 points April 30, 2009 17-5 +10...

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