Activity 6: Spectroscopy

Transcription

1 Activity 6: Spectroscopy Reference: Kenkel (p ) and arris, (hp 17-21) Turn in a hardcopy of this activity. If you do not, you will not receive credit. Dropbox submission is not valid. / Score Name (last) (first) For all spectral analysis, show your complete analysis and provide evidence for your proposed structure. Basic Spectroscopy 1. Determine the index of hydrogen deficiency (also known as the unsaturated index) for the following- a) 8 7 N b) 4 4 BrN 2 c) N 2 2 DU = = 6 DU = = 3 DU = = 12 Index of hydrogen deficiency also called unsaturated index = π bonds + rings DU = n 2 i (v i 2) R +DB = X 2 + N +1 where X is a halogen 2 2. alculate the molecular formulas for possible compounds with molecular masses of 136, using the Rule of Thirteen. You may assume that only other atom present in each molecules are carbon and hydrogena) A compound with two oxygen atoms: 136/13 = = 10 + (0.46)13 = = b) A compound with two nitrogen atoms and one oxygen atom: N N 2 c) A compound with five carbon atoms and four oxygen atoms: In the rules of 13, first establish a base formula : M 13 = n + r 13 Where M is the molar mass and n + r is the number of hydrogens The base formula is n n+r UV-Vis Molecular Absorption Spectroscopy 3. Draw the structural formulas that show a UV maximum at- Use the following site to predict the λ max for absorption wavelength for conjugate systems a) An acid, l 2 shows a UV with λ max = 242 nm. 2,4 dichlorobenzoic acid or 3,4-dichlorobenzoic acid b) An aldehyde 8 12, absorbs in the UV with λ max = 244 nm. 2-methyl-1-cyclohexenecarboxaldehyde section 7.13 λ max = Predict the UV maximum for each of the following substances a) 3 3 b) 3 N 3 alculate 249 nm observe 249 nm alculate 275 UV-Vis observe 274 nm 1

2 InfraRed Spectroscopy 5. In each of the following part of the molecular formula is given. Deduce the structure that is consistent with the infrared spectrum. There may be more than one answer ,5-diméthylanisole a) 9 7 N innamonitrile b) 2

3 Nuclear Magnetic Resonance Spectroscopy 6. Determine the organic structure for the following. Provide as much evidence as possible for the structure you proposed.. a) The following NMR spectra are of mono-substituted aromatic hydrocarbon compound with the formula The chemical shift in the aromatic region is found between 7.1 and 7.3 ppm. Proposed a reasonable structures. b) Below are the NMR spectra of two isomeric carboxylic acid, 35l2. Proposed reasonable structures. l l 3

4 7. Using a correlation table for 13 chemical shifts, calculate the δ for the carbons for the following chemicals. 3 a) 3 b) c) Mass Spectroscopy 8. The mass spectrum of an unknown halo-organic liquid shows a molecular ion peak at m/e = 78 with a relative intensity of Suggest a reasonable chemical formula and structure given the relative intensities of the isotopic peaks are as follow- m/e: = Relative intensities: = DU = = = = 43 da Formula: 3 7 l MW = 78 and 80 Note peak 79: A+ 1 (0.79/23.6) * 100 = 3.3 % 3 carbons Note peak 80: A+2 (7.55/23.6) 100 = 32% g Notice the A+2 is 33% intense compare to the molecular ion peak. This reflects the fact that chlorine contains 3 times as much of the 35 l isotope as the 37 l one. That means that there will be 3 times more molecules containing the lighter isotope than the heavier one. Plausible formula: 3 3 carbons 36 amu l 1 chlorine 35 amu Rule of 13: 78/13 = 6, Number of extra hydrogen = = 78 Substitue 3 + for l DU = = = = 43 da Analysis : l = 3 7 l Proposed formula: 3 7 l MW = 78 and 80 4

5 Atomic Absorption Spectroscopy 9. The chromium in an aqueous sample was determine by pipetting 10.0mL of the unknown into each of five 50.0-mL volumetric flasks. Various volumes of a standard containing 12.2 ppm r were added to the flask, followed by diluting each solution to volume.. Unknown, ml Standard, ml Absorbance a) Plot the data b) Derive an equation for the relationship between absorbance and volume of standard. c) alculate the standard deviation for the slope and the standard deviation about regression in (b) d) alculate the ppm r in the sample e) alculate the standard deviation of the results in (d) For Standard Addition - A x A x+s = Where, [X] " i V Rearranging this equation yields, A 0 + V % s [S] f + [X] x+s $ ' = A x + A " x V % [S] s f # & [X] i $ ' i # & A x - Absorbance for analyte A x+s - Absorbance for sample + standard [X] i - oncentration inital of the analyte [S] f - oncentration final of the standard [X] f - oncentration final of the analyte V 0 - Volume initial of unknown V s - Volume added from standard V 0 V 0 A = ml b) s m = c) s r = d) 28.0 ppm r e) 0.22 ppm r " V Graph A 0 + V % " s V % x+s $ ' vs [S] s # V i $ ' yields [X] 0 & # V i as the x - intercept 0 & V T = 50.0 ml, V 0 = 10.0 ml, V s = variable, s = variable, x = to be determine 5

6 Fluorescence Spectroscopy 10. Quinine is one of the best known fluorescent molecules, and the sensitivities of fluoroimeter are often specified in terms of the detection limit for this molecule. The The structure of quinine is shown. Predict the part of the molecule that is most likely likely to behave as the chromophore and fluorescence center. 3 onjugation in the fused aromatic, naphthalene. N N 11. The reduced form of nicotinamide adenine dinucleotide (NAD) is an important and highly fluorescent coenzyme. It has an absorption maximum at 340nm and an emission maximum at 465 nm. Standard solutions of NAD gave the following fluorescene intensities: onc NAD, (umol/l) = Relative intensities, I = a) onstruct a calibration curve for NAD. See graph below b) ompute the least-squares regression parameters of a linear equation for the plot in part (a) See below c) alculate the standard deviation of the slope and the intercept. See below d) An unknown exhibits a relative fluorescence of alculate the concentration of NAD micromol/l e) alculate the relative standard deviation for the result in part (d) f) alculate the relative standard deviation for the result in part (d) if the reading of was the mean of three measurements. 12. Why is spectrofluoroimetry potentially more sensitive than spectrophotometry? igh sensitivity a- Absorbance detects small differences between incident and transmitted light. Many photons must be absorbed to give detectable absorbance. b- Fluorescence is an absolute measurement. Theoretically, a single fluorescing photon can be detected with, for example, a scintillation counter. In practice, stray light severely limits sensitivity. Nonetheless good fluorescence methods can be 10 to 1000 fold more sensitive than good absorbance methods. c- bserving fluorescence is similar to observing stars. ne sees faint light against a dark background. As in astronomy stray light can limit detection sensitivity. This is why fluorescent objects are usually photographed in hoods. 6

7 Spectroscopy ombination Spectral Analysis 13. Use the information provided to propose a reasonable organic structure for the halogenated alkane l 7

8 14. The compound has the formula 48. The 1 NMR spectrum shows a triplet at 9.8 ppm and the signal at 2.4 ppm is a triplet of doublet. Suggest a reasonable structure for the compound.. Error formula should be The compound has the formula 362. The UV spectrum of the compound shows no maximum above 205 nm. The carbon NMR spectrum shows peaks at 14, 60 and 161 ppm. The peak at 161 ppm appears as a possible peak in the DEPT-90 spectrum