<p> Exam 1 Spring 2003 Chem 3372 1. Describe the characteristic infra red absorption frequencies that would allow you to distinguished between the following pairs of compounds. Give approximate values in cm-1 and shapes where appropriate in your answer. a. O and O Conjugated ketone 1680 cm-1 unconjugated 1710</p><p> b Cyclohexanol and cyclohexene (gift) 3100 cm-1 (unsat C-H broad peak at 3300 cm-1 rounded at bottom small peak at 1640 cm-1</p><p>O O OH and c. HO map of US C=O at 1710 cm-1 broad peak at 3300 cm-1 rounded at bottom and C=O at 1710 cm-1</p><p>2. Using a 60 MHz spectrophotometer, a student observes the following absorption doublet, J = 8 Hz at  ppm a. How many herz from the TMS peak is the absorption? _300 Hz______b. Where would the peak be located (in herz and ppm) in the 600 MHz spectrum? _____3000____Hz _____5 ppm. c. What would be the j value in the 600 MHz spectrum of this sample? _____8______Hz? 3. Describe precisely how you would use 1H NMR spectra to distinguish between the following pairs of compounds. You should compare splitting patterns and/or chemical shifts in your answer.</p><p>O O O and a, O</p><p>Conjugated ester IR around 1710 cm-1 non-conjugated ester IR around 1740 cm-1</p><p>Chemical shift of CH2 group in ethyl in conjugated ester is further downfield thatthat in the non- conjugated ester</p><p>Me Me b. C OMe C Me C and Me OMe IR - not much help H H2 Me 2 NMR signals - two singlets one singlet (OMe), double (CH2(, multiplet, (CH) and doublet Me2C</p><p>4-NMR signals</p><p>O H</p><p> c. O</p><p>IR =1680 cm-1 (conj.ketone) ir = 1710 cm-1 non-conjugated aldehydic hydrogen 2750 cm-1</p><p>4. Using mass spectroscopy state precisely how one could distinguished between the above pair. (hint: Think Benzyl) CH3 Cl Cl As shown on next page 4-chloro-1-methyl forms stable M-1 peak containing chlorine; will have peaks at 125 and 127 in a ratio of 3:1, respectively. On the other hand 4-chlorobenzyl chloride losses chlorine to give 4-methylbenzyl cation which does not contain chlorine atom and thus no chlorine isotope peaks. Chlorobenzyl cation</p><p>CH3 CH2 -H</p><p>Cl Cl</p><p>M = 126 (35Cl) M-1 = 125 m/z 37 M+2 = 128 (Cl ) M+1 =127 ratio of 3:1 since it is chlorine ratio of 3:1 since it is chlorine</p><p>4-methylbenzyl cation</p><p>CL Cl</p><p>H3C H3C</p><p>M = 126 (35Cl) M-35 peak at m/z = 91 (Cl37) M+2 = 128 no chlorine isotopes peaks ratio of 3:1 since it is chlorine 5. Which elements other thanC and H are present in molecules with the following MS peaks.</p><p> a. P and P+ 2 peaks of approximate intensity. _____Bromine______</p><p> b. P and P + 2 peak in the ratio of 3:1, respectively. __Chlorine______</p><p> c. P peak with molecular mass of 121. ___Nitrogen______</p><p> d. A small P+2 peak but slightly larger than the P+1 peak. __Sulfur______</p><p>6 Predict the multiplicity (the number of peaks as the result of splitting) for each underline proton in the following compounds.</p><p>H CH3 doublet</p><p>Cl H Cl H3C H doublet H O</p><p> septet singlet</p><p>7. Predict the number of nmr signals for the following compounds 1,2-dibromoethane __one______CH3CH(CH3)CH2OH __four______</p><p>ClCH2CH(Br)CH3 _four______Me-CH=CH2 (CAREFUL) __four______HNote:O CH2O is a diasterotopic proton Since CH(Br)CH3) is chiral 8. Write a rough NMR spectrum for above compound. A signal around delta 9.5 for CHO that is split into a triplet by the adjacent CH2 group. A signal delta 5 that is split into a doublet by the CHO. Both splittings have the same J value. A signal around 4 for the CH next to O that is split into a septet by the adjacent two methyl groups A signal around 2 adjacent to the CH group that is split into a double. J values to the above two signals are the same. </p>