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A student attempted to create the following conjugated compound via a double elimination. The mass spectrum showed the correct compound had been formed. And yet, the IR spectrum should definitive evidence that the wrong compound was formed. Answer using the bold letters. What do you think was formed ? Which functional group? A alcohols B alkyne C ketone D another functional group What peak was lacking from the IR spectrum that should have been there E 1500-1800cm1 F 2000-2500cm-1 G 3000-3500cm1 What peak was present in the IR spectrum that confirmed the wrong product H 1500-1800cm1 I 2000-2500cm1 J 3000-3500cm1 Br 2eq NaOH

1. Report your 1H NMR analyses here. Fill in the table. ( inserted below)  2. Draw structure and math the peaks with the protons.  3. Explain any special features:                IR: conjugation, resonance, H-bonding, etc.                 1HNMR: enantiotopic, diastereotopic, H-bonding, cis-                trans, coupling constants, etc.

c. Can the carboxylic acid derivative shown as a reactant in part a be used  make an anhydride? d. Can product A be used to make a carboxylic acid? e. What type of carboxylic acid derivative iss product A? f. Using IR spectroscopy, how would you determine the reaction in part a produced A? (Compare the expected IR spectra for the reactant and product structures to provide the best answer. Be specific.)

24. How can IR spectroscopy help to monitor the following reaction? OH e PCC a. What structural features distinguish the reactant from the product? b. What structural features would NOT help in distinguishing the two structures? C. What would you look for in the IR data of both structures? d. How would you know when the reaction was complete?

2. READ THE DIRECTIONS TO EACH QUESTION CAREFULLY - not following the directions means you get the question wrong! 3. FORMAT YOUR ANSWERS AS DIRECTED-formatting your answers incorrect means you get the question wrong! 4. Below are the IR regions and NMR chemical shifts table for use with the spectroscopy problems: ¹H NMR 12 11 10 R C=O ... OH R 'H 9 8 7 H ppm 6 OH C=C-H 5 NH H-C-X H-C-O -OH -NH X = F, Cl, Br (i.e. electronegative atom) 3 2 H-C-N H-C-S -CH3 -CH₂- C=C-H H-C-C=O -CH H-C-C=C HC-0 0 4000 DIO small range range of values broad peak =C-H N-H 3250- 3300 broad with spikes -3300 3500 -O-H broad-3300 -3400 O-H -CN usually strong N-H CO =C H H 2730- 2820 3000- 2 peaks 3100 1 H -C-H 2850- 2960 O 0 -C-O-H broad-3000 3000 -CEN 2200 -CICH ← 2200 2500 wavenumber, cm-1 2000 C H.C. N н 1680 0.00 0.00 0.00 H 1600- 1660 O C. OR 2 :ں 1600 1500 1000

5. The proxluct of the reaction below gives the IR spectrum shown. Although you are unfamiliar with the rcaction, usc your knowledge of IR spectroscopy to predict a likely product. (Note: the number of carbon atoms in the product is the same as in the starting material.) Page 4 OH Al(OiPr)3 acetone no 4000 c hle belveen sall rlate cas. e Ged LIFE