Cyclooctatetraene (2R)-2-phenylbutane benzyl chloride cycloocta-1,3,5,7-tetraene

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1 CM S05 Practice Material for xam 1 This is a compilation from relevant problems from last spring s exam 1&2. This is too long! 1. (4 pts) Draw 2 of the following 3: Cross one out or graded in order. N - meta-nitrotoluene sec-butylbenzene (1 3)-1,4-Diphenyl-1,3-butadiene 2. (4 pts) Name 2 of the following 3 with an IUPAC or other non-ambiguous name. Indicate R/S and /Z. Cross one out or graded in order. Cl Cyclooctatetraene (2R)-2-phenylbutane benzyl chloride cycloocta-1,3,5,7-tetraene 3. (20 pts) lectrophilic addition of to dienes. a. 1-Methyl-1,3-cyclohexadiene undergoes both 1,2 and 1,4 addition with. Draw products from reaction with alkene A and then alkene B.: A B A reacts C C - C - C - C - b. 1,4-Dimethyl-1,4-cyclohexadiene also reacts with. Predict the result from reaction: No allyl possible one mole only

2 2 4. (14 pts) Styrene C 8 8 is a conjugated alkene that only produces one product when reacted with. a. Provide structures and arrows to show the mechanism for the reaction below. Provide necessary LP s. C 2 C - Draw styrene and arrow(s) b. Answer these questions: Why is the 1,4 product not formed? C C 1,4 addition product. eaks aromaticity. What is the name of the special position that the cation holds in the intermediate? Benzyl cation Finish the three important resonance structures that show the unusual stability of this cation: C C C C C 5 (10 pts) Given: pka s: N 4 = 9; N 3 = 33;; Cl= --6; 2 = 16. Benzoic Acid, C has pka ~4 and is mostly insoluble in water. ne way to make benzoic acid soluble is to make the salt anion by removing the acidic from the part of the acid. (a) Show the reaction of benzoic acid with any base to make the benzoic acid anion in large quantities and predict the Keq. - - Keq= Benzoic Acid Base of your choice Benzoic Acid anion Conjugate Acid of Base Na used in the CM235 lab. Also, N 3 N 2 - To recover benzoic acid, what acid would be effective in turning the benzoic acid anion back into benzoic acid? Cl As used in the CM235L.

3 6. (10 pts) Below are the three orbitals for the allyl cation C C C C 2 y 3 2 nodes 2 C C C 2 y 2 1 node 2 C C C 2 y 1 0 nodes (a) Color in the p orbitals for ψ 1, ψ 2 ψ 3 Note that ψ 2 has no p orbital on the middle carbon = non bonding -- more stable than anitbonding. (b) Based on the diagram above: ψ 1 has 0 node(s). ψ 2 has 1 node(s). ψ 3 has 2 node(s). (c) Fill in the Molecular orbital diagram with the proper number of pi electrons for the allyl cation system. (d) Based on the picture above, which carbon would you expect to have more degree of positive charge, the C or the C 2? Why? Middle C has a M representation where there is no electron density. lectron density comes from electrons in RBITALS. 7. (18 pts) Diels-Alder Reaction: (a) The following reactions do not occur as written. iefly explain what is wrong with the reactions: Diene not conjugated C 2 CN CN Benzene would not want to break aromaticity.

4 b. The following adducts were prepared by Diels Alder reaction Draw structures for the diene and dienophile. new pi bond former pi bond 4 C 2 C 2 lost pi bond diene C 2 dienophile new pi bond C 2 diene dienophile 8. (8 pts) Classify as aromatic A or not aromatic NA: Si 2 P C - N N Aromatic or NA NA A A NA # of pi electrons (5 pts) xplain briefly why you cannot easily tell the difference between the following two molecules in the 1 NMR. Both have 1 resonances that are singlets between 2 3 ppm. ow could you tell the difference? 13 C NMR would show 2 peaks for hexamethylbenzene, but 3 peaks for hexaacetylbenzene. C= comes near 200 ppm. IR shows C=0 for the hexaacetylbenzene.

5 10 18 pts) Draw resonance structures for the intermediate from the electrophilic addition of to anisole in the para and meta positions. 5 C3 a. Is the addition of in the para or meta position more favorable? BRIFLY explain. Para more favorable than meta because extra resonance structure. Intermediate lower in energy, TS lower for para faster reaction. b. In a kinetics experiment, it was observed that added to anisole 10,000 times faster than to benzene. BRIFLY explain. Anisole is DG pumps electron density into the benzene ring. Benzene more reactive towards 11..(10 points) Provide reagents for the transformations below. A B N - C N - ortho Mg t 2 N 2 N - 1. D 2. acid workup Mg N - A ( ) 2 CCl, AlCl 3 B N 3 2 S 4 C NBS

6 6 12 (6 pts) xplain why the strategies below cannot make the targets: N - 1. Na, aq, RT N - 2. acid workup target molecule a. not made NAS Nitro group must be in the o, p position to stabilize Meisenheimer intermediate. C 2 2 (one mole) Fe 3 b. not made 2 reacts with alkene. 2 not reactive enough to react with benzene breaks aromaticity. 13. (18 pts) Use the following list of reagents to carry out the transformations. Place the letter in the box. There may be more than one way. XPRSS PTIN. You can do all 3 for 2 points each step R choose 2 and cross out one for 3 points each step/ C 2 target molecules C 2 a. KMn 4, 3 b. Cl AlCl 3 c. Cl, AlCl 3 d. 2, Fe 3 e. C 2 C 2 C 2 Cl, AlCl 3 f. NaN 2, N 3 g. Na(aq) C h. 2, Pd i. 2 S 4 (fuming) j. N 3, 2 S 4 k. NBS, peroxides l. C 2 Cl, AlCl 3 m. Cl, Sn n. Na, 2, RT o. Na fusion (heat), aqueous acid 1 C 2 Cl, AlCl 3 2 KMn 4 N - Nitration first does not work! 3 N 3, 2 S 4 1 BuCl, AlCl 3 2 N 2 N 3, 2 S 4 C 3 N 2 3 Sn, Cl 1 C 2 C 2 C()Cl, AlCl , Fe 3 2, Pd

7 14. (10 pts) Draw out a structure that will fit the following NMR spectra. To receive full credit, assign all NMR peaks by drawing an arrow to the peak from the structure. Formula = C C 2 C 2 For extra credit assign as many peaks in the IR as you can. sp 2 C- sp 3 C- C= C=C C- The 13 C NMR is on the next page. You don t have to assign it.

8 8 C 2 C C= pts) Draw out a structure that will fit the following NMR spectra. To receive full credit, assign all NMR peaks by drawing an arrow to the peak from the structure. Formula = C For extra credit assign as many peaks in the 13 C NMR as you can.

9 9 The IR spectrum is here. You don t have to assign it. sp 2 C- sp 3 C- C=C C-