Organic Chemistry Peer Tutoring Department - UCI Sites

Organic Chemistry Peer Tutoring Department Chem 51BUniversity of California, Irvine Professor GuanRachel Bauer ([email protected]) https://sites.uci.edu/ochemtutors/Jisoo Kim ([email protected])

Midterm 1 Review Key

1. Rank each species in order of increasing leaving group ability. (7.52 3rd Edition)

a. -OH, F-, -NH2 c. Br-, Cl-, I-

-NH2 < -OH < F- Cl- < Br- < I-

b. H2O, -NH2, -OH d. NH3, H2S, H2O-NH2 < -OH < H2O NH3 < H2O < H2S

Leaving group ability increases as the basicity decreases, which is why the weakest bases arefavored for a, b, and d. For c, leaving group ability increases down the periodic table, as atomicsize increases, and still following the basicity trend. The weaker the base (or stronger theconjugate acid), the better the leaving group.

2. Rank each of the species in order of increasing nucleophilicity. (7.55 3rd Edition)

a. CH3-, -OH, -NH2 c. F-, Cl-, -OH in acetone

-OH < -NH2 < CH3- Cl- < F- < -OH

b. H2O, -SH, -OH in CH3OH d. F-, Cl-, HS- in CH3OHH2O < -OH < -SH F- < Cl- < HS-

When no solvents are involved, nucleophilicity decreases across a row of the periodic table.When a polar protic solvent is involved, nucleophilicity increases down a column of the periodictable. In a polar aprotic solvent, nucleophilicity decreases down a column of the periodic table.

3. Draw the product of each SN2 reaction and indicate stereochemistry. (7.24 3rd Edition)

a.

b.

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SN2 reactions are bimolecular, one step reactions, that only occur via backside attack, whichresults in the inversion of stereochemistry. The strong nucleophiles attack the carbon and theleaving group (the halide) leaves to form the product.

4. Draw the products of each SN1 reaction and indicate stereochemistry when necissary.(7.68 3rd Edition)

a.

b.

SN1 reactions are two step reactions that involve a carbocation intermediate allowing frontsideand backside attack to occur, which results in a mixture of products. The stereochemistry flipsonly on the center where the substitution occurs, but not the other stereocenters like in part B,with that extra methyl group.

5. For each reaction, use the identity of the alkyl halide and nucleophile to determine whichsubstitution mechanism occurs. Then determine which solvent affords the faster reaction. (7.393rd Edition)

a.





SN1 reaction: Tertiary halide; weak nucleophile. Thus, water is the favored solvent as it is polarprotic which favors SN1.

b.

SN1 reaction: tertiary halide; weak nucleophile. Thus, methanol is the favored solvent, beingpolar protic.

c.

SN2 reaction: primary halide; strong nucleophile. DMF is favored as an aprotic solvent.

d.

SN2 reaction: Secondary halide, but strong nucleophile. HMPA is favored as a polar aproticsolvent.

6. What alkyl halide and nucleophile are needed to prepare each compound? (7.42 3rd Edition)a. c.





b. (CH3)3CCH2CH2SH d.

When determining starting material from a product, cut between the nucleophile and the adjacentCarbon bond IF it’s a single bond. Then add a halide to the removed Carbon and balance thecharge with the remaining nucleophile by adding sodium to make it a salt. This was done forparts a-c. However, part d had a triple bond which cannot be “cut”, so we move to the nearestsingle bond, and that triple bonded Carbon becomes our nucleophile. In addition, for these typesof problems we give the longest carbon chain the halide so that our nucleophile is not superbulky.

7. Nicotine can be made when the following ammonium salt is treated with Na2CO3. Draw astepwise mechanism for this reaction. (7.75 3rd Edition)

Deprotonation occurs first, as the nucleophile has a positive charge. Then the nucleophileproceeds via SN2 and attacks the carbon with the halogen bonded to it for an intramolecular





reaction which results in the formation of a ring. The resulting Nitrogen has a positive charge onit again, so another deprotonation step occurs to form the neutral product.

8. How does each of the following changes affect the rate of an E1 reaction? (8.16 3rd Edition)a. Doubling [RX] d. Changing the leaving group from Cl- to Br-

Doubles the rate Rate increasesb. Doubling [B:] e. Changing the solvent from DMSO to CH3OH

No change Rate increasesc. Changing the alkyl halide from

(CH3)3CBr to CH3CH2CH2BrRate decreases

For E1 reactions, more substituted halides react faster, only the concentration of the halideaffects the rate (not the base concentration), weaker bases are favored, better leaving groups reactfaster, and polar protic solvents are favored.

9. What is the major stereoisomer formed in each reaction? (Elimination product only) (8.39 3rdEdition)

a.

b.

For elimination reactions, the trans isomer is always favored as it is more stable than a cis isomerdue to steric hindrance. In this problem, zaitsev’s rule does not apply, so only one product in thatsense is possible





10. Which elimination reaction in each pair is faster? (8.43 3rd Edition)a. b. c.

Regardless of whether it’s an E2 or E1 reaction, tertiary alkyl halides are the fastest reacting sofor a and b, those are the fastest. For part c, looking at the base strength, we can see that this willbe an E2 reaction and thus will be faster with a polar aprotic solvent, like DMSO.

11. Explain the following observation. Treatment of alkyl chloride A with NaOCH2CH3 yieldsonly one product B, whereas treatment of A with a very dilute base in CH3CH2OH yields amixture of alkenes B and C, with C predominating. (8.45 3rd Edition)





With the strong base – OCH2CH3, the mechanism is E2, whereas with the dilute (weak) base,the mechanism is E1. E2 elimination proceeds with an antiperiplanar arrangement of H and X.This just means that the hydrogen and leaving group are on the same plane at 180 degrees fromeach other. In this case, there is only one hydrogen trans to Cl, so the disubstituted alkene isformed (B). In the E1 mechanism there is no requirement for elimination to proceed with antiperiplanar geometry because once the carbocation is formed, there is no sense ofantiperiplanar—everything becomes flat! Because the major product is always the moresubstituted alkene (think about the effect of hyperconjugation), C is the major product under E1conditions.

12. Taking into account anti periplanar geometry, predict the major E2 product formed from eachstarting material. (8.47 3rd Edition)





a.

b.





c.

d.

For E2 reaction to proceed, the H and X must be coplanar and anti of each other to satisfy therequired antiperiplanar geometry. If the example does not present the anti-periplanar position,rotate the bond into another conformation until the H and X are antiperiplanar. One way to checkwhether they’re antiperiplanar is to convert into chair conformation. See if beta-H and X aretrans-diaxial for E2 to proceed. If not, add a chair flip to make them on axial positions.

13. Draw the major product formed when (3R)-1-chloro-3-methylpentane is treated with eachreagent: (a) NaOCH2CH3; (b) KCN; (c) DBU. (8.61 3rd Edition)





First, look at your substrate. If it is a primary alkyl halide, then it is almost certainly going to beSN2 with a good nucleophile. Although sterics aren’t an issue for E2, more substituted halideswill react faster (think about the transition state and the energy of the barrier). For E2 to befavored, it needs a strong, non-nucleophilic base, a beta-hydrogen, and a good leaving group.

14. Draw a stepwise, detailed mechanism for the following reaction. (8.62 3rd Edition)

Determine the reagents. We have a tertiary halide reacting with a relatively weakbase/nucleophile. This satisfies the condition for both SN1 and E1. Besides, SN1 and E1





reactions are always in competition. The first step is identical (formation of C+). In this case, awater molecule is used as a nucleophile to produce two different substituted products dependingon the side of attack. We also have a formation of an alkene product by a weak base Br- attackingthe beta-hydrogen.

15. Name each ether and epoxide (9.41 3rd Edition)

For naming ethers, name both alkyl groups bound to the oxygen and arrange them alphabeticallywith the word “ether” at the end. For symmetrical ethers, use the prefix “di-”; epoxides can benamed in 3 ways: epoxyalkanes- use 2 numbers to prioritize the atoms bonded to the O and add“epoxy” to name the epoxide as a substituent; oxiranes- put the O atom at position 1 and the first





subtitudent at position 2; alkene oxides- mentally imagine the epoxide oxygen with a doublebond and add the word “oxide”.

16. Although alcohol V gives a single alkene W when treated with POCl3 and pyridine, threeisomeric alkenes (X–Z) are formed on dehydration with H2SO4. Draw a stepwise mechanism foreach reaction and explain why the difference occurs. (9.56 3rd Edition)





With POCl3 (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has ahydrogen, only one product is formed and no carbocation arrangement occurs in this reaction asit is operated by E2 mechanisms. With H2SO4, the mechanism of elimination is E1. A 2°carbocation rearranges to a more stable form of 3° carbocation, which has three pathways forelimination.





17. Identify the reagents (a–h) needed to carry out each reaction. (9.78 3rd Edition)

18. Propranolol, an antihypertensive agent used in the treatment of high blood pressure, can beprepared from 1-naphthol, epichlorohydrin, and isopropylamine using two successivenucleophilic substitution reactions. Devise a stepwise synthesis of propranolol from thesestarting materials. (9.79 3rd Edition)





NaH is a strong base, perfect for the deprotonation process. The base reacts with 1-naphthol togive an alkoxide ion. Step 2 describes SN2 reaction substituting an alkoxide ion from step 1 tothe epichlorohydrin as primary halide leaves. Finally, an isopropylamine molecule acts as a baseto attack less substituted carbon of epoxide. Epoxide opens up readily as it has high reactivity tothe open ring and relieves ring strain.



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Organic Chemistry Peer Tutoring Department - UCI Sites