CHAPTER 9 HW SOLUTIONS: ALCOHOLS + ETHERS

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1 CAPTER 9 W SLUTNS: ALCLS + ETERS ALCL + ETER NMENCLATURE. Give the UPAC name for each compound. nclude cis/trans or R/S stereochemistry if necessary. Structure C Name -methyl--heptanol -t-butylcyclopentanol (2R,S)--methyl- 2-pentanol Structure Name 4,5-dimethyl--decanol (S,2S)- 2-chlorocycloheptanol trans-4-(-methylbutyl) cyclohexanol (can t use R,S because it s achiral) 2. Give the UPAC or common name for each compound. nclude cis/trans or R/S stereochemistry if needed. Structure C 2 C 2 C Name hexyl isopropyl ether or -isopropoxyhexane dicyclohexyl ether cyclopentyl propyl ether or propoxycyclopentane (no R,S- not chiral) Structure Name (R)-4-cyclopropoxyoctane (S,4R)- 4-t-butoxycyclooctanol,6-diethoxy-2-methylheptane Page

2 REVEW F ALCL SYNTESES. Provide the starting alkyl halide and reagents needed in order to produce each alcohol through a substitution reaction. Na C X C S N 2 X Na S N 2 2 X d. S N X 2 S N WLLAMSN ETER SYNTESS 4. What is the purpose of the sodium hydride (Na) in the following reaction? C C C Na C (hydride) is a strong base and removes the from the alcohol (acid-base). This converts C (poor nucleophile) into C (good nucleophile), so the reaction is faster. 5. Give the curved arrow mechanism for the following reactions. Na C C C C C C C acid-base S N 2 C C Na C Na C + C acid-base C E2 because 2 o or o RX + C Page 2

3 6. Give the major organic product for the following reactions. Na C C d. Na Na C C 2 C 2 e. Na C Na E2 f. Na 7. Using the Williamson Ether Synthesis, show a synthetic route (complete with reagents) that efficiently produces each ether below. C Na C C C Na C Both methods work. Na C C 2 Below doesn't produce the ether well because E2 is the main pathway with a 2 o RX. C C 2 Na C C + 2 o RX. E2 C 2 C 2 C Na C 2 C 2 C Na This doesn't work well: E2 occurs instead. 8. Provide the reagents needed to complete each reaction. C C 2 C 2 C S N Na C C 2 X Williamson Ether Synth. C 2 C Page

4 NTRAMLECULAR REACTNS 9. Give the curved arrow mechanism for the following reaction. Na acid-base ntramolecular S N 2 0. Give the major organic product of each reaction. 5 Na Na 4 2 DEYDRATN REACTNS. Give the curved arrow mechanism for each reaction. nclude the Lewis structure of the acid in your mechanism. con. 2 S 4 S S E con. 2 S 4 S 2 S 4 con. P 4 P 2 P E2 o dont form carbocations Page 4

5 2. For the following reaction, Draw the curved arrow mechanism. con. 2 S 4 S Step 2 Step 2 S 4 Step Use the mechanism to identify two reasons why acid is a catalyst in the dehydration reaction. Note: acid is any strongly acidic source, such as 2 S 4, +, or R 2 +. Acid accelerates the reaction by turning a bad leaving group ( ) into a good leaving group ( 2 ). This makes the catalytic pathway have a lower activation barrier. The acid is not consumed in the reaction. t is used in step, but is regenerated in step. Draw the energy diagram. E Alkene is higher energy than alcohol reactant. 2 Step Acid-base reaction is favorable (check pkas). Draw all probable dehydration products for these reactions, including stereoisomers. Then decide which should be the major product and briefly explain your answer. con. 2 S 4 most substituted (di) and trans has fewer repulsions, so lowest energy. con. P 4 Trisubsituted is lower energy than disu con. 2 S 4 Δ No other possibility. d. con. P 4 Trisub lower E than disu Major put bulkiest groups opposite. (n truth all 4 trisub are similar E.) Page 5

6 YDRDE AND ALKYL SFTS 4. Draw the intermediate formed after each mechanistic step. C C 5 4 C 6 5. What is a possible driving force (or reason) for the rearrangement in: Problem 4a? A secondary carbocation is converted into a tertiary carbocation through the rearrangement, which is more stabilized by hyperconjugation. The motivation for the shift may be to lower the energy of the carbocation. Problem 4c? C C C 2 C C C C C C 4 5 A four-membered ring is converted into a five-membered ring through the rearrangement, which has less ring strain. The motivation for the shift may be to relieve the ring strain (lower the energy of the system). 6. Give the curved arrow mechanism for each reaction. nclude the Lewis structure of the acid. 6 2 d C C 7 C C Reaction con. 2 S 4 S 2 -shift 2 or S 4 S con. 2 S 4 Δ 2 -shift another -shift S 4 Page 6

7 con. P 4 P 2 C methyl shift C 2 P 4 d. con. 2 S 4 Δ S 4 S 2 alkyl shift (ring expansion) = 7. dentify which of the reactions W-Z would synthesize,-dimethylcyclopentene most efficiently (with the fewest competing products). Then explain why the other routes are less efficient. W X BEST Y b a con. 2 S 4 KC(C ) KC(C ) minor + + major Reaction Y would most efficiently produce,- dimethylcyclopentene as there are no significant alternative products. There is only one type of beta-hydrogen, so the E2 reaction can make only one possible product. (Also S N 2 does not compete much when using such a bulky base.) n Reaction W, two different beta-hydrogens are present, con. 2 S Z 4 + leading to a likely mixture of the products shown. This would split likely the yield, lessening the quantity of,-dimethylcyclopentene obtained. Also this mechanism proceeds through a carbocation, so rearrangements could further divert the yield. n Reaction X, a bulky base would more likely remove a beta-hydrogen from position a, creating the wrong alkene isomer as the major product. Removal of a beta-hydrogen from position a is still possible, but the intended alkene would be a minor product. n Reaction Z, the carbocation generated in this mechanism is neighboring a quaternary center, so rearrangements involving a methyl shift are very likely, producing the wrong product.,-dimethylcyclopentene Page 7

8 ALCL REACTN WT X 8. Give the curved arrow mechanism for each reaction. Reaction 2 S N 2 C C C 2 C C hydride shift C S N C ethyl shift 2 (same as above) P, S 2, AND TSYLATE REACTNS 9. Fill in the boxes with the organic product from each reaction. Na C C Ts KN py. N Ts N N N S 2 KN N Page 8

9 CMBNED ALCL REACTNS 20. Give the major organic product for each reaction. Consider plausible rearrangements. i. C S py. C Ts S 2 j. KC 2 C C 2 C don t form carbocations, so no rearrangement occurs k. Ts, py. NaN N d. P l. P NaCN CN Double S N 2 e. S 2 Ts, py. m. Na Single S N 2 f. C Ts py. C Ts n. C C g. Na D o. alcohol goes through S N 2, so inversion occurs D h. P Page 9

10 2. The synthesis of the product shown in reaction L (call it product Q) is best achieved by this method. Explain why methods M- will not effectively synthesize product Q. (L) P NaS C (M) NaS C (N) 2 S 4 2 S C () Ts, py. NaS S C Q Best synthesis of Q: Double S N 2 gives the right stereochemistry of the S group (P inverts the center, S inverts the center again). Reaction (M) doesn t work: is not a good leaving group for S N 2 reactions. Reaction (N) is not efficient: the reaction goes through a flat carbocation intermediate, so attack of 2 S gives both isomers (with S out and S back). Carbocations also often rearrange, and a significant product might be with the S on the same carbon as the methyl group. E may also compete. Reaction () doesn t product Q: single S N 2 produces the wrong isomer. EPXDE REACTNS 22. Give the curved arrow mechanism for the following reactions. Na 2 + C C 2 C 2 C + C C 2 C C C 2 C 2 C 2 C KC 2 C DMF C C 2 C C 2 C C 2 C C 2 Page 0

11 2. Concerning the following two reactions: a b C NaS 2 S C C C Explain the regioselectivity of the reactions. The first reaction involves a negatively charged nucleophile (good Nu) so reacts S N 2-style at the least hindered side of the epoxide, at site The second reaction involves protonation of the epoxide before addition of the δ+ nucleophile ( δ+ ). Since the LG is better now, the LG has already partially left (S N -like). There are partial positive charges on the two carbons of the epoxide, C but the more substituted side ( vs. 2 ) stabilizes a partial carbocation better. δ+ Thus there is a greater d + on the more hindered side, which is where the nucleophile attacks. Explain the stereoselectivity of the reactions. Both reactions occur with inversion of the reacting carbon, S N 2-style, with backside attack. Even though the protonated epoxide in the second reaction has a partial carbocation, it is not a full carbocation. The reaction mechanism is really in between S N (strong d + on the C) and S N 2 (backside attack). 24. Give the major organic product for each reaction. ndicate if a racemic mixture is formed. C C 2 K C C 2 e. C + C C C + (C ) 2 C racemic f. C K 2 C g. d. NaCN 2 CN h. C C 2 C C Page

12 CMBNED ALCL + EPXDE REACTNS 25. Give the major organic product for these reactions. Consider plausible rearrangements and indicate if a racemic mixture is formed. C 2 C Na C C 2 C 2 C e. C 2 C Na 2 f. racemic Ts g. Na C C con. 2 S 4 or Ts or C C C C d. P C C h. C KCN 2 C CN Page 2