CAPTER 23 W: ENLS + ENLATES KET-ENL TAUTMERSM 1. Draw the curved arrow mechanism to show the interconversion of the keto and enol form in either trace acid or base. trace - 2 trace 3 + 2 + E1 2 c. trace - 2 d. trace 3 + 2 2 + Page 1
2. Draw all possible enol tautomers for each compound (ignore E/Z stereoisomers). Keto form Enol Tautomers 3. Draw the main enol form for ketones A and B, then answer the questions. internal -bonding A B Keto form of A Main Enol form of A Keto form of B Main Enol form of B Under what reaction conditions are the keto and enol forms at equilibrium? When exposed to an aqueous solution and trace acid or base. Explain why the keto form of A is largely favored at equilibrium over the enol form. Most keto forms are favored over the enol forms because a C= tends to be stronger than a C=C bond (so enthalpy favors the keto form). Note: this is the normal trend (applies to most systems). c. Explain why the enol form of B is largely favored at equilibrium over the keto form. Certain enols are favored over the keto forms, as in this example. This is because: The C= and C=C in the enol are strong because they re conjugated. There is extra stabilization of the system due to the ability to have internal hydrogen bonding. Page 2
4. Draw the main enol forms for each compound below. Then rank compound A-C in order from lowest to highest percentage of enol present at equilibrium. Explain your ranking. (could also draw di-enol) or Keto A Enol A Keto B Enol B Keto C Enol C The enol of B is not conjugated with the carbonyl, so this system follows the typical trend: the keto form is favored, as C= are stronger than C=C. B should have the least # of enol. Both A and C produce conjugated enols, so they will have more enol relative to B. The enol of C also has the ability to internally hydrogen bond, so it would be extra stabilized relative to the enol of A. Least enol: B < A < C Most enol ALPA SMERZATN 5. An optically active solution of compound C (R isomer) loses its optical activity when exposed to either aqueous acid or base. Explain this phenomenon, including a curved arrow mechanism for the loss of optical activity in both aqueous acid and base. C C 3 Exposure of this chiral substance to aqueous acid or base allows it to equilibrate with a flat intermediate (either the enol in acid or enolate in base). These intermediates can react to reform the keto form, but because they are flat they can add the alpha to either face of the alkene, leading to C or its enantiomer. ver time, the system will create equal amounts of each enantiomer as neither is lower energy Flat ENL and the creation of 2 compounds + + is favored by entropy. Therefore, a racemic mixture is formed which is no longer optically active. C 3 Enantiomer 1 C 3 Enantiomer 1 C 3 C 3 (E/Z mix) Flat ENLATE C 3 C 3 2 2 C 3 Enantiomer 2 C 3 (E/Z mix) C 3 Enantiomer 2 Page 3
6. The cis ketone D is isomerized to the trans ketone E with aqueous Na. A similar isomerization does not occur with the cis ketone F. Explain the difference in reactivity. Na 2 Na 2 No reaction C(C 3 ) 3 C(C 3 ) 3 D E F C(C 3 ) 3 Na D E F The cis ketone D has one group axial and another equatorial. Base allows for it to equilibrate with a flat enolate, which can then add the alpha back in such a way that makes the ketone equatorial. This allows for conversion of D into E, which is a lower energy system because all groups are equatorial. The cis ketone F is already in its lowest energy state, with both groups equatorial. Therefore, there is no motivation to isomerize to the trans, as that would be higher energy. ALPA ACDTY 7. Explain the differences in acidity. Use resonance structures with your explanation. Acetaldehyde has a pk a of 17, while acetone has a pk a of 19. Acetaldehyde s alpha is more acidic than acetone s because it has a lower energy conjugate base. 3 C C 2 C C 2 C C The extra methyl in acetone is a EDG, and destabilizes the negative charge in the conjugate base (makes the anion more negative). 3 C C C 3 2 C C C 3 2 C C C 3 Methyl acetate (pk a 25) is less acidic than acetone (pk a 19). The ester s alpha is less acidic than the ketone s. The 3 C C C 3 2 C C 2 C C C 3 C 3 ester has a higher energy conjugate base because the R group is a stronger 3 C C C 3 2 C C 2 C C C 3 C 2 C C C 3 3 EDG (through resonance) than a methyl (through hyperconjugation, a weaker effect). The ester s conjugate base is more destabilized, making it less acidic. Page 4
8. Rank each set of compounds in order from least to most acidic. Explain your ranking, using resonance structures with your explanation. Least acidic 3 < 2 < 1 Most acidic C The conjugate base of 3 is least stabilized, as 3 there is resonance involving only one carbonyl, 1 2 3 making 3 the least acidic. The other s CB s have resonance involving 2 carbonyls, so the charge is more delocalized and the system is lower in energy. owever, 2 is less acidic than 1 because the ester s R group is a strong EDG and destabilizes the anionic CB somewhat. Conj. base of 1 Conj. base of 3 Conj. base of 2 C 3 C 3 C 3 C 3 4 5 6 Conj. base of 4 Conj. base of 5 Conj. base of 6 Least acidic 4 < 5 < 6 Most acidic The conjugate base of 6 is the most stabilized because it has a negative charge delocalized on two oxygen atoms (whereas the CB of 4 and 5 have a negative charge partially on carbon; being less electronegative, carbon does not stabilize the charge as well). The conjugate base of 5 is more stabilized than 4 because it does not have a second methyl EDG like 4. 9. Rank the hydrogen atoms at each position (a-c) from least to most acidic. Explain your answer. a b c Least acidic c < a < b Most acidic b is the most acidic because it has the lowest energy CB (it is resonance stabilized involving 2 carbonyls, which allows the negative charge to be delocalized on both oxygen atoms). c is the least acidic because the CB involves resonance with only one carbonyl (least stabilized). a is more acidic than c because the CB is more delocalized: there is resonance stabilization from both the carbonyl and the aromatic ring (Note: the aromatic contribution is not as stabilizing as a second carbonyl like with b, because the resonance structures place the negative charge on carbon, not oxygen). Page 5
ALPA ALGENATN 10. Give the curved arrow mechanism for these reactions. 2 Na 2 Na + C 3 (s) alpha-halogenation nucleophilic substitution C C 11. ow many equivalents of Br 2 are needed to react completely with one equivalent each of 2-butanone and Na? 2-butanone has 5 alpha atoms, so it would react fully with 5 equivalents of Br 2. 12. Give the major organic product of each reaction, assuming excess reagent in each case. Br 2 Br c. 2 Na K Na Br 2 d. Na 2 Page 6
ALPA ALKYLATN 13. Give the curved arrow mechanism for each alkylation reaction. LiN-iPr 2 (LDA) C 3 Br C 3 N 3 C Br C 3 LiN-iPr 2 (LDA) C 2 =CC 2 N 14. Explain why reaction G gives good yields of the product shown, while the yields are not as good with reaction. n reaction G, the alpha between the carbonyls is so LDA much more acidic than the other alpha s, that LDA G: C 3 reacts there exclusively. There will be no mixture of C 3 enolates formed, so only one product (better yields). n contrast, the alpha s in reaction are of similar LDA : acidity, meaning LDA will deprotonate either side and C 3 C 3 create a mixture of enolates. The yields will not be as good in reaction because the yield will be split among alkylation products from the different enolates. 15. Give the major organic product of each reaction. 1 equiv. LDA C 3 C 2 c. LDA (1 equiv.) C 2 1 equiv. LDA C 3 C 3 S 2 C 3 d. LDA (1 equiv.) Br Page 7
16. Give the curved arrow mechanism that shows how two alkylation products are formed in this reaction. LDA + C 3 N 3 C 3 C 17. Treatment of ketone J with 1 equivalent of LDA followed by C 3 C 2 did not form the desired alkylation product K. J K 3 C 2 C What product might have formed instead? Draw the compound in the space above. Devise a successful synthesis of J to K that uses 1 equivalent of LDA. TMS TMS a) LDA b) C 3 C 2 TMS TBAF K (Alcohol could also be protected with an ester (using RC) as ester alpha are less acidic than ketone alpha, so LDA will form enolate still where we want it. Esters can be deprotected with aqueous acid or base). MALNC ESTER AND ACETACETC ESTER SYNTESES 18. When forming an enolate of the compound below, explain why the base NaC 3 is preferred over Na. C 3 Ester hydrolysis: Na C C 3 3 C 3 C3 C 3 C 3 + C 3 C3 + C 3. Na may react as a nucleophile (reacting at the carbonyl) instead of a base (reacting with the alpha ). Reaction of the Na at the carbonyl can cause hydrolysis of the ester. When NaC 3 is used instead, if it reacts with the carbonyl the ester will remain unchanged. Page 8
19. Draw the curved arrow mechanism for each reaction sequence. You can skip the ester hydrolysis part of the mechanism, but do draw the decarboxylation and keto-enol tautomerism steps. Et NaEt C 3 C 2 Br Et c. +, 2, heat Et Et Et Et Et Et Et Alpha alkylation C 2 Br Ester hydrolysis C 3 - C 2 + 2 Decarboxylation Enol-keto tautomerism Me Me NaMe C 2 c. +, 2, heat Me Me Me Me Me Me C 2 Alpha alkylation - C 2 Decarboxylation + Enol-keto tautomerism Me 2 Ester hydrolysis product Page 9
20. Use the malonic ester synthesis (starting material= RCC 2 C 2 R) or the acetoacetic ester synthesis (starting material= C 3 CC 2 C 2 R) to synthesize each compound. a Key: start with malonic ester (diester) if product is a carboxylic acid, and acetoacetic (ester/ketone) if product is a methyl ketone. (- C 2 ) Et a) NaEt Et b) Br Et Et +, 2 heat b (- C 2 ) Et a) NaEt b) C 2 Br Et +, 2 heat c Me Me a) NaMe Me b) Br a) NaMe Me b) Br Me Me +, 2 heat d Me a) NaMe a) NaMe b) C 3 Me b) Br Me +, 2 heat product Page 10