APTER 22 W: 2 DERIVATIVES MELATURE 1. Give the name for each compound (IUPA or common name). Use R/S naming where needed. Structure 2 3 1 3 ame 2,2-dimethylpropanoyl chloride (R)-3-methylpentanoyl bromide Structure 1 2 3 4 5 ame Ethanoic anhydride or Acetic anhydride 3-methylpentanoic anhydride 2. Give the name for each ester. Structure 1 2 3 4 5 ame yclohexyl butanoate 4-methylpentyl ethanoate or 4-methylpentyl acetate Ethyl benzoate 3. Give the name for each amide. Structure F F 2 3 3 ame 5,5-difluoropentanamide -propylethanamide or -propylacetamide 4-bromo-,dimethylbutanamide Page 1
4. Give the name for each compound, using cis/trans, E/Z or R/S naming where needed. Structure ame omoethanoic anhydride or omoacetic anhydride Benzyl ethanoate or Benzyl acetate Benzoic anhydride Structure 3 1 4 2 F 3 1 2 2 3 ame Trans-2-pentenamide or (E)-2-pentenamide -butyl-3-fluoro-methylpropanamide 2-ethoxybutyl benzoate Structure 5 4 3 2 1 I 3 1 2 3 4 5 ame 3-ethyl-2-oxopentanoyl iodide Methyl 6,6,6- trichlorohexanoate,-diethyl-4- phenylpentanamide SPETRSPY 5. For each pair of isomers (A+B, then +D), give two methods that IR spectroscopy can be used to distinguish between the isomers. Give specific wavenumbers where their IR spectra would differ. A B 3 A would have a large broad blobby peak in the IR at 3300 cm -1 representing the stretch, while B would not. The = stretching mode would absorb at ~1715 cm -1 for A (ketone) and 1735 cm -1 for B (ester). 3 2 3 D D would have 2 moderate-weak peaks around 3400 cm -1 representing the - stretching modes, while would lack these peaks. The = stretching mode would be at ~1680 cm -1 for (amide) and ~1715 cm -1 for D (ketone). 6. Acetic anhydride has two strong signals in its IR spectrum, at 1827 and 1755 cm -1. What do these signals specifically represent? They are two modes of stretching involving the carbonyls (symmetric and asymmetric stretch). = symmetric stretches = asymmetric stretches Page 2
7. In each set are listed wavenumbers for the carbonyl stretching modes in each compound s IR spectrum. Explain in detail why in each set, one carbonyl absorbs IR at a higher wavenumber than the other. int: draw resonance structures. Acetyl chloride v = 1799 cm -1, Ethyl acetate v = 1743 cm -1 Ethyl acetate has a lower = wavenumber than acetyl chloride because it s = has more singlebond character. The second resonance structure shown for ethyl acetate contributes to the resonance hybrid significantly, and in that resonance structure the = is a -. Therefore, the = is weakened, lowering the wavenumber. The second resonance structure shown for acetyl chloride does not contribute to the hybrid much (poor orbital overlap in the pi system of the 2p orbitals of the = and the 3p orbital of the ). Therefore, the acetyl chloride = is more double bond like, stronger, and has a higher wavenumber. Butanoic acid v = 1717 cm -1, Butanamide v = 1660 cm -1 The second resonance structures shown for each contribute significantly to the resonance hybrid, but the amide s second resonance structure contributes the most. This is because nitrogen is less electronegative than oxygen, so accommodates a positive charge better (or oxygen destabilizes a positive charge more). Therefore, the amide s carbonyl has more single-bond character (as the resonance with the = as a - contributes more), which is a weaker = and therefore has a lower wavenumber. 8. Predict which carbonyl would absorb infrared radiation at a higher wavenumber. Explain in detail, using appropriate structures. 2 2 Both esters have this sort of resonance contribution: R R The aromatic ester also has additional resonance shown below. As these extra resonance structures have the = as a - bond, the aromatic ester s = has more single-bond character. This causes the aromatic ester s = to be weaker and absorb IR at a lower wavenumber. etc. 9. The 1 MR of DMF has 3 signals at 2.93, 3.03 and 8.00 ppm. Explain why 3 signals are observed. 3 3 3 3 DMF The 2 nd resonance structure shown is a significant contributor for an amide, so the - bond has significant double-bond character. This restricts rotation around the - bond, and causes the 3 s to be in different environments (one is cis to the oxygen, the other is trans). Page 3
10. Predict the number of signals expected in both the 1 and 13 MR spectra of each compound. 2 1 MR: 5 signals ( 2 are different due to amide resonance: 2 signals for 2 ) 1 MR: 6 signals (Each ethyl is different due to amide resonance) 13 MR: 5 signals (ne carbonyl, 4 aromatic) 13 MR: 7 signals REATIVITY TREDS 11. Explain the trends in reactivity toward nucleophilic acyl substitution, using appropriate structures with your answer. Acetyl bromide is more reactive than methyl acetate. Me Me Reactivity in acyl substitution can be described by the energy of the carbonyl species. The ester (methyl acetate) is resonance stabilized (second resonance structure shown above contributes significantly), causing it to start at a lower energy. This causes it to have a higher E a and be less reactive. The acid bromide (acetyl bromide) is not very resonance stabilized (second resonance structure shown does not contribute much as there is poor orbital overlap between the -2p and the -3p), so the acid bromide is a high energy carbonyl species is more reactive. Acetic anhydride is more reactive than methyl acetate. δ+ δ+ Me Me The ester is resonance stabilized, which lowers its initial energy and therefore reactivity. The anhydride is not as resonance stabilized (and is therefore at a higher initial energy and more reactive), as the 2 nd + 3 rd resonance structures shown do not contribute that much. They place a positively charged oxygen atom next to a strongly δ+ center (the carbonyl). c. Ethyl acetate is more reactive than -methylacetamide. Both compounds have contributing resonance structures that stabilize the carbonyl and lower their reactivity (shown above). The 2 nd resonance structure is a larger contributor for the amide than the ester, because nitrogen is less electronegative than oxygen and so can handle a positive charge better. The amide therefore has a lower initial energy and is less reactive. 3 3 Page 4
AID ALIDE AD AYDRIDE REATIS 12. Draw the curved arrow mechanism for each reaction. excess 3 2 1 2 3 3 3 (Steps 2+3 can be reversed) 3 3 2 1 2 3 3 3 3 (Steps 2+3 can be reversed) 13. Give all organic and inorganic products of these reactions. 3 2 + e. 2 2 py. f. py. c. 3 2 py. 2 3 g. 2 3 d. 2 3 h. excess 3 2 3 3 3 2 2 3 Page 5
ARBXYLI AID REATIS 14. Give the curved arrow mechanism for each reaction. + S 3 para-toluenesulfonic acid ( p-ts) 3 + 3 3 3 3 + 3 3 3 3 3 p-ts + + 2 R or 2 R (starting material) 15. Give the major organic product of these reactions. 3 2 2 S 4 2 3 c. 2 S 4 cat. + d. 3 p-ts 3 16. The Fischer esterification reaction does not always produce good yields. What specific methods can be used to drive the reaction to completion? Le hâtelier s Principle can be used: use an excess of reactant (either alcohol or carboxylic acid) or remove the product (perhaps water through a dehydrating agent). Page 6
17. What reagents are needed to synthesize each compound using a Fischer Esterification reaction? 3 + 3 + + (cat.) + + (cat.) ESTER REATIS 18. Give the curved arrow mechanism for each reaction. K 2 + 3 2 S 4 2 + 3 + 2 3 3 2 3 3 + 3 2 Page 7
19. Give the major organic product of these reactions. 3 a (aq) c. Et K 2 2 S 4 2 d. p-ts 2 20. Explain why ester hydrolysis using acid generally has an equilibrium constant of ~1, while ester hydrolysis using base has an equilibrium constant greater than 1. Ester hydrolysis under acidic conditions have a K~1 because there is little difference in bond strengths of the reactants and products ( 2 is similar to R; ester is just as resonance stabilized as a carboxylic acid). R R R R + 2 + + R R + R + R Ester hydrolysis under basic conditions have a K >1 (are favorable), as the reaction involves a transfer of charge from a localized system ( ) to a delocalized system (resonance stabilized carboxylate ion). 21. Give the curved arrow mechanism for this reaction. Mg 3 +, 2 + product 3 3 22. omplete the sequence by filling in the necessary reagents for each synthetic step. S 2 3 pyridine 3 a) 3 2 2 Mg b) +, 2 Page 8
MBIED REATIS 23. Give the major organic product of these reactions. S 2 f. 2 S 4 2 S 4 2 g. 3 a 2 c. 2 S 4 3 2 h. 3 2 S 4 d. S 2 3 2 3 i. p-ts e. a 2 j. 2 S 4 2 AMIDE AD ITRILE REATIS 24. Give all organic and inorganic products of these reactions. 2 con. a 2 + a + + 3 c. con. + 2 2 + con. + 2, heat + 4 + d. con. a 2, heat + a + + 3 Page 9
25. Give the curved arrow mechanism for each reaction. 2 con. a 2 + 3 2 2 2 + 3 con. + 2 3 + 2 2 + 2 3 c. con. a 2, heat + 3 looks like a carboxylic acid 2 = 2 2 2 + 3 AMIDE Page 10
26 continued d. 3 con. + 2, heat 3 + 4 + 3 + 3 3 2 2 3 + 3 3 3 3 2 2 Protonated Amide 2 2 2 + 3 3 3 3 3 + 4 + YDRIDE REATIS 26. Give the major organic products of these reactions. Assume excess inorganic reagent in each. LiAl 4 DIBAL- f. +, 2 +, 2 + 2 LiAl 4 +, 2 2 g. LiAl 4 +, 2 2 2 c. 2 DIBAL- +, 2 h. 3 LiAl 4 +, 2 d. DIBAL- +, 2 i. 2 LiAl 4 +, 2 2 e. LiAl 4 +, 2 j. DIBAL- +, 2 Page 11
SYTESIS 27. Devise a synthesis than converts 3-bromo-1-propene into 3-butenoic acid. (ote the extra carbon in the product; must make a - bond.) Method 1 Method 2 a Mg 2 c. +, 2 con. +, 2 heat 28. Design a synthesis to perform the following transformations, showing all reagents and synthetic intermediates (no mechanisms are necessary). S 2 a) LiAl 4 b) +, 2 ote: product has 1 extra carbon; need to make a - bond a) DIBAL a - Method 1 b) +, 2 Method 2 a) Mg a) LiAl 4 P b) 2 b) +, 2 c) +, 2 c. 2 2 heat Diels-Alder (h 16) +, 2 Page 12
d. 2 e. 2 a) DIBAL - b) +, 2 2 + Imine formation (h 21) 2 KMn 4 -, heat a) S 2 b) 3 2 Benzylic oxidation (h 19) f. 2 3 3 a) 3 2 Mg a) a 2 3 + (cat.) 3 b) +, 2 b) 3 2 I Williamson Ether Synthesis (h 9) Variation: a) S 2 a) 3 2 Mg 2 3 b) R, py. R b) 3 2 X g. Variation Al 3 2 Fe 3 ab 4 Et a) Mg b) c) +, 2 or etc. or + Page 13
h. 3 a) DIBAL - a) MgX r 3 3 b) +, 2 b) +, 2 + i. +, 2 r 3, + 3 P= 2 Wittig Rxn (h 21) Page 14