Page 1 C HAPTER 22 HW: CO 2 H D ERIVATIVES NOMENCLATURE 1. Give the name for each compound (IUPAC or common name). Use R/S naming where needed. Structure Name 2,2-dimethylpropanoyl chloride (R)-3-methylpentanoyl bromide Structure Name Ethanoic anhydride or Acetic anhydride 3-methylpentanoic anhydride 2. Give the name for each ester. Structure Name Cyclohexyl butanoate 4-methylpentyl ethanoate or 4-methylpentyl acetate Ethyl benzoate 3. Give the name for each amide. Structure Name 5,5-difluoropentanamide N-propylethanamide or N-propylacetamide 4-bromo-N,N- dimethylbutanamide Cl O Br O CH 3 1 2 3 O O O O O O 1 2 3 4 5 O O O O 1 2 3 4 5 O O F NH 2 O F O N H N O Br CH 3 H 3 C
Transcript
Page 1
CHAPTER 22 HW: CO2H DERIVATIVES
NOMENCLATURE
1. Give the name for each compound (IUPAC or common name). Use R/S naming where needed.
Structure
Name 2,2-dimethylpropanoyl chloride (R)-3-methylpentanoyl bromide
Structure
Name Ethanoic anhydride or Acetic anhydride 3-methylpentanoic anhydride
2. Give the name for each ester.
Structure
Name Cyclohexyl butanoate 4-methylpentyl ethanoate or 4-methylpentyl acetate
Ethyl benzoate
3. Give the name for each amide.
Structure
Name 5,5-difluoropentanamide N-propylethanamide
or N-propylacetamide 4-bromo-N,N-
dimethylbutanamide
Cl
O Br
OCH312
3
O
OO O
O O
123
4 5
O
O
O
O1
23
45
O
O
F NH2
OFO
N
H
N
O
Br
CH3
H3C
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4. Give the name for each compound, using cis/trans, E/Z or R/S naming where needed.
Structure
Name Bromoethanoic anhydride or Bromoacetic anhydride
Benzyl ethanoate or Benzyl acetate
Benzoic anhydride
Structure
Name Trans-2-pentenamide or (E)-2-pentenamide
N-butyl-3-fluoro-N-methylpropanamide
2-ethoxybutyl benzoate
Structure
Name 3-ethyl-2-oxopentanoyl
iodide Methyl 6,6,6-
trichlorohexanoate N,N-diethyl-4-
phenylpentanamide
SPECTROSCOPY
5. For each pair of isomers (A+B, then C+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 would have a large broad “blobby” peak in the IR at 3300 cm-1 representing the OH stretch, while B would not.
• The C=O stretching mode would absorb at ~1715 cm-1 for A (ketone) and 1735 cm-1 for B (ester).
• D would have 2 moderate-weak peaks around 3400 cm-1 representing the N-H stretching modes, while C would lack these peaks.
• The C=O stretching mode would be at ~1680 cm-1 for C (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).
BrO
O O
Br O
O O
O
O
O
NH2F N
O
CH3
12
3 O
O
O
12
34
O
O
I12
34
5Cl
OCH3Cl
Cl
ON
O
12
34
5
OO
OH
O
CH3A B
NCH3
O
CH3
NH2
O
C D
O
OO
O
OO
C=O symmetric stretches
C=O asymmetric stretches
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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. Hint: draw resonance structures. a. Acetyl chloride v = 1799 cm-1, Ethyl acetate v = 1743 cm-1
Ethyl acetate has a lower C=O wavenumber than acetyl chloride because it’s C=O has more “single-bond character.” The second resonance structure shown for ethyl acetate contributes to the resonance hybrid significantly, and in that resonance structure the C=O is a C-O. Therefore, the C=O 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 C=O and the 3p orbital of the Cl). Therefore, the acetyl chloride C=O is more double bond like, stronger, and has a higher wavenumber.
b. 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 C=O as a C-O contributes more), which is a weaker C=O and therefore has a lower wavenumber.
8. Predict which carbonyl would absorb infrared radiation at a higher wavenumber. Explain in detail,
using appropriate structures.
Both esters have this sort of resonance contribution:
The aromatic ester also has additional resonance shown below. As these extra resonance structures have the C=O as a C-O bond, the aromatic ester’s C=O has more “single-bond character.” This causes the aromatic ester’s C=O to be weaker and absorb IR at a lower wavenumber.
9. The 1H NMR of DMF has 3 signals at 2.93, 3.03 and 8.00 ppm. Explain why 3 signals are observed.
The 2nd resonance structure shown is a significant contributor for an amide, so the C-N bond has significant “double-bond character.” This restricts rotation around the C-N bond, and causes the CH3’s to be in different environments (one is cis to the oxygen, the other is trans).
Cl
O
Cl
O
O
O
O
O
OH
O
OH
O
NH2
O
NH2
O
O R
O
O R
OO
O
O
O
H N
O
CH3
CH3
H N
O
CH3
CH3DMF
O
O
O
O
etc.
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10. Predict the number of signals expected in both the 1H and 13C NMR spectra of each compound.
1H NMR: 5 signals (NH2 are different due to amide resonance: 2 signals for NH2)
1H NMR: 6 signals (Each ethyl is different due to amide resonance)
13C NMR: 5 signals (One carbonyl, 4 aromatic)
13C NMR: 7 signals
REACTIVITY TRENDS
11. Explain the trends in reactivity toward nucleophilic acyl substitution, using appropriate structures with your answer. a. Acetyl bromide is more reactive than methyl acetate.
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 Ea 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 C-2p and the Br-3p), so the acid bromide is a high energy carbonyl species is more reactive.
b. Acetic anhydride is more reactive than methyl acetate.
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 2nd + 3rd 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 N-methylacetamide.
Both compounds have contributing resonance structures that stabilize the carbonyl and lower their reactivity (shown above). The 2nd 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.
O
NH2 N
O
Br
O
Br
O
OMe
O
OMe
O
O
OO
O
OO
O
OO
OMe
O
OMe
O
δ+ δ+
O
O
O
O
N
O
N
OCH3
H
CH3
H
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ACID HALIDE AND ANHYDRIDE REACTIONS
12. Draw the curved arrow mechanism for each reaction.
13. Give all organic and inorganic products of these reactions.
a.
O
Clexcess
NH3
O
NH2
O
Cl NH3
O
NH2
O
Cl
N HH H
NH3
O
Cl
NH H
(Steps 2+3 can be reversed)
1 2 3
b.O
O
O CH3OH
N
O
OCH3
O
O
O O
OCH3
CH3OH
O
O
O
OH CH3 N
O
OCH3
H(Steps 2+3 can be reversed)
1 2 3
a. H2OO
Cl
O
OH+ HCl e.
O
O
OH2O
O
OH2
b.OH
O
Cl
py.
O
O
NH
Cl
f.Ph
O
O
O
PhOH
py. Ph
O
O
O
O
PhNH
c.
Br
OCH3CH2OH
py. OCH2CH3
O
NH
Br
g.NH2
O
O
O HN
O
NH3O
O
NH2O
Cl
d. NH
O
NH3 Cl
h.excess
CH3CH2NHCH3
O
O
O O
N
O
OCH3
CH3CH2NH2CH3
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CARBOXYLIC ACID REACTIONS
14. Give the curved arrow mechanism for each reaction.
15. Give the major organic product of these reactions.
16. The Fischer esterification reaction does not always produce good yields. What specific methods can
be used to drive the reaction to completion?
Le Châtelier’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).
a.O
OH
S OH
OO
+
para-toluenesulfonic acid ( p-TsOH)
CH3OHO
OCH3
O
OCH3
O
OH
H+ O
OH
H
CH3OHOH
HOO
CH3
HCH3OH
OH
HOOCH3
H+
O
HOOCH3
HH
O
OCH3
H O
OCH3
HCH3OH
b. p-TsOH
HO OH
O
O
O
HO OH
O H+
HO OH
OH
O
OH
OH
HROH
(starting material)
O
OH
OH H+
O
OH
OH2
O
OH
ROH or H2O O
O
a.CH3CH2OH
H2SO4OH
O
OCH2CH3
Oc.
H2SO4
OHOH
OO
O
b.cat. H+
Ph OH
O
OH O
O
d.CH3OH
p-TsOHOHO OH3CO
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17. What reagents are needed to synthesize each compound using a Fischer Esterification reaction?
ESTER REACTIONS
18. Give the curved arrow mechanism for each reaction.
a.O
OCH3
O
OH+ CH3OH + H+ (cat.)
b. O
O
OHHO
O
+ H+ (cat.)
a.O
OOH
O
O+
H2O
KOH
OO
OH
O
OOH O O
OH
OO
O
H
OCH3
O
H2O
H2SO4OH
O
+ CH3OHb.
Ph OCH3
O
OH2
H+
Ph OCH3
OH
Ph OCH3
HO OH
H OH2
Ph OCH3
HO OH
H+
Ph O
HO OHCH3
HPh OH
OH
Ph OH
OH OH2
Ph OH
O
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19. Give the major organic product of these reactions.
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 (H2O is similar to ROH; ester is just as resonance stabilized as a carboxylic acid).
Ester hydrolysis under basic conditions have a K >1 (are favorable), as the reaction involves a transfer of charge from a localized system (OH–) to a delocalized system (resonance stabilized carboxylate ion).
21. Give the curved arrow mechanism for this reaction.
22. Complete the sequence by filling in the necessary reagents for each synthetic step.
a.NaOH (aq)
OCH3
O
O
Oc.
KOH
H2O
OEtO OO
b.H2O
H2SO4O Ph
O
HO Ph
Od.
H2O
p-TsOH
OO
HOO
HO
a.
b. H+, H2O
O
OCH3
OHMgBr
O
OCH3
O
OCH3
O O H+
product
OH
O
Cl
O
OCH3
O OHSOCl2 CH3OH
pyridine
a) CH3CH2CH2MgBr
b) H+, H2O
R OR
OH+
+ H2OR OH
O+ ROH
R OR
O+ OH
R O
O+ ROH
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COMBINED REACTIONS
23. Give the major organic product of these reactions.
AMIDE AND NITRILE REACTIONS
24. Give all organic and inorganic products of these reactions.
a.SOCl2
OH
O
Cl
Of.
O
OHHO
H2SO4
O
O
O
Ob.
H2O
H2SO4 O
OHg.
OCH3
O
H2ONaOH
O
O
c.O
OH CH3CH2OH
H2SO4 O
Oh.
OH H3C
O
OH
H2SO4
O
O
d.
O
OH b. CH3NH2
a. SOCl2O
NH
CH3
OHO
OHi.
p-TsOH
O
O
e.O O
H2ONaOH
O O
j.O
O
H2O
H2SO4
OH
O
OH
a.
O NH2
con. NaOH
H2OO O
+ Na+
+ NH3 c. N
Ocon. H+
H2ONH2
O
HO+
b.CN con. H+
H2O, heat + NH4+
O OHd. CN
con. NaOH
H2O, heat CO
O
+ Na+
+ NH3
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25. Give the curved arrow mechanism for each reaction.
a.con. NaOH
H2O+ NH3
NH2
O
O
O
+ NH3NH2
O
O
O
OH NH2O
O
HNH2
O
OH
b.NH
Ocon. H+
H2OHO NH3
O
NH
O
HO NH3
O
H+
HO NH2
OH
NH
OHOH2 NH
OHO
HH OH2
NH
OHOH
H+
N
OHOH
HH
C con. NaOH
H2O, heatc.
O
O + NH3
N
PhC NHO–
PhC
N
OH
HOH
PhC
N
O PhC
N
O
HOH
PhC
NH2
O
PhC
N
OH
–OHH
looks like a carboxylic acid
=Ph NH2
O
AMIDE
–OH
H H
Ph NH2
O OH
PhC
O
OH
NH2
PhC
O
O+ NH3
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26 continued
HYDRIDE REACTIONS
26. Give the major organic products of these reactions. Assume excess inorganic reagent in each.
d.
CH3
CN
con. H+
H2O, heat
CH3
O
OH + NH4+
CH3
OH
O + NH4+
H+
CH3
CN
CH3
CN
H
OH2CH3
CN
H
O
H
HOH2
CH3
C
N H
OH
H+
CH3
C
N H
OH
H
CH3
C
NH2
O
H2O
H
CH3
CNH2
OH
OH
HOH2
CH3
CNH2
OH
HO H+
Protonated AmideCH3
CNH3
OH
HO
CH3
O NH3
OH
H
a.a. LiAlH4
b. H+, H2O
O
O+ PhOH
H
OHH
f.a. DIBAL-H
b. H+, H2O
O
O
O
H
b.a. LiAlH4
b. H+, H2ONH2
O
NH2
g.O
CNa. LiAlH4
b. H+, H2O
OH
CH2NH2
c.a. DIBAL-H
b. H+, H2ONH2
O
H
O
h.
a. LiAlH4
b. H+, H2O
O
OCH3
OH
d.CN a. DIBAL-H
b. H+, H2O H
O
i.NH2
O
OHO
a. LiAlH4
b. H+, H2O NH2
OH
e.OH
O a. LiAlH4
b. H+, H2O OHj.
a. DIBAL-H
b. H+, H2OCN
O
H
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SYNTHESIS
27. Devise a synthesis than converts 3-bromo-1-propene into 3-butenoic acid.
(Note the extra carbon in the product; must make a C-C bond.)
28. Design a synthesis to perform the following transformations, showing all reagents and synthetic intermediates (no mechanisms are necessary). a.
b.
Note: product has 1 extra carbon; need to make a C-C bond
