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This the solution manual for art of Writing Reasonable Organic Reaction Mechanisms
Typology: Exercises
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1.1. The resonance structure on the right is better because every atom has its octet.
1.2.
NMe (^2)
NMe2 NMe (^2)
NMe 2 NMe 2 NMe (^2)
the second structure is hopelessly strained
Ph
Ph
sp H 3 C CH (^3)
sp 2 sp 3 sp
sp 2
sp 2 sp^2
sp 2
all sp^2 all sp^2
sp 3
sp 3
sp 2 sp 3 sp 3
sp 3 sp 3
sp 2 sp^
2
sp 2 sp 3 sp 2 sp^2
sp
1.4. The O atom in furan has sp^2 hybridization. One lone pair resides in the p orbital and is used in resonance; the other resides in an sp^2 orbital and is not used in resonance.
1.5.
(a) No by-products. C(1–3) and C(6–9) are the keys to numbering.
Ph O
Ph H
4 5 6 7 8 9 10
11
12
13
12
13
10
9 8 2
1
3
(^4 )
7 6 11
(b) After numbering the major product, C6 and Br25 are left over, so make a bond between them and call it the by-product.
1 2 3 4 5 6
7
(^98)
10 11
12 13 14 15 16 17 18
Br
OMe
OMe
Br N
OMe
Br
19
20 21
22 23
Br Br
24 25 (^1) Me Br 2 3 4 5
6
7
8
10
9 19
18
17
(^15 ) 24 14 13 20
(^1211 )
1.6. (a) Make C4–O12, C6–C11, C9–O12. Break C4–C6, C9–C11, C11–O12. (b) Make C8–N10, C9–C13, C12–Br24. Break O5–C6, C8–C9.
1.7. PhC≡CH is much more acidic than BuC≡CH. Because the p K b of HO–^ is 15, PhC≡CH has a p K a ≤ 23 and BuC≡CH has p K a > 23.
reaction on O reaction on N
(b) Esters are lower in energy than ketones because of resonance stabilization from the O atom. Upon addition of a nucleophile to either an ester or a ketone, a tetrahedral intermediate is obtained for which resonance is not nearly as important, and therefore the tetrahedral product from the ester is nearly the same energy as the tetrahedral product from the ketone. As a result it costs more energy to add a nucleophile to an ester than it does to add one to a ketone.
(c) Exactly the same argument as in (b) can be applied to the acidity of acyl chlorides versus the acidity of esters. Note that Cl and O have the same electronegativity, so the difference in acidity between acyl chlorides and esters cannot be due to inductive effects and must be due to resonance effects.
(d) A resonance structure can be drawn for 1 in which charge is separated. Normally a charge-separated structure would be a minor contributor, but in this case the two rings are made aromatic, so it is much more important than normal.
(e) The difference between 3 and 4 is that the former is cyclic. Loss of an acidic H from the γ C of 3 gives a structure for which an aromatic resonance structure can be drawn. This is not true of 4.
(f) Both imidazole and pyridine are aromatic compounds. The lone pair of the H-bearing N in imidazole is required to maintain aromaticity, so the other N, which has its lone pair in an sp^2 orbital that is perpen- dicular to the aromatic system, is the basic one. Protonation of this N gives a compound for which two
equally good aromatic resonance structures can be drawn. By contrast, protonation of pyridine gives an aromatic compound for which only one good resonance structure can be drawn.
(g) The C=C π bonds of simple hydrocarbons are usually nucleophilic, not electrophilic. However, when a nucleophile attacks the exocyclic C atom of the nonaromatic compound fulvene, the electrons from the C=C π bond go to the endocyclic C and make the ring aromatic.
non-aromatic aromatic
(h) The tautomer of 2,4-cyclohexadienone, a nonaromatic compound, is phenol, an aromatic compound.
(i) Carbonyl groups C=O have an important resonance contributor C+ –O–. In cyclopentadienone, this resonance contributor is antiaromatic.
[ Common error alert: Many cume points have been lost over the years when graduate students used cyclohexadienone or cyclopentadienone as a starting material in a synthesis problem!]
(j) PhOH is considerably more acidic than EtOH (pKa= 10 vs. 17) because of resonance stabilization of the conjugate base in the former. S is larger than O, so the S(p)–C(p) overlap in PhS –^ is much smaller than the O(p)–C(p) overlap in PhO–. The reduced overlap in PhS–^ leads to reduced resonance stabiliza- tion, so the presence of a Ph ring makes less of a difference for the acidity of RSH than it does for the acidity of ROH.
(k) Attack of an electrophile E +^ on C2 gives a carbocation for which three good resonance structures can be drawn. Attack of an electrophile E+^ on C3 gives a carbocation for which only two good resonance structures can be drawn.
(f)
Acidity increases as you move down a column in the periodic table due to increasing atomic size and hence worse overlap in the A–H bond
(g)
CO (^) 2Et CO (^) 2Et
The anion of phenylacetate is stabilized by resonance into the phenyl ring.
(h)
EtO2 C CO 2 Et EtO2 C CO 2 Et
Anions of 1,3-dicarbonyl compounds are stabilized by resonance into two carbonyl groups
(i)
The anion of 4-nitrophenol is stabilized by resonance directly into the nitro group. The anion of 3-nitrophenol can't do this. Draw resonance structures to convince yourself of this.
(j)
O (^) More electronegative atoms are more acidic than less electronegative atoms in the same row of the periodic table
(k)
Ph CH 3 Ph H
C(sp) is more acidic than C(sp 3 ), even when the anion of the latter can be delocalized into a Ph ring.
(l)
O O The anion of the latter cannot overlap with the C=O π bond, hence cannot delocalize, hence is not made acidic by the carbonyl group.
(m)
this C atom
this C atom
The C(sp^2 )–H bond on the upper atom is the plane of the paper, orthogonal to the p orbitals of the C=O bond, so the C=O bond provides no acidifying influence. The C(sp 3 )–H bonds on the lower atom are in and out of the plane of the paper, so there is overlap with the C=O orbitals.
(a) Free-radical. (Catalytic peroxide tips you off.) (b) Metal-mediated. (Os) (c) Polar, acidic. (Nitric acid.) (d) Polar, basic. (Fluoride ion is a good base. Clearly it’s not acting as a nucleophile in this reaction.) (e) Free-radical. (Air.) Yes, an overall transformation can sometimes be achieved by more than one mechanism. (f) Pericyclic. (Electrons go around in circle. No nucleophile or electrophile, no metal.) (g) Polar, basic. (LDA is strong base; allyl bromide is electrophile.) (h) Free-radical. (AIBN tips you off.) (i) Pericyclic. (Electrons go around in circle. No nucleophile or electrophile, no metal.) (j) Metal-mediated. (k) Pericyclic. (Electrons go around in circle. No nucleophile or electrophile, no metal.) (l) Polar, basic. (Ethoxide base. Good nucleophile, good electrophile.) (m) Pericyclic. (Electrons go around in circle. No nucleophile or electrophile, no metal.)
1 2
3 4 5 6
7 Br CO 2 Me
MeO OMe CO 2 Me H
Bu MeO^ OMe 3 SnH cat. AIBN
1 2 (^4 ) (^5) + Bu3 SnBr
7 6
(b) Ag+^ is a good Lewis acid, especially where halides are concerned, so polar acidic mechanism is a
each equivalent of dibromoethane. One of the CO 2 Et groups from cyanoacetate is missing in the product and is replaced by H. The H can come from EtOH or HOH, so the CO 2 Et is bound to EtO or HO. The two products differ only in the location of a H atom and a π bond; their numbering is the same. Made: C2–C5, C2'–C6, C2'–C3, C1'–OEt. Broken: C1'–C2', C5–Br, C6–Br.
1 2
3
5 6
EtO2 C CN NaOEt, EtOH; 1/2 BrCH 2 CH2Br
EtO2 C CN +
4
1 (^2 )
4
(^5 )
2' 3' 4'^ EtO2^ C^ OEt
1'
(g) Polar under acidic conditions. The enzyme serves to guide the reaction pathway toward one particular result, but the mechanism remains fundamentally unchanged from a solution phase mechanism. The Me groups provide clues as to the numbering. Made: C1–C6, C2–C15, C9–C14. Broken: C15–O16.
1
(^32)
4
(^5 6 )
9 10 11 12 13
14
15 16^ Me Me
Me Me
Me
enzyme
Geranylgeranyl pyrophosphate A Taxane
Me Me
Me
Me
Me
1 (^32)
4
5 6
(^7 )
9
10 11
13 12
14 15
(h) Two types of mechanism are involved here: First polar under basic conditions, then pericyclic. At first the numbering might seem very difficult. There are two CH 3 groups in the starting material, C5 and C16, and two in the product. Use these as anchors to decide the best numbering method. Made: C1– C14, C2–C12, C12–C15. Broken: C3–C12, O7–Si8.
Me 3 SiO
Li ; warm to RT; aq. NaHCO (^3)
1 2 (^43)
5
(^876)
9
10 11
12
13 14 15
16 5 4 6
7
9 10
3 2
1
11
12
13
14 15 16
(i) The carboxylic acid suggests a polar acidic mechanism. Made: C2–C7, C2–O3, C4–O6. Broken: O3–C4.
Bn N C Ph
Me H
Me (^) Bn N H
O Ph
Me Me
(^1 )
3
4 5
6
7
(^8 1 )
3
8
7
6 4
5
(j) Free-radical mechanism (AIBN). Both Br7 and Sn11 are missing from the product, so they are probably connected to one another in a by-product. H12 appears connected to C10 in the product, as C is the only C that has a different number of H’s attached in S.M. and product. Made: C1–C9, C2–C6, Br7–Sn11. Broken: C6–Br7.
Ph
Br Bu 3 SnH cat. AIBN (^1) Ph
2 3
4 5
(^6 ) 8 9 10
11
10 H (^) H H
9
8
5
6
(^34)
2 1 +^ Bu^3 SnBr
11 7
(k) No acid or base is present, and the reaction involves changes in π bonds. This is a pericyclic mechanism. Use C8 with its two Me groups as an anchor to start numbering. Ozone is a symmetrical molecule, but the middle O is different from the end O’s; it’s not clear which O in ozone ends up attached to which atom in the product. However, it is clear where O4 ends up, as it remains attached to C3. Made: C1–O11, C1–O4, C2–O9, C2–O10. Broken: C1–C2, O9–O10.
1 2 (^43)
5 6
7
8
9,10,11 1 8 7 6
3 5
2 4
10
11
9
(l) Polar mechanism under basic conditions. Again, use C11 with its two Me groups as an anchor to start numbering. C7 remains attached to C8 and O6 in the product. C2 leaves as formate ion; the two O’s attached to C2 in the S.M. remain attached to it in the formate product. O4 is still missing; it’s probably lost as H 2 O, with the two H’s in H2O coming from C8. Made: C5–C8. Broken: C2–C7, O3–O4, O4– C5, C5–O6.
aq. NaOH
1 2
3
4 5 6
8
9 10
11 7
10 9
7
8
5
6
2 1,3 4
1.1. The resonance structure on the right is better because every atom has its octet.
1.2.
NMe (^2)
NMe2 NMe (^2)
NMe 2 NMe 2 NMe (^2)
the second structure is hopelessly strained
1.8. The OH is more acidic (p K a ≈ 17) than the C α to the ketone (p K a ≈ 20). Because the by-product of the reaction is H 2 O, there is no need to break the O–H bond to get to product, but the C–H bond α to the ketone must be broken.
reaction on O reaction on N
(b) Esters are lower in energy than ketones because of resonance stabilization from the O atom. Upon addition of a nucleophile to either an ester or a ketone, a tetrahedral intermediate is obtained for which resonance is not nearly as important, and therefore the tetrahedral product from the ester is nearly the same energy as the tetrahedral product from the ketone. As a result it costs more energy to add a nucleophile to an ester than it does to add one to a ketone.
(c) Exactly the same argument as in (b) can be applied to the acidity of acyl chlorides versus the acidity of esters. Note that Cl and O have the same electronegativity, so the difference in acidity between acyl chlorides and esters cannot be due to inductive effects and must be due to resonance effects.
(d) A resonance structure can be drawn for 1 in which charge is separated. Normally a charge-separated structure would be a minor contributor, but in this case the two rings are made aromatic, so it is much more important than normal.
(e) The difference between 3 and 4 is that the former is cyclic. Loss of an acidic H from the γ C of 3 gives a structure for which an aromatic resonance structure can be drawn. This is not true of 4.
(f) Both imidazole and pyridine are aromatic compounds. The lone pair of the H-bearing N in imidazole is required to maintain aromaticity, so the other N, which has its lone pair in an sp^2 orbital that is perpen- dicular to the aromatic system, is the basic one. Protonation of this N gives a compound for which two
3
Cl 3 C
inductive electron-withdrawing effect of F is greater than Cl
(b)
In general, AH +^ is more acidic than AH
(c)
EtO
Ketones are more acidic than esters
(d)
Deprotonation of 5-membered ring gives aromatic anion; deprotonation of 7-membered ring gives anti-aromatic anion.
(e)
The N(sp 2 ) lone pair derived from deprotonation of pyridine is in lower energy orbital, hence more stable, than the N(sp 3 ) lone pair derived from deprotonation of piperidine.
(f)
Acidity increases as you move down a column in the periodic table due to increasing atomic size and hence worse overlap in the A–H bond
(g)
CO (^) 2Et CO (^) 2Et
The anion of phenylacetate is stabilized by resonance into the phenyl ring.
(h)
EtO2 C CO 2 Et EtO2 C CO 2 Et
Anions of 1,3-dicarbonyl compounds are stabilized by resonance into two carbonyl groups
(i)
The anion of 4-nitrophenol is stabilized by resonance directly into the nitro group. The anion of 3-nitrophenol can't do this. Draw resonance structures to convince yourself of this.
(j)
O (^) More electronegative atoms are more acidic than less electronegative atoms in the same row of the periodic table
(k)
Ph CH 3 Ph H
C(sp) is more acidic than C(sp 3 ), even when the anion of the latter can be delocalized into a Ph ring.
