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- Predict the relative acidities of cycloheptatriene versus 1,3,6-heptatriene. Explain.
H
H
H
H
H
H
H
H
H
H
etc
etc
Cycloheptatrienyl anion has 8 electrons in a contiguous cyclic array. This is a 4n system and is
antiaromatic. Even though the electrons can be delocalized it will still not form as readily as
heptatrienyl anion which also has 8 electrons but is not antiaromatic. 1,3,6-heptatriene should be
more acidic.
- Predict the relative acidities of cyclopentadiene versus 1,4-pentadiene. Explain.
+ H
etc
etc
H
H
Cyclopentadienyl anion has 6 electrons in a contiguous cyclic array. This is a 4n+2 system and is
aromatic and will form more readily than the pentadienyl anion. Therefore cyclopentadiene and
the stronger acid.
- Predict the relative acidities of propene versus 1,3-pentadiene. Cyclopropene versus propene.
Explain.
1.3-pentadiene is more acidic than propene due to higher resonance stabilization of the conjugate
base, particularly the middle structure representing a di-allylic character.
H
H
CH 2
CH 2
CH
H C 2
H C 2
base
base
Examination of the HMO diagrams for each carbanion species reveals that the conjugate base of
propene contains a single filled bonding orbital and a single filled non-bonding orbital. Meanwhile,
the conjugate base of 1,3-pentadiene, contains two filled bonding orbitals and a single filled non-
bonding orbital. The key observation is that the difference in energy of the corresponding occupied
propene and pentadiene orbitals is greater for the pentadiene/pentadienyl system.
The cyclopropene with be less acidic since the loss of proton generates antiaromatic system. See
problem 2.
+ H
etc
H
H
H + H H
H
O O
sp
2
?
In order to delocalize the electrons and form an enolate the a carbon must nominally rehybridize
to sp
2
from sp
3
. The preferred bond angles for sp
2
is 120
o
so there is an increase in angle strain
when the enolate forms. This will not be the case for 3-pentanone which should be a stronger
acid.
- Various studies indicate the orbitals used to make carbon-carbon bonds of cyclopropane have
about 80% p character, i.e. sp
4
. What does this suggest about the carbon orbitals used to make
CH bonds and the acidity of cyclopropane versus cyclohexane? (Cylcohexane has about the
same pKa as propane.)
The total amount of s and p orbitals available for bonding (hybridization) is constant. If more p is
used for one orbital less is available for others. (To minimize angle strain, cyclopropane uses more p
for C-C bonds to reduce bonding angle which will minimize strain and optimize overlap, atomic
orbitals will point at each other better). This leaves less p for making C-H bonds and will have sp
2
like hybridization. This will also make the CH bonds more acidic (compared to cyclohexane) since
the electrons in the carbanion will be left in an orbital with more s character.
- Show the product and mechanism for the following reaction.
O (^) O Br 2
/base O^ O
H H H
most acidic (pKa?)
Br-Br
O (^) O
H Br
Br
O (^) O
Br Br
Bromination occurs at most acidic site. It will probably be difficult to stop at monobromination
since the product will be more acidic than the starting material.
- Which compound has more enol at equilibrium? Explain and show enols.
O
O (^) O
O
Only the acyclic enol can adopt a conformation that allows internal hydrogen bonding.
OH
O
O O
H
- Show how benzoic acid can be made from a carbanion.
Br
Mg
ether
O
O
MgBr
+ CO
2
MgBr
H 3
O
O
OH
- Show how the following transformation can be completed.
Br (^) CO 2
H
- Show the mechanism for the following reaction.
HO
S
N
O O O
Br 2
/ Base
Br S
N
O O
The most acidic proton is removed followed by decarboxylation. The carbanion is delocalized on
the SO 2 N group which assists the decarboxylation.
HO
S
N
O O (^) O
Br 2
/ Base
O
S
N
O O (^) O
CO 2
H 2 C
S
N
O (^) O Br Br
H 2
C
S
N
O (^) O
Br
Br
Cl
Li +
or
- Show the mechanisms and products of the following reactions.
Nucleophilic substitution followed by hydrolysis of the acetal product.
Li
EtO
EtO
OEt
H
EtO
OEt
OEt
H
2 ) H
1 )
EtO
OHEt
H
EtO
HOEt
H
H 2
O
H 2 O
EtO
OH 2
H
EtO
OH
H
EtO
OH
H
H
EtO
OH
H
H
O
H
+H
- Show the mechanisms and products of the following reaction
Ortho attach is inhibited by methyl group which cannot leave like a proton.
OH
HCCl 3
CCl 3
CCl 2
H CCl 2
OH-
O
CHCl 2
OH-
O
Cl
Cl
H
O
Cl
Cl
H
OH-
O
Cl
HO
H
O
Cl
HO
H
OH-
O
O H
- Show the mechanism and product of the following reaction.
EtO
O
Cl
NaOEt
O
Epoxide product: this is the Darzens reaction see 10.5.3.
- Show the mechanisms and products of the following sequence of reactions.
OH
CHCl 2
NaOH
H
2
O
H+
OH
CH 2
Cl
OH
Na
O
CH 2
Cl
O
H 2
C
Cl
O
CH 2
OH
Na
O
CH 2
OH
OH
CH 2
OH
H
- Show mechanisms to account for the following.
OH
NaOH
CO 2
OH
KOH
CO 2
O
CO 2
O
CO 2
Sodium ion is smaller with higher charge density and greater electronegativity bind to anion and
CO2 to favor reaction at ortho position. Potassium ion does not coordinate as tightly allowing
raection to occur at para position where there is less steric congestion.
OH O O
O
CO 2
C
O
O
Na
H
OH-
OH- OH
CO 2
H 2 O
H+/H 2
O
OH
CO 2
H
O
H+
Same as first step for 24 except bromination occurs at ethyl group since the enol on that side is more
stable and the reaction is faster.
- Show a mechanism to account for the following results for this reaction and explain. At low
bromine concentrations racemization of starting material is faster than bromination. At high
bromine concentration no racemization of starting material is observed.
Ph
O
Et
Me
H
OH-, Br 2
Ph
O
Et
Me
Br
racemic product
At low bromine concentration, k-1[H 2 O] >> k 2 [Br 2 ], and the proton goes back on faster the
bromine and either side of the carbanion can be protonated leading to racemization of starting
material. At high bromine concentration, k-1[H 2 O] << k 2 [Br 2 ], and the bromination is so faster
there is not time to reprotonate so racemization of starting material can be observed when
stopping the reaction before completion. The product is expected to be racemic under both
conditions.
Ph
H
O
OH-
Br-Br
O
Br
O
Br
k k^2 1
k
Ph
H
O
H
2
O
OH-
k 1 k
- If the reaction is done with deuterated water and OD-, mechanistically what does it mean if
the rate of H-D exchange is equal to the racemization of starting material? If the rate of H-D
exchange is faster then the rate of racemization of starting material? If the rate of H-D
exchange is slower than to the rate of racemization of starting material?
Ph
O
Et
Me
H
OH-, Br 2
Ph
O
Et
Me
Br
racemic product
H
O
OD-
HOD
O (^) D
O
D
O
k rac
= k ex
then D added on each side equally
k ex
k rac
then HOD or D 2
O adds D from same side faster than opposite side
k rac
k ex
then HOD must add H to opposite side faster than D.
