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May 1, 2024
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Chem 236, S2024
Problem Set 3
M. Burke, J. Berger, S. Roth, D. Bergandine
a) Draw the S
N
2 product for the reference reaction, as well as for the two aminoalkylbromides.
b) Draw a reaction coordinate diagram for each reaction on one plot. Clearly label each starting material, transition state and product. Make sure to accurately reflect the ΔΔ
G
‡
of each transition state in your reaction coordinate diagram.
1) Researchers at Florida State University were interested in studying the rate of intramolecular S
N
2 reactions of aminoalkylbromides with varying change lengths. In order to quantify this trend, the authors calculated the Gibbs Free Energy of the transition state for a simple reaction between ethylamine and 1-bromopropane. After normalizing this value to zero, the authors could caulate the change in Gibbs Free Energy (
ΔΔ
G
‡
) for each reaction compared to the reference and use that as a metric for determining the rate.
NH
2
Br
N
H
H
2
N
Br
HN
H
2
N
Br
N
H
-HBr
-HBr
-HBr
Reactant
ΔΔ
G
‡
NH
2
Br
0 (this is the reference reaction)
H
2
N
Br
0.34
H
2
N
Br
-2.93
Reaction A
Reaction B
Reaction C
For this problem, the actual heights of each TS do not matter so long as they are correctly placed relative to each other based on the ΔΔ
G
‡
, which represents the change in Δ
G
‡
from the reference
1 point per S
N
2 product, 3 points total
2 points for RCD (+1 for general structure, +1 for correct transition state differences)
Question 1: 12 points total
Chem 236, S2024
Problem Set 3
M. Burke, J. Berger, S. Roth, D. Bergandine
c) Predict the relative rates of each reaction (1 = fastest, 3 = slowest). Explain your ranking using the data provided. Why can we use ΔΔ
G
‡
as a metric for rate in this context?
d) The authors then wanted to investigate the effects of different alkyl substituents on the ΔΔ
G
‡
of the reaction. They found that increasing the steric bulk on the methylene beta to the amine led to a greater change in ΔΔ
G
‡
. For each of substrates below, clearly draw the transition state of the S
N
2 reaction. Please make sure to correctly capture the three dimensionality of the carbon being attacked, as well clearly depict which bonds are being formed and broken.
e) Explain the functional impact that increasing steric bulk on the beta methylene has on the rate of reaction. What does increasing the steric bulk do to the molecular geometry of the substituted methylene carbon?
H
2
N
ΔΔ
G
‡
Br
H
3
C
CH
3
H
2
N
Br
H
3
C
CH
3
H
2
N
Br
H
3
C
CH
3
CH
3
CH
3
-9.03
-9.75
-11.06
1 = Reaction C
, 2 = Reaction A
, 3 = Reaction B
Since all starting materials are roughly the same in energy, the less energy required to reach the transition state, the faster the reaction.
Me
Me
N
H
H
Br
H
H
N
H
H
Br
H
H
N
H
H
Br
H
H
Me
Me
Me
Me
Me
Me
The functional impact is increasing rate of ring formation. This is known as the "Thorpe-Ingold Effect"
More steric bulk on the methylene carbon decreases the bond angle between the two smaller substituents. When those substitutents are “pushed” together it promotes cyclization.
As the nucleophile approaches and the leaving group leaves, the carbon being attacked planarizes so that all groups are in plane and 120 degrees from each other. the nucleophile and leaving group are 180 degrees opposite each other
+2 (+1 correct order, +1 explanation connencting activation energy to rate)
+3 (+1 per transition state)
+2 points (+1 function, +1 explanation)
2) Authors were interested in studying the effects of different solvents on the reaction between p
-nitrophenylacetate and different hydroxymate nucleophiles.
O
O
CH
3
O
2
N
p-nitrophenylacetate
R
O
N
H
HO
NaOH
O
O
2
N
Na
R
O
N
H
O
H
3
C
O
Benzohydroxamic Acid (BHA)
R
Acetohydroxamic Acid (AHA)
H
3
C
O
Salicylhydroxamic Acid (SHA)
OH
Chem 236, S2024
Problem Set 3
M. Burke, J. Berger, S. Roth, D. Bergandine
a) Using curved arrows, draw the mechanism for the reaction between p-nitrophenylacetate and a generic hydroxamate nucleophile, clearly showing all intermediates. Label which species is the nucleophile and which is the electrophile. Using resonance structures, explain why the p-
nitrophenoxide is a good leaving group. (hint: this reaction proceeds via a carbonyl addition-elimination mechanism).
R
O
N
H
O
-
OH
H
R
O
N
H
O
O
O
CH
3
O
2
N
O
O
CH
3
O
2
N
O
HN
R
O
O
O
2
N
Na
R
O
N
H
O
H
3
C
O
nucleophile
electrophile
O
O
2
N
O
N
O
N
O
O
O
O
negative charge on oxygen can be delocalized into the ring, and the nitro group is strongly electron withdrawing by induction.
"tetrahedral intermediate"
4 points:
+3 mechanism (+1 correct nucleophile, +1 correct electrophile, +1 tetrahedral intermediate)
+1 resonance
Question 2: 10 points
b) The authors observed the following trends for the rate of reaction of each nucleophile in different solvents. Classify the solvents based on their polarity and whether or not they are protic or aprotic. Using this classifications, explain the trends observed for the rate of reaction of each nucleophile. Be sure to discuss the impact of the R group.
c) Predict the order for rates of reaction for each of the above nucleophiles in the following solvents: water and n-hexanes. Explain why you ranked the compounds in the order you did.
d) Explain the functional importance of the R group and how it affects of the rate of reaction.
Chem 236, S2024
Problem Set 3
M. Burke, J. Berger, S. Roth, D. Bergandine
All three solvents are polar aprotic. However, the DMF and DMSO can participate in hydrogen bonding, causing the SHA (which has a hydroxyl group, an H-bond acceptor and donator) to solubilize better and thus have a faster rate. The most nonpolar (acetonitrile) leads to the fastest rate for the most nonpolar R group (benzene in BHA).
Water: SHA > AHA > BHA Hexanes: BHA > AHA > SHA. The most nonpolar R group will favor the nonpolar solvent, and vice versa.
The importance of the R group is that it can tune the solubility of each compound towards a specific solvent. The more soluble the nucleophile is in the reaction, the more it can interact with the electrophile in solution, leading to a faster rate.
3 points:
+0.5 per assignment of solvents (x3)
+1.5 explanation
+2 (+1 per solvent)
+1 explanation
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When the following ester is treated with lithium lodide in DMF, a carboxylate ion is obtained:
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