Transcript Proposing a mech

Minimum criteria a proposed
mechanism should meet
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It must be consistent with all of the experimental data.
It must make experimentally testable predictions that, if not
verified, would prove it false.
If several mechanisms are consistent with the known data,
prefrerence is given to the least complicated one.
In any multistep mechanism, individual steps should be
unimolecular or bimolecular.
Each step in a mechanism should be energetically
favorable.
Each step in a mechanism should be chemically
“reasonable”.
Where possible, ad hoc additions to a mechanism as
devices to explain away inconsistencies with experimental
facts should be avoided.
Proposing a Mechanism
Consider the following substitution reactions:
-
CH3 Br + OH
-
(CH3 )3 CBr + OH
1
CH3 OH + Br
(CH3 )3 COH + Br
rate = k CH3 Br
-
-
1
OH
1
rate = k (CH3 )3 CBr
Observation: In both substitution reactions, a C-Br bond is being broken in the
reactant, a C -O bond is being formed to create product, and bromide is being
expelled.
Question: Do both substitution reactions share a common m echanistic pathway
given that they share all of the above transformations?
Answer: No. Why not?
Proposing a mechanism utilizing the
experimental rate law
Consider the following substitution reactions:
-
CH3 Br + OH
-
(CH3 )3 CBr + OH
CH3 OH + Br
1
-
(CH3 )3 COH + Br
rate = k CH3 Br
-
-
1
OH
1
rate = k (CH3 )3 CBr
Observation: In both substitution reactions, a C-Br bond is being broken in the
reactant, a C -O bond is being formed to create product, and bromide is being
expelled.
Question: Do both substitution reactions share a common m echanistic pathway
given that they share all of the above transformations?
Answer: No. Why not?
Appr oach: As a first approximation, use the experimentally derived rate law to
identify what is transpiring in the slow step (rds) of the m echanism .
Proposing a mechanism utilizing the
experimental rate law - cont’d.
Overall:
CH3 Br + OH-
CH3 OH + Br
-
1
rate = k CH3 Br
-
1
OH
Procedure: Use the species that appear in the rate law as the reactants in the rds
of the mechanism . U se the orders of the species as the coefficients in the rds. C om pare
what you have at this point to the overall transformation. What m ore needs to be done?
Answer:
(rds)
1 CH3 Br + 1 OH-
1 CH3 OH + 1 Br
-
The above transform ation is a one step transformation in which both reactants in the
overall reaction participate in the slow step (rds).
Overall:
-
(CH3 )3 CBr + OH
(CH3 )3 COH + Br
-
1
rate = k (CH3 )3 CBr
Note: A lthough hydroxide is a reactant in the overall transformation, it does not
participate in the slow step (rds) of the mechanism . O nly the C-Br bond of the other
reactant undergoes cleavage.
Step #1:
Question:
(rds) 1 (CH3 )3 CBr
?
How exactly does the C-Br bond break?
Hint: To create the desired product, a C-O bond needs to form. What would m ake
carbon undergo attack by the hydroxide ion that's been waiting in the wings (so to
speak)?
Proposing a mechanism utilizing the
experimental rate law - cont’d.
Overall:
-
(CH3 )3 CBr + OH
(CH3 )3 COH + Br
-
1
rate = k (CH3 )3 CBr
Note: A lthough hydroxide is a reactant in the overall transformation, it does not
participate in the slow step (rds) of the m echanism. Only the C-Br bond of the other
reactant undergoes cleavage.
Step #1:
Question:
(rds) 1 (CH3 )3 CBr
?
How exactly does the C-Br bond break?
Hint: To create the desired product, a C-O bond needs to form. What would make
carbon undergo attack by the hydroxide ion that's been waiting in the wings (so to
speak)?
(rds) 1 (CH3 )3 CBr
1 (CH3 )3 C+ + 1 Br-
Answer:
Step #1:
Question:
What m ust happen next to complete the overall transformation?
Proposing a mechanism utilizing the
experimental rate law - cont’d.
-
Overall:
(CH3 )3 COH + Br
(CH3 )3 CBr + OH
-
1
rate = k (CH3 )3 CBr
Note: A lthough hydroxide is a reactant in the overall transformation, it does not
participate in the slow step (rds) of the m echanism. Only the C-Br bond of the other
reactant undergoes cleavage.
Step #1:
Question:
(rds) 1 (CH3 )3 CBr
?
How exactly does the C-Br bond break?
Hint: To create the desired product, a C-O bond needs to form. What would make
carbon undergo attack by the hydroxide ion that's been waiting in the wings (so to
speak)?
1 (CH3 )3 C+ + 1 Br-
Answer:
Step #1:
(rds) 1 (CH3 )3 CBr
Question:
What m ust happen next to complete the overall transformation?
Answer:
Step #2:
+
(fast) 1 (CH3 )3 C + 1 OH-
1 (CH3 )3 COH
Acceptable m echanism
Step #1:
(rds) 1 (CH3 )3 CBr
Step #2:
+
(fast) 1 (CH3 )3 C + 1 OH
Overall:
-
1 (CH3 )3 CBr + 1 OH
1 (CH3 )3 C+ + 1 Br1 (CH3 )3 COH
1 (CH3 )3 COH + 1 Br
-
Proposing a mechanism - Example #1
1
Overall rxn:
2 NO 2 Cl
2 NO 2 + 1 Cl2
Question: Can this be a one step mechanism?
Answer:
No. Why not?
rate = k NO 2 Cl
Proposing a mechanism - Example #1
1
Overall rxn:
2 NO 2 Cl
2 NO 2 + 1 Cl2
rate = k NO 2 Cl
Question: Can this be a one step m echanism ?
Answer:
No. Why not?
Approach: As a first approximation, let the experimentally derived rate law
dictate what is transpiring in the rds of the m echanism .
(rds)
1 NO 2 Cl
?
Proposing a mechanism - Example #1
1
Overall rxn:
2 NO 2 Cl
2 NO 2 + 1 Cl2
rate = k NO 2 Cl
Question: Can this be a one step mechanism?
Answer:
No. Why not?
Approach: A s a first approxim ation, let the experim entally derived rate law
dictate what is transpiring in the rds of the mechanism.
(rds)
Answer:
1 NO 2 Cl
Let the reactant in the rds decompose to form one of the products
in the overall rxn.
(rds)
Question:
Hint:
?
1 NO 2 Cl
1 NO 2 + 1 Cl
What next?
Does atomic chlorine appear in the overall rxn?
Proposing a mechanism - Example #1
Approach: A s a first approxim ation, let the experim entally derived rate law
dictate what is transpiring in the rds of the mechanism.
(rds)
Answer:
1 NO 2 Cl
Let the reactant in the rds decompose to form one of the products
in the overall rxn.
(rds)
Question:
?
1 NO 2 Cl
1 NO 2 + 1 Cl
What next?
Hint:
Does atomic chlorine appear in the overall rxn?
Answer:
U se atom ic chlorine as a reactant in the next step.
(fast)
1 Cl + 1 NO 2 Cl
1 NO 2 + 1 Cl2
Acceptable m echanism
(rds)
(fast)
1 NO 2 Cl
1 Cl + 1 NO 2 Cl
Overall rxn:
2 NO 2 Cl
1 NO 2 + 1 Cl
1 NO 2 + 1 Cl2
2 NO 2 + 1 Cl2
Proposing a mechanism - Example #2
Overall rxn:
1 NO + 1 CO2
1NO 2 + 1 CO
2
Observation:
rate = k NO 2
0
CO
Question:
What is the significance of the fact that the experimentally derived
rate law is zero order in CO?
Proposing a mechanism - Example #2
Overall rxn:
1 NO + 1 CO2
1NO 2 + 1 CO
2
Observation:
rate = k NO 2
0
CO
Question:
What is the significance of the fact that the experim entally derived
rate law is zero order in CO?
Answer:
CO is not a reactant in the rds of the m echanism.
Approach:
Keep in m ind that all steps in a mechanism are balanced chemica
equations. Use this fact and the fact that the rate law is second order in nitrogen
dioxide to obtain the rds.
Proposing a mechanism - Example #2
Overall rxn:
1 NO + 1 CO2
1NO 2 + 1 CO
2
Observation:
rate = k NO 2
0
CO
Question:
What is the significance of the fact that the experim entally derived
rate law is zero order in CO?
Answer:
CO is not a reactant in the rds of the m echanism.
Approach:
Keep in m ind that all steps in a mechanism are balanced chemical
equations. Use this fact and the fact that the rate law is second order in nitrogen
dioxide to obtain the rds.
Acceptable m echanism
(rds)
1NO 2 + 1 NO 2
1 NO + 1NO 3
(fast)
1NO 3 + 1 CO
1 CO2 + 1NO 2
1NO 2 + 1 CO
1 NO + 1 CO2
Overall rxn:
Proposing a mechanism - Example #3
Overall rxn:
Observation:
2 NO + 1 O2
rate = k NO
2 NO 2
2
1
O2
Question:
Can this be a one step mechanism?
Answer:
No. Why not?
Proposing a mechanism - Example #3
Overall rxn:
2 NO + 1 O2
2 NO2
2
rate = k NO
Observation:
1
O2
Question:
Can this be a one step mechanism?
Answer:
No. Why not?
Approach:
In order to avoid a step in the mechanism that invokes
a three-body collision, one step needs to be a rapid equilibrium (a reversible
reaction in which the rate of the forward reaction equals that of the reverse
reaction).
(fast)
(rds)
Overall:
Accetable mechanism
kf
1 NO + 1 O2
1 NO3
kr ev
1 NO3 + 1 NO
2 NO + 1 O2
kr ds
2 NO2
2 NO2
Proposing a mechanism - Example #3
Overall rxn:
2 NO + 1 O2
Observation:
rate = k NO
(fast)
(rds)
2 NO 2
2
1
O2
Accetable m echanism
kf
1 NO + 1 O2
1 NO 3
k r ev
1 NO 3 + 1 NO
2 NO 2
kr ds
Overall:
2 NO + 1 O2
Pr oof:
rate = kr ds NO 3
kf NO
kf NO
1
1
kf NO
1
NO
1
1
= k r ev NO 3
1
O2
kr ev
rate = kr ds
1
O2
2 NO 2
1
= NO 3
1
O2
1
NO
k r ev
2
rate = k' NO
O2
1
1
Proposing a mechanism - Example #4
Overall equation:
Expt. #
NO
o
2NO(g) + 1 Cl2 (g)
Cl2
o
2 NOC l(g)
Measured initial rate
-6
1
0.100M
0.100M
2.53x10 M/sec
2
3
0.100M
0.200M
0.200M
0.100M
5.06x10 M/sec
-5
1.01x10 M/sec
4
0.300M
0.100M
2.28x10 M/sec
Determine:
rate = k NO
-6
-5
x
Cl2
y
Find the numerical value of the specfic rate constant, k.
Propose a plausible m echanism to account for the above transformation.
Proposing a mechanism - Example #4
NO
Expt. #
o
Cl2
Measured initial rate
o
-6
0.100M
0.100M
3
0.100M
0.200M
0.200M
0.100M
2.53x10 M/sec
-6
5.06x10 M/sec
-5
1.01x10 M/sec
4
0.300M
0.100M
2.28x10 M/sec
1
2
-5
Solve for x using data from experim ents #3 & #1:
-6
x
x
1.01x10 M/sec
0.200M
(2.00) = 3.99
=
0.100M
2.53x10-6 M/sec
(
)
x= 2
Solve for y using data from experim ents #2 & #1:
-6
y
y
5.06x10 M/sec
0.200M
=
(2.00)
= 2.00
0.100M
2.53x10-6 M/sec
(
)
2
NO
rate = k
y= 1
1
Cl2
To solve for the num erical value of k, use the data from any one of the experim ents:
-6
2
1
From Expt. #1: 2.53x10 M/sec = k(0.100M) (0.100M)
-6
k=
2.53x10 M/sec
(0.100M)2 (0.100M)1
=
-3
2
-1
2.53x10 (M sec)
Proposing a mechanism - Example #4
Overall equation:
Observation:
(fast)
(rds)
Overall:
Pr oof:
2 NO + 1 Cl2
rate = k
NO
2 NOCl
2
Cl2
1
Accetable m echanism
kf
1 NO + 1 Cl2
1 NOCl2
kr ev
1 NOCl2 + 1 NO
2 NO + 1 Cl2
2 NO Cl
rate = kr ds NOCl2
kf NO
kf NO
1
1
Cl2
Cl2
1
1
NO
1
1
= kr ev
NOCl2
1
1
= NOCl2
kr ev
rate = kr ds
2 NOCl
kr ds
kf NO
1
Cl2
1
NO
kr ev
2
rate = k' NO
1
Cl2
1
A Mechanism Exercise
Consider the following overall reaction:
2 H2 + 2 NO
1 N 2 + 2 H2 O
Observation: When the concentration of NO is fixed and that of hydrogen is doubled,
the rate of reaction doubles. When the concentrations of NO and hydrogen are both
doubled, the rate of reaction increases by a factor of eight.
Question: Which of the following mechanisms is(are) consistent with the
experimental rate law?
1 N 2 + 2 H2 O
Mechanism #1:
(rds)
2 H2 + 2 NO
Mechanism #2:
(fast)
(rds)
2 NO
1 N 2 O2 + 1 H2
1 N 2 O2
1 H2 O2 + 1 N 2
(fast)
1 H2 O2 + 1 H2
2 H2 O
Mechanism #3:
1 H2 + 1 NO
(rds) 1 H2 NO + 1 NO
(fast)
1 N 2 O + 1 H2
(fast)
1 H2 NO
1 H2 O + 1 N 2 O
1 H2 O + 1N 2