Page 303 - Advanced Organic Chemistry Part A - Structure and Mechanisms, 5th ed (2007) - Carey _ Sundberg

284 are reversible, so the potential reversibility of each step must be taken into account.
A completely reversible mechanism is as follows:
CHAPTER 3
Structural Effects on O O
Stability and Reactivity (1) – k 1
H OH
2 5
PhCCH 3 + OC H k PhCCH 2 – + C 2 5 deprotonation
O –1 O – O
k 2
(2) PhCCH 2 – + PhCH=O PhCHCH CPh addition
k –2 2
O – O fast OH O
(3) C 2 5 k 3 proton equilibration
H OH +
–
PhCHCH 2 CPh PhCHCH CPh + OC H
2 5
k –3 2
OH O O
k
(4) – 4 PhCH=CHCPh + OH + C H OH dehydration
–
PhCHCH 2 CPh + OC 2 H 5 2 5
k –4
Because proton transfer reactions between oxygen atoms are usually very fast, Step 3
can be assumed to be a rapid equilibrium. With the above mechanism assumed, let us
examine the rate expression that would result, depending upon which of the steps is
rate determining. If Step 1 is rate controlling the rate expression would be
–
Rate = k [PhCOCH ][ OC H ]
2 5
3
1
Under these conditions the concentration of the second reactant, benzaldehyde, would
not enter into the rate expression. If Step 1 is an equilibrium and Step 2 is rate
controlling, we obtain the rate expression
–
Rate = k [PhCOCH ][PhCHO]
2
2
which on substituting in terms of the prior equilibrium gives

–
Rate = k K [PhCOCH ][ OC H ][PhCHO]
2 5
2 1
3
since
–
–
[PhCOCH ] = K [PhCOCH ][ OC H ]
1
3
2
2 5
where K is the equilibrium constant for the deprotonation in the first step. If the final
1
step is rate controlling the rate is
OH
–
Rate = k [PhCHCH COPh][ OC H ]
4
2
2 5
The concentration of the intermediate PhCHOHCH COPh I can be expressed in
2
−
terms of the three prior equilibria. Using I for the intermediate and I for its conjugate
base and neglecting [EtOH], since it is the solvent and will remain constant, gives the
following relationships:
I OEt I
−
−
K = and I = K
3
I 3 −
OEt
−