THE “BLUE” DIMER: WATER OXIDATION CATALYSTPresented By: Margo Roemeling

Mentor: Dr. James K. Hurst

THE BASIS FOR LIFE Photosynthesis is the basis for all

aerobic life on Earth. The process uses water and carbon

dioxide as a source of electrons to make sugar and oxygen as a bi-product.

It involves the use of biocatalysts and energy from sunlight.

PHOTOSYNTHESIS AND THE OXYGEN EVOLVING COMPLEX

Pigments

Oxygen Evolving Complex (O.E.C.)

Electron Acceptor

OXYGEN EVOLVING COMPLEX MECHANISM

The O.E.C. is the water oxidation center of PSII

Has a 4 Mn metal active center

Di-oxo bridges 2 terminal waters Uses 4 photons to lose 4

e- and 4 H+ from waters Upon reaching the S4

state, O2 is given off and 2 waters are taken up to bring it back to its most reduced state.

Light is flashed on the photo reaction center, and every four flashes, O2 is given off.

“ARTIFICIAL BIOSYNTHESIS”

“Artificial biosynthesis” attempts to mimic photosynthetic reactions in simpler systems

In the growing energy crisis, it has become exceedingly important that we find new alternative fuels to replace fossil fuels.

Using these systems, we could mimic the way photosynthesis converts sunlight to energy, and convert sunlight into usable fuel energy.

Artificial biosynthesis could yield hydrogen fuel as well as alcohol fuels.

A SIMPLER SYSTEM

2H2O

O2 + 4H+

WOC

photo-active

element

4[e-]

2H2 (or CH2O + H2O)

4H+ (+ CO2)

hn

THE “BLUE DIMER”

The “blue dimer” is a very effective catalyst for water oxidation (like the O.E.C.)

Water oxidation mechanisms of blue dimer are analogous to the water oxidation mechanisms in the O.E.C.

STRUCTURE

2 Ruthenium metal centers

2 Terminal waters

Oxygen 3 Bipyridine

ligands

“BLUE DIMER” VS. O.E.C.

{3,3}-[(bpy)2Ru(OH2)]2O4+ PSII

“BLUE DIMER” MECHANISM

2 Ru metal active center

Mono-oxo bridge 2 terminal waters Loses 4 e- and 4 H+

from water Upon reaching {5,5}

oxidation state, gives off an O2 and takes up 2 waters bringing it back to its most reduced state.

{3,3}

{3,4}

{4,4}

{4,5}

{5,5}

e-, H+

e-, H+

e-, H+

e-, H+

THE CRUCIAL STEP

Ru O Ru(bpy)2

O OHOH

(bpy)2

{5,5} 4+

Ru O Ru(bpy)2

O OH

(bpy)2

4+

OH

{4,4}

Ru O Ru(bpy)2

OH2 OH2

(bpy)2

4+{3,3}

H2O

O-O

• The two ruthenyl groups are structurally situated to allow joint addition of water to form the peroxo-bound intermediate

• Once formed, the intermediate species is unstable and internal electronic rearrangements lead directly to the final products ({3,3} and O2).

A SIMILAR SYSTEM

RuL33+

RuL32+*RuL3

2+

S2O82-

2SO42-

hv

2H2OWOCn+4

O2 + 4H+WOCn

Net: 2S2O82- + 2H2O 4SO4

2- + 4H+ + O22h

2-4 cycles

Electron acceptor

“Pigment”, Photoreaction center

Water Oxidation Center

THE REACTION: PERSULFATE

S2O8 is often used as a reactant (electron acceptor) to study catalyzed water oxidation by redox-active metal ions.

S2O82-

2SO42-

SURPRISING RESULTS

Persulfate reacts thermally with the blue dimer and partially oxidizes it.

2{3,3} + S2O82- 2{3,4} + 2SO4

2-

We need to understand this reaction and its relationship to the overall photocatalytic system.

QUESTION

Does this reaction involve direct reaction between persulfate and {3,3}? (1) S2O8

2- + {3,3} {3,4} + SO42- + SO4

.-

(2) SO4.- + {3,3} fast 2{3,4} + 2SO4

2-

Net: S2O8 + 2{3,3} 2{3,4} + 2SO42-

Or is it indirect?

S2O82- 2SO4

.-

SO4.- + {3,3} fast {3,4} + SO4

2-

HYPOTHESIS

We can use kinetics to distinguish between these reactions. If direct,

Rate = K[S2O8][{3,3}]

If indirect,Rate = K[S2O8]

METHODS To study the kinetics of

the reaction, a special instrument is used.

Because the reactions are very fast in basic solution, we use a stopped-flow machine.

2 Syringes, one for each solution.

Solutions are quickly mixed and absorption is measured for 100 seconds as reaction is progressing.

STOPPED-FLOW TRACE• This trace

from the stopped-flow machine shows the exponential decay of the reaction which tells us that it is first order.

PH DEPENDENCE

The reaction is very fast in basic solution and very slow in acidic solution.

THE RATE LAW

From the stopped-flow data, we can get the rate law.

If Rate = k[{3,3}][S2O8], this means that it is first order in {3,3}

So, we can treat [S2O8] as a constant making the new rate law: Rate = k’[{3,3}]

FIRST ORDER REACTION

0 100 200 300 400 5000.00

0.05

0.10

0.15

0.20

0.25

k (s

-1)

[S2O82-] (M)

5/8/2010

{3,3} + S2O

8

2- in

2 mM borate, pH 9.2

• Using Rate = k’[{3,3}], we can graph k’ vs. [S2O8]

• And if the reaction is first order in {3,3} like we predicted, we should see a straight line.

FURTHER TESTS• To further test our

hypothesis of direct reaction, we tested the reaction at various ionic strengths.

• If there was direct reaction between S2O8 and {3,3} we would see that as ionic strength increases, the rate of the reaction decreases.

TEMPERATURE DEPENDENCE• Testing the

temperature dependence allowed us to look at the activation energy barrier of the reaction

FUTURE

In the future, we plan to use computer modeling simulations to further study the kinetics of the reaction.

ACKNOWLEDGEMENTSHoward Hughes Medical Institute

Dr. James K. Hurst

Dr. Kevin Ahern

The Beckman Lab Group