3) Copper forms stable four-coordinate complexes with 1,10-phenanthrolines. 1,10

Question

3) Copper forms stable four-coordinate complexes with 1,10-phenanthrolines. 1,10-phenanthroline (phen) 2,9-dimethyl-1,10-phenanthroline (dmp) 2,9-di-tert-buty-1,10-phenanthroline for [Cu(dmp)2 and a. Explain how you expect the Cu(II)(1) reduction potentials [Cu(phen)2]2, to differ (dmp = 2,9-dimethyl-1 , 10-phenanthroline). b. Explain how you expect the reduction potential of the Culbis(2,9-di-tert-butyl-1,10- phenanthrolinecomplex to compare with those in part (a). Arrange the three complexes from most positive to most negative reduction potential. Which one is hardest to reduce?

Explanation / Answer

First of all, Let the complex of Cu with phen be C1 , with dmp be C2 and with the last one be C3.

Reduction potential indicates the tendency to copper to get reduced (gain electron) and changes its oxidation state from +2 to +1.

In all these complexes N of phenanthrolines forms coordinate bond with Cu(II).

More positive the reduction potential, higher will be the tendency to get reduced and vice versa.

steric hinderance increases from C1 to C2 to C3 as C3 contain bulky tertiary butyl groups at 2 and 9 position whereas C2 contain methyl groups and C1 contain only H at position 2 and 9 .

Large size rings destabilise the complex with Cu(II) and stabilze the Cu(I) state because of steric effects. These large size rings cannot acheive planarity with respect to one another in Cu(I) state, Thus minimizes the steric hinderance and forms stable complexes.

Out of C1, C2, C3 ;most stable is C1 so it will be difficult to reduce as compared to C2 and C3. Also we can say that C3 is easier to reduce because after reduction the complex contain Cu(I) which is the stable one.

So order of their tendency to get reduced will be :

C3 > C2 > C1

Order of reduction potential:

C3 > C2 > C1

C3 has most positive reduction potential whereas C1 has most negative reduction potential.

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