Problem 1

Compounds of divalent platinum with the general formula

\ce{PtX2(amine)2}

(X =

\ce{Cl2}

\ce{SO4^{2-}}

, malonate, etc.) have attracted great scientific interest because of their antitumour activity. The best-known clinically used compound is

\ce{PtCl2(NH3)2}

, in which platinum has square-planar coordination; it has two geometrical isomers of which only one is antitumour-active. 1.1 Sketch the spatial structures of the two possible isomers. 1.2 How many isomers does

\ce{PtBrCl(NH3)2}

have? Sketch all of them. 1.3 The amine ligands may be replaced by one ligand with two donor atoms (N), e.g. 1,2-diaminoethane (en). Show that

\ce{PtBrCl(en)}

has only one stable structure. 1.4 In aqueous solution the compounds can isomerise through dissociation of a ligand and transient replacement by water;

\ce{Cl^-}

and

\ce{Br^-}

are replaced relatively easily, the amine ligands only with difficulty.

\ce{PtCl2(en)}

reacts with

\ce{Br^-}

(molar ratio 1 : 2) at room temperature. Which compounds form and in what proportion? (Pt–Br and Pt–Cl bonds are equally strong; neglect hydrolysis.) 1.5 Using a chemical equilibrium equation, show why hydrolysis of

\ce{PtCl2(NH3)2}

hardly occurs in blood but does occur inside cells (Cl− concentration is high in blood, low in cells). 1.6 After hydrolysis in the tumour cell a reactive platinum ion bearing two

\ce{NH3}

ligands binds to cellular DNA at a guanine N atom; a second bond to a guanine of the same strand can then form. Show by calculation which of the two isomers of 1.1 can form this bond. (Pt–N distance = 210 pm; distance between DNA bases = 320 pm.)

Step 3 of 4: Hydrolysis equilibrium

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\ce{PtCl2(NH3)2 <=> PtCl(H2O)(NH3)2^+ + Cl^- <=> Pt(H2O)2(NH3)2^{2+} + 2Cl^-}

Analysis

Stepwise loss of $\ce{Cl^-}$ produces the aqua species. In blood ( $[\ce{Cl^-}] \approx 0.1$ M) Le Chatelier pushes the equilibria left and the drug stays inert; inside cells the low $\ce{Cl^-}$ pulls the equilibria right, liberating the reactive aqua-platinum ion that attacks DNA.