Chemistry Labs

Problem 2

Square-planar Pt(II) complexes substitute with retention of stereochemistry, and the rate of substitution of X by Y depends on the ligand trans to X (trans effect): CNX−>HX−>NOX2X−,IX−>BrX−,ClX−>pyridine,NHX3,OHX−,HX2O\ce{CN-} > \ce{H-} > \ce{NO2-}, \ce{I-} > \ce{Br-}, \ce{Cl-} > \text{pyridine}, \ce{NH3}, \ce{OH-}, \ce{H2O}. (a) Draw all stereoisomers of square-planar Pt(py)(NHX3)BrCl\ce{Pt(py)(NH3)BrCl} (py = pyridine). (b) Write reaction schemes, via intermediates, to prepare each stereoisomer of [Pt(NHX3)(NOX2)ClX2]X−\ce{[Pt(NH3)(NO2)Cl2]-} in water starting from PtClX4X2−\ce{PtCl4^{2-}}, NHX3\ce{NH3}, NOX2X−\ce{NO2-}, exploiting the trans effect. (c) Substitution follows Rate=kS[MLX3X]+kY[Y][MLX3X]\mathrm{Rate} = k_S[\ce{ML3X}] + k_Y[\ce{Y}][\ce{ML3X}]; for [py]≫[Pt][\ce{py}] \gg [\ce{Pt}], Rate=kobs[Pt]\mathrm{Rate} = k_{obs}[\ce{Pt}]. For the displacement of ClX−\ce{Cl-} by pyridine in methanol at 25 °C: [py][\text{py}] = 0.122, 0.061, 0.030 mol dm−3^{-3} give kobsk_{obs} = 7.20×10−47.20 \times 10^{-4}, 3.45×10−43.45 \times 10^{-4}, 1.75×10−41.75 \times 10^{-4} s−1^{-1}. Find kSk_S and kYk_Y and decide which pathway dominates at [py]=0.10[\text{py}] = 0.10 mol dm−3^{-3}. (d) A gold nanoparticle of diameter 13 nm carries 90 oligonucleotide groups, 98 % bound to a Pt(IV) complex; a 1.0 cm3^3 vessel contains 1.0×10−61.0 \times 10^{-6} mol dm−3^{-3} in Pt (density of Au = 19.3 g cm−3^{-3}). Calculate the masses of Pt and Au used.
Step 1 of 3: Isomers and trans-effect routes
3 isomers of Pt(py)(NHX3)BrCl;cis:PtClX4X2−→NHX3[Pt(NHX3)ClX3]X−→NOX2X−cis-[Pt(NHX3)(NOX2)ClX2]X−;trans:PtClX4X2−→NOX2X−[Pt(NOX2)ClX3]X2−→NHX3trans-[Pt(NHX3)(NOX2)ClX2]X−\text{3 isomers of } \ce{Pt(py)(NH3)BrCl};\quad \text{cis}: \ce{PtCl4^{2-} ->[NH3] [Pt(NH3)Cl3]- ->[NO2-] cis\text{-}[Pt(NH3)(NO2)Cl2]-};\quad \text{trans}: \ce{PtCl4^{2-} ->[NO2-] [Pt(NO2)Cl3]^{2-} ->[NH3] trans\text{-}[Pt(NH3)(NO2)Cl2]-}
Analysis

Pt(py)(NHX3)BrCl\ce{Pt(py)(NH3)BrCl} has 3 isomers (each ligand can be trans to one of the three others). For [Pt(NHX3)(NOX2)ClX2]X−\ce{[Pt(NH3)(NO2)Cl2]-}: adding NHX3\ce{NH3} first to PtClX4X2−\ce{PtCl4^{2-}} gives [Pt(NHX3)ClX3]X−\ce{[Pt(NH3)Cl3]-}; the incoming NOX2X−\ce{NO2-} substitutes the Cl trans to Cl (Cl labilises trans position weakly, NOX2\ce{NO2} enters cis to NHX3\ce{NH3}). Conversely NOX2X−\ce{NO2-} first gives [Pt(NOX2)ClX3]X2−\ce{[Pt(NO2)Cl3]^{2-}}; the strong trans-effector NOX2\ce{NO2} directs NHX3\ce{NH3} trans to itself, giving the trans isomer.