Question

Consider the given complex ion $[Pt(bn)_2]^{2+}$, where bn stands for 2, 3-diamino butane. Find the value of n (where n is the total number of stereoisomers).

Answer 1

6

Answer 2

The complex ion is $[Pt(bn)_2]^{2+}$, where bn stands for 2,3-diaminobutane. Platinum(II) typically forms square planar complexes. The ligand 2,3-diaminobutane is a bidentate ligand.

First, let's analyze the stereochemistry of the ligand, 2,3-diaminobutane: The structure of 2,3-diaminobutane is CH$_3$-CH(NH$_2$)-CH(NH$_2$)-CH$_3$. It has two chiral centers (the two carbon atoms bonded to the -NH$_2$ groups). Like tartaric acid, 2,3-diaminobutane exists in three stereoisomeric forms:

1. (2R,3R)-2,3-diaminobutane (chiral): Let's denote this as L$_{\text{RR}}$.
2. (2S,3S)-2,3-diaminobutane (chiral): This is the enantiomer of (2R,3R)-bn. Let's denote this as L$_{\text{SS}}$.
3. meso-2,3-diaminobutane (2R,3S or 2S,3R) (achiral): This form has an internal plane of symmetry. Let's denote this as L$_{\text{meso}}$.

Now, we consider the formation of the square planar complex $[Pt(bn)_2]^{2+}$ using these different forms of the ligand. We need to find all possible distinct stereoisomers.

Let's list the possible combinations of the ligands and the resulting complex's chirality:

1. Complex with two (2R,3R)-bn ligands: $[Pt(\text{L}_{\text{RR}})_2]^{2+}$ Since L$_{\text{RR}}$ is a chiral ligand, the complex formed by two such ligands will be chiral. (1 distinct isomer)

2. Complex with two (2S,3S)-bn ligands: $[Pt(\text{L}_{\text{SS}})_2]^{2+}$ Since L$_{\text{SS}}$ is the enantiomer of L$_{\text{RR}}$, this complex will be chiral and is the enantiomer of $[Pt(\text{L}_{\text{RR}})_2]^{2+}$. (1 distinct isomer, forming an enantiomeric pair with the above)

3. Complex with two meso-bn ligands: $[Pt(\text{L}_{\text{meso}})_2]^{2+}$ Since L$_{\text{meso}}$ is an achiral ligand, the complex formed by two such ligands will be achiral. (1 distinct isomer)

4. Complex with one (2R,3R)-bn and one (2S,3S)-bn ligand: $[Pt(\text{L}_{\text{RR}})(\text{L}_{\text{SS}})]^{2+}$ When a complex contains a pair of enantiomeric ligands in a square planar geometry, it possesses an internal plane of symmetry, making the complex achiral (a meso compound). (1 distinct isomer)

5. Complex with one (2R,3R)-bn and one meso-bn ligand: $[Pt(\text{L}_{\text{RR}})(\text{L}_{\text{meso}})]^{2+}$ This complex contains one chiral ligand (L$_{\text{RR}}$) and one achiral ligand (L$_{\text{meso}}$). The presence of the chiral ligand makes the overall complex chiral. (1 distinct isomer)

6. Complex with one (2S,3S)-bn and one meso-bn ligand: $[Pt(\text{L}_{\text{SS}})(\text{L}_{\text{meso}})]^{2+}$ This complex contains one chiral ligand (L$_{\text{SS}}$) and one achiral ligand (L$_{\text{meso}}$). This complex will be chiral and is the enantiomer of $[Pt(\text{L}_{\text{RR}})(\text{L}_{\text{meso}})]^{2+}$. (1 distinct isomer, forming an enantiomeric pair with the above)

By summing up all the distinct stereoisomers from these combinations: Total number of stereoisomers = 1 (from case 1) + 1 (from case 2) + 1 (from case 3) + 1 (from case 4) + 1 (from case 5) + 1 (from case 6) = 6.

The value of n (total number of stereoisomers) is 6.