2009 sub handout (dragged)

Upload: anhonest

Post on 05-Apr-2018

216 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/31/2019 2009 Sub Handout (Dragged)

    1/2

    Substitution Reactions Dr L.L. Wong, Michaelmas Term, 3rd

    Year_______________________________________________________________________________________________________________________________________________

    _______________________________________________________________________________________________________________________________________________

    Page 11 of 14

    Substitution Reactions of Square Planar Complexes

    Common metal oxidation states with square planar complexes. All have d8 configuration.

    Ni2+

    Rh+ Pd2+

    Ir+ Pt2+ Au3+

    Most details have come from studies on Pt2+ complexes. The reaction

    occurs with retentionof configuration and has a general two-term rate law because of two parallel

    mechanisms the solvent pathway and the nucleophilic pathway.

    "d[ML3X]

    dt= (ks + ky[Y])[ML3X]

    The mechanisms:

    The two vacant coordination sites in square

    planar complexes mean that the mechanism is

    almost invariably associative. The reactions show

    large and negative values of "S and "V.

    The incoming nucleophile donates its electrons

    into the empty metal pzorbital.

    The ks term arises from the solvent pathway in

    which a solvent molecule displaces the leaving

    group in a slow reaction followed by rapid

    substitution of the coordinated solvent molecule by

    the incoming nucleophile. The solvent does not

    appear in the rate law.

    The ky term is the direct attack by the incoming

    group, hence the second order rate law.

    Both mechanisms proceed via 5-coordinate

    intermediates which have been detected for Rh+

    and Ni2+ but not for Pt2+.

    Reactivity at different metal centres will depend

    on the readiness with which the centres move

    away from the square planar geometry.

    Hence the rate order: Ni2+ > Pd2+ > Pt2+.

    With Au3+ the higher charge will attract the

    incoming nucleophile, and Au3+ > Pt2+. Fig. 1: The parallel solvent andnucleophilic pathways.

    M

    L

    XL

    L

    + Y M

    L

    YL

    L

    + X

    M

    L X

    LL

    M

    L X

    LL

    + S

    S

    M

    L X

    L

    L

    S

    M

    L X

    LLY

    + Y

    M

    L S

    LL

    M

    L X

    L

    L

    Y

    + Y

    M

    L Y

    LL

    M

    L Y

    LL

    X

    M

    L S

    L

    L

    Y

    - S

    Reactant

    Product

    - X

    - X

    Substitution Reactions Dr L.L. Wong, Michaelmas Term, 3rd

    Year_______________________________________________________________________________________________________________________________________________

    _______________________________________________________________________________________________________________________________________________

    Page 12 of 14

    The rate law:

    For the exchange reaction:

    the plot of kobs = ks + ky[Y] shows that the

    intercept is effectively zero in the non-polar

    solvent hexane, i.e. the solvent pathway does

    not operate and the slope gives ky. In the polar

    solvent methanol, the rate is independent of

    [Y], i.e. the solvent pathway dominates.

    The entering group:

    For the substitution reaction:

    the plot of kobs = ks + ky[Y] shows that the

    entering ligand has a significant effect on

    the rate, in contrast to octahedral

    complexes. The activity trends parallel

    the nucleophilicity of the incoming ligand.

    Hence:

    I > Br > Cl >> F

    PR3 >> NR3

    R2S >> R2O

    Overlap and solvation factors.

    The leaving group:

    In the reaction

    Pt(dien)X+ + Py ! Pt(dien)Py2+ + X

    the lability varies by 106 in the order:

    NO3 > H2O > Cl

    > Br > I >> N3 > SCN > NO2

    > CN

    This sequence broadly follows ligand basicity and ML bond strength, and suggests that bond

    breaking must be an important factor even in an overall associative mechanism.

    In general, a metal-halide bond is more labile than a MN bond. Hence all other things being

    equal a PtCl bond will undergo substitution faster than a PtNH3 or PtPy bond.

    Pt

    PEt3

    Cl Cl

    NHEt2

    *NHEt2Pt

    PEt3

    Cl Cl

    NHEt2*

    0#0.5#

    1#1.5#

    2#2.5#

    3#

    0# 0.05# 0.1# 0.15# 0.2# 0.25#

    Pt

    Py

    Cl Cl

    Py

    Pt

    Py

    Cl Y

    Py

    Y

    0#0.5#

    1#1.5#

    2#2.5#

    3#3.5#

    4#

    0# 20# 40# 60# 80# 100# 120#

    in hexane

    in methanol

    [Y] = [Et2NH] / M

    105kobs / s1

    103kobs / s1

    SCN

    I

    [Y] / mM

    Me2S

    BrN3

    NH3

  • 7/31/2019 2009 Sub Handout (Dragged)

    2/2