sigma-Bound Ligands: Alkyls, Aryls, Hydrides & Agostic Interactions

Line-by-Line Solution:

(a) Equilibrium Constant for Homolysis at $298\text{ K}$:

1. For $\text{CH}_3\text{Mn}(\text{CO})_5$:

• $\Delta H^\circ = +155\text{ kJ/mol} = 155,000\text{ J/mol}$
• $\Delta S^\circ = +125\text{ J/(mol}\cdot\text{K)}$

2. For Ethane ($\text{CH}_3-\text{CH}_3$):

• $\Delta H^\circ = +377\text{ kJ/mol} = 377,000\text{ J/mol}$

(b) Physical Explanation of Kinetic Lability vs. Thermodynamic Stability:

• Both complexes have extremely small equilibrium constants for homolysis ($K_\text{hom} \ll 10^{-20}$), meaning neither undergoes spontaneous homolytic cleavage at room temperature at an observable rate.
• However, transition metal-alkyls decompose not by homolysis, but through low-barrier concerted pathways such as $\beta$-hydride elimination or reductive elimination, where $\Delta G^\ddagger < 60-80\text{ kJ/mol}$.
• In contrast, ethane has no vacant low-energy orbitals or accessible oxidation states; its only decomposition pathway is homolytic cleavage, which possesses an insurmountable activation barrier ($E_a \approx 377\text{ kJ/mol}$).
• Therefore, transition metal-alkyls are thermodynamically stable with respect to homolysis, but kinetically labile due to concerted intramolecular reaction channels.

Line-by-Line Solution:

(a) Evaluation of $\beta$-Hydrogen Availability and Geometry:

1. $L_n M-\text{CH}(\text{CH}_3)_2$ (isopropyl):

• Possesses six $\beta$-hydrogens attached to two primary carbons.
• Secondary alkyl group with high steric congestion and multiple pathways for syn-coplanar alignment.
• Extremely rapid $\beta$-elimination; least stable.

2. $L_n M-\text{CH}_2\text{CH}_3$ (ethyl):

• Possesses three $\beta$-hydrogens on a primary carbon.
• Undergoes smooth, rapid $\beta$-elimination via a low-energy planar four-membered transition state.

3. $L_n M-\text{C}_7\text{H}_{11}$ (1-norbornyl):

• Possesses bridgehead $\beta$-hydrogens.
• $\beta$-elimination would generate a double bond at a bridgehead carbon, violating Bredt's rule and incurring severe ring strain.
• Highly stable; decomposes only at elevated temperatures.

4. $L_n M-\text{CH}_2\text{C}(\text{CH}_3)_3$ (neopentyl):

• The $\beta$-carbon is quaternary; zero $\beta$-hydrogens.
• $\beta$-elimination is chemically impossible.

5. $L_n M-\text{CH}_2\text{Si}(\text{CH}_3)_3$ (trimethylsilylmethyl):

• Silicon occupies the $\beta$-position; zero $\beta$-hydrogens.
• Thermally robust.

6. $L_n M-\text{CH}_3$ (methyl):

• No $\beta$-carbon; zero $\beta$-hydrogens.
• Highly stable toward $\beta$-elimination.

(b) Ranking of Increasing Kinetic Stability:

The first two undergo rapid $\beta$-elimination, while the remaining four are kinetically robust.

Line-by-Line Solution:

(a) Structural Assignment from Coupling Constants:

1. Rhodium-103 is a $100\%$ abundant spin-$1/2$ nucleus ($I = 1/2$). Coupling to $^{103}\text{Rh}$ splits the hydride resonance into a doublet with $^1J(\text{Rh}-\text{H}) = 28\text{ Hz}$.
2. The two equivalent phosphorus-31 nuclei ($I = 1/2$) further split the signal into a triplet with $^2J(\text{P}-\text{H}) = 14\text{ Hz}$.
3. In transition metal coordination chemistry:

• Trans phosphine-hydride coupling: $^2J(\text{P}-\text{H})_\text{trans} = 90 - 160\text{ Hz}$.
• Cis phosphine-hydride coupling: $^2J(\text{P}-\text{H})_\text{cis} = 10 - 30\text{ Hz}$.

1. The observed value of $^2J(\text{P}-\text{H}) = 14\text{ Hz}$ falls directly within the cis-coupling range.

• Conclusion: Both triphenylphosphine ligands occupy positions mutually cis to the hydride ligand.

Line-by-Line Solution:

(a) Failure of the Diamagnetic Term $\sigma_d$:

1. The Lamb formula for the diamagnetic shielding of an $s$-electron at distance $r$ from a nucleus is:

1. For an isolated hydrogen atom with a $1s$ electron ($a_0 = 0.529$ Å):

1. Even for a hypothetical free hydride ion $\text{H}^-$ with two electrons, the maximum diamagnetic shielding cannot exceed $\approx 26\text{ ppm}$ relative to a bare proton.
2. Relative to the standard reference TMS (tetramethylsilane, which has $\sigma \approx 31\text{ ppm}$), a shift of $\delta = -40\text{ ppm}$ corresponds to an absolute shielding of:

1. Because $\sigma_d$ from the hydrogen electron cloud is at most $26\text{ ppm}$, diamagnetic shielding of the proton itself is completely incapable of explaining shifts below $\delta = 0\text{ ppm}$.

(b) Buckingham-Stephens Paramagnetic Shielding Formulation:

1. The transition metal center possesses filled non-bonding $d$-orbitals (e.g., $t_{2g}$ in $O_h$, such as $d_{xy}, d_{yz}, d_{xz}$) at energy $E_0$ and low-lying empty $d$-orbitals (e.g., $e_g^*$ or $p_z$) at energy $E_n$.
2. An external magnetic field $B_0$ directed along the $z$-axis mixes the ground state $|0\rangle$ with excited states $|n\rangle$ via the angular momentum operator $\hat{L}_z$:

1. This magnetic mixing induces a circular current of electrons within the metal $d$-orbitals around the metal center, generating an orbital magnetic dipole moment $\boldsymbol{\mu}_M$ at the metal atom:

where $\chi_{vv}$ is the van Vleck paramagnetic susceptibility of the metal center.

1. The dipolar magnetic field produced by this metal-centered magnetic moment at the adjacent hydride proton (located at distance $R$ along the $z$-axis) is given by the classical dipole field equation:

1. For a hydride ligand situated along the $z$-axis ($z = R$), the induced secondary magnetic field at the proton is:

Because the orbital current is paramagnetic with respect to the metal, $\boldsymbol{\mu}_M$ is oriented to produce an opposing magnetic field in the local equatorial plane, but an enforcing positive shielding field directly along the bond axis.

1. The resulting long-range shielding constant at the hydride nucleus is:

Because $E_n - E_0 > 0$, this contribution is strictly positive and inversely proportional to $R^3$.

1. Because the $M-\text{H}$ bond distance is exceptionally short ($R \approx 1.5 - 1.7$ Å), $1/R^3$ is very large, contributing $+15$ to $+45\text{ ppm}$ of positive magnetic shielding to the proton, producing the extreme negative chemical shifts observed experimentally.

Line-by-Line Solution:

(a) Elementary Reaction Mechanism:

1. Starting complex: $(\text{PPh}_3)_2\text{Pt}(\text{Et})_2$ is a 16-electron square planar $d^8$ $\text{Pt}(\text{II})$ complex.
2. Although it possesses only 16 electrons, square planar geometry has all 4 in-plane coordination sites occupied. For the ethyl group to achieve a syn-coplanar four-membered transition state, the Pt center requires an open coordination site in the coordination plane.
3. Step 1: Reversible Dissociation of Phosphine:

generating a 14-electron, 3-coordinate T-shaped intermediate $[(\text{PPh}_3)\text{Pt}(\text{Et})_2]$.

4. Step 2: $\beta$-Hydride Elimination:

5. Step 3: Reductive Elimination of Ethane:

Applying the steady-state approximation to the 14-electron intermediate:

The overall rate of decomposition is:

This matches the empirical rate law with $K = \frac{k_{-1}}{k_2}$.

(b) Inhibition by Excess Triphenylphosphine:

• Excess $\text{PPh}_3$ drives the equilibrium back toward the 4-coordinate 16-electron complex via Le Chatelier's principle ($k_{-1}[\text{PPh}_3] \gg k_2$).
• Denying the complex a vacant coordination site completely halts the $\beta$-hydride elimination pathway.

(c) Stoichiometry of Products:

• Products: exactly one equivalent of ethene gas, one equivalent of ethane gas, and platinum(0) species.

Line-by-Line Solution:

(a) Unambiguous Structural Assignment from $^1J(H-D)$:

1. In classical metal dihydrides, the two hydrogen atoms are bonded directly to the metal center and not to each other ($d(H-H) > 1.6$ Å). Spin-spin coupling between the two sites through the metal center is small:

1. In non-classical dihydrogen complexes, the $H-D$ single bond remains intact ($d(H-D) \approx 0.8-1.0$ Å). In free $\text{HD}$ gas, $^1J(\text{H}-\text{D}) = 43.2\text{ Hz}$.
2. Coordinated dihydrogen complexes characteristically display:

1. The observed value of $^1J(\text{H}-\text{D}) = 29.5\text{ Hz}$ represents unequivocal proof that the hydrogen-deuterium bond is intact.

• Assignment: The complex is a non-classical dihydrogen complex, $[(\text{dppe})_2\text{Ru}(\eta^2-\text{H}_2)]$.

(b) Internuclear $H-H$ Distance Calculation from $T_{1,\text{min}}$: Given:

• $T_{1,\text{min}} = 18\text{ ms} = 0.018\text{ s}$
• Spectrometer frequency $\nu = 500\text{ MHz}$

Substitute into the dipole-dipole relaxation formula:

1. Compute the ratio:

1. Compute the sixth root:

Let $y = 3.60 \times 10^{-5}$:

• $\ln(y) = \ln(3.60) + \ln(10^{-5}) = 1.2809 - 11.5129 = -10.232$
• $\frac{\ln(y)}{6} = \frac{-10.232}{6} \approx -1.7053$
• $\exp(-1.7053) \approx 0.1817$

1. Compute $r_{HH}$:

• Conclusion: The internuclear distance is $r_{HH} = 1.06$ Å. This is elongated relative to free $\text{H}_2$ ($0.74$ Å) due to $\pi$-backbonding, but well within the non-classical dihydrogen regime ($r_{HH} < 1.15$ Å), confirming the non-classical coordination mode.

Line-by-Line Solution:

(a) Electron Counting:

• Titanium is in Group 4 ($n_v = 4$).
• Oxidation state: Three chlorides ($-3$) and one ethyl ($-1$) $\implies \text{Ti}(\text{IV}) (d^0)$.
• Without Agostic Interaction:
• $VEC = 4 (\text{Ti}) + 3 \times 1 (\text{Cl}) + 1 (\text{Et}) = 8$ valence electrons (drastically sub-octet, highly electron-deficient).
• With Agostic $\text{C}-\text{H}\cdots\text{Ti}$ Interaction:
• The $\beta$-$\text{C}-\text{H}$ bond acts as a 2-electron donor into a vacant titanium $d$-orbital ($L$-type):
• $VEC = 8 + 2 = \mathbf{10\text{ valence electrons}}$.
• The agostic interaction partially alleviates the extreme electronic deficiency of the $d^0$ metal.

(b) Molecular Orbital Origin of Spectroscopic Signatures:

1. Three-Center Two-Electron ($3c-2e$) Orbital Overlap:

• The filled $\sigma(\text{C}-\text{H})$ bonding orbital donates into the empty $d_{z^2}/d_{xz}$ hybrid orbital of $\text{Ti}(\text{IV})$.
• This dative interaction depopulates electron density from the $\text{C}-\text{H}$ internuclear bonding region.

2. Bond Order Depletion:

• The formal $\text{C}-\text{H}$ bond order drops from $1.0$ to $\approx 0.5-0.6$.
• Depletion of $\sigma(\text{C}-\text{H})$ electron density weakens the force constant $k_{CH}$, shifting the stretching frequency from $2950\text{ cm}^{-1}$ down to $2500\text{ cm}^{-1}$ ($\Delta \nu = -450\text{ cm}^{-1}$).

3. Reduction in $^1J_{CH}$:

• The Fermi contact term for spin-spin coupling between carbon-13 and proton is directly proportional to the $s$-electron density at both nuclei and the $C-H$ bond order:

• Transferring electron density out of the $\sigma(\text{C}-\text{H})$ orbital reduces the effective bond order $P_{CH}$, decreasing the coupling constant from $140\text{ Hz}$ to $65\text{ Hz}$.

(c) Reduction of Activation Barrier in Ziegler-Natta Insertion:

1. Migratory insertion requires coordinating an incoming ethylene molecule and migrating the alkyl chain onto ethylene via a four-centered transition state.
2. The agostic interaction pre-organizes the alkyl ligand into an orientation where the $\alpha$- and $\beta$-carbons are already bent toward the metal coordination plane ($ngle(\text{Ti}-\text{C}-\text{C}) \approx 85-95^\circ$ instead of tetrahedral $109.5^\circ$).
3. When ethylene coordinates, the weak agostic interaction (bond energy $\approx 40-60\text{ kJ/mol}$) is displaced easily without requiring ligand dissociation.
4. During the migration step, the developing transition state is stabilized by a continuous agostic interaction between the metal and the migrating carbon's hydrogen atom, smoothing the electronic potential energy surface and reducing the activation energy $\Delta G^\ddagger$ by $20-35\text{ kJ/mol}$.

Line-by-Line Solution:

(a) Thermodynamic Cycle for Hydricity: Consider the following thermodynamic steps in solution:

1. Deprotonation of the Hydride:

2. Two-Electron Oxidation of the Conjugate Base:

3. Formation of Hydride Ion from Proton and Electrons:

Summing steps 1, 2, and 3 yields the net hydride dissociation:

The absolute hydricity $\Delta G_{\text{H}^-}^\circ$ is:

where $E_{\text{avg}}^\circ = \frac{E^\circ(M^{n+1}/M^n) + E^\circ(M^n/M^{n-1})}{2}$.

(b) Numerical Calculation of Absolute Hydricity in Acetonitrile: In acetonitrile solvent:

• Free energy contribution from acidity:

• Using the standard thermodynamic benchmark established by Daniel DuBois:

In SI units ($ ext{kJ/mol}$):

Given:

• $\text{p}K_a = 22.0$
• $E^\circ = -0.80\text{ V}$

Compute:

• Conclusion: The absolute hydricity of the cobalt complex is $\mathbf{304.3\text{ kJ/mol}}$ ($\approx 72.7\text{ kcal/mol}$). Lower values of $\Delta G_{\text{H}^-}^\circ$ denote stronger hydride donors; this complex is a potent hydride donor capable of reducing carbon dioxide ($\Delta G_{ ext{H}^-}^\circ(\text{HCOO}^-) = 314\text{ kJ/mol}$) to formate.

Line-by-Line Solution:

(a) Equilibrium Positions from Potential Minimization: The potential energy function along the $H-H$ internuclear separation coordinate $r$ is:

Equilibrium points correspond to the roots of the first derivative:

For a symmetric double well ($c = 0$):

The roots are:

• $r_0 = 0$ (unstable local maximum: transition state barrier separating the two states).
• $r_1, r_2 = \pm \sqrt{\frac{b}{2a}}$.

Taking the physical branch $r > 0$, when an asymmetric term $c \ne 0$ is introduced (representing electronic bias toward one state):

• Minimum 1 ($r_1 \approx 0.85$ Å): Non-classical dihydrogen complex.
• Minimum 2 ($r_2 \approx 1.65$ Å): Classical dihydride complex.
• The barrier height between the two states is:

(b) Inelastic Neutron Scattering (INS) and Rotational Tunneling:

1. Dihydrogen coordinated side-on ($M-\eta^2-\text{H}_2$) acts as a two-dimensional quantum rotor hindered by a twofold or fourfold potential barrier $V(\phi) = \frac{V_2}{2}(1 - \cos 2\phi)$.
2. Because neutrons have wavelengths comparable to molecular bond lengths ($1-2$ Å) and possess zero charge, they scatter directly off atomic nuclei without selection rules.
3. If the potential is a single well (pure dihydrogen), INS spectra exhibit discrete rotational transitions between quantized rotor states ($J=0 \to J=1$, para-to-ortho hydrogen transition) at low energy ($0.5 - 5\text{ meV}$).
4. If a low-barrier double well exists, coherent quantum tunneling between the two wells splits each degenerate vibrational level into a doublet with an energy separation $\hbar \Omega$ that is extraordinarily sensitive to isotopic substitution ($H/D$).

(c) WKB Semiclassical Tunneling Splitting Derivation: Under the Wentzel-Kramers-Brillouin (WKB) approximation, the tunneling probability $P$ through the potential barrier $V(r)$ between classical turning points $r_a$ and $r_b$ is:

where $\mu = \frac{m_H}{2}$ is the reduced mass of the $\text{H}_2$ oscillator. The quantum tunneling frequency $\Omega$ (and corresponding energy splitting $\Delta E = \hbar \Omega$) is given by:

where $\omega_0$ is the classical attempt frequency within the well.

• Isotope Effect on Tunneling:

Because $\mu_D = 2\mu_H$, the exponent increases by $\sqrt{2} \approx 1.414$. Consequently:

This colossal quantum tunneling isotope effect observed in low-temperature INS unambiguously verifies the double-well topology of the dihydrogen-to-dihydride reaction coordinate.