Stable carbenes and carbene analogues of the p-block elements (i.e. low oxidation state chemistry)

Question 1

Inert pair effect

Answer 1

Arises from increasing s-p separation as you go down the group due to the increased penetration of the ns orbital over the np orbital
s electrons stabilised by increasing nuclear charge
Also orbitals become more diffuse down the group so bonds get weaker - the energy released from bonding needs to be enough to be able to utilise the low-lying s electrons

Question 2

Carbenes

Answer 2

Neutral compounds featuring a divalent carbon with only 6 electrons in its valence shell
i.e. R2C:

Question 3

Carbene geometry

Answer 3

Triplet carbenes can be either linear or bent (singlet always bent)
Linear = sp hybrid i.e. bonds to substituents are formed using sp orbitals
Bent = sp2 hybrid i.e. bonds to substituents are formed using sp2 orbitals

Question 4

Triplet carbene

Answer 4

Spins paired (electrons in different orbitals)

Question 5

Singlet carbene

Answer 5

Opposite spins (electrons in the same sp2 orbital, diamagnetic)
Higher energy than triplet

Question 6

Why are bent triplet carbenes more stable than linear triplet carbenes?

Answer 6

Draw MO diagram

Question 7

When does a singlet ‘carbene’ become the ground state rather than a triplet?

Answer 7

If the 3a1-1b1 (i.e. HOMO-LUMO) separation is > 2 eV (45 kcal mol-1)

Question 8

Difference between the ground states of carbenes and group 14 metallylenes

Answer 8

The increasing s-p separation down the group leads to increasing stability of the singlet state down the group
This is a result of the inert pair effect - electrons want to remain paired up in the low energy orbitals of high s character

Question 9

DeltaE(ST) for H2M (M = group 14 element)

Answer 9

= E(triplet) - E(singlet)
C -14
Si +16.7
Ge +21.8
Sn +24.8
Pb +34.8

Question 10

Why does Et2Sn exist as a hexamer rather than a monomer?

Answer 10

Sn-Sn single bonds are sufficiently strong to allow the valence 5s2 electrons in tin to rehybridise to form bonds
(Et2Sn)6 is the thermodynamically favoured product of the reaction

Question 11

How can isolation of monomeric dialkyl tin compounds be favoured?

Answer 11

Through the introduction of steric bulk around the tin (i.e. altering the kinetics of the reaction)
Using a bulky R group will introduce kinetic protection around the tin centre and therefore result in a higher barrier towards condensation of the monomer units
i.e. will stabilise the low oxidation state

Question 12

Examples of common bulky ligands than can provide kinetic stabilisation

Answer 12

Draw

Remember ‘amide’ ligands are 1 electron donors (ie. NH2/NR2)

Question 13

How to rationalise the singlet ground state of heavier group 14 metallylenes

Answer 13

Can’t just simply rationalise using the inert pair effect - because the s orbitals are involved in bonding with the substituents (lowest energy bonding orbital - 2a1 orbital)
Rationalise instead through mixing of the a1 orbitals (second order Jahn-Teller effect)
The sigma* orbital reduces in energy down group 14 because the bonding gets weaker (i.e. smaller sigma/sigma* gap)
This reduction in energy leads to mixing of the HOMO (3a1) and the sigma* orbital (4a1) because they are of the same symmetry
This increases the s-character of the HOMO (because the sigma* orbital is the antibonding combination of s orbitals)
Therefore the HOMO becomes lower in energy (increasing HOMO/LUMO gap)
This mixing effect increases down the group as orbitals get more diffuse and bonds get weaker

Question 14

How can the HOMO/LUMO gap be narrowed?

Answer 14

By increasing the angle between the substituents (e.g. by using bulky substituents)
i.e. moving away from bent back towards linear
The overlap of orbitals in 3a1 (HOMO) becomes less efficient, increasing the HOMO energy and narrowing the HOMO/LUMO gap
Get a red-shifted absorption (longer wavelength/lower energy absorption)

Question 15

Inductive effects on HOMO/LUMO gap

Answer 15

The 3a1 HOMO can be stabilised by using ligands with electronegative substituents e.g. O, N
Therefore increasing HOMO/LUMO gap and stabilising singlet state over highly reactive triplet state

Question 16

Mesomeric effects on HOMO/LUMO gap

Answer 16

The LUMO can be destabilised by pi-donor ligands
i.e. there is an interaction between the empty p orbital on the central element and the pi-donor lone pairs, destabilising the LUMO
(see NHC stability stuff too)
Therefore increasing the HOMO/LUMO gap

Question 17

Why is NHC stable?

Answer 17

Because it is a singlet rather than triplet carbene
N-C-N angle approx 95 degrees which favours the 3a1 stabilising interaction
The donor N centres are electronegative so stabilise the HOMO overall
System is completely planar (N and C all sp2) so N lone pairs in p orbital can form a pi-type interaction with the empty p(y) orbital on the C, thus stabilising the system overall

Question 18

Synthesis of stable compounds of group 14 M(II) (i.e. NHMs where M = C, Si, Ge, Sn, Pb)

Answer 18

Can use MCl2 as a starting material (only available for the heavier group 14 elements)
Use MCl4 for C/Si
React with dilithiated amide ligand
Then have to reduce from M(IV) to M(II) for MCl4 starting materials
Can do this using K

Flashcard

Question 19

Carbene analogues in groups 13-16

Answer 19

ISOELECTRONIC! All 6 electrons
Therefore
Group 13 = -
Group 14 = neutral
Group 15 = +
Group 16 = 2+

Question 20

Synthesis of group 13 carbene analogues

Answer 20

Draw

Question 21

Reactivity of group 13 and 14 ‘carbene analogues’

Answer 21

Oxidative addition (2 electron oxidation) i.e. oxidative addition of M(II) across X-Y bond to give M(IV)
Lewis base character/acting as a nucleophile
Flashcards