4c

Question 1

metal = Carbon : what can the two things be called

Answer 1

a carbene or alkylidenes

Question 2

carbenes and alkylidenes diff in what

Answer 2

they differ in reactivity
theyre both M=C tho

Question 3

C=O ligands should theoretically have what happen

Answer 3

nuc attackssss

Question 4

how do we make a carbene

Answer 4

Mo with 6CO,, and a MeLi nuc attacks one CO

u then have the 5CO - Metal - carbonyl with Me

the O then bonds to Li and u get 5CO - Metal = C - OLi and - Me

this then reacts with MeLi again and instead of OLi we get OMe and Me as the Metal=C subs

Question 5

what’s different between carbenes and alkylidenes that cause their reactivity differences

Answer 5

carbenes have a singlet carbon that sigma bond into the metals orbital

the metal also has a singlet that donates into the nonbnfing orbital of C.

we therefore have sigma donation and pi bb.

the carbene is a two electron donor // acceptor ligand.

Question 6

carbene singlet drawing is similar to what

Answer 6

to CO

Question 7

alkylidene isn’t a singlet,, it’s a what

Answer 7

it’s a triplet carbon

aka u have two separate spin up electrons in the lhs and the top orbital.

the lhs one sigma donates to the metal which also only have one spin down one

there’s also a pi bond type thing from the single electrons on the top

Question 8

alkylidene triplet carbon is similar to what bonding

Answer 8

the bonding in single and double bonds

aka sigma and then the pi part

Question 9

how do we normally make alkylidenes

Answer 9

alpha elimination of alkyl complexes

Question 10

n2 example

Answer 10

M - alkene

like ethene

Question 11

n3 examples

Answer 11

m - )> CH

with CH2 on both sides

Question 12

n5 example

Answer 12

pentane wirh a circle in the middle and the M being bonded to all of the Carbons

Question 13

n6 example

Answer 13

benzene where M is bonded to all

Question 14

n4 example

Answer 14

cyclobutadiene

the box with 2 double bonds in it

Question 15

can u pls describe the bonding interaction between n2 and a metal also known as DCD bonding model

Answer 15

u have the metal with a petal pointing to the right

u then have ur alkene on the side so the pi cloud is close to the metal petal and the other pi cloud is on the other side.

the pi cloud close to the petal donates electron density into it

sigma donation from the homo into metal

Question 16

can u describe pi back bonding of metal and alkene

Answer 16

metal has the d orbitals
alkene has the pi* orbitals

the d on metal pi backbonds into pi* of alkene by donating electron density into them

Question 17

what’s the bond order of the alkene when pi back bonding occurs

Answer 17

it’s between one and 2

bc obvs backbonding weakens it a bit

Question 18

what’s a better pi acceptor,, CO or CC

Answer 18

CO is the better pi acceptor

Question 19

when backbonding occurs into the alkene what happens,, aka what evidence do we have that backbonding occurs

Answer 19

• CC bond is longer
• bond order is reduced
• sp3 character occurs
• vibrational frequency is reduced

Question 20

in n2 aka alkenes,, what’s the angelo bonding thing we can look for and what orbitals match each one

Answer 20

u have pi and pi*

the pi is the bonding interaction between the two p orbitals : S, Pz and dz2 can overlap here for sigma bonding.

the pi* is the pi bb interaction where the p are out of phase. the px and dzx match here

Question 21

common alkene complexes

Answer 21

cyclooctene (octane with a double bond)

norbornene ( the folded card with a triangle on top and a double bond on one of the sides.)

Question 22

alkenes wre what type of ligands

Answer 22

alkenes are weak alkenes

Question 23

even if an alkene ligand isn’t stable,, what can happen for it to be stabilised

Answer 23

alkene ligand forming a complex can reduce the ring strain of the ligand as some sp3 character forms due to pi bb

Question 24

what effect brings stability to alkene ligands

Answer 24

chelate effect brings stability

Question 25

MAC stands

Answer 25

metal alkene complexes

Question 26

describe // explain fluxionality in mac

Answer 26

the alkenes can spin round like propelles around the metal n2 axis. this gives averaging and gives one broad peak in NMR bc the rotation is faster than the NMR timescale in low temps such as -20 we see two peaks bc there are 2 H env. the one closer to the metal and the one further away ( aka think of pi clouds)

Question 27

what’s a n3 ligand

Answer 27

an allyl aka like a M - )> type thing

Question 28

what’s n3 made up of

Answer 28

it’s made up of a n1 and a n2

Question 29

if it’s n1 or n3 depends on what

Answer 29

depends on the 18e- rule and if the complex needs 3 more e- or 1 more e-

Question 30

can the n1 and n3 interconvert

Answer 30

yesss they can interconvert to give fluxional behaviour

Question 31

can u explain n3 bonding diagrams plsss,, the angelo one

Answer 31

we use this > but n form so it’s a little mountain. upside down v basically. u have all bonding p orbitals which is pi ( s, pz and dz2 overlap here) this is filled with 2 e- then u have the non bonding one i think we’re u just have the p orbitals on the sides and these don’t match up ( px and dx2 overlap here) this has 1 e- then u have pi* where none of the p match up aka all antibonding interactions, the py and dzy overlap here) this has no electrons

Question 33

py and dzy exp

Answer 33

these are the ones where part of it is into the page and part of it is coming out of the page

Question 34

how do we synthesise n3 allyls

Answer 34

[M] (low os) + alkene with X (halogen) —> metal bonded to X and alkene n1 end. —> metal bonded to x and n3 or metal bonded to X and XMg alkene —> we lose MgX2 to give us metal bound to n1 of alkene then this gives us metal bonded to n3

Question 35

what controls the synthesis of n3

Answer 35

other ligands control the regioselectivity // stereochem

Question 37

n4 butadiene angelo bonding diagram

Answer 37

pi, pi, homo pi*, lumo pi* pu and pu are both filled u have all in phase 1 AB interaction 2 AB interactions 3 AB interactions s, py and dz2 overlap px and dxz overlap py and dzy overlap dxy overlap u have donation from homo to the metal. loss of e- density reduces bonding in bonding interactions HOMO: losing electron density therefore bonding interactions are weakened ( side ones) LUMO: gains electron density but obvs 2 AB and one B,, the middle is strengthened bc this one is bonding bht the sides are AB and therefore get weakened.

Question 38

when we see CO as a ligand what must we thing of when we see it with a alkene

Answer 38

it’s a rlly good pi acceptor ligand meaning it takes electron density away from the metal. this means that the alkene will sigma donate to the metal but the metal won’t backbond as much into the alkene this means that the bonds won’t be altered that much!! they’re similar lengths as if they were a free alkene

Question 39

what should we think when we see a Cp ligand in a complex with an alkene

Answer 39

we should think that the Cp is a rlly good pi donor. this means that it pi donates to the metal a lot,, meaning the metal is very electron rich. there’s also sigma donation from the alkene to the metal. the metal will backbond into the alkene bc it’s electron rich and will weaken the lumo ab orbitals (side ones) these will lengthen and weaken the bonding orbital on the alkene homo will be strengthened

Question 40

what can backbonding manipulate

Answer 40

backbonding can manipulate bond lengths

Question 41

describe n4 cyclobutadiene

Answer 41

a box with 2 double bonds in it it’s very unstable bc of ring strain it has diradical character 4 pi electrons so not aromatic ,, antiaromatic acc. dimerises above -200*C

Question 42

how can a cyclobutadiene be stabilised

Answer 42

it can be stabilised via backbonding when it forms part of a complex. aka with CO bonds it’s very stable

Question 43

angelo bonding diagram of n4 cyclobutadiene

Answer 43

u have 2e- in one orbital,, then 2 orbitals with 1 e- each,, then 1 orbital with no electrons u have bonding where they’re all in phase,, s, pz and dz2 overlap here u have 2 degenerate ones where there’s 2 bonding and 2 antibonding,, px, dzx, py and dyz overlap here. then u have the completely anti phase one where dx2-y2 overlaps with.

Question 44

with backbonding into n4 cyclobutadiene what can happen

Answer 44

bc of diradical character,, with enough bb we can put 2e- into the cyclobutadiene making it 6e- and similar to benzene!!!

Question 45

describe n5

Answer 45

cyclopentadienyl Cp!!!! it’s very stable. it’s a pentagon with a circle in the middle ITS A METALLOCENE and can even half coordinate with only some of its doibke bonds if needed to fit with 18e- rule.

Question 46

what’s CP*

Answer 46

pentagon with me on every carbon

Question 47

describe n6

Answer 47

arene it’s a benzene need to think of when we can add tho bc it’s 6e- donating and we always look for 18e- rule

Question 48

the H on an n6 is what

Answer 48

it’s acidiccccc.

Question 49

when an arene aka n6 coordinated what happens

Answer 49

there’s a change in reactivity

Question 50

Cr(CO)3 is what

Answer 50

it’s an ewg!!!