pi bonding in phosphazines and onwards // lecture 7 Flashcards
how many pi electrons are there in cyclophosphazine
there’s 6 pi electrons
one from each n and p
describe nitrogen in cyclophosphazines
sp2 hybridised
there’s 1e- in the p orbital as in cyclophosphazine,, each atom gives 1 e-!!
the lone pair is in a sp2 hybrid
describe P in cyclophosphazine
it’s sp3 hybridised
forms 4 bonds and there’s an e- in each hybrid bond. (in the sigma bonding framework)
there’s one more e- tho!! and we want to figure out where this one e- is located
what are the 2 types of pi bonding in cyclophosphazine
there’s in plane pi bonding
out of plane pi bonding
what type of pi bonding was borazine and benzene
out of plane pi bonding!!!
what’s the diff between in plane and out of plane pi bonding
deciding whether to use the P 3d orbitals
or the sigma* orbitals
ON THE P
what do we need to think of when deciding whether to use the 3d or sigma* orbitals on P
the energy
the 3d orbitals are probs too high in energy to be involved in bonding.
sigma* orbitals may be lower in energy and used in any pi bonding used in the ring system.
PF5 hybridisation used and explanation - think about how many orbitals each hybridisation gives and how many bonds the molecule has
so sp3 forms 4 bonds,, and this is one less than PF5 with 5 bonds
so it has to be sp3d1 hybrid
bc we need one more orbital and this must come from d as all of the p and s are used already
PF5,, how many F environments
theres 2 diff environments
axial and equitorial
spin of P
i. = 1/2
spin of f
i = 1/2
if P couples to the axial F’s in PF5 u gettttt
a triplet
bc 2x2x0.5 + 1 = 3
if P couples to the equitorial F’s u get ,,, in PF5
2x3x0.5+1 = 4
a quartet
coupling is stronger and the bonds are thereofre shorter.
so PF5 in nmr givesss
a quartet of triplets
the equitorial split it 3 times (quartet)
then the axisl (diff env) split each peak twice : triplet
what happens in cold temps to PF5
at cold temps u can tell a difference between the axial and equitorial fluorines
in room temp they look like 5 equivalent fluorines.
if the d orbitals in PF5 were too high in energy, how would we explain the bonding w/o them
we say P is sp2 hybridised
3 orbitals bond to the equitorial F’s
left over p orbital to bond to the axial F’s.
the F’s use their s orbital to bond; its a circle like in H
3 MO for FPF,, THE AXIAL ATOMSSSSSSS
bonding
shaded o, shaded p, unshaded p, unshaded o (in phase with each side of the p orbital so theyre in the same phase so theyre bonding)
non bonding: shaded o,,, node at p bc any way u draw it, 1 will be inphase and 1 will be out of phase,, shaded o.
antibonding: shaded o,, unshaded p, unshaded p, shaded o. nodes between the p and the o’s
each bond has 2e-,, so u get 4e- from the FPF so u fill the bonding and non bonding AO’s
by o’s we mean the s orbital but thats the shape of the s orbital so im gonna say o.
PFP is what type of bond
3c-4e
3 atoms but 4e- !!!
normally its 2c-4e-
in PFP,, in PF5,, theres 2e- in bonding and 2e- in nonbonding,, so how many e- are holding the PFP together
only 2e-!!!
bc 2e- are non bonding
bond order = 1/2
so axial bonds are weaker and longer. thats why the coupling constant is smallerrrrrr!! they make the triplet after the quartet.
phosphines are what type of ligands
theyre acceptor ligands
so where do they accept into
phosphines are pi acceptor ligands,, where can the e- go
either the d orbitals of the metal from the P:,, then back into the P’s d orbitals via back bonding
or the sigma* orbitals of the P molecule from the metals d orbital
pi acceptor ability of P with d orbital explanation
RRR P:
P has a lone pair
this can be donated to the transition metal’s d orbitals (dx^2-y^2)
this is the bonding interaction!!
same thing as amines,, sigma donation of lone pair to the metal.
what do pi accepting mean when it comes to P being a pi acceptor ligand
it means that P can donate its lone pair to the metals d orbital,, but it can also accept e- into the Ps d orbitals
wait wait wait!! explain pi acceptor ability of phosphines WITH d orbitals
phosphines = P RRR
p: –> metal d orbital (dx2-y2)
this is a bonding interaction
then meta d orbital –> P d orbitals
back bonding
d orbitals are available for P bc its in the 3rd period. it can accept e- here.
draw the clover and RRR bonds.
pi acceptor ability of P without d orbitals
so the bonding interaction is the same bc we use the p: … the p’s lone pair to donate the lone pair to the metals d orbitals.
so bonding interaction : RRRP: –> metal (dx2-dy2)
backbonding:
accept e- density fromthe metals d orbital into the P-R sigma* orbital ( not shaded and shaded)
when we think about P’s pi acceptor abilities we need to think offff
bonding interaction ,, same for both with // without d orbitals
the back bonding interactionnn
different depending on if d or sigma* is lower in energy.
when explaining the pi acceptor ability of P,, what is the accepted method of backbonding
the use of their sigma* // antibonding orbitals
usually due to them having a lower energy
when u donate e- into the sigma* of PR what happens
u weaken the bond between that R and P
but u strengthen the bond between P and the metal
PF5 and phosphines is diff to
phosphazines
N is phsophazines
sp2
P in phosphazines is
sp3
in phosphazine,, the N is sp2 but where is the lone pair
the lone pair is exooooo
aka its not in the p orbital its in the sp2 hybrid orbitals
d orbitals on P can lead tooo
in plane // out of plane pi bonding
out of plane being similar to borazine and benzene
using the P d orbitals in cyclophosphines
we say the axis coming out of the page is the z axis
we get 6 MO frpm 6 AO.
3/6 = antibonding
2/3 = strongly antibonding
1/3 = weakly antibonding
3/6 = bonding
1/3 = weakly bonding
2/3 = strongly antibonding.
for pi bonding we use dxz and dyz orbitals on P and p orbitals on the N.
bonding interaction: phase on top of d orbitals matches that of p orbitals!!!
U GET A NODE AT P BC NOTHING WILL MATCH ITS PHASE: MOST BONDING MO STILL HAS A NODE // either a N or P doesnt rlly matter.
node makes smt less stable + unstable
this isnt fully conjugated bc we have a node!!! not truley aromatic bc we have a node at the most bonding MO. less stabe than borazine// benzene
why is cyclophosphazine less stable then borazine and benzene
bc even the most bonding MO have a node in them
if u draw the d orbitals that look like p’s // wings from birds eye view (dzx and dzy) and then the p orbital on N which looks like a circle,, u cant make all of them be in phase!! there will be a node on either N or P!!
and a node will weaken an interaction and leads to a degree of unstability.
the more unstable MO of cyclophosphazine uses what d orbital
it uses the dyz orbital
which means the d orbital wings arent going along with the bonds but are perpendicular to them
this gives antibonding interactions outside the system but bonding interactions inside the ring.
better orbital overlap on the inside + bonding MOs so its majority bonding. still weaker tho bc u have antibonding on the outside.
in plane pi bonding of cyclophosphazines
N: is donated to the empty dx2-y2 orbital on P.
aka they back donate to P, into its d orbital
in plane pi bonding bc u can see the p orbitals and d orbitals for what they rlly are.
dumbbell and clover shaped,, in plane bc the orbitals are flat in the plane
so the N lone pair is pointing out the system and is donated into one of the d orbital petals (dx2-y2) orbitalllll. STRENGTHENS THE N-P BOND IN THE RING: SHORTENS THE P-N BONDS!!!
the strength of the NP bond in cyclophosphazines in plane bonding aka where u can see every orbital for what it is,,, the NP bond strength depends on what
the R grouyp u have attached onto P
bc this affects on hpow much the N lone pair can backbond ]
EWG = more back donation as P is more (+)
perfectly planar if F is used as u get more back donation as P will be more delta (+)
PN bond length decreases
which enhances out of plane pi bonding?
advantages of in plane pi bonding and using d orbitals
accurately describes geometries: plananr pn rings with short NP bonds
d orbitals are too high in energy to be used in bonding
explaining bonding in cyclophosphazines without using d orbitals,, explanation of the ring and advantages//disadvantages
we use the ionic model
N is more electronegative so P donates 1e- to N.
each P will be (+)
5-1e- = 4 so sp3 hybrid
N will be (-)
2 lone pairs available now!!
so u draw N with 2 lone pair orbitals
sp3 hybridised
advantage: no d orbitals
disadvantage: N is sp3 hybridised and 2 lone pairs,, so the ring should look like a cyclohexane ring! but the ring is acc planar!!
negative hyper conjugation:
donation of a pi orbital // filled orbital // lone pair into a sigma* orbital
eg in an ester,, one of the oxygens lone pairs that are in plane,, they can donate lone pair to the carbonyls// C-O sigma* orbital: we want this as it stabilises the energy of the lone pair.
whats the anomeric effect - think of the O in anOmeric
when the cyclohexane with an O instead of a C and an OH as a substituent is preferred to have the OH as axial.
when the o is adjacent as it allows the lone pair to bc donated to the C-OH sigma* which stabilises the energy of the lone pair. and none bonding lone pair becomes bonding by being donated into the sigma*.
what is needed for the anOmeric effect to occur
the lone pair needs to be donated to the sigma* bond but the sigma* bond between at least one electronegative atom
sigma* is more stable near electronegtative groups!! non electronegative groups are higher in energy. O or F to lower the sigma* energy so it matches the lone pair better.
explain the bonding in cyclophosphazenes using negative hyperconjugation - out of plane so p orbitals going up and looking in an angle
- start with ionic model
- lone pairs on N in the p orbital
- N: donated into sigma* on P,, the bond between P and X (substituent)
- if X is more EWG, the sigma* will be lower in energy,,
is nonbonding or bonding higer in energy
nonbonding is higher in energy
whe nthe lone pair (nonbonding) is donated to the sigma*,, its energy is lowered as it becomes bonding.
negative hyperconjugation in cyclophosphazine in in plane bonding
- N: is sticking out and pointing outside the cyclic molecule
- the N: is in the sp2 hybrid orbital
- can donate into P-N sigma* orbitals in the ring
aka the bonds that make up the actual ring.
in in plane bonding,, the N: is in what orbital
its in the pz orbital
in out of plane bonding: where is the N:
its in the sp2 hybrid
out of plane bonding in cyclophosphazines,, where is the N: and where is it donated to
N: is in the sp2 hybrid
its donated into the NP sigma*
in plane bonding,, where is the N: and where is it donated
its in the pz orbital
its donated into the P-X sigma* orbital.
X being the substituent on P.
if X is a EWG,, we like this bc it reduces the energy of the sigma* bond.
