lecture 1b Flashcards
1
Q
describe methyl lithium and why this is the correct structure
A
MeLi is made up of 4 Me and 4Li and these are bridged together in a cube type shape. with alternating Me or Li on each corner
it’s made up of linked (CHLi)4 tetramers in the solid state
2
Q
connectivity of C in MeLi
A
connectivity is 6
it’s bonded to 3 H’s and 3 Li’s
3
Q
describe trimethyl aluminium
A
dimeric structure with symmetrical methyl bridges
u have 2 Me’s bonded to an aluminium (terminal Me’s) the Al is then bonded to 2 Me’s which connect it to the other Al. which is also bonded to 2 more terminal Me’s!!!
4
Q
what does the NMR of 1H and 13C show at room temp when we look at trimethyl aluminium
A
we would see 1 methyl signal!!!
bc there’s interconversion between the bridging and terminal Me ligands (via mono bridged species)
5
Q
what does the NMR of 1H and 13C show of trimethyl aluminium in (-40*C)
A
it shows 2 signals (2:1)
bc interconversion slows down at low temps and u can fully see the different structures u couldn’t see at higher temps.
6
Q
describe borane
A
BH3
7
Q
describe diborane
A
2 terminal H’s bonded to B ,, bridged by 2H’s,, then another B then 2 terminal H’s
8
Q
how do we synthesise diborane // borane
A
NaBH4 + I2 —>
2BH4- + I2 —> 2’BH3’ + H2(g) + 2I
‘BH3’ bc it’s acc diborane and not borane
9
Q
how can we tell diborane was formed and not borane
A
bc there’s a BH asymmetric and symmetric bond stretch seen in IR spectra at around 2500
then there’s an unknown at 1500-2000
the IR spectra is not conclusive that it is diborane
10
Q
% abundance of 10B
A
20%
11
Q
% abundance of 11 B
A
80%
12
Q
chance of B2H6 (diborane) being completely 11B
A
0.8 x 0.8 = 64%
13
Q
chance of B2H6 being made up of one 10B and one 11B
A
0.8 x 0.2 (x2) bc it could be the other way around aswell
32%
14
Q
chance of B2H6 being made jo of two 10B
A
0.2 x 0.2 = 4%
15
Q
if we add all the %’s for the different isotopes of B in B2H6 what do we get
A
we get 100%!!!
aka diborane must be made up of 2B’s!!
16
Q
a borane dimer can look likeeee
A
like ethane
like diborane (the bridging H’s)
17
Q
going back to the IR spec,, how do we now know it’s def diborane
A
bc there’s a BH bridging stretch (the previously unknown one at 1500-2000)
and a BH terminal bond stretch at 2000
18
Q
why do we ignore 10B in nmr
A
bc it has a spin of i=3
this means it has a bigger quadrupole moment and it’s hard to see bc it’s broad and problematic
19
Q
what B do we focus on when looking at NMR of diborane
A
we look at 11B
which has a spin of i = 3/2
20
Q
1H NMR of BH3 explained
A
H nmr so look at the protons // Hs
- 1 H environment!!
- one peak gets split by 11B and it’s different spin states (3/2,1/2,-1/2,-3/2)
- so u get a quartet (2x1x1.5+1) with a 1:1:1:1 ratio!!
21
Q
11B NMR of BH3
A
- B nmr so look at the B
- 1 B environment
- bonded to 3Hs ,, i = 1/2
- 2x3x0.5+1 = quartet
- but 1:3:3:1 ratio bc the peak gets split 3 times for the 3 different H’s
22
Q
1H nmr of B2H6 (ethane one)
A
- look at H’s bc H nmr
- 1 environment
- coupled to 1 B
- B i=3/2
- 1 peak split by 1 B and it’s 4 spin states to give a quartet with a 1:1:1:1 ratio.
bc 2x1x1.5+1!!
23
Q
11B nmr of B2H6 (ethane one)
A
- B nmr so look at B
- 1 B environment
- coupled to 3 H’s
- H i = 1/2
- 2x3x0.5+1
gives a quarter with a 1:3:3:1 ratio
24
Q
1H environment of B2H6 (bridging one)
A
- look at H’s
- 2 different H environments (terminal and bridging)
- ‼️‼️‼️‼️‼️‼️‼️ go over again
25
11B nmr of B2H6 (bridging structure)
- 1B environment
- coupled to H’s (2T and 2B)
TERM: 2x2x0.5+1 = triplet (1:3:1)
BRIDGING: 2x2x0.5+1 (triplet 1:3:1)
triple of triplets. bc 1 B environment and not 2. (so can’t be 2 different signals)
26
H nmr of diborane
look at the H’s
for some reason we only looked at the bridging ones
they’re bonded to 2Bs
2x2x(3/2)+1 = 7 so a septet
27
looking at the bridging H when we look at 1H nmr what do we see for diborane
okay so the H is coupled to 2B’s
so 2x2x(3/2)+1 = 7 or septet
it’s also coupled to 4H’s
2x4x0.5+1 = 5 or quintet
so u get a septet of quintets
bc it’s coupled to 2 different things u would get a diff cc this is why the septet stuff splits into quintets.
28
does valence bond theory work for diborane + explain
nope
u have 8 bonds in diborane
but only 12 valence e-
(3 + 3 + 6)
the BHB bridging is a 3c 2e bond
29
what does the conventional valence bond theory require
it requires 2e- for every bond
2C, 2e bond
30
is diborane fit the conventional valance bond theory how many electrons would it have
it would have 16e-
bc it’s made up of 8 bonds
and 2e- in each bond
31
why is diborane called electron deficient
bc conventional valence bond theory says u need 2e- in each bond
but diborane only has 12 e- ,, not 16e-
32
boron hydrides and anionic boranes general formula
BnHm
BnHmx-
both are electron deficient.
new theories of chemical bonding were formulated to explain their structures.
33
explain homolytic cleavage of diborane
homolytic cleavage so u attack different B’s
bc the o’s on homo are separated.
one base attacks one B and the bridging BH bond is cleaved onto the other B.
then the other base attacks the different B and also cleaves off the bridging BH H onto the original B.
homolytic cleavage gives us a base on both B’s
34
heterolytic cleavage
this is where the base attacks the same B on the diborane.
base attack one B and cleaves off the H
then another base attacks the same B and cleaves off another H.
so there’s 2 bases on one B.
35
what does heterolytic cleavage give
bases on the same B!!!!
it gives a cation and an anion.
[BH2L2]+ [BH4]-
36
what does homolytic cleavage give us
Bases on the same B
[BH3L1] [BH3L1]
37
when the base is large,, what cleavage is expected
Homolytic cleavage
to reduce steric hinderance
38
when a small base is used what cleavage is expected
small base gives heterolytic cleavage
the Bases attack the same B as there’s not much steric hinderance.
39
reacting B2H6 with BMe3 givesss
a scramble
but BMe3 is too big to bridge the B’s so it can only go in the terminal position.
remember to think of cis and trans products bc there’s wedges and dashes.
u just keep on adding Me till there’s no more terminal H’s.
40
if ur reacting 2 things together and they give 6 different product possibilities,; how many of everything do u have
u have 8 species on solution (scramble and reactants)
and 6 new products (scramble products)
41
how B and Br is different to B and F bonding : talk about lewis acid strength and orbitals
BF: both have small 2p orbitals that are not diffuse. the F reduces the B’s lewis acid abilities as there’s a strong pi sigma overlap bc the orbitals are both small.
B cannot accept things as readily into its LUMO (2p)
BBr,, B has a 2p but Br has a 4p which is larger and more diffuse. this size difference reduces the sigma and pi overlap between B and Br. it weakens the pi bonding meaning the B’s p orbital is emptier. it’s a stronger lewis acid than with BF
42
how do we form higher nuclearity clusters - aka more metals in a cluster
thermolysis of diborane.
heating diborane under anaerobic conditions leads to higher nuclearity clusters
43
what happens when u heat us B2H6 with no oxygen
- loss of H2
- loss/ gain of BH units
- loss / gain of BH3 units
44
how were boron hydrides originally prepared
adding acids to metal borides
45
boron hydrides were interesting due to what
their potenital use as rocket fuel
46
structures of many boron hydrides are based on what
based on polyhedra
47
name some polyhedra boron hydrides structures correspond to
octahedron ( square with 2 points)
octahedron with one missing vertex
48
what are the BH polyhedra usually looking like
BH units at the verttices
sometimes bridging Hs if ur missing a vertex
49
what hydridisation do we say the B in BH is
sp
meaning theres 2 SP hybrid orbitals
and 2 unybrid p orbitals
50
the 2 unhubrid p orbitals in BH can be used for what
for cluster bonding
51
what are wades rules based on
based on MO theory
52
what are wades rules used to do
predict the structure of boron hydride species by working out the nukber of pairs of e- used for cluster bonding
53
what are PSEPs
polyhedral skeletal electron pairs
the number of pairs of e- used for cluster bonding
54
wades rules and counting e- : how many e- does BH donate ,, how many does a H bonate ,,, how does charge affect the e- count
donates 2e-
aka one pair
H donates 1 e-
u remove // add electrons based on the charge,, add if anionic and subtract if cationic
55
once youve counted how many e- u have in wades rules,, whatdo u do
u divide it by 2
and the answer is the amount of PSEPS u haveeee
56
in wades rules,, if u have a structure with n pseps,, what is the structure based on
its based on the shape that has n-1 vertices
aka 7PSEPs = based on 6 vertices = octahedron , bc this is the 6 vertex polyhedron
57
describe the different polyhedral shapes u have in wades rules
- trigonal bipyradmidal: triangle with 2 points: 5 vertices, so 6 pairs of e- ,, u get 3 diff isomers of it
- octahedron ,, square with 2 points,, 6 vertices,, 7 PSEPs,, 2 isomers
- pentagonal bipyramid ,, pentagon with 2 points,, 7 vertices,, 8 PSEPs ,,
58
once we know what polyhedral shape we are gonna have due to wades rules,, what do we do
count the number of BH units u have
if the number of units u have = number of vertices = closo
if u have one less unit than = nido
if u have less units than = arachno ,, make sure u skip 2 veertices adjacent to one another
59
when it comes to nido and arachno,, what do u need to remeber to do
add bridging hydrogens around the open face.
60
the more open a structure is,, aka u have less units than vertices,, what does that make the structure
it makes the stricture more reactive bc its more open
so arachno is more reactive than nido which is more reactive than closo.
61
when u have a closo octahedron with BH units,, how many peaks will u see in B11[H] NMR
u see one peak ,, bc theyre all in the same env if u just rotate the structure a little bit.
62
nido BH gives how many B environments
it gives us 2 B environments!!!
the cube ones and then the one at the point
63
octahedron when u remove 2 vertices
u get a butterfly structure
aka 2 triangles back to back
with H's bridging ,, and 2 H's coming from each end point
64
how do u form carboranes
u take a borane and an alkyne and high temp
65
what else can we use on carboranes
u can use wades rules
66
whats a carborane
a borane but where some B's have been turned into C's
67
how do we make carboranes,, whats the specific temp too
borane and alkyne and 500-600 degrees
68
what does bh contribute to cluster bonding
2 e-
3 orbitals
69
what is B-H isoelectronic to
its isoelectronic to CH+
70
how many e- do BH and CH+ have
they have 4 e- each
and are 2e fragments for cluster bonding
71
C in CH+ aka carboranes has what type of hybridisation
it has sp hybridisation
same has B in BH
meaning it has 2 sp orbitals and 2 unhybridised p orbitals
72
what is the isolobal principle and what does it relate to
the isolobal principal relates to the BH and CH+ fragments
they are said to be isolobal bc theyre both sp hybrids and said to behave // react in the same way
- they have the same number of e-
- their frontier orbitals have the same symmetry
- their frontier orbitals are of similar energy
frontier being homo and lumo
73
bc theyre isolobal what can be done with clusters with BH and CH
u can still use wades rules on them
74
describe the hybridisation of BH
B = sp
sp points out and overlaps with s orbital of H
u then have 2 p and another sp
and these are non bonding orbitals
75
okay describe a hexagon type with with BH on the corners,, like which way are the orbitals of BH facing
sp ,, the bigger lobe points in
then theres P orbitals along the hexagon ,, not pointing in or out,, just pointing along the edge
then u have the other P whicih is kinda giving out of plane pi bonding,, as in its
76
if we have 6 BH fragments what else do we have
6 sp type orbitals
therefore 6 radial orbitals
we therefore also have 12 tangential orbitals,, 12 P type orbitals
77
what happens to the 6 sp and 12 p orbitals on B6H6 clusters
they overlap and this overlap forms MOs!!
bonding, antibonding and nonbonding
u have 7 bonding ]11 nonbonding and antibonding =!!
78
in B6H6 we have 7 bonding mo's,, so how many pseps do we need to fill them
we need 7 PSEPs
and this will be 7-1 so based on octahedron. 6 vertices which need 7 PSEPs
79
a n vertex pollyhedron has ow many MOs and how many PSEPs do these require
n vertices
so n + 1 MO's
so n +1 PSEPS
80
when u have a closeo structure,, adding more e- does what
it makes us have a nido structure
aka the e- go into the antibonding orbital and therefore favoures bond breaking making nido and arachno structures.
81
how do we make phosphboranes
thermolysis of a 2:1 mixture of (iPr2N)BCl-P(TMS)2. and (iPR2N)BCl2
yields (BR')3P2 where R is (iPr2N)
82
BR is a what e- structure,, wades
its a 2 e- structure so when u use ades rules u saiy one BR gives 2 e-
83
how many electrons does P give u when u use wades rules
it gives u 3 electrons - number of valence electrons
84
wades rules for (BR')3P2
12e- so 6 PSEPS
6-1 = 5 so based on trigonal bipyramidal
needs 5 and has 5 so closo trigonal bipyramidal
85
trigonal bipyramidal gives how many isomers
3
top and bottom
top and middle
middle and middle
86
BH is also isolobal with whattttt
M(CO)3 fragment
where M is a metal
87
how do we find the number of e- in a M(CO)3
we need to calculate the number of d electrons it has
then subtract 6 ( 2e- for every CO)
88
when ur giving a percent abundance of an elemtn,, how do u find how mamny elements is in the molecule
u do the % of the thing) x ( the molecular weight of the total product
///// the m of that element
89
when ur findigng the formula of a product cluster ffrom percent mass and all that what do we need to look out for
we need to make sure the % is adding to 100
90
when u have 2 carbons in ur fragment,, and u have diff isotopes u can make by putting the C's in diff positons,, what do u want to do
u want to put the C's as far away as possible
so for octahedron closo u want them in the 2 furthest point.
91
main group cluster,, describe white phosphorus,
P4
P gives 3 e-
4 x 3 = 12
12/2 = 6
6-1 = 5 so trigonal bipyramidal ,, needs 5 but has 4
so trigonal bipyramidal but nido,, so tetrahderal
92
whwhat happens when u reduce group 14 atoms withh sodium with ligand en
u get isolated large main group elements
93
whats en a ligand,,
NH2 ethene NH2 ,, its a stabilsiing ligand which when mixed with a group 14 element and sodium,, u would isolate the main group cluster that is large
94
okay what if u have a pyramid type shape
this is tetrahedral geometry
95
what is the triangle with 2 caps
trigonal bipyramidal
96
describe tricapped trigonal prysmatic ,
toblerone with a cap at top and the sides
97
describe monocapped square-antiprismatic
twisted cube with a pyramid // cap at the top
98
describe bicapped square antiprismatic
twisted cube with 2 pyramids up and down
99
recuction of group 15 elements,, what doesnt woerk here and why
wades rules dont work here bc goup 15 elements are e- rich
100
when do wades rules work
on electron deficient thingsss
101
reduction of group 25 elements leads to what
leads to open structures, M7^3-
102
what can be used to describe group 15 bonding
valence bond theory!!!
103
what group 15 element forms cationic clusters and therefore can use wades rules
Bi!!!!
Bi5 3+
Bi8 2+
104
what does an organometallic need to have
it needs a direct M-carbon bond
it needs to be matal and a carbon directly attached
105
in gas phase,, what does a BeH2 have shape wise and describe the MO
its linear!!!
think that Be outer orbital is an S
the H orbital is also S
so u have 3 S orbitals
they can be bonding, where theyre all in phase and one colour
then they can be nonbonding where theres a bonding and non bonding interaction cancel eachother out ,, white white black
then u have antibonidng where theres 2 out of phase interactions,, white black white
u have 4 e- in total so ur filling two sigma orbitals. not sigma and pi somehow
so the e- arent found in between the bonds,, theyre found in a massive orbital that surrounds HBeH!!!
so theyre in a bubble and the e- are also in the bubble
this is bc Be is electron deficient and u cant get the full octet rule,, so it just kinda forms a bubble with 2e- in it.
106
whats cool about lithium organometallics
Li - C are normally more covalent and more stable than larger metals
107
group 1 organometallics,, describe them
u have lithium which has the most covalent character and more stable
the other elements are said to have more ionic character and are more reactive and difficult to isolate
108
2 methods of synthesising group 1 organometallics
- direct addition of the metal with alkyl or aryl halide , MeLi!!!
aka CH3Br + 2Li --> (ET2O + 20 degrees) to give CH3Li + LiBr
- reaction of strong lithium containing base with an acidic C-H bond, LiC5Me5
C5Me5H + n-BuLi --> (THF, -78*C) C5Me5Li + n-BuH.
109
whats smt cool about lithium alkyls we talked about last time but then whats aslo kinda sad about them
cool thing = C-Li bonds are more covalent and more stable than other ones
sad thing = they cant respect and reach the octet rule from simple 2 centre 2 e- bonds!!
bridging ligands and oligomeric strutures are common in solution ans in solid state
110
bc Li alkyls cannot reach the full octet rule using 2 centre 2 e- bond,, what must they do
they must use bridging ligands
or form oligomers
111
describe a MeLi tetramer ,, (CH3Li)4
u have Li on some vertices and CH3 on other vertices
the H's kinda stick out from the side
this distorted cube shape this is the tetramer
112
whats a tetramer tho
a structure made up of 4 units
113
okay so the (LiCH3)4 tetramer can also be describes as,, and this can be used to decribe its bonding
Li tetrahedron,, aka u have like a weird 4 point cheese slice looking thing.
and Li is SP3 hybridised so u have the SP3 ligand pointing in thr same direction ,,, these can be all in phase,,, out of phase or nonbonding and cancel eachother out!!
3 black and facing out, all the same size
1 blacj and 1 white with no middle one
2 black and 1 massive white one in the middle
when these all overlap with eachother they cancel eachother out
this combo is the same symmetry as a CH3 sp3 ligand ,, with the larger lobe being in phase wirth the 3 all black Li SP3 orbitals facing it,, these overlap to give bonding and antibonding MOs
theres 1 e- for each CH3 and theres 1 e- for each Li so u have 8e- each!! ,, and bc u have 4 faces that counts as 2e- per face. therefore its a 4C-2e bond
u only need 2e- to fill low lying orbitals,, and u have 4 faces so u need 8e- so 2e- per face and 8e- in total
114
in the MeLi thing,, theyre both sp3 hybrids so how many e- do they contribute
they contribute 1 e- each
and bc u have (CH3Li)4 u have 8e- and only 2e- are needed to fill the low lying orbitals,, which we know bc the MO,, and u have 4 faces so 4x2=8
so we need 8e- ,, which we have
8e- / 4 faces = 2e- per face
each face has 4 atoms that make it up
so u have a 2e - 4c bonddd
its a very reactive moelcule. and u have an empty sp3 obital bc we only use 3.
115
how can the (LiMe3)4 structure be proven and describe what occurs for each one
using 7Li or 13C NMR
for 7Li we look at the i value of 13C!! we expect a quartet bc we assume the Li is bonded to 3/4 of the C!!
- however the acc NMR gives us a septet,, with half the J value we assume,
THIS IS BC WE ENRICHED THE SAMPLE WITH 50% 13C,, WHIHC MEANS THAT NOT ALL THE COUPLING IS TO 13C,, WHIHC MEANS THAT WE WONT ALWAYS GET THE SPLITTING WE ASSUME WE ARE GONNA GET.
AKA IF IT COUPLES TO NO 13C U GET A SINGLET,, IF IT COUPLES TO 1 13C U GET A DOUBLET, IF IT COUPLES TO 2 13C U GET A TRIPLET, IF U COUPLE TO 3 13C U GET A QUARTET,,
HOWEVER ALL OF THESE OVERLAP TO GIVE THE SEPTET WE SEE IN NMR,, SO ITS THE AVERAGE NMR IS U COUPLE TO DIFF 13C,, BC U ONLY ENRICH IT BY 50%,, NOT 100%.
u can then do a 13C nmr using 6Li,, at185k and 299k aka at a cold and hotter temperature.
and from the diff nmr wr can see that thre will be 1C bonded to 3Li so we expect to see a septet,, bc the i value of 6Li is 1. and we do see this in the low temp 13C nmr,,, but then we also see 9 peaks at high temp,, bc the CH3 decides to jump to each face meaning it couples to all 4 Li,, meaning u get 9 peaks!!! and it moves around faster than the nmr can sense so yhh u get a septet or a 9 peak peak!!
116
his iswhat happens when u mix a group 2 metal and a RX
u get a schlenk equilibrium
MgX2 + MgR2 <---> 2RMgX.
117
what happens when u go down the group of group 13,, aka the B, Al, Ga, Ln, Tl thing
u get a lower oxidation state.
u go from 3+ at the top of the group to +1 at the bottom of the group!!
due to the inert pair effect and the fact that the s electrons are less likely to engage in hybridisation,, bc thereyre better at shielding and penetration close to the nucleus,, making them less likely to occur in hybridisation.
118
describe the shape of BMe3
the shape of BMe3 is a monomer,, aka it doesnt form a dimer or bridge
its just a B with Me attached to it
119
how d u form BR3,, u get a gringard reagent and an ether.
this gives u BF3 and BR3 and a salt and ether.
3RMgX + OEt2 --> BF3 + BR3 + 3MgXF + Et2O
120
describe the structure of Al2Me6
its a dimeric structure
meaning theres bridging CH3
so u have 2 Al and u have 2 bridging CH3 and then u also have 2CH3 coming from the Al ,, and the angle between these is 123* which is more similar to a sp2,, trigonal planar shape instead of sp3 but its calm.
u have symmetrical methyl bridges : the C of these is pentavalent
121
in the Al2Me6 structure,, what do we seen in room temp 1H and 13C NMR at room and -40* temps
at room temp u have one methyl signal
at -40 u have 2 methyl signals
this is bc u have bridging and terminal CH3
meaning theyre in different environments,, but in reality the bridging Al-C bonds are breaking and a terminal C is then bonding and being turned into a bridging one.
this is happening rlly fast at room temp so the 2 different C's are averaged into one. so u eg tone Me peak.
at cooler temps this process slows down,, and then u can acc see the differences between the terminal and the bridging Me!!!
and this is why u get 2 different peaks for the different environments!!!
122
in the Al2Me6 dimer with the bridging Me things,, what hybridisation is the Al
its sp3 hybridisation,, meaning u have 4 total sp3 orbitals,, they use 2 of these to bond to the terminal CH3
then they use another sp3 to bond with the other Al and the bridging Me,, they,, aka the Al do this by just pointing their sp3 orbitals upwards,, and these can be bonding or antibonding or nonbonding obvs,, if theyre doing the bonding interaction then the sp3 of Al are the same phase,, and so is the sp3 CH3. when its antibonding the Al sp3 are the same phase and the CH3 sp3 is the opposite phase.
when u have non bonding the Al sp3 are opposite phaes and the sp3 of the C is the same phase as one of them!!! which is how the interactions cancel out and u get non bonding instead of bonding or antibonding.
so yesss
123
when u have the Al2Me6 structure,, what type of bond is the Al-C-Al bond
its a 3C, 2e bond!! bc theres only 2e- between the whole 3 atoms things
124
in the Al2Me3 structure we said thats they're sp3,, but how else can they be described // explained as
they can be explained as sp2
with the sp2 hybrid orbitals forming the 2c-2e bond
and the P orbitals either being bonding or non bondoing but overlapping with the sp3 of the C to give u the 4c-2e bond
125
amide in inorganic chem =
R3N- this is an amide,,, in inorganic chem,, when u have dimers,, what does the strength of the bridging atoms depend
depends on their electronegativity // pi electron donation ,
aka the lone pair of things like R2N, Or, Cl or Br allow it to bond to the Al
then things like alkynes and phenyl rings can use their pi electrons to do so.
then u just have ur Me and other alkyls at the back.
R2N> OR > Cl > Br >>> alkyne > Ph > Me.
the more electronegtaive smt is,, the worse it is at being a bridging group bc its not gonna want to donate its lone pair to the Al.
126
when u have a trans bent structure,, what can u use to help stabilise it
u can use bulky ligands like MeS to thermodynamically and kinetically stabilise it.
127
Sn=Sn bonds are also known as what
they're known as distannylenes
