lecture 1b Flashcards
describe methyl lithium and why this is the correct structure
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
connectivity of C in MeLi
connectivity is 6
it’s bonded to 3 H’s and 3 Li’s
describe trimethyl aluminium
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!!!
what does the NMR of 1H and 13C show at room temp when we look at trimethyl aluminium
we would see 1 methyl signal!!!
bc there’s interconversion between the bridging and terminal Me ligands (via mono bridged species)
what does the NMR of 1H and 13C show of trimethyl aluminium in (-40*C)
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.
describe borane
BH3
describe diborane
2 terminal H’s bonded to B ,, bridged by 2H’s,, then another B then 2 terminal H’s
how do we synthesise diborane // borane
NaBH4 + I2 —>
2BH4- + I2 —> 2’BH3’ + H2(g) + 2I
‘BH3’ bc it’s acc diborane and not borane
how can we tell diborane was formed and not borane
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
% abundance of 10B
20%
% abundance of 11 B
80%
chance of B2H6 (diborane) being completely 11B
0.8 x 0.8 = 64%
chance of B2H6 being made up of one 10B and one 11B
0.8 x 0.2 (x2) bc it could be the other way around aswell
32%
chance of B2H6 being made jo of two 10B
0.2 x 0.2 = 4%
if we add all the %’s for the different isotopes of B in B2H6 what do we get
we get 100%!!!
aka diborane must be made up of 2B’s!!
a borane dimer can look likeeee
like ethane
like diborane (the bridging H’s)
going back to the IR spec,, how do we now know it’s def diborane
bc there’s a BH bridging stretch (the previously unknown one at 1500-2000)
and a BH terminal bond stretch at 2000
why do we ignore 10B in nmr
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
what B do we focus on when looking at NMR of diborane
we look at 11B
which has a spin of i = 3/2
1H NMR of BH3 explained
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!!
11B NMR of BH3
- 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
1H nmr of B2H6 (ethane one)
- 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!!
11B nmr of B2H6 (ethane one)
- 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
1H environment of B2H6 (bridging one)
- look at H’s
- 2 different H environments (terminal and bridging)
- ‼️‼️‼️‼️‼️‼️‼️ go over again
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)
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
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.
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
what does the conventional valence bond theory require
it requires 2e- for every bond
2C, 2e bond
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
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-
boron hydrides and anionic boranes general formula
BnHm
BnHmx-
both are electron deficient.
new theories of chemical bonding were formulated to explain their structures.
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
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.
what does heterolytic cleavage give
bases on the same B!!!!
it gives a cation and an anion.
[BH2L2]+ [BH4]-
what does homolytic cleavage give us
Bases on the same B
[BH3L1] [BH3L1]
when the base is large,, what cleavage is expected
Homolytic cleavage
to reduce steric hinderance
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.
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.
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)
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
how do we form higher nuclearity clusters
thermolysis of diborane.
heating diborane under anaerobic conditions leads to higher nuclearity clusters
