Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

inorganic y2 / s1 > lecture 1b > Flashcards

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

connectivity of C in MeLi

A

connectivity is 6

it’s bonded to 3 H’s and 3 Li’s

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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!!!

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

describe borane

A

BH3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

describe diborane

A

2 terminal H’s bonded to B ,, bridged by 2H’s,, then another B then 2 terminal H’s

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

% abundance of 10B

A

20%

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

% abundance of 11 B

A

80%

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

chance of B2H6 (diborane) being completely 11B

A

0.8 x 0.8 = 64%

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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%

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

chance of B2H6 being made jo of two 10B

A

0.2 x 0.2 = 4%

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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!!

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

a borane dimer can look likeeee

A

like ethane
like diborane (the bridging H’s)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
Q

11B nmr of B2H6 (bridging structure)

A
  • 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
Q

H nmr of diborane

A

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
Q

looking at the bridging H when we look at 1H nmr what do we see for diborane

A

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
Q

does valence bond theory work for diborane + explain

A

nope
u have 8 bonds in diborane
but only 12 valence e-
(3 + 3 + 6)

the BHB bridging is a 3c 2e bond

29
Q

what does the conventional valence bond theory require

A

it requires 2e- for every bond
2C, 2e bond

30
Q

is diborane fit the conventional valance bond theory how many electrons would it have

A

it would have 16e-

bc it’s made up of 8 bonds
and 2e- in each bond

31
Q

why is diborane called electron deficient

A

bc conventional valence bond theory says u need 2e- in each bond

but diborane only has 12 e- ,, not 16e-

32
Q

boron hydrides and anionic boranes general formula

A

BnHm
BnHmx-

both are electron deficient.
new theories of chemical bonding were formulated to explain their structures.

33
Q

explain homolytic cleavage of diborane

A

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
Q

heterolytic cleavage

A

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
Q

what does heterolytic cleavage give

A

bases on the same B!!!!

it gives a cation and an anion.
[BH2L2]+ [BH4]-

36
Q

what does homolytic cleavage give us

A

Bases on the same B
[BH3L1] [BH3L1]

37
Q

when the base is large,, what cleavage is expected

A

Homolytic cleavage

to reduce steric hinderance

38
Q

when a small base is used what cleavage is expected

A

small base gives heterolytic cleavage

the Bases attack the same B as there’s not much steric hinderance.

39
Q

reacting B2H6 with BMe3 givesss

A

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
Q

if ur reacting 2 things together and they give 6 different product possibilities,; how many of everything do u have

A

u have 8 species on solution (scramble and reactants)

and 6 new products (scramble products)

41
Q

how B and Br is different to B and F bonding : talk about lewis acid strength and orbitals

A

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
Q

how do we form higher nuclearity clusters

A

thermolysis of diborane.

heating diborane under anaerobic conditions leads to higher nuclearity clusters

inorganic y2 / s1 (10 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions