part 1 year 2 how do we make Flashcards
what does borazine cosist of
B and N cyclic skeleton
how do u make a borazine with Ph on the B and 1Me on the N
u need
3Me2NH2 + 3 BPhCl ā> the structure
u also need to remove the 6HCl when u make it
sp2 C building block makes what
cage: fullerenes
2D sheet: graphite
an sp3 C building block makes what
cage: P4 aka white phosphorus
2D sheet: black phosphorus, P.infinity
describe Pinfinity aka black phosphorus
u have a cyclohexane type thing but with P
and u have bonds equitorially to other Pās and lone pairs axially
huckels rule
flat // planar molecule
conjugated double bonds
4n+2 pi electrons
cyclic
a neutral B with 3 bonds is what hybridisation
its sp2
when u have an O and B adjacent to one another,, why would they both be sp2 hybridised even if they have single bonds
bc back donation can occur
the lone pair from O can gi into the empty B orbital giving them both sp2.
how can we say that a normal B with 3 bonds is sp2 hybridised
bc it has 3 bonds meaning 3 bonding interactions and no lone pairs and an empty p orbital
3 bonds and an empty p orbital is characteristically sp2!!!
a dative bond is treated like
a dative bond is treated the same as a sigma bond
so it still counts as a normal bond
B with 4 single bonds is
sp3 hybridised
bc sp3 = 4 bonding interactions
B with 2 single bonds and one double bond is
sp2
bc still 120 bond angles
if theres a N with 3 sigma bonds and a lone pair,, why could it not be sp3
look for if the lone pair can be conjugated in the molecule,, if it can,, its sp2.
things to look at to see if smt is sp3, sp2, etc
- how many bonds // lone pairs does it have
- can the lone pairs be put into a empty p orbital // conjugated into the system
- look at the bond angles
when u have a ring with double bonds and the substituents are a phenyl,, when ur choosing whether its aromatic or not,, do i inlude the phenyl e- in the equation
nopeee
u just use the e- in the ring
are BN // borazine rings flate
yes
when is something antiaromatic
when its planar, conjugated and cyclic but has 4n pi electrons instead of 4n+2 electrons
when we draw the lowest energy mo of smt that is aromatic with C, N and B atoms in the ring what do we need to think of
think of the large energy diagram thing
with the cicles representing the atoms but remmeber that for benzene all the circles are the same size bc they have the same coefficient bc they have the same electronegativity
but when u have N C and B
KF and cyclophosphazing
- activates P by making is more electropositive and so addition will be geminal
cyclophosphazine and 4M2eNH
only 2 get added to mop up the HCl formed by making a salt
non geminal bc N: goes into P centre which deactivates it.
trans bc of steric and theyre large
u also make a salt with the X and the rest of the amine
2Me2NH2 (+) Cl(-) this is the salt u form
lets say the question says u have 1:1 ratio of two things what do we need to look for
we need to see if we have a compound of N and a compound with another element.
we need to see if the other element is in the same group as B
if it is ur gonna make a ring with the B compound and N.
ur gonna keep one of the substituents on the B molecule that it comes with,, the N is also gona. ahve one of the substituents it comes with
1 peak in nmr means what
there can be multiple of that element,, but theyre in the same environment