I'm just here and YouTube Studios is
closed so I have no way to shoot more
content and I'm almost out of backlog
releases so I was trying to brainstorm
some other ways to make content from
home and I came up with this idea of
doing some ochem practice I came across
some old practice exams from a CSUN
ochem course from when I used to tutor
over there and I thought why not just go
through some of these so I'm gonna make
a few videos with a little bit of ochem
practice for you guys so let's go ahead
and and jump right into this one so we
have this substrate and we want to know
what's going to happen under these
conditions so we have this organo
cuprate right so we've got an organic
OOP rate and this is as such we've got
two ethyl substituents right and copper
bares a negative charge and then we've
got a lithium counter ion so I do have a
tutorial on organic cuprates that this
looks unfamiliar you can go and check
that out but what these do this
essentially just delivers an alkyl
nucleophile so we're gonna be able to do
this let's say this ethyl group right
here will go and just do sn2 and kick
off the bromine so when we see organic
cuprates
it's it's a little confusing because we
see two of that of that alkyl group but
whatever it is one of them is going to
act as a nucleophile so what do we end
up with we've got the same molecule so
that's the that's the carbon that was
attacked and then we need to have two
more carbons on there
so these are the two additional carbons
that came from that organic cooperate so
that's that first one okay so this one's
pretty easy
whenever we see magnesium we know what
we are going to be generating a grignard
reagent if we see some alkyl halide in
this case we better now key
bromide and so all we're gonna do is
we're going to insert magnesium into
that carbon halogen bond so we're here
and then instead of Br we're gonna put
mg right there and then we're gonna put
BR right there
so just one quick point notice how I put
mg here and then BR rather than mg BR we
do want to show that the magnesium is
covalently bound to the carbon right
this is the this is now the partial -
carbon that is going to go and attack a
carbonyl but we do want to put mg right
here and then BR out there because mg is
in between BR and carbon so that's just
preparation of a grignard reagent okay
so now we are doing bromination this is
radical bromination right we know that
we need some light of a particular
frequency to promote home hemolysis of
the bromine molecule right we know that
we are gonna under these conditions
we're gonna get bromine radicals and
that is going to where does gonna allow
us to brominate an alkane right and so
what we need to know is that the bromine
radical is highly regioselective because
of its enhanced ability over the
chlorine radical and so we're gonna be
looking at this it is going to attack
specifically the most substituted
location on the molecule because that
will that will provide the most stable
radical intermediate and so we know that
whenever we do radical bromination we
always get the most substituted alkyl
bromide that we can and so because we
have two tertiary carbons we can just
pick either of them let's just let's
just put the br here let's just say that
right so we're going to add that bromine
atom right there and that's going to be
the answer for that one
okay moving on we have this alkene
substrate and we are going to
drogyn eight over in this case a plea
diem catalyst and so we know that when
we do this we're hydrogenating we're
hydrogenating the PI bond we're gonna
put two hydrogen atoms on there and so
all we have to do is get rid of the PI
bond that's all we're gonna do there so
that is the case they're alkynes we also
can hydrogenate alkynes and were there
were the case that we would be producing
chiral centers then we would have to
know that it is a syn addition but in
this case it's not relevant so we just
get rid of the PI bond and that's all
there is to it okay just a couple of
multiple choice here so number one cyclo
butane is less stable than cyclohexane
primarily due to so we've got some
options here remember cyclobutane it
looks like that and cyclohexane looks
like this so what do we know about these
molecules well we do know that
cyclohexane is an extremely stable
cyclic molecule because all of the sp3
centres obey a perfect tetrahedral
geometry these are all 109 whoops 109
point five degree bond angles right so
this is there's no strain on any of the
carbon atoms any of the carbon-carbon
bonds or carbon hydrogen bonds in this
molecule
whereas for cyclo butane which is
essentially square what do we know about
the angles in a square well those are 90
degree angles right and 90 degree angles
that that forces certain electron
domains to be closer together than they
would be under you know these are still
sp3 centers right any sp3 Center will
prefer to have tetrahedral geometry in
109.5 degrees bond angles so this is
angle strain right the the the the the
lower stability of cyclobutane is
attributed to these compressed bond
angles so it's not torsional strain that
would be like if you were looking at a
Newman projection and you had some
gauche interactions or something like
that that's a kind of steric strain and
it's not steric strain
in general it is it is due to the
compression of these bond angles beyond
what they would prefer to be so that
would be angle strength now a molecular
orbital can be occupied by and so we
don't even care this is a molecular
orbital any orbital any orbital can be
occupied by a maximum of two electrons
right there can be two there can be zero
one or two electrons in any orbital
whether atomic orbital hybrid orbital
molecular orbital and the must have if
there are two electrons they must have
opposite spin right we know that because
the Pollack's exclusion principle no two
electrons in a system can have the same
set of quantum numbers and so in the
same orbital they have three out of the
four quantum numbers identical but that
fourth one the spin quantum number will
be opposite so that does not violate the
Pauli exclusion principle so any orbital
maximum two electrons covalent bond
hemolysis forms and we just saw an
example of this when we saw the radical
bromination of of that alkane right that
would be like this this is covalent bond
hemolysis right hemolysis involves the
single headed electron pushing arrows
which denote the motion of one electron
at a time this is different from what we
usually see in most mechanisms which is
heterosis where right let's say you have
right some alcohol let's say a base
takes this proton now notice the
double-headed arrow this would be
heterosis both of the electrons in this
bond go to oxygen so hemolysis is
different and this affords us with in
this case bromine radical so hemolysis
we want to associate that with radicals
okay now we have on I you pack this is a
pretty easy one let's go ahead and
identify the longest carbon chain we
know that when we're doing IU pact we
always want to identify the parent chain
which has to be the longest carbon chain
that
one of the rules of IU pack so we got to
go up this way we got to go up here
because that is a longer carbon chain
remember don't just take whatever's in
the in this in this line here left to
right if there's a longer chain you got
to use that so that's gonna go up there
now we got a number this is this going
to be numbered left to right or right to
left well we need the substituents occur
the first substituent occurring soonest
so we got to go left to right because
that gives us this methyl on two so we
go 1 2 3 4 5 6 7 so we do have a heptane
because we have a heptane that's we know
it's not that right and so did we number
this the right way right we've got two
options that that are the difference is
if you numbered it the right way or not
so we know we have 2 5 dimethyl if we
numbered it incorrectly we would have
gotten 3 4 5 6 so that's the trick
answer there we don't want that 2 5
dimethyl heptane we need the methyls
occurring sooner oh and that's it for
the first one there so I'm gonna do a
bunch more of these and stay tuned for
those thanks for watching guys subscribe
to my channel for more tutorials support
me on patreon so I can keep making
content and as always feel free to email
me professor Dave explains at gmail.com
