ok guys today's practice problem is about nomenclature, we've got an alkane
we've got an alkene, we've got an alkyne, and a cycloalkane, so let's go ahead
and give these the correct IUPAC name and if you're having trouble with this
go ahead and check out, I have several lectures on IUPAC nomenclature
ok so let's go ahead and name these one at a time, remember that when we're
generating an IUPAC name for a compound we have to do a couple of things
the first thing we have to do is identify the parent chain, the parent
chain is always going to be the longest carbon chain in the molecule, you need to
be careful because we have a tendency to just say okay there's the parent chain
but actually it's gonna be longer if we go up here so here's the parent chain
from here to here
the second thing we have to do is then number that parent chain. if it's a
linear molecule it's either gonna be left to right or right to left. remember
that if all we're dealing with is alkyl and halogen substituents which all have
the same priority we are simply going to choose the side that has the first
substituent occurring sooner, so if we're starting over here we have a methyl
group on carbon 2, if we're starting over here we have a methyl group on
carbon 3, so that means we're gonna number this left to right so we're gonna
go 1, 2, 3, 4, 5, 6, 7, 8 so that's the first step we identify the parent chain and then we
select a direction in which we're gonna number so now that we've done that we
have to identify all the substituents. so let's identify the substituents we
have a methyl there, we have an ethyl there, we have two fluoros and we have a
methyl there. now completely separately from the algorithm we used to number the
parent chain, a completely different algorithm we use is the one to decide
the order in which we list the substituents and that is going to be in
alphabetical order, and so what we need to do here
is we've got an ethyl, we've got methyls, and we've got
fluoros so the first thing alphabetically is ethyl, and then we just list the
substituents in alphabetical order where they occur, so on carbon three we've got
an ethyl so that's going to be three ethyl. now on carbon four we have two
fluoros so that is going to be 4,4 we do have to list both of those
and we are going to say difluoro, first of all thats fluoro not flouro, I see
that common error all the time, and 4,4 we do have to list where
both of those are occurring even if they are on the same
atom, so that there is no ambiguity. and lastly we do disregard the prefixes in
terms of alphabeticity so difluoro still occurs after ethyl because
it's F for fluoro, not D for difluoro, so E for ethyl, F for fluoro, and then next we
have 2,6 dimethyl and then we go directly into the parent molecule
with no space which is going to be an octane, so 3-ethyl-4,4-difluoro-2,6
dimethyloctane is going to be the name of that one. now for this one here we have
two functional groups that tend to take greater priority over alkyl and
halogen in terms of numbering, however the one that takes the greater priority
is the hydroxyl. hydroxyl has priority over alkyne which has priority over
alkene which has priority over alkyl and halogen substituents in terms of
numbering the carbon chain. so we know that the carbon chain is gonna be 1, 2, 3, 4, 5, 6
like that but we're deciding to go left to right or right to left on the basis
of the position of the hydroxyl, so we're going to number right to left because
that will have the hydroxyl occurring sooner. ok now we list the
substituents except for hydroxyl and alkene because those are both modifying
the suffix of the molecule, we will save those for the end. so what we've got, we've got
a promo group, B for bromo
comes first, 6-bromo and then 4-isopropyl, now comes the tricky part
because both the double bond and hydroxyl are trying to modify the suffix
of the molecule, so this is what we end up doing, we have to say first of all
it's six carbons and we say 5-hexen, and we drop the e on the ene because
we need this e right there to indicate the position of the double bond
or an alkene but then we drop off that last e because there's more to go and
we say 3-ol
because we need to know that it is an alcohol and the position that the
hydroxyl group is on. 6-bromo-4-isopropyl-5-hexen-3-ol so there's that one
over here we've got an alkyne and we know that we're gonna number this parent
chain so as to have the triple bond occurring soonest, so once again
left-to-right or right-to-left it will happen sooner from the right so let's go
1, 2, 3, 4, 5, 6, 7 so let's realize that these are carbos, these are sp hybridized carbons, a
very very common error is to not realize when drawing an alkyne in line notation
we know that carbons participating in triple bonds are sp hybridized which
therefore have linear molecular geometry so that's why we draw them like this
they have to be, in order to be geometrically accurate we do draw them
like this, that is a carbon, that is a carbon, we have a seven carbon alkyne
as well as a methyl group so we're just gonna list that methyl real quick so
that is going to be 6-methyl and then now we list the presence of the triple
bond as well as where it occurs, it begins at carbon 3 so that is
3-heptyne, hept because it is a seven carbon alkyne and then yne for
the triple bond, 6-methyl-3-heptyne. now lastly we have this cyclohexane
substituted cyclohexane, now these are numbered a little bit differently than
the linear molecules if we recall because it is not about left to right
and right to left
with a linear molecule we've got two options, this way or that way but with
the ring this is cyclical, there's no given
starting point so the algorithm we use to number a cyclohexane ring
or any ring, is a little bit different, we are going to assign values to the
carbons in the ring so as to give all of the substituents occurring at the
lowest set of locants, so basically we've got twelve options, any of these
six could be carbon one and then from there we could go clockwise or
counterclockwise, so we've got two options there, two there, so
12, we've got 12 possibilities and so we're gonna pick one of those twelve that
occurs that gives the substituents occurring on the lowest set of locants
and so for example let's say we chose this was carbon one and then we went
clockwise it'd be 1, 4, 5 right over the other way it be 1, 3, 4. 1, 3, 4 is lower than 1, 4, 5 so
that's preferential but the lowest possibility is here, if we choose this to
be carbon one and then go clockwise, 1, 2, 4 is the lowest set of locants, there is no
other numbering scheme that you could choose that will have those occurring on
a set of numbers that will add up to be lower than seven. this is the preferred
numbering scheme and so once again because these are, so alkyl and
halogen all have equal priority so it's arbitrary how we, there's not one group
that takes priority over another
if there was a hydroxyl group that would take priority and that would be carbon
one, it would be implied, but here they're all halogen so that's how we
number it and so then we just list in alphabetical order so that's going to be
4-bromo-2-chloro-1-iodocyclohexane so that's the correct IUPAC nomenclature
for all four of these compounds
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