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5.2 How to Assign R and S Configuration | Absolute Configuration | Organic Chemistry

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assigning r and s in this lesson we're

going to talk about what are termed

absolute configurations now these are a

designation given to a

chiral center now this is the second

lesson in a whole chapter on isomers and stereochemistry

stereochemistry

in the first chapter part of what we

learned was what a chiral center was and

how to recognize them in a molecule

and that they exist in two different

forms that are mirror images of each other

other

in this lesson we're going to learn that

there's a system of rules called the

khan ingold prelog system

and they help us give a designation to

those two different forms of a chiral

center as either

r or s so that we can distinguish them

one from

another now if this is your first time

joining me my name is chad and welcome

to chad's prep

my goal here for the channel is to help

make science both understandable

and maybe even enjoyable now this is

part of my new organic chemistry

playlist i'll be releasing these lessons

weekly throughout the 2020-21 school year

year

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all right so let's get right into this

here so absolute configurations

assigning rns

we'll look at some chiral centers here

and as a reminder chiral centers have four

four

different groups attached and so in this

case we can see one right here this

carbon right here

is bonded to a chlorine it's also bonded

to a hydrogen that's not drawn in that

would be a

dashed bond and that hydrogen is

technically right behind that wedge bond

the dash is always right behind the wedge

wedge

so so again this chiral center right

here this carbon is bonded to a chlorine

a hydrogen then it's bonded to the

carbon of a methyl group and the carbon

of an ethyl group so those carbons are

definitely different now the first part

of your con angled prelog system says

you're going to give the four things

that chiral centers attach to a

designation of

or a priority we should say of one two

three and four and you first start those

priorities by

uh according to atomic number here so

out of hydrogen chlorine carbon and

carbon chlorine's got the highest atomic number

number

so we're gonna give him a designation of

one hydrant's got the low atomic number

we're gonna give him a designation of

four and from there if you've got a tie

like we do with these two carbons

then you'll take a deeper look and see

what are those carbons bonded to

so in this case they both have a bond

coming from the chiral center itself we

don't recount that

so but then you say what other three

atoms are we bonded to well this one

right here

is one of the three hydrogens and so we

just list those kind of hhh

now the next one here it's bonded to a

carbon right here

and then it's bonded to these two

hydrogens right here

and so in this case it's a carbon and

two hydrogens and you want to list those

in descending order of atomic number so

carbon first and then hydrogen

and then hydrogen and so when comparing

these two carbons right here

i'm actually comparing these three to

these three but you're not actually

comparing them all at the same time

you're comparing them until you find the

first point where they're different

and the first place where the different

carbon beats hydrogen once again

according to

atomic number and so as a result we'd

give this carbon right here a

designation of two

this one a designation of three and

again the hydrogen being the low atomic

number got the designation of four

now from there you're then going to make

a circle from one to two to three

now some people say go one two to three

to four but don't really worry about

four just one to two to three is suffice

it to say

and in this case we're going to go around

around

and we can see that this is clockwise

now i don't like saying clockwise i like

saying a right-handed turn like if you

were turning a steering wheel this would

be a right-handed turn

and that's actually where r and s come

from so r comes from the latin

word rectus which means right so s comes

from latin word sinister which

means left and so a right-handed turn is

going to

go with r and a left-handed turn is

going to go with s as we'll see here

so but they had to give us just a little

more than just that so

because it depends on which side of the

molecule you're looking at so notice if

i make a

i'm going to put my hands up do me a

favor right on the other side of the

camera put your hands up

so i'm waiting for you come on yep there

you go put your hands up now i'm going

to make a right-handed turn

so and if you're putting your hands up

and matching my hands look and you

should be making a

left-handed turn but notice again i was

making a right-handed turn

and so they had to give you some

perspective here they couldn't just tell

you make a right or left-handed turn

they had to tell you which side of the

molecule you're looking on because in

this case if i look at this molecule

from this side it looks like a right hand

hand

turn but if i could crawl in behind the

board and look at the molecule from the

other side it would actually end up

looking like a left-handed turn

so going around this way instead and so

what they said is that when you

look at your turn whether it be left

hand or right-handed

your number four priority group has to

be going away from you in the dashed position

position

that's kind of the designation they gave

and so this is indeed a right-handed turn

turn

and as long as your number four party is

in the back going away from you

then right-handed turn is going to mean

r cool so what if

your lowest priority group is not in the

dashed position

and that's we're going to deal with in

this next example so in the in this next

example chlorine is still going to be

number one so this carbon's still going

to be number two and this carbon is

still going to be number three

and once again that hydrogen that's not

is going to be right there and that's

going to be priority number

four and so now we have a problem i make

my circle from one to two to three

and for me it still looks like a

right-handed turn and it should look

like a right hand turn to you

the problem is we're looking at this

molecule from the wrong side

they said no no no your number four

priority is not going away from you it's

coming out towards you

what you need to do is actually flip

this molecule over

so that that hydrogen is actually going

away from you well we're not actually

going to flip it over

what we're going to do is just take

advantage of the fact that what we just learned

learned

that if from this perspective which is

exactly 180 degrees the wrong

perspective it looks like a right-handed turn

turn

then if actually again i crawled in

behind the board

and was looking at it from the other

side that hydrant would be going away

from me

and it would look like a left-handed

turn and so the idea is this then

if your number four priority rather than

being a dash where it's supposed to be

if it's a wedged bond it's just opposite

of the way it looks

so it looks like a right-handed turn

from this side which is the wrong side

from the correct side it would have

looked like a left-handed turn and this

corresponds to

the s isomer instead

okay so last possibility though and this

is the worst one of them

is what if your lowest priority group is

not a dashed bond

not a wedge bond but it's one of the two

bonds that's in the plane of your paper

in this case in the plane of the board

so well that kind of sucks and i'm gonna

warn you

it's the worst case scenario and it's

the one that students struggle with the

most and

there's a couple of different approaches

for how to look at this now i'm gonna

give you the approach that

i look at it with and then i'm gonna

recommend that you don't look at it that

way just yet

but the way i look at this is instead of

looking at this head on

so i go through assign my priorities one two

two

three and four so same same notice it's

the same groups we had before

these other two examples and what i do

is instead of looking at it head on

i come off to the side here and look at

it from the side with the hydrogen going

away from me

and so as a result the chlorine here on

a wedge bond is out over here

out in front of the board so number two

is up in the plane of the board and then

number three

is back behind the board down below at

that methyl group so one

two three one two three one two three

one two

three and that's a left-handed turn for

me with the h going away from me

and that means this is going to be s

so however students often struggle with

the three-dimensionality of this early

on in ocam

and so as a result maybe this is not the

best way now

for students who have had a year of

organic chemistry and coming back and

reviewing some organic chemistry

this is actually what i recommend they

do so but for sophomores who are taking

and experiencing

organic chemistry for the first time

this might not be the best approach

so we're going to take a little bit

different view of this now one of the

things we can do

is we could take the other bond that's

in the plane and just kind of

rotate that bond around until this

hydrogen was either on the dash or the wedge

wedge

and that's one way to approach and it's

commonly taught that way but students

screw that up all the time so i'm

actually not going to teach that way

either but you can do that as long as

you're just rotating a bond it's the

same molecule

but what i'm going to do and recommend

is something a little bit different

now in the last lesson we learned that a

chiral center is an example of a

stereocenter if you take

any two groups and you trade them places it

you get the opposite version of that

chiral center so well in this case

that's going to mean that

r turns to s and s turns to r and so

what i'm going to do

is have the hydrogen trade places with

another group

and technically i could have a trade

place with either the wedge or the dash

and and i

put it in a configuration where i could

assign it like one of these two so but

usually i just trade it with the dash

and go from there so

in this case what we're going to do is

we're going to leave the ethyl group alone

alone

and we're going to leave the wedged chlorine

chlorine

alone so however i'm gonna where the

hydrogen used to be that's where i'm

gonna put the methyl group

and where the methyl group used to be

that's where i'm gonna put the hydrogen

now we have to keep in mind here that

these are not the same molecule

if you have any two groups exactly two

groups trade place on a chiral center

it automatically inverts it and it's the

opposite form so these are not the same

they're opposites

and again this one's still hard to

assign but this one's not that number

four priorities in the back that's easy

to assign

and so chlorine is number one carbon's

number two this methyl group is number three

three

and one two three hydrogen's four so but

that's a right handed turn

and this is r and if having substituted

you know having two group swap places

gives me the inverted stereochemistry

where it's r

then the original just like we already

had said must indeed be

s so that's kind of the way i look at it

so again i didn't rotate a bond here if i

if all you do is rotate a bond you still

have the same molecule i didn't rotate a bond

bond

i had two groups and i detached them and

then reattached them

in the opposite places and again if you

have two groups swap places on a chiral center

center

you get the opposite version but once

again it was really easy to assign this

one right-handed turn

lowest priority in the back it was r

which means that the original must have been

been

s cool if you've already got this down

in one of the other two ways

fantastic and again looking at it from a

different perspective so you can get the

number four priority going away from you

really is the fastest way but again

this last one i find is just the one

that students tend to screw up the least

when taking organic chemistry as sophomores

sophomores

all right so now we're going to look at

a couple of little more difficult

examples and

professors are notorious for going over

rather simplistic examples in class and

then asking much harder ones on the exam so

i'm gonna try and ask some questions

here that might be a little more

representative of what you might

encounter on an exam here

in your organic chemistry course so in

this case this first one's definitely

got a chiral center right here

it's the only chiral center in the

molecule so and that

chiral carbon right there is bonded to

four different things now three to those

four things are all

carbon atoms so but not drawn in here

there is a

fourth bond in this case the fourth

group is a hydrogen

and then comparing carbon to hydrogen

the hydrogen definitely is going to have

the lowest priority

so that's going to be priority number

four now you'll find that a lot of

organic molecules your

lowest priority is a hydrogen and a lot

of students get in the habit of being

like well just make sure the

you know find your hydrogen well it

doesn't have to be hydrogen just

whatever your lowest priority is whether

it's height or not

so that's kind of again how they define

things and when it's in the back

that's when a right-handed turn means

are and a left-handed turn means s

so life is good here getting your having

your your number four priority already be

on the dashed position that's where we

want it but assigning priority for the

others we got a three-way tie with

carbon-carbon and carbon

in this case similar to what we've done

before we'll then look at what three

additional atoms are these carbons

bonded to well this carbon right here is

bonded to this carbon

he's also bonded to these two hydrogens

and so we'll list this as carbon

hydrogen hydrogen listed in descending order

order

of atomic number so for this carbon down here

here

he's bonded to the sulfur and then he

also is bonded to a couple of hydrogens

not drawn in

and so we'll list this as sulfur

hydrogen hydrogen

and then finally this carbon right here

he's got a double bond to an oxygen here

a single bond to an oxygen here

but he's not bonded to three additional

atoms he's only bonded to two additional atoms

atoms

and so to take into this into account

what you do when you've got a double

bond is you actually count each bond separately

separately

so this counts as one bond oxygen this

counts as a second bond auction

and then we have yet another bond to an

oxygen and so we'll list this out as oxygen

oxygen

oxygen oxygen those first two oxygens

being accounted for by the fact that

it's a double bond

to a single oxygen this way we're always

comparing three things

now we get here and we can see that

there are some differences here

and a lot of students will ask me hey

chad what's better three oxygens or a

sulfur and two hydrogens who's going to

get the higher priority

so and that's not the right question the

con angled pre-log system says

when you're going to compare this now

you just look for the first place where

they're different

and so what i'm really comparing here is

oxygen to sulfur

and then on this one to carbon and it's

there's no ties anymore and so in this case

case

sulfur's got the highest atomic number

out of those three and so he's going to

cause the carbon he was attached to

to be priority number one so oxygen is

the second highest atomic number so

it's going to cause the carbon he was

attached to to be priority number two

and then finally the lowest of the three

is the carbon and so the carbon he's

attached to

is gonna be number three and then we'll

make our circle one two three

so in this case that is a left-handed turn

turn

and a left-handed turn when your number

four priority in the back

corresponds to s so this is the s

isomer here or we'd say that this

molecule is in the s configuration or this

this

chiral centers in the s configuration

cool so got one trickier example out of

the way and this next one's gonna get

even a little trickier from there so

we take a look at this one here we've

got a single chiral center in this

molecule as well and that's

right there and in this case just like

the last one

that chiral carbon is bonded to three

other carbon atoms and then he's also

bonded to a hydrogen and that hydrogen

will infer

must be to a wedge bond and that

hydrogen once again is going to be

priority number

four so and once again your number four

priority doesn't have to be hydrant it

often will but it doesn't have to be

so but wherever it is we can make some

determinations here because it's on a

wedged bond

that means if you see a right-handed

turn from this side

that really corresponds to a left-handed

turn if you crawled in behind the board

and so a right-handed turn from this

perspective with your number 4 as a wedge

wedge

doesn't mean r it means s and a

left-handed turn would mean r

so not s cool so just want to

kind of put that out there before we

start here but assigning priorities for

these three carbons now is going to be a

little bit of a pain in the butt

so i'm going to start with this bottom

one because he's kind of the easiest and

in addition to being monitored this

carbon he's bonded to these two

hydrogens and so we're just going to

list carbon hydrogen

hydrogen again a descending order of

atomic number

now this carbon right here is bonded to

one hydrogen and then he's got a double

bond to that carbon right there

and so with a double bond to that carbon

we'd count it as one bond of carbon

second bond to carbon bond to hydrogen

and so we'd list him out

as carbon carbon hydrogen where these

two carbons are actually both

that carbon right there all right so

then we move over to this carbon right

here and he's also bonded to

one hydrogen but he's bonded to this

carbon and he's bonded to this carbon as well

well

and so we'll list him out as carbon

carbon hydrogen

as well and we see one point of

difference so these two there's no point

of difference but

these two compared to this one we're

really comparing the hydrogen here

to the carbon right here and so this

hydrogen is definitely going to

lose if you will and so we knew these

were going to be priorities one two and

three and we just figured out that

therefore he's going to have to be

at least the carbon he's attached to is

going to be priority number

three but we've still got a tie between

these two carbons

even based on what they're bonded to so

then we have to go further in the chain

and say well then

what are these carbons bonded to like

this one is represents these two

or these two i guess represent this one

so and then these two carbons are these

two and say well what are they bonded to

as well

let's do the one on the right first so

this one right here this carbon right here

here

let's draw this in with yet another

color is bonded to two hydrogens

and that is bonded to this carbon and so

we list it out as

carbon hydrogen hydrogen and then this

carbon right here is bonded to two hydrogens

hydrogens

and it's also bonded to that same carbon

as well and so its three things are

still going to be carbon

okay so taking this a little further

then so this carbon right here again

is both of these carbons so and things

are going to get a little bit funny so

he's bonded to two hydrogens

but he's not bonded to anything else at all

all

so what do we do because there's got to

be three things

and so what they do in this case is they

count the pi bond

back to the previous atom and so we'll

actually count a bond back to the

previous carbon

as part of that pi bond so instead of

just saying h it'll actually be

along with this carbon yet again and so

this is going to be c

h h and this one also is going to be chh

because they're

again representing both that carbon and

lo and behold this is not looking good

because we still

have a tie so now we have to do the

green carbons

well in this case the green carbon was

the previous one that was blue and it's done

done

he's already been taken care of we have

nowhere else to go with this we're done

assigning designations you can't go back

up the chain you'd have to continue

further out so

there's nothing else to do here but for

this one

we're now looking at the green carbon

and the green carbon here

and then depending on who you talk to is

bonded to one other carbon so

when we're coming from this carbon over

here out to here he was bonded to this

carbon and these two h's

so and when we were coming from this

carbon out to here he was bonded to this

carbon in two h's

and so regardless of which one you look

at here he's gonna be bonded to a carbon

and two h's so we had further

designations to go we have nothing to

compare that to

and they therefore these are definitely

going to be nothing to compare it to

and so as a result took all this work to

finally realize

that this was priority number one and

this was priority number two

and then we'll make our circle one two three

three

and we can see that once again this is a

left-handed turn

and as long as your four priorities in

the back a left-handed term means s

but in our case our number four priority

is the wedge bond that's coming out of

the board

and so a left-handed turn is going to mean

mean

r instead we're looking at the molecule

from the wrong side and if we crawled in

behind the board

what looks to us like a left-hand turn

from this side would look like a

right-handed turn from the other side

cool so these are a couple of the

trickier examples i could give you

so and again this might be more

representative of some of the harder

ones you might see

on an exam so now we're going to take a

brief look at

naming molecules that have chiral

centers and first thing you want to do

is just kind of name it as if you didn't

know it had any chiral centers and

things of this sort and we'll find out

we just have to add a designation to it

at the beginning of the name so

in this case longest continuous chain

here would be the one

the only carbon chain all the carbons

essentially and we'd number it from

right to left

so and we'd see that we've got a

chlorine at position two and the parent

chain is gonna be called

hexane for a six carbon chain and so we

would name this

hexane now the problem is that name is

not sufficient because while that is

indeed two chlorohexane so

is that that's also two chlorohexane

but these are not the same molecule

they're enantiomers and so somehow we

can't give

two molecules exactly the same name as

they're not the same molecule

and so that's why we got to come up with

a way of including the r and the s

in the name so let's get him out of here

so in this case

we'll go through and assign priorities

once again and for the chiral center

that is right here he's the guy who

bonded to four different groups

chlorine's number one the hydrogen

that's in the back

will definitely be number four and then

between these two carbons this carbon's

bonded to

three h's this carbon right here is

going to be bonded to this carbon and

two h's

and we see that carbon is going to be

hydrogen and so this will be priority

number two

this will be priority number three and

if we go from one to two to

three we'll see that that's a

right-handed turn and a right-handed

turn when your four priorities in the back

back

corresponds to r so we know this is the

r configuration

and how do we put that in the name well

that's always going to go at the

beginning of the name and it's going to

go at the beginning of the name

in parentheses and so here we'll just say

say

r 2-chlorohexane so

when you have just a single chiral

center of your molecule just put rs in

parentheses at the beginning of the name

done now if you've got more than one

chiral center

not only will you have to stick the

designation at the beginning but you'll

also have to give me the carbon location

where it's located assuming

in most cases in pretty much every case

you're likely to see it's going to be on

a carbon

so once again if we numbered this it

would be one

two three four five six and so notice

this is my

numbering for naming it not any kind of

priorities or anything like this and

so in this case we would call this two

chloro alphabetical order here if you

recall so two

let's save myself a little room i'll

drop down here so two

and once again we're going to save some

room for some parentheses

at the beginning but this molecules

actually got two

chiral centers so it's got the one at

two here with the chlorine

on it it's got four different groups but

this one also is going to be

and so we'll assign priorities

separately here so for this carbon right

here chlorine is definitely number one

there once again is a hydrogen attached

that's definitely going to be number four

four

and between these two carbons this one's

bonded to three h's

this one right here is bonded to a

carbon up here a carbon over here and

then one h that's not drawn in so it'd

be c

c h and he's definitely going to win

then so one

and then two three and so if we go

around our circle

i can see that this is a right-handed

turn and it's with the number four

priority in the back

which means it is r

okay so so far so good and rather than

trying to muck up this diagram

even further i'm just going to redraw it

one more time

and so our other chiral center is this one

one

right here he's bonded to four different

groups as well and this one's going to

be a little more challenging now

he is bonded to a hydrogen and that hydrogen

hydrogen

is on a wedged bond but he's going to be

priority number four

because the other three atoms are all

carbons carbon

carbon carbon it'll be a three-way tie

but hydrogen's definitely gonna be

priority number four

and once again because it's a wedged

bond now it'll be opposite of normal so

a right-handed turn

would actually correspond to s and a

left-handed term would correspond to r

because we're looking at the molecule

from the wrong side all right to compare

these three carbons

so this carbon over here is bonded to a

chlorine a carbon

and then also to one hydrogen so a

chlorine a carbon

and then to that hydrogen right there so

we list it as chlorine

carbon hydrogen so this carbon over here

in addition to being bonded to

those two hydrogens is bonded to a

carbon as well and so we list it as carbon

carbon

hydrogen hydrogen and then finally

obviously that carbon right there is

just bonded to all three of the h's of

that methyl

group so h h h

and we see the point of difference here

and that point of difference

chlorine is going to be carbon which is

going to be

hydrogen and so as a result the carbon

the chlorine is bonded to is going to be number

number

one this carbon over here is going to be

number two and this carbon up here is

going to be number

three and as we go from one to two to three

three

it's a right-handed turn and once again

because our four priority is on the

wedge instead of a right-handed term meaning

meaning

r it's actually going to mean s

so now when we go to our designations a

lot of students just want to do this

they want to come out here and be like well

well

there it is it's rs or sr whatever

so well it turns out that's not good

enough because then i don't know which

one is which and things of a sort and so

what you do out here is kind of put them

in numerical order

and in this case the chiral center with

the chlorine is at position

two so on our parent chain and so we're

going to say

and he gave a designation of r so we're

going to say 2r

and then you put a comma so and then the

chiral center

at that position 3 ends up being s and

so then you'll say

3 s cool and we only have to worry about

this when you have more than one chiral

center if you only have one chiral

center don't include the number just say r

or just say s it's when you have

multiple designations to give out here

that you gotta start including locations

where it just gets confusing on which

one you're talking about

and so this is 2r 3s 2-chloro 3-methyl

hexane now if you found this lesson

helpful consider giving me a like and a share

or if you just feel sorry for like short

bald guys consider giving me a like and

a share for that as well

if you've got questions feel free to

leave them in the comment section i'm on

there pretty religiously

and if you're looking for practice

problems or you want the study guides

that go with this

check out my premium course on chatsprep.com

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YouTube Transcript:5.2 How to Assign R and S Configuration | Absolute Configuration | Organic Chemistry

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5.2 How to Assign R and S Configuration | Absolute Configuration | Organic Chemistry