Stereochemistry - R S Configuration & Fischer Projections - Free Educational videos for Students in K-12

Stereochemistry - R S Configuration & Fischer Projections - By The Organic Chemistry Tutor

Transcript


00:0-1	in this video , we're going to talk about steroid
00:02	chemistry . We're going to talk about how to assign
00:05	Rs configuration to Cairo centres , including fisher projections and
00:10	a lot of other examples as well . So let's
00:12	start with this one , assign a R . Or
00:14	as configuration to each Carlson as shown below . So
00:18	first , what is a carol center ? A Cairo
00:21	center or a Cairo carbon is a carbon atom that
00:25	has four different groups attached to it . In this
00:28	case , These are the four different groups . Now
00:33	, in order to assign that carol center A .
00:36	R . S configuration , we need to rank those
00:39	four groups using the con single pre log process grouping
00:46	will one , it's going to have the highest priority
00:50	and the way we determine priorities based on atomic number
00:54	. So looking at these four atoms bro mean chlorine
00:58	, carbon and hydrogen , which of these four atoms
01:01	have the highest atomic number based on the periodic table
01:05	. So if you're using the periodic table , you'll
01:07	find that grooming has the highest atomic number . So
01:12	this is going to be group number one . It
01:13	has the highest priority chlorine has the second highest atomic
01:17	number relative to bring me . So that's gonna be
01:21	group # two . Carbon has an atomic number of
01:24	12 hydrogen has the lowest atomic number of all of
01:27	the elements in the product table . It has an
01:30	atomic number of one . So it's always going to
01:32	be a group . And before now , before we
01:36	can count it , We need to make sure a
01:39	group # four is in the back or it's in
01:41	the hatch wedge , which it is in this case
01:45	. So we can count from one , 2 ,
01:46	2 , 2 , 3 . So notice that we're
01:49	rotating in the counter clockwise direction . Whenever that happens
01:54	, you need to assign an S . Configuration if
01:58	grouping before is in the back . So that's the
02:01	configuration for this particular carl center . Now let's focus
02:05	on the Cairo center on the right . And if
02:07	you want to pause the video to try that example
02:10	, feel free to do so this burning adam ,
02:13	we know it's gonna be number one chlorine is going
02:16	to be too . The method group will be group
02:19	number three and ages four , so counting it from
02:22	1 to 2 to three . This time we are
02:25	rotated in a clockwise direction . So this is going
02:28	to have the our configuration . Now , looking at
02:34	these two molecules , what is the relationship between them
02:39	? Would you describe them as constitutional , ice hammers
02:43	and nancy Morales . Die stormers , miso compounds .
02:47	How would you describe these two compounds ? Now notice
02:52	that these two molecules are mirror images of each other
02:57	, therefore they are known as . And the and
03:01	commerce and consumers are a special type of stereo iceman
03:10	. A stereo I smell like any other . I
03:12	simmer have the same chemical formula . They are connected
03:15	the same way , but notice that the way the
03:18	atoms are arranged in space , it's a range of
03:22	space differently . This carbon and that carbon they're connected
03:26	to the same for groups . But the way those
03:29	four groups are arranged in space , they're not the
03:31	same here . Chlorine is on the left side here
03:34	, it's on the right side . So they are
03:35	arranged in space differently , which makes it a certain
03:38	type of steroid iceberg , die streamers and consumers cis
03:44	trans geometric iceman's . Those are all sterilizers . But
03:49	I know trimmers , they're mirror images of each other
03:52	and that's how you can distinguish them from the customers
03:58	. Another feature of an answer is that they have
04:00	the opposite configuration at the karl center here , the
04:03	configuration . We determine it to be us here .
04:06	It's our Mhm . Now let's work on some more
04:10	examples for the sake of practice . So go ahead
04:14	and assign the R and S configuration to each Kyle
04:20	center . I'm going to draw . So let's put
04:25	an ethyl group here . Let's add a hydrogen a
04:32	broom in adam and let's put a hydroxy group .
04:37	So go ahead and determined the configuration at this carbon
04:42	. And so while you do that , I'm gonna
04:44	write up another one . Yeah , so let's start
05:09	with the first one . The first thing we need
05:11	to do is determine which group has the highest priority
05:15	. So we know it's going to be booming next
05:18	. If we compare oxygen to carbon and hydrogen ,
05:22	we know oxygen is going to have a higher atomic
05:24	number . The atomic number for uh oxygen is eight
05:28	for carbon at six . So this is going to
05:30	be number two , three . And then hydrogen is
05:32	always for the hydrogen is in the back , which
05:35	is good . So we're going to just go from
05:37	1-2 to 3 . And so for this one ,
05:40	we have the r configuration , we're rotating clockwise or
05:46	in the direction of a clock . Now for the
05:49	next one , hydrogen is going to be grouping before
05:52	here we have oxygen , carbon , carbon oxygen wins
05:56	. So that's gonna be number one . Next notice
06:00	that these two carbons , they have the same atomic
06:03	number . So if we move on to the next
06:06	group here , we have carbon versus hydrogen carbon winds
06:10	. So the ethnic group has a higher priority than
06:13	the method group . So if we counted from 1
06:18	to 2-3 , notice that were rotated in a clockwise
06:22	direction , But group # four , it's not in
06:26	the back , it's in the front , it's not
06:28	on the dash . This time it's on the solid
06:31	wedge . So because it's on a solid wedge or
06:35	it's in the front coming out of the paper ,
06:38	we need to rotate it . So just switch the
06:41	configuration and it's equivalent to putting it back to the
06:45	it's putting group before in the back . So it
06:48	was all right . But now it's s so that's
06:51	the configuration for this carl center . So anytime H
06:54	is in the front simply counted from 1 to 23
06:57	And then reverse your answer number two . Name the
07:02	following compounds using the R . S . System for
07:05	stereo is MERS . So first let's identify the carl
07:10	center . So let's focus on its carbon . That
07:13	carbon has a burning hydrogen , a method group and
07:17	an ethyl group . So it has four different groups
07:19	which makes the Cairo center . Now let's rank those
07:22	groups , Bromine has the highest atomic number . That's
07:25	gonna be number one . We know S . O
07:28	. Has a higher priority than metal . So this
07:31	will be too that's gonna be three . Age is
07:33	gonna be four . So going from 1 to 2
07:37	to 3 . We can see that It's going in
07:41	our direction . And when you count from 1-2 to
07:44	3 , Ignore four . Now ask you counted .
07:49	We're going to focus on four . four is in
07:52	the front , it's not in back which means we
07:54	need to reverse it so it's going counter clockwise .
07:58	Which means the configuration is s let's put the S
08:04	. Configuration here . Now in order to name it
08:09	, we need to count the carbons in the parent
08:12	chain . So we're gonna count in this direction .
08:15	So we have a four carbon parent chain . We're
08:18	gonna be dealing with beauty . Yeah . So first
08:22	we need to write the configuration of the carl center
08:25	which is s we're going to close the in parentheses
08:29	and then it's going to be we have a blooming
08:31	on carbon . To so to Bruno and for four
08:35	cardinal cane that's going to be butane . So it's
08:38	s to burma butane , that's how we can name
08:42	this particular stereo Eisenberg . Now let's move on to
08:46	the next one . So here we have to Cairo
08:51	centers and you can check it . Here we have
08:57	one group 23 relative to this carbon . This is
09:01	the fourth group . If we focus on the second
09:04	carl center , this is one group to the ethyl
09:08	stree And that's the 4th group . So let's focus
09:14	on the first carol center . This one . So
09:18	we're comparing a chlorine atom , hydrogen , carbon ,
09:21	carbon . So if we look at the first item
09:25	that's connected to that carol center , we see that
09:28	chlorine has the highest atomic number . We don't look
09:31	at growing because it's not the first atom attached to
09:35	this Carlson . So we look at it from a
09:38	once one basis . So this chlorine has a higher
09:42	atomic number than that carbon . So this is going
09:45	to be group # one . Now , for group
09:49	No two , We know HS # four . So
09:52	we got to compare these two . Here , we
09:56	have a carbon and carbon . And then after that
09:59	carbon is a hydrogen here , there's a broom in
10:02	this is going to win . So this entire group
10:06	is group number three , which makes the matter group
10:14	. Actually take that back . This is to this
10:17	is # three . This has a higher priority than
10:19	the method group . So now we can count it
10:25	from 1-2 to 3 . And so that's going to
10:29	be the our configuration . So let's put our here
10:38	and now let's focus on the next Cairo center .
10:44	So bro , main hydrogen carbon , carbon , brahman
10:48	is going to have the highest atomic number . So
10:50	that's one , eight should be four . So now
10:53	let's compare these two . So those two carbon atoms
10:56	are the same . Next we have a chlorine versus
10:58	a carbon . The client is going to win .
11:02	So this whole group will be group number two And
11:05	the effort group is gonna be number three . So
11:09	going from 1 to 2-3 , we're going in our
11:13	direction , but age is in the front , we're
11:15	gonna have to reverse it , making it S so
11:24	this is us now that we've assigned the configuration to
11:30	each carol center . We can name this theory I
11:32	smart , so we need to number it from left
11:36	to right . So the substitutions will be on 2
11:39	, 3 as opposed to 34 . Now on carbon
11:49	to we have the R configuration . So this is
11:51	going to be to our And on carbon three we
11:54	have the s . configuration . So this is going
11:57	to be three s . Now we need a comment
12:00	between them and then we need to put the bromo
12:03	and the chlor a group in alphabetical order . So
12:06	B comes before . See this is going to be
12:09	three burma , It's on carbon three chlorine and carbon
12:13	too . So we're gonna have to claro And for
12:17	five carbon out cain , that's going to be painting
12:22	. So this is to R three S . Three
12:24	bergamo to chloral pending . So that's how we can
12:27	name at this particular stereo iceberg . Now , here's
12:33	the question for you how many stereo ISIS members are
12:36	possible if you have one carl center and how many
12:40	are possible if you have to carl centers . So
12:45	if you have one Carl Center , you can have
12:51	the S . Ice summer or you can have the
12:54	iceman . So there's two possibilities . The equation that
12:58	will give you the number of steroids members is to
13:00	to the end where n . Is the number of
13:02	carl centers . So in this case and is one
13:05	to the first power is too . So for one
13:08	Carl Center , There are two possible stereo I summers
13:12	. Now , what if we have to carol centers
13:18	? How many stereo I Summers are possible ? Well
13:23	, it's going to be an S2 , so it's
13:26	gonna be two squared , two squared is two times
13:29	two , which is four . So we have four
13:32	possibilities . The first possibility is if both carl centers
13:36	are are the second one is if one is our
13:39	I mean if the first ones are in the second
13:41	one is S or if the first ones asked the
13:44	second ones are Or if they're both fast . So
13:47	those are the four possibilities . Or the four sterilizers
13:51	that are possible if you have to Kyle centers .
13:55	So the number of steroids MERS is going to be
13:58	to to the end carol centers . Number three .
14:02	The chemical structure of cholesterol is shown below . How
14:06	many stereo customers are possible for this molecule , so
14:11	feel free to take a minute and try this problem
14:14	. So this right here is a carl center .
14:17	We have an O . H . Group , there
14:19	is an invisible hydrogen , but now it's invisible and
14:22	then this side is different than that side . This
14:27	side is closer to the double bond . This other
14:29	side has some other group . So these two sides
14:34	are different , which makes this a Carlson . Now
14:40	this carbon is also a car center has four different
14:42	groups . Here we have a method group here is
14:46	a secondary carbon , a tertiary carbon and this tertiary
14:50	carbon . This tertiary carbon is close to the double
14:53	bond . This one is not so they're different which
14:57	makes this a Cairo center here is another carl center
15:05	, this is one group , this is another ,
15:07	this one's another and there is a hydrogen attached to
15:10	it and we could put that hydrogen here . This
15:19	carbon here is not Cairo . Whenever you see a
15:22	carbon with two bonds and that's it . It's a
15:25	ch two Carden , which means it doesn't have for
15:28	different groups because these are the same . So that
15:31	is not a Cairo carbon . So it's best to
15:35	focus on the tertiary carbons and the coronary carbons .
15:40	This is a tertiary carbon , It's attached to three
15:43	other carbons and there's hydrogen attached to it . So
15:47	that is another carl center , here's another tertiary carbon
15:55	. It has three other carbons , all of which
15:57	are different . And there's a hydrogen here too .
16:04	Now this carbon here is coronary . It's attached to
16:08	four different carbon atoms . So that's another Cairo carbon
16:17	. This tertiary carbon is carl too . There's a
16:19	hydrogen here and this carbon is tertiary and its Cairo
16:28	, This is the group one , this is another
16:30	group , This is another group . And then the
16:33	hydrogen is the 4th group . So let's put that
16:42	hydrogen on the hatch wedge . Now this carbon here
16:46	is Treasury , but it's not carbon because It doesn't
16:50	have four different groups . These two method groups are
16:54	the same . That's cholesterol has a total of 1
17:00	, 2 , 3 , 4 , 567 eight Kyle
17:06	centers . So because cholesterol has eight carol centers ,
17:10	it's gonna have to to the eight stereo prisoners To
17:14	to the eight is 2-4 times 2-4 , four plus
17:17	four is eight . If you multiply 24 times two
17:20	times two is four times two is eight times two
17:22	is 16 . You got 16 times 16 which is
17:26	256 . So this is a number of steroids murders
17:31	that are possible for this particular molecule . Number four
17:36	determine the absolute configuration of each carl center and name
17:40	each Fischer projection shown below when dealing with fisher projections
17:45	. It's easy to determine the carl center . So
17:49	we have to carol centers , which means that we
17:51	have four possible sterilizers for this particular structure . Now
17:57	, before we determine the absolute configuration of each car
17:59	center , let's talk about fish projections . So let's
18:03	say if we have this particular Fischer projection , let's
18:10	make this an ethnic group . And let's say this
18:14	is hydrogen . This is ohh , we need to
18:17	realize is that the groups on the horizontal part of
18:21	the official projection there in the front . So you
18:25	can redraw the structure like this if you want .
18:29	They're coming out of the page . And so they're
18:31	on the solid wedge . Yeah , these two groups
18:38	they're on the dash or the hatch wedge . So
18:43	they're going into the page . You can view them
18:48	as being in the back . So make sure you
18:51	understand that . Now hydrogen is typically on the side
18:54	, which means that as a group number four ,
18:56	you're gonna have to reverse it when determining the configuration
19:00	. So keep that in mind anytime google before he's
19:04	in the front , you need to reverse it .
19:08	So let's begin by determining the configuration of this carol
19:10	center . Yeah , the hydroxy group is going to
19:14	be group number one oxygen beats these two carbon atoms
19:19	now comparing those two carbon atoms , they're the same
19:22	. But here we have a booming here is the
19:24	hydrogen . So this group has a higher priority than
19:28	the method group . So this will be group number
19:30	two , The metal will be three , and then
19:33	hydrogen is always for if it's there , So going
19:38	from 1 to 2-3 , we're going in a clockwise
19:42	direction . So that's our but h remember h when
19:46	it's in the horizontal part , it's like being on
19:48	the solid wedge ages in the front , so we
19:51	got to reverse it , so we're gonna get s
19:57	so we have the S configuration for that carl center
20:03	Now , for the next one . So we're going
20:05	to focus on this carol . Senate Bro . Mean
20:09	, is gonna be group # one has the highest
20:11	priority . Next were compared to method group with this
20:15	group . So we have carbon , it's a carbon
20:19	and then hydrogen to oxygen oxygen wins . So this
20:24	entire group is going to be number two . Methodist
20:27	three , H is four , So going from 1
20:31	to 2-3 , it looks like we're going in the
20:34	S direction but if we reverse it it's going to
20:37	be our so we have the R . Configuration here
20:44	. Mhm . So now that we've assigned the absolute
20:49	configuration to each car center , we can go ahead
20:52	and name this particular molecule . So the hydroxyl group
20:57	is going to have priority over the broom in groups
21:00	. So we're gonna count it in a way that
21:03	we give the hydroxy group the lower number . We
21:06	want to give it a two instead of three .
21:07	So we're gonna count in this direction . So on
21:15	carbon to we have the S . Configuration . So
21:18	this is going to be to us and on carbon
21:22	three we have the R . Configuration . So three
21:24	are Yeah and then We have a brahmin of carbon
21:30	three . So this is gonna be three bromo any
21:33	alcohol is can be part of the parent name and
21:36	it's all carbon too . So it's gonna be too
21:38	beating all because we have a four carbon chain .
21:43	So two as three R three bromo to be ,
21:45	you know that's how we can name this particular fisher
21:47	projection . Now let's try another example . Go ahead
21:58	and name this fisher projection and determine the configuration at
22:02	each cow center , just like we did before .
22:17	So let's start with this one , Brahman is gonna
22:20	be a group # one . This group with a
22:22	chlorine atom , we know that's going to be too
22:25	Method Astri Ages four , so going from 1 to
22:28	2 to three , ignoring four , this is going
22:31	in the counter clockwise direction sets s but age is
22:35	in the front , so we need to reverse it
22:38	so we're gonna get arm For the 1st 1 .
22:45	Now for the second carol center Chlorine is gonna be
22:49	group # one . This entire group , It's number
22:52	two , if we compare carbon to carbon , it's
22:55	a tie at any carbon tube roaming roaming winds .
22:59	So this is number two ethel's number three ages four
23:04	. So going from 1 to 2 to three ,
23:06	skipping four , it appears to be art , but
23:09	H is in the front , so we reverse it
23:11	and we get the s configuration . So now that
23:18	we have the configuration at both carl centers , we
23:21	can name it . So we don't want to count
23:23	in this direction because this will be three and that
23:27	will be four . Rather , we want to count
23:29	in this direction . So this will be too and
23:33	this will be three . So we have a five
23:36	carbon chain , so we didn't live painting . But
23:39	first let's put let's focus on the configuration . So
23:45	the configuration is our at carbon too . So we're
23:48	gonna have to art and then it's US At carbon
23:52	three . So 3 of us . And then we
23:58	have a br on carbon too . And we need
24:01	to put it in alphabetical order B comes before See
24:05	. So this is going to be to Bruno .
24:07	And then we have a cl on carbon three .
24:10	So three chloral And then for the five carbon chain
24:14	that's going to be repenting . So that's the nomenclature
24:18	or the I . U . P . Economic nature
24:19	for this particular fisher projection . It's two R three
24:22	S two bromo three chloral panting . Now , let's
24:25	work on a more challenging problem . So let's say
24:32	this carol center has an ethnic group , a method
24:37	group in the front . And let's put a chlorine
24:43	group in the back . All right . And the
24:49	hydrogen go ahead and assign the configuration to the Cairo
24:55	carbon . Is it R . S . Now ,
24:58	this problem is different because hydrogen group number four is
25:02	neither in the front nor is in the back .
25:06	So , how do you assign the configuration in this
25:09	situation ? Well , let's begin . We know that
25:12	chlorine has the highest atomic number . So that's going
25:14	to be group number one , estelle has more priority
25:18	than metal . So , with those two methods three
25:22	ages four . In a situation like this , there's
25:26	a technique that you could use to assign the configuration
25:30	, whatever group is in the back , put it
25:32	in a circle . So I'm going to put one
25:34	in the circle and you can put a small sub
25:37	script B to indicate that what ? Seven in a
25:39	circle ? It's in the back . Sometimes it could
25:42	be in the front . So , but for now
25:45	, whatever is in a circle . And this problem
25:46	is in the back . Now , the other numbers
25:51	2 , 4 and three , you want to arrange
25:54	them in the form of a triangle . Now ,
25:57	if you notice too is at the top . So
25:59	let's put that at the top . Four is at
26:03	the bottom left with respect to So let's put that
26:05	here . three is on the bottom right Now ,
26:10	what we're gonna do is we're going to rotate this
26:12	molecule in such a way that number four will be
26:16	at the top or to us . So we're going
26:18	to rotate it 120° clockwise . So one is still
26:25	in a circle . Four has replaced too . Two
26:29	is now with three ones and three years were formless
26:34	. Now the next thing we're gonna do is we're
26:35	going to flip the molecule as we flip it .
26:39	two and 3 will move to the top Group .
26:42	Number four is going to expel group number one .
26:46	So now four is in a circle , one is
26:49	beneath it And then three and 2 or at the
26:52	top . Now four group before it's in the back
26:57	because what's in a circle represents what's in the back
27:01	. And so we can count from 1 to 2-3
27:04	. And this gives us the assessment . So the
27:08	configuration of this carol center is s So that's the
27:12	technique that you could use whenever group # four is
27:16	not in the front or in the back .

Summarizer

DESCRIPTION:

This video provides an overview of the stereochemistry of organic compounds and defines what exactly a chiral carbon center is. This video also shows you how to assign R and S configuration to a chirality center whenever the hydrogen (4th group) atom is in the front, back, or neither. It also discusses nomenclature - how to name organic compounds using R and S absolute configuration. It provides a lot of examples and practice problems.

OVERVIEW:

Stereochemistry - R S Configuration & Fischer Projections is a free educational video by The Organic Chemistry Tutor.

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