Resonance Structures - Free Educational videos for Students in K-12

Resonance Structures - By The Organic Chemistry Tutor

Transcript


00:00	in this lesson , we're going to talk about resonant
00:02	structures . How can we draw the resonance structure for
00:10	this ? A lilic , carbo , catomine . The
00:16	first thing you should do is you should start the
00:20	arrow from the double bond and it's going to flow
00:22	towards the positive charge . Electrons will flow from a
00:26	region of high negative charge towards the region of low
00:31	negative charge or from negative to positive . And so
00:35	the arrow , it's always going to go from the
00:38	nuclear filic region of the molecule to the electra filic
00:41	region of the molecule or ion in this case .
00:48	So the resident structure for this , a little cat
00:50	iron looks like this . The double bond is going
00:54	to move to the right side and it's going to
00:57	trade places with the positive charge . Since this carmen
01:01	lost pi electrons , it now has a positive charge
01:07	. This carbon lost a pie bon but regained it
01:10	on the right side . So it doesn't have the
01:12	positive charge now , which of these structures is the
01:18	major resonance contributor ? Is it the one on the
01:21	left or the one on the right ? Both of
01:25	these are primary , a little carbon credits . So
01:29	they're equally stable . So there is no major or
01:32	minor residents contributor . In this example , the residents
01:37	hybrid is actually a blend of these two resonance structures
01:42	. So the pie bon is shared among the three
01:45	carbon atoms and so is the positive charge . It's
01:50	actually shared between These two carbon atoms . So they
01:57	both carry a partial positive charge or you could say
02:00	half Of a positive charge if he divided by two
02:11	. Yeah , go ahead and try this problem .
02:18	Yeah , draws many residents structures as you can and
02:22	identify which one is the major residents contributor or which
02:27	resident structure is the most stable . So the pie
02:32	bon is going to move towards the carbon catomine and
02:36	a positive charge is going to jump to places ,
02:40	or two carbon atoms towards the pipe on . So
02:44	the positive charge will not be here . And so
02:49	that's one resident structure that we can draw and we
02:52	can draw another one . Mhm . So the plus
03:02	charge is going to jump two carbons to the left
03:05	. So it's that one that carbon . So in
03:09	our first example we have a primary carbon canteen or
03:12	a little caramel candy . And in the second example
03:15	it's tertiary And in the 3rd 1 it's secondary .
03:20	So therefore This one is the major residents contributor because
03:26	a tertiary , a little carbon canteen is more stable
03:29	than a primary or secondary . A little Karpel Captain
03:35	. So recall that tertiary carbon headlines are more stable
03:38	than secondary ones , which are more stable than primary
03:41	ones . So if we compare , let's say ,
03:47	a tertiary carbon canteen with a primary caramel candy .
03:53	It's important to understand that the method groups our electron
03:58	donating groups relative to hydrogen . As a result ,
04:03	they can donate electron density to the carbon headline by
04:08	means or through the sigma bond , which is known
04:11	as the inductive effect or through the overlap of atomic
04:14	orbit does , which is known as hyper congregation .
04:19	And so a carbon headline with more method groups or
04:23	more carbon atoms attached to it . It's going to
04:25	be more stable Than one that has less . Now
04:31	, let's move on to our next example . In
04:38	this example , we have a Benz Ilic carbo Katie
04:42	. It drives many resonance structures as possible and identify
04:45	which structure is the major resonance contributor . So we
04:51	can move the double bond towards the positive charge And
04:56	just like before the positive charge is going to jump
04:58	two carbons towards that to move on . And so
05:06	we have this resident structure and then we can move
05:11	this double bond here , Given us our 3rd resident
05:25	structure , and then we can move this double bond
05:28	and that location . So this is a primary carbon
05:42	canteen or technically Ben's Ilic carbon copy . This one
05:47	is secondary , this is secondary , and this is
05:50	secondary . So this one is going to be the
05:53	minor resonance contributor because it's the least stable . Now
05:58	these are equally stable , so they form part of
06:03	the major resonance contributor . Now , going back to
06:12	the last problem , there's one more resonant structure that
06:15	we can draw out of the common ones . There's
06:19	a lot of uncommon resonant structures that we can draw
06:21	. So just keep that amount . I prefer to
06:23	focus on the common ones . So we could take
06:26	this double bond , move it to the plus charge
06:30	and we'll get this answer . It's similar to the
06:37	original answer , but notice that the double bonds in
06:39	the benzene ring have been shifted . Now let's move
06:44	on to our next example , draw the resident structure
06:51	of the benzene molecule . All we can do here
06:55	is simply rotate the pie bahn's and so this is
07:05	the resident structure for benzene . And that's all we
07:09	can do here . Now what about for the car
07:18	box , leyte island . What resonance structure can we
07:21	draw ? Go ahead and try that problem . So
07:26	what we can do is we can take a lump
07:28	here from the negatively charged oxygen , use it to
07:32	create a pie bon and then break this pylon .
07:39	And so that's going to give us this particular structure
07:49	. Now these two structures are identical and so they're
07:53	equally stable . There is no major or minor resonant
07:57	structure in this example . Now let's move on to
08:07	the next example . So let's say if we have
08:11	a sulfur atom instead of an auction item , identify
08:17	the major resonance contributor in this case . So we
08:22	could follow the same pattern in order to draw the
08:24	resident structure . So the question is , is it
08:39	better to put a negative charge on the oxygen atom
08:44	or on the sulfur atom ? Now granted oxygen is
08:49	more electro negative than sulfur . So that fact favors
08:53	oxygen . However , sulfur is bigger than oxygen .
09:00	And let's say if we have a small atom versus
09:03	a larger atom , if we were to make each
09:07	adamant ion by giving it a negative charge , which
09:11	charges more stable a big guy on is more stable
09:16	than a small line , assuming they have the same
09:18	charge , because this has more volume where it can
09:23	stabilize that negative charge . So this negative charges you
09:27	can think of , It has been diluted over a
09:29	larger surface area , this one here , it's more
09:33	concentrated and so larger ions can are more stable than
09:39	small ions . The bigger the atom the better it
09:42	can stabilize a negative charge . So therefore it's better
09:46	to put the negative charge on the sulfur atom and
09:48	then on an oxygen atom . So this is going
09:51	to be the major residents contributor on the left .
09:56	This one here is the minor resonance contributor . Here's
10:01	another example , let's say we have an alcohol but
10:07	adjacent to the O . H . Group , there's
10:08	a positive charge . Draw the resonant structure and identify
10:13	the major business contributor . So we know the electrons
10:18	will flow from a region of high negative charge towards
10:21	the region of low negative charge . So the arrow
10:25	is going to point towards like from the lone pair
10:28	but towards the positive charge . And so we can
10:31	use that lone pair to create a pi bond .
10:34	And so now the oxygen has one lone pair as
10:37	opposed to two because this carbon atom , it gained
10:40	a pie bon , the positive charge on that carbon
10:44	atom has been neutralized . The oxygen lost the lump
10:47	here . So now it has a positive formal charge
10:51	. So is it better to put a positive charge
10:55	on an oxygen atom or on a carbon atom ?
11:00	Now oxygen is more electoral negative than carbon . And
11:04	so as a result it's better to put the positive
11:07	charge on a carbon atom . However , it turns
11:10	out that the structure on the left is the major
11:13	residence contributor and the one on the I mean this
11:17	is the right side , not the left side ,
11:18	the one on the left is the minor resident contributor
11:23	. So why is this the major resident contributor ?
11:26	If we have a positive charge on an election negative
11:29	atom , when we know it's better to put a
11:31	positive charge on a more electoral positive atoms like carbon
11:37	. The answer has to do with the arctic rule
11:41	. So if you look at the structure on the
11:43	left , the carbon atom , it doesn't obey the
11:46	oxide rule , it doesn't have eight electrons . It
11:49	has three bonds , no lone pairs , so it
11:51	has six electrons . If we look on the right
11:56	side , this carbon atom has four bonds . So
11:59	it has eight electrons . It satisfies the octet rule
12:04	. The oxygen also satisfies the octet rule . It
12:07	has three bonds that six electrons plus a lump here
12:10	. So it has eight electrons on the left side
12:15	. The oxygen satisfies the octet rule has one bond
12:18	plus the O . H . Bond . That's two
12:20	that's four electrons plus two lone pairs . So it
12:23	has eight . So the reason why this is the
12:26	minor resonance contributor is because carbon does not obey the
12:30	octet rule when it has a positive charge . And
12:34	so that's why you'll see that when dealing with resident
12:37	structures , it's better to put the positive charge on
12:40	the oxygen atom as opposed to the carbon atom ,
12:44	to satisfy the octet rule . So electro negativity is
12:48	not a priority in this example .

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