[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.77,0:00:02.63,Default,,0000,0000,0000,,- [Instructor] The S N\N1 and S N 2 reactions
Dialogue: 0,0:00:02.63,0:00:04.35,Default,,0000,0000,0000,,involve leaving groups.
Dialogue: 0,0:00:04.35,0:00:06.36,Default,,0000,0000,0000,,Let's look at this pKa table
Dialogue: 0,0:00:06.36,0:00:09.10,Default,,0000,0000,0000,,to study leaving groups in more detail.
Dialogue: 0,0:00:09.10,0:00:10.95,Default,,0000,0000,0000,,On the left, we have the acid.
Dialogue: 0,0:00:10.95,0:00:14.21,Default,,0000,0000,0000,,For example, hydroiodic acid, HI,
Dialogue: 0,0:00:14.21,0:00:17.60,Default,,0000,0000,0000,,with an approximate pKa of negative 11.
Dialogue: 0,0:00:17.60,0:00:19.63,Default,,0000,0000,0000,,Remember, the lower the pKa value,
Dialogue: 0,0:00:19.63,0:00:21.31,Default,,0000,0000,0000,,the stronger the acid.
Dialogue: 0,0:00:21.31,0:00:24.93,Default,,0000,0000,0000,,So on this table with the\NpKa value of negative 11,
Dialogue: 0,0:00:24.93,0:00:28.13,Default,,0000,0000,0000,,hydroiodic acid is the strongest acid
Dialogue: 0,0:00:28.13,0:00:29.66,Default,,0000,0000,0000,,and the stronger the acid,
Dialogue: 0,0:00:29.66,0:00:32.11,Default,,0000,0000,0000,,the more stable the conjugate base.
Dialogue: 0,0:00:32.11,0:00:33.88,Default,,0000,0000,0000,,So the conjugate base to HI
Dialogue: 0,0:00:33.88,0:00:36.21,Default,,0000,0000,0000,,is I minus the iodide anion.
Dialogue: 0,0:00:37.20,0:00:40.22,Default,,0000,0000,0000,,And since this is the conjugate\Nbased to the strongest acid,
Dialogue: 0,0:00:40.22,0:00:42.57,Default,,0000,0000,0000,,this is the most stable base.
Dialogue: 0,0:00:42.57,0:00:43.96,Default,,0000,0000,0000,,Let me write that down here.
Dialogue: 0,0:00:43.96,0:00:47.66,Default,,0000,0000,0000,,This is the most stable base on the table,
Dialogue: 0,0:00:47.66,0:00:49.93,Default,,0000,0000,0000,,which means that the iodide anion
Dialogue: 0,0:00:49.93,0:00:51.53,Default,,0000,0000,0000,,is an excellent leaving group
Dialogue: 0,0:00:51.53,0:00:54.14,Default,,0000,0000,0000,,because it is very stable.
Dialogue: 0,0:00:54.14,0:00:56.11,Default,,0000,0000,0000,,Next, we have hydrobromic acid,
Dialogue: 0,0:00:56.11,0:00:58.68,Default,,0000,0000,0000,,approximate pKa of negative nine.
Dialogue: 0,0:00:58.68,0:01:01.21,Default,,0000,0000,0000,,So the conjugate base\Nwould be the bromide anion,
Dialogue: 0,0:01:01.21,0:01:04.17,Default,,0000,0000,0000,,so also a stable conjugate base.
Dialogue: 0,0:01:04.17,0:01:06.50,Default,,0000,0000,0000,,So therefore, a good leaving group.
Dialogue: 0,0:01:06.50,0:01:09.83,Default,,0000,0000,0000,,For HCl, it's the chloride anion,
Dialogue: 0,0:01:09.83,0:01:11.71,Default,,0000,0000,0000,,also a good leaving group.
Dialogue: 0,0:01:11.71,0:01:14.15,Default,,0000,0000,0000,,So you see these halide anions
Dialogue: 0,0:01:14.15,0:01:17.42,Default,,0000,0000,0000,,as leaving groups all the\Ntime in organic mechanisms.
Dialogue: 0,0:01:17.42,0:01:18.59,Default,,0000,0000,0000,,Let me write this down here.
Dialogue: 0,0:01:18.59,0:01:20.13,Default,,0000,0000,0000,,So these are all examples
Dialogue: 0,0:01:20.13,0:01:21.03,Default,,0000,0000,0000,,of
Dialogue: 0,0:01:21.03,0:01:22.06,Default,,0000,0000,0000,,good
Dialogue: 0,0:01:22.06,0:01:23.11,Default,,0000,0000,0000,,leaving
Dialogue: 0,0:01:23.11,0:01:24.51,Default,,0000,0000,0000,,groups.
Dialogue: 0,0:01:24.51,0:01:27.62,Default,,0000,0000,0000,,Next, let's look at this\Nacid on the left here.
Dialogue: 0,0:01:27.62,0:01:30.78,Default,,0000,0000,0000,,This is p-Toluenesulfonic acid
Dialogue: 0,0:01:30.78,0:01:33.10,Default,,0000,0000,0000,,with a pKa value of negative three,
Dialogue: 0,0:01:33.10,0:01:34.94,Default,,0000,0000,0000,,so it's still pretty acidic.
Dialogue: 0,0:01:34.94,0:01:38.25,Default,,0000,0000,0000,,The conjugate base to\Nthis is on the right here
Dialogue: 0,0:01:38.25,0:01:41.56,Default,,0000,0000,0000,,and we call this anion a tosylate group.
Dialogue: 0,0:01:41.56,0:01:42.55,Default,,0000,0000,0000,,Let me write this down.
Dialogue: 0,0:01:42.55,0:01:45.22,Default,,0000,0000,0000,,This is called a tosylate group.
Dialogue: 0,0:01:46.15,0:01:48.84,Default,,0000,0000,0000,,And since it's kind of a bulky group,
Dialogue: 0,0:01:48.84,0:01:51.38,Default,,0000,0000,0000,,instead of drawing this out all the time,
Dialogue: 0,0:01:51.38,0:01:53.60,Default,,0000,0000,0000,,you often see OTs written.
Dialogue: 0,0:01:53.60,0:01:54.43,Default,,0000,0000,0000,,So, OTs,
Dialogue: 0,0:01:55.75,0:01:56.58,Default,,0000,0000,0000,,like that,
Dialogue: 0,0:01:56.58,0:01:57.54,Default,,0000,0000,0000,,and you could put a negative charge
Dialogue: 0,0:01:57.54,0:01:59.96,Default,,0000,0000,0000,,on the oxygen here if you wanted to.
Dialogue: 0,0:01:59.96,0:02:01.89,Default,,0000,0000,0000,,So you'll see the tosylate group
Dialogue: 0,0:02:01.89,0:02:05.72,Default,,0000,0000,0000,,function as a leaving\Ngroup in many reactions.
Dialogue: 0,0:02:07.12,0:02:09.10,Default,,0000,0000,0000,,Let's look at an example of another acid.
Dialogue: 0,0:02:09.10,0:02:13.66,Default,,0000,0000,0000,,So if I move down here to\NH3O+, the hydronium ion,
Dialogue: 0,0:02:13.66,0:02:15.95,Default,,0000,0000,0000,,with a pKa value of negative two.
Dialogue: 0,0:02:15.95,0:02:19.33,Default,,0000,0000,0000,,The conjugate base to H3O+ is H2O
Dialogue: 0,0:02:19.33,0:02:22.45,Default,,0000,0000,0000,,and water is also a good leaving group.
Dialogue: 0,0:02:22.45,0:02:24.46,Default,,0000,0000,0000,,So let's go back up here to the topic
Dialogue: 0,0:02:24.46,0:02:27.38,Default,,0000,0000,0000,,and we can see that all the\Nacids that we talked about
Dialogue: 0,0:02:27.38,0:02:29.40,Default,,0000,0000,0000,,have negative pKa values,
Dialogue: 0,0:02:29.40,0:02:31.82,Default,,0000,0000,0000,,so negative 11, negative\Nnine, negative seven,
Dialogue: 0,0:02:31.82,0:02:34.57,Default,,0000,0000,0000,,negative three, and negative two.
Dialogue: 0,0:02:35.55,0:02:37.10,Default,,0000,0000,0000,,And notice all of the conjugate bases
Dialogue: 0,0:02:37.10,0:02:39.25,Default,,0000,0000,0000,,are good leaving groups.
Dialogue: 0,0:02:39.25,0:02:40.68,Default,,0000,0000,0000,,So you can say that if an acid
Dialogue: 0,0:02:40.68,0:02:43.46,Default,,0000,0000,0000,,has a negative value for the pKa,
Dialogue: 0,0:02:43.46,0:02:47.26,Default,,0000,0000,0000,,the conjugate base will\Nbe a good leaving group.
Dialogue: 0,0:02:47.26,0:02:50.41,Default,,0000,0000,0000,,Let's look at another example of an acid.
Dialogue: 0,0:02:50.41,0:02:51.93,Default,,0000,0000,0000,,So, water.
Dialogue: 0,0:02:51.93,0:02:54.85,Default,,0000,0000,0000,,Water's pKa value is positive 15.7,
Dialogue: 0,0:02:56.57,0:02:59.28,Default,,0000,0000,0000,,so it's not a very strong acid.
Dialogue: 0,0:02:59.28,0:03:00.68,Default,,0000,0000,0000,,The conjugate base to water
Dialogue: 0,0:03:00.68,0:03:03.01,Default,,0000,0000,0000,,is the hydroxide anion, OH-,
Dialogue: 0,0:03:03.100,0:03:05.60,Default,,0000,0000,0000,,and this is a bad leaving group.
Dialogue: 0,0:03:05.60,0:03:08.06,Default,,0000,0000,0000,,So hydroxide ion is a bad leaving group
Dialogue: 0,0:03:08.06,0:03:10.64,Default,,0000,0000,0000,,and that's because water\Nis not a strong acid.
Dialogue: 0,0:03:10.64,0:03:12.31,Default,,0000,0000,0000,,Look at this value for the pKa,
Dialogue: 0,0:03:12.31,0:03:13.47,Default,,0000,0000,0000,,positive 15.7.
Dialogue: 0,0:03:15.30,0:03:18.06,Default,,0000,0000,0000,,So if we look at ethanol,\Nsimilar story here.
Dialogue: 0,0:03:18.06,0:03:21.60,Default,,0000,0000,0000,,So ethanol has a pKa value of positive 16.
Dialogue: 0,0:03:21.60,0:03:25.09,Default,,0000,0000,0000,,So the ethoxide anion is\Nnot a good leaving group,
Dialogue: 0,0:03:25.09,0:03:28.51,Default,,0000,0000,0000,,so this pKa values are in the positive
Dialogue: 0,0:03:28.51,0:03:33.18,Default,,0000,0000,0000,,and these conjugate bases\Nmust not be very stable
Dialogue: 0,0:03:33.18,0:03:35.40,Default,,0000,0000,0000,,which means they are bad leaving groups.
Dialogue: 0,0:03:35.40,0:03:36.39,Default,,0000,0000,0000,,Let me write that down here.
Dialogue: 0,0:03:36.39,0:03:40.06,Default,,0000,0000,0000,,So these are examples\Nof bad leaving groups.
Dialogue: 0,0:03:43.23,0:03:45.20,Default,,0000,0000,0000,,Both S N 1 and S N 2 reactions
Dialogue: 0,0:03:45.20,0:03:46.91,Default,,0000,0000,0000,,need good leaving groups.
Dialogue: 0,0:03:46.91,0:03:50.33,Default,,0000,0000,0000,,However, the S N 1 reaction\Nis even more sensitive.
Dialogue: 0,0:03:50.33,0:03:52.05,Default,,0000,0000,0000,,Let's look at tert-Butyl chloride.
Dialogue: 0,0:03:52.05,0:03:55.90,Default,,0000,0000,0000,,Let's say it's reacting\Nvia an S N 1 mechanism.
Dialogue: 0,0:03:55.90,0:03:58.30,Default,,0000,0000,0000,,The first step should be\Nloss of leaving group.
Dialogue: 0,0:03:58.30,0:04:00.83,Default,,0000,0000,0000,,So these electrons come\Noff onto the chlorine.
Dialogue: 0,0:04:00.83,0:04:03.18,Default,,0000,0000,0000,,We would form the chloride anion
Dialogue: 0,0:04:03.18,0:04:06.42,Default,,0000,0000,0000,,which has a negative one formal charge.
Dialogue: 0,0:04:06.42,0:04:08.80,Default,,0000,0000,0000,,We just saw on our pKa table
Dialogue: 0,0:04:08.80,0:04:11.98,Default,,0000,0000,0000,,that the chloride anion is\Na stable conjugate base.
Dialogue: 0,0:04:11.98,0:04:15.08,Default,,0000,0000,0000,,So therefore, this is\Na good leaving group.
Dialogue: 0,0:04:15.08,0:04:18.15,Default,,0000,0000,0000,,We're taking a bond away\Nfrom the carbon in red,
Dialogue: 0,0:04:18.15,0:04:22.66,Default,,0000,0000,0000,,so the carbon in red gets\Na plus one formal charge
Dialogue: 0,0:04:22.66,0:04:26.40,Default,,0000,0000,0000,,and we form a tertiary\Ncarbocation as well.
Dialogue: 0,0:04:26.40,0:04:29.31,Default,,0000,0000,0000,,Since this is the rate determining step
Dialogue: 0,0:04:29.31,0:04:31.36,Default,,0000,0000,0000,,of our S N 1 mechanism,
Dialogue: 0,0:04:31.36,0:04:34.42,Default,,0000,0000,0000,,the formation of our stable anion,
Dialogue: 0,0:04:34.42,0:04:36.65,Default,,0000,0000,0000,,this formation of a good leaving group
Dialogue: 0,0:04:36.65,0:04:39.96,Default,,0000,0000,0000,,helps the S N 1 mechanism occur.
Dialogue: 0,0:04:39.96,0:04:42.29,Default,,0000,0000,0000,,Next, let's look at this alcohol here.
Dialogue: 0,0:04:42.29,0:04:43.59,Default,,0000,0000,0000,,If we approach it the same way
Dialogue: 0,0:04:43.59,0:04:45.22,Default,,0000,0000,0000,,as we did in the previous problem
Dialogue: 0,0:04:45.22,0:04:47.82,Default,,0000,0000,0000,,and we said, "Okay, first step\Nis loss of the leaving group
Dialogue: 0,0:04:47.82,0:04:50.59,Default,,0000,0000,0000,,"and these electrons come\Noff onto the oxygen."
Dialogue: 0,0:04:50.59,0:04:53.11,Default,,0000,0000,0000,,Think about what leaving group that is.
Dialogue: 0,0:04:53.11,0:04:55.85,Default,,0000,0000,0000,,That would be the hydroxide ion
Dialogue: 0,0:04:55.85,0:04:58.12,Default,,0000,0000,0000,,which we know from our pKa table
Dialogue: 0,0:04:58.12,0:05:01.03,Default,,0000,0000,0000,,is not a good leaving group.
Dialogue: 0,0:05:01.03,0:05:02.94,Default,,0000,0000,0000,,So the hydroxide ion
Dialogue: 0,0:05:02.94,0:05:07.11,Default,,0000,0000,0000,,is not as stable of an\Nanion as the chloride anion.
Dialogue: 0,0:05:07.11,0:05:09.15,Default,,0000,0000,0000,,So the chloride anion\Nis a good leaving group.
Dialogue: 0,0:05:09.15,0:05:11.59,Default,,0000,0000,0000,,The hydroxide anion is\Na bad leaving group.
Dialogue: 0,0:05:11.59,0:05:14.94,Default,,0000,0000,0000,,So that's not the first\Nstep of this mechanism.
Dialogue: 0,0:05:14.94,0:05:17.45,Default,,0000,0000,0000,,We need to make a better leaving group
Dialogue: 0,0:05:17.45,0:05:20.49,Default,,0000,0000,0000,,and you can do that by\Nhaving a proton source.
Dialogue: 0,0:05:20.49,0:05:23.47,Default,,0000,0000,0000,,Let's say we have a source of\Nprotons, an acid in solution,
Dialogue: 0,0:05:23.47,0:05:25.48,Default,,0000,0000,0000,,so let's say there's an H+ here.
Dialogue: 0,0:05:25.48,0:05:28.22,Default,,0000,0000,0000,,The first step would be\Nto protonate our alcohol,
Dialogue: 0,0:05:28.22,0:05:30.23,Default,,0000,0000,0000,,so our alcohol is gonna act as a base
Dialogue: 0,0:05:30.23,0:05:32.25,Default,,0000,0000,0000,,and pick up a proton.
Dialogue: 0,0:05:32.25,0:05:33.95,Default,,0000,0000,0000,,Let's draw the results of that.
Dialogue: 0,0:05:33.95,0:05:36.02,Default,,0000,0000,0000,,We have our ring.
Dialogue: 0,0:05:36.02,0:05:37.70,Default,,0000,0000,0000,,Let's put in that methyl group.
Dialogue: 0,0:05:37.70,0:05:41.64,Default,,0000,0000,0000,,And now our oxygen is\Nbonded to two hydrogens.
Dialogue: 0,0:05:41.64,0:05:44.82,Default,,0000,0000,0000,,There's still a lone pair\Nof electrons on this oxygen
Dialogue: 0,0:05:44.82,0:05:47.14,Default,,0000,0000,0000,,which give the oxygen a plus,
Dialogue: 0,0:05:47.14,0:05:49.48,Default,,0000,0000,0000,,which gives the oxygen a\Nplus one formal charge.
Dialogue: 0,0:05:49.48,0:05:52.78,Default,,0000,0000,0000,,So the electrons here\Nin magenta, let's say,
Dialogue: 0,0:05:52.78,0:05:55.59,Default,,0000,0000,0000,,pick up this proton to form this bond.
Dialogue: 0,0:05:55.59,0:05:57.85,Default,,0000,0000,0000,,Now we're ready for\Nloss of a leaving group
Dialogue: 0,0:05:57.85,0:06:01.73,Default,,0000,0000,0000,,because if these electrons\Ncome off onto the oxygen now,
Dialogue: 0,0:06:01.73,0:06:04.38,Default,,0000,0000,0000,,we form water as a leaving group.
Dialogue: 0,0:06:04.38,0:06:05.86,Default,,0000,0000,0000,,Let me draw that in here.
Dialogue: 0,0:06:05.86,0:06:08.36,Default,,0000,0000,0000,,So here is the water molecule.
Dialogue: 0,0:06:09.36,0:06:12.81,Default,,0000,0000,0000,,Let me highlight those electrons in blue.
Dialogue: 0,0:06:12.81,0:06:17.58,Default,,0000,0000,0000,,These electrons come off onto\Nthe oxygen then we form water
Dialogue: 0,0:06:17.58,0:06:19.22,Default,,0000,0000,0000,,and we know from our pKa table
Dialogue: 0,0:06:19.22,0:06:22.30,Default,,0000,0000,0000,,that water is a good leaving group.
Dialogue: 0,0:06:22.30,0:06:25.80,Default,,0000,0000,0000,,We're taking a bond away\Nfrom this carbon in red,
Dialogue: 0,0:06:25.80,0:06:28.94,Default,,0000,0000,0000,,so we're also gonna form\Na tertiary carbocation.
Dialogue: 0,0:06:28.94,0:06:30.79,Default,,0000,0000,0000,,Let me draw that in here.
Dialogue: 0,0:06:30.79,0:06:31.93,Default,,0000,0000,0000,,Here's our ring.
Dialogue: 0,0:06:31.93,0:06:33.38,Default,,0000,0000,0000,,Here's our methyl group.
Dialogue: 0,0:06:33.38,0:06:37.85,Default,,0000,0000,0000,,A plus one formal charge\Non the carbon in red.
Dialogue: 0,0:06:37.85,0:06:40.52,Default,,0000,0000,0000,,So by thinking about your pKa values,
Dialogue: 0,0:06:40.52,0:06:43.77,Default,,0000,0000,0000,,you can determine the\Nstability of the conjugate base
Dialogue: 0,0:06:43.77,0:06:45.36,Default,,0000,0000,0000,,and therefore, if a leaving group
Dialogue: 0,0:06:45.36,0:06:48.11,Default,,0000,0000,0000,,is a good leaving group\Nor a bad leaving group
Dialogue: 0,0:06:48.11,0:06:51.82,Default,,0000,0000,0000,,and that helps you out when\Nyou're drawing mechanisms.