1 00:00:00,030 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,840 Commons License. 3 00:00:03,840 --> 00:00:06,840 Your support will help MIT Open Courseware continue to 4 00:00:06,840 --> 00:00:10,520 offer high-quality educational resources for free. 5 00:00:10,520 --> 00:00:13,390 To make a donation or view additional materials from 6 00:00:13,390 --> 00:00:17,430 hundreds of MIT courses, visit MIT Open Courseware at 7 00:00:17,430 --> 00:00:18,680 ocw.mit.edu. 8 00:00:22,580 --> 00:00:23,070 Hi. 9 00:00:23,070 --> 00:00:23,700 I'm Brian. 10 00:00:23,700 --> 00:00:26,730 We're going to be going over problem number 13 of the fall 11 00:00:26,730 --> 00:00:29,030 2009 final exam. 12 00:00:29,030 --> 00:00:32,650 I like to think of the theme of this problem as polymers. 13 00:00:32,650 --> 00:00:34,130 So that's the overarching concept. 14 00:00:34,130 --> 00:00:36,130 But I'm going to give you some things I think are important 15 00:00:36,130 --> 00:00:39,240 to review and to know before actually meaningfully 16 00:00:39,240 --> 00:00:40,980 attempting this problem. 17 00:00:40,980 --> 00:00:44,370 So there's three things that I that I sort of thought were 18 00:00:44,370 --> 00:00:45,080 very important. 19 00:00:45,080 --> 00:00:48,000 The first one is polymerization processes. 20 00:00:48,000 --> 00:00:50,730 So we learned in class and in the book as well that there's 21 00:00:50,730 --> 00:00:54,120 two ways that we've covered that you can create a polymer. 22 00:00:54,120 --> 00:00:56,610 And that is through either addition polymerization or 23 00:00:56,610 --> 00:00:58,520 condensation polymerization. 24 00:00:58,520 --> 00:01:01,150 And I'll talk about that more in a second. 25 00:01:01,150 --> 00:01:03,770 The second thing you want to know is basically molecular 26 00:01:03,770 --> 00:01:05,930 weight and how it applies to polymers. 27 00:01:05,930 --> 00:01:09,150 And you want to know about elastomers for this problem. 28 00:01:09,150 --> 00:01:10,660 If you don't know about elastomers then you'll have 29 00:01:10,660 --> 00:01:12,370 trouble with the last part of the problem. 30 00:01:12,370 --> 00:01:15,000 So that's kind of what I would review. 31 00:01:15,000 --> 00:01:17,400 And then once you've got that under your belt give the 32 00:01:17,400 --> 00:01:18,990 problem a try. 33 00:01:18,990 --> 00:01:20,060 So we're going to go and we're going to 34 00:01:20,060 --> 00:01:21,310 start the problem now. 35 00:01:23,600 --> 00:01:27,850 We're given in the problem the 6-aminohexanoic acid, which 36 00:01:27,850 --> 00:01:29,560 I've drawn here. 37 00:01:29,560 --> 00:01:32,170 And we're also given at the beginning of the problem the 38 00:01:32,170 --> 00:01:34,300 structure of nylon 6. 39 00:01:34,300 --> 00:01:38,130 This is the actual reaction that's used to create nylon. 40 00:01:38,130 --> 00:01:41,300 So what I'm going to do is I'm going to show you-- 41 00:01:41,300 --> 00:01:43,990 or ask the question of how do we create nylon from 42 00:01:43,990 --> 00:01:46,850 6-aminohexanoic acid. 43 00:01:46,850 --> 00:01:50,350 And that's actually going to answer part A and part B to 44 00:01:50,350 --> 00:01:52,350 this problem. 45 00:01:52,350 --> 00:01:54,410 So to create a polymer-- 46 00:01:54,410 --> 00:01:57,060 in this case we're talking about the polymer nylon 6-- 47 00:01:57,060 --> 00:01:59,940 you actually have to add together many mers. 48 00:01:59,940 --> 00:02:03,900 So merge is the term we used to denote the single unit 49 00:02:03,900 --> 00:02:09,380 which is repeated n times to create the poly mer-- 50 00:02:09,380 --> 00:02:10,410 polymer. 51 00:02:10,410 --> 00:02:16,180 So in this case the mer, the individual unit of nylon 6, is 52 00:02:16,180 --> 00:02:18,650 6-aminohexanoic acid. 53 00:02:18,650 --> 00:02:19,590 And that's this. 54 00:02:19,590 --> 00:02:21,920 That's the same thing right there. 55 00:02:21,920 --> 00:02:23,990 So we're going to add these together and we're going to 56 00:02:23,990 --> 00:02:26,720 start creating our long chain of nylon 6. 57 00:02:26,720 --> 00:02:30,440 Remember, nylon 6 a polymer is a really, really long chain. 58 00:02:30,440 --> 00:02:33,040 Think of it as spaghetti, very long spaghetti. 59 00:02:33,040 --> 00:02:35,620 It isn't just two molecules that react. 60 00:02:35,620 --> 00:02:37,680 So let's just start somewhere. 61 00:02:37,680 --> 00:02:38,750 It's got to begin somewhere. 62 00:02:38,750 --> 00:02:40,870 You have to have two molecules at some point in the beginning 63 00:02:40,870 --> 00:02:42,590 reacting with each other. 64 00:02:42,590 --> 00:02:44,460 So let's look at our two molecules of 65 00:02:44,460 --> 00:02:45,970 6-aminohexanoic acid-- 66 00:02:45,970 --> 00:02:49,340 I have to keep looking back at how to say that-- 67 00:02:49,340 --> 00:02:51,980 so let's take a look and we have to think about now our 68 00:02:51,980 --> 00:02:55,260 two possible processes or routes towards polymerization. 69 00:02:55,260 --> 00:02:57,860 There's addition and there's condensation. 70 00:02:57,860 --> 00:03:01,230 Now addition, as you probably read and heard, requires there 71 00:03:01,230 --> 00:03:04,550 to be a double bond present between two carbon atoms. The 72 00:03:04,550 --> 00:03:07,160 reason for this is because usually have a free radical 73 00:03:07,160 --> 00:03:09,830 coming in, which will then initiate. 74 00:03:09,830 --> 00:03:12,950 That's why oftentimes the free radical molecule or free 75 00:03:12,950 --> 00:03:14,690 radical provider is called the initiator. 76 00:03:14,690 --> 00:03:18,650 You need an initiator to come in to break that double bond 77 00:03:18,650 --> 00:03:20,330 and then create another free radical. 78 00:03:20,330 --> 00:03:23,260 Which then lets the process continue along in 79 00:03:23,260 --> 00:03:24,810 a chain like that. 80 00:03:24,810 --> 00:03:27,890 If we look at our molecule here, we don't actually have 81 00:03:27,890 --> 00:03:30,280 any double bonds between two carbon atoms. We have a double 82 00:03:30,280 --> 00:03:33,630 bond with an oxygen but that's not sufficient for an 83 00:03:33,630 --> 00:03:36,300 addition-type reaction. 84 00:03:36,300 --> 00:03:39,720 For a condensation reaction, what that basically implies is 85 00:03:39,720 --> 00:03:44,840 two mer units, or two things coming together to react and 86 00:03:44,840 --> 00:03:46,630 they give off a byproduct. 87 00:03:46,630 --> 00:03:49,390 So that's how I remember condensation, very often the 88 00:03:49,390 --> 00:03:50,660 byproduct is water. 89 00:03:50,660 --> 00:03:55,200 You can have other byproducts, like HCl or any other sort of 90 00:03:55,200 --> 00:03:57,160 small molecules can be given off. 91 00:03:57,160 --> 00:04:00,260 But most times what we'll see in 3091 will be water given 92 00:04:00,260 --> 00:04:02,620 off in condensation, so it's sort of intuitive to call it 93 00:04:02,620 --> 00:04:03,990 condensation. 94 00:04:03,990 --> 00:04:07,570 In this problem, if we look at this, we have a molecule that 95 00:04:07,570 --> 00:04:11,090 has the Hs on one side and has Os on the other side, we can 96 00:04:11,090 --> 00:04:13,830 actually almost see this immediately as being a 97 00:04:13,830 --> 00:04:15,170 condensation reaction. 98 00:04:15,170 --> 00:04:18,010 What's going to happen is we're going to have this O 99 00:04:18,010 --> 00:04:19,360 reacting with the positive end of this 100 00:04:19,360 --> 00:04:21,020 molecule, which has Hs. 101 00:04:21,020 --> 00:04:23,450 And we're going to have an H2O liberated and given off. 102 00:04:23,450 --> 00:04:24,940 We're going to have the new molecule formed. 103 00:04:24,940 --> 00:04:26,760 Let me draw the molecule for you. 104 00:04:26,760 --> 00:04:28,010 Let me write it out. 105 00:04:59,830 --> 00:05:02,490 So we've just begun forming our polymer and this is a 106 00:05:02,490 --> 00:05:04,120 polymer which has an n of 2. 107 00:05:04,120 --> 00:05:06,180 It has 2 mer units building it. 108 00:05:06,180 --> 00:05:08,350 Notice I've kept the ends here. 109 00:05:08,350 --> 00:05:11,090 The positive and the negative here. 110 00:05:11,090 --> 00:05:14,030 In reality what's going to happen is we have a big vat of 111 00:05:14,030 --> 00:05:17,940 this 6-aminohexanoic acid and they're 112 00:05:17,940 --> 00:05:18,940 going to keep reacting. 113 00:05:18,940 --> 00:05:21,960 So we might have another one of these molecules float over 114 00:05:21,960 --> 00:05:23,790 and react with this O. 115 00:05:23,790 --> 00:05:26,800 Or conversely, we could have another one of these molecules 116 00:05:26,800 --> 00:05:29,150 come over and react with the H. 117 00:05:29,150 --> 00:05:30,960 So this thing is going to keep building. 118 00:05:30,960 --> 00:05:33,410 And the way these polymerization processes begin 119 00:05:33,410 --> 00:05:37,470 is the control of some parameter of the system. 120 00:05:37,470 --> 00:05:39,570 It could be temperature, pressure you can add something 121 00:05:39,570 --> 00:05:40,990 in to sort of initiate it. 122 00:05:40,990 --> 00:05:43,000 But that's how this is going to start. 123 00:05:43,000 --> 00:05:45,800 So we've actually sort of answered part a and b. 124 00:05:45,800 --> 00:05:48,890 We've come to the conclusion that this is obviously a 125 00:05:48,890 --> 00:05:50,340 condensation reaction. 126 00:05:50,340 --> 00:05:51,965 Because we're going to give off-- 127 00:05:51,965 --> 00:05:56,570 I left that out there, see if you guys caught it-- 128 00:05:56,570 --> 00:05:58,460 H2O. 129 00:05:58,460 --> 00:06:04,070 So we're losing this O and two of these Hs and 130 00:06:04,070 --> 00:06:05,320 we're forming water. 131 00:06:05,320 --> 00:06:06,880 And that's the byproduct. 132 00:06:06,880 --> 00:06:09,080 And a byproduct immediately should tell you we've got 133 00:06:09,080 --> 00:06:10,390 condensation going on. 134 00:06:10,390 --> 00:06:12,720 So condensation polymerization, part A. 135 00:06:12,720 --> 00:06:16,160 Part B, we've actually already done to sort of logic out the 136 00:06:16,160 --> 00:06:17,410 answer to part A. 137 00:06:17,410 --> 00:06:19,420 This is the answer to part B. 138 00:06:19,420 --> 00:06:20,660 Part B is just the reaction. 139 00:06:20,660 --> 00:06:22,150 I'll read it very specifically. 140 00:06:22,150 --> 00:06:24,380 Write the reaction that converts two molecules of 141 00:06:24,380 --> 00:06:28,800 6-aminohexanoic acid into a dimer of nylon 6. 142 00:06:28,800 --> 00:06:31,350 So we have two molecules forming the dimer. 143 00:06:31,350 --> 00:06:33,500 You are going to have trimer, and it goes on and on. 144 00:06:33,500 --> 00:06:36,420 The most common mistake on this problem was to give us 145 00:06:36,420 --> 00:06:39,530 the full repeating units. 146 00:06:39,530 --> 00:06:41,400 Let me just show you what that would look like. 147 00:06:41,400 --> 00:06:42,290 And then we'll move on with the problem. 148 00:06:42,290 --> 00:06:46,000 So the most common mistake on this problem was to not have 149 00:06:46,000 --> 00:06:46,990 these end units. 150 00:06:46,990 --> 00:06:50,070 And instead say, oh it's created this huge, long chain. 151 00:06:50,070 --> 00:06:51,720 So I'm going to write it as follows. 152 00:06:57,820 --> 00:07:01,940 And this is the symbol to imply that it repeats. 153 00:07:01,940 --> 00:07:03,190 Likewise over here. 154 00:07:09,340 --> 00:07:10,950 So people would give us this. 155 00:07:10,950 --> 00:07:12,430 And that's not correct. 156 00:07:12,430 --> 00:07:14,070 And the reason that's a problem is because not only do 157 00:07:14,070 --> 00:07:15,690 you lose points on this part, you also lose 158 00:07:15,690 --> 00:07:16,880 points on the next part. 159 00:07:16,880 --> 00:07:21,180 So part C is now asking us to calculate the molecular weight 160 00:07:21,180 --> 00:07:23,190 of a molecule given a certain n. 161 00:07:23,190 --> 00:07:25,160 n is the degree of polymerization. 162 00:07:25,160 --> 00:07:28,140 So in this case, n equals 2, we've got a dimer. 163 00:07:28,140 --> 00:07:31,680 In part C, we have n equals-- 164 00:07:31,680 --> 00:07:33,230 surprisingly-- 165 00:07:33,230 --> 00:07:38,760 3,091 You'll see that's a recurring theme in this class. 166 00:07:38,760 --> 00:07:42,420 So we have n equals 3,091 And the question is, how do we 167 00:07:42,420 --> 00:07:43,890 approach the molecular weight problem? 168 00:07:43,890 --> 00:07:48,220 Well we're told that we have 3,091 one of these mer units 169 00:07:48,220 --> 00:07:50,710 making up this polymer chain. 170 00:07:50,710 --> 00:07:51,950 So let me just write this down. 171 00:07:51,950 --> 00:07:54,950 We're going to form an equation. 172 00:07:54,950 --> 00:07:58,640 If we take the mass of the full chain and divide it by 173 00:07:58,640 --> 00:08:02,790 the mass of a single mer unit we should be able to extract 174 00:08:02,790 --> 00:08:04,280 the number of units in the chain. 175 00:08:04,280 --> 00:08:06,420 So let me write that out for you. 176 00:08:06,420 --> 00:08:09,820 So the total molecular weight of the molecule, divided by 177 00:08:09,820 --> 00:08:13,580 the weight of a particular mer unit. 178 00:08:13,580 --> 00:08:16,755 OK so we're going to put that as mer. 179 00:08:19,460 --> 00:08:20,710 And so this is easy. 180 00:08:20,710 --> 00:08:22,750 Let's first identify what our mer unit is. 181 00:08:22,750 --> 00:08:24,340 And this is why it was easy get 182 00:08:24,340 --> 00:08:25,390 tripped up on this problem. 183 00:08:25,390 --> 00:08:27,530 Because if you made a mistake on the part B, you have 184 00:08:27,530 --> 00:08:28,950 trouble on part C. 185 00:08:28,950 --> 00:08:30,710 What is our mer unit? 186 00:08:30,710 --> 00:08:33,960 So in our mer unit, what we're talking about is a chain which 187 00:08:33,960 --> 00:08:40,360 has 3,091 of these linked pieces here. 188 00:08:40,360 --> 00:08:41,580 So let's look at this. 189 00:08:41,580 --> 00:08:42,460 What do we have? 190 00:08:42,460 --> 00:08:43,490 Here's our mer unit. 191 00:08:43,490 --> 00:08:44,740 Let me circle it. 192 00:08:47,350 --> 00:08:51,700 We have 6 carbons. 193 00:08:51,700 --> 00:08:56,510 We have 11 hydrogens, 1 nitrogen and 1 194 00:08:56,510 --> 00:08:58,650 oxygen in this mer unit. 195 00:08:58,650 --> 00:09:00,860 We're not going to take the full, complete end unit. 196 00:09:00,860 --> 00:09:01,970 We're not counting this O. 197 00:09:01,970 --> 00:09:03,990 We're not counting those 2 Hs. 198 00:09:03,990 --> 00:09:08,630 We're dropping off 18 grams per reaction into water. 199 00:09:08,630 --> 00:09:09,600 So this is our mere unit. 200 00:09:09,600 --> 00:09:12,880 Let's calculate the mass of our mer unit. 201 00:09:12,880 --> 00:09:14,472 So it's going to be-- 202 00:09:14,472 --> 00:09:17,890 let me write this here-- 203 00:09:17,890 --> 00:09:20,260 we're looking for molecular weight. 204 00:09:20,260 --> 00:09:22,220 We're going to divide that by the mass of the mer unit. 205 00:09:22,220 --> 00:09:26,200 Let me just write out the full line of what it would be. 206 00:09:26,200 --> 00:09:31,830 Approximately 6 times 12 plus 11 times 1. 207 00:09:31,830 --> 00:09:34,610 So this is 6 carbons, times the mass of carbon. 208 00:09:34,610 --> 00:09:36,890 11 hydrogens times the mass of hydrogen. 209 00:09:36,890 --> 00:09:40,450 We're going to have 1 oxygen times the mass of oxygen. 210 00:09:40,450 --> 00:09:42,660 And we're going to have the nitrogen as well. 211 00:09:42,660 --> 00:09:43,940 So let me add that down here. 212 00:09:43,940 --> 00:09:46,760 1 nitrogen times the mass of nitrogen. 213 00:09:46,760 --> 00:09:47,280 I rounded. 214 00:09:47,280 --> 00:09:47,730 That's OK. 215 00:09:47,730 --> 00:09:50,590 As long as you get the idea right. 216 00:09:50,590 --> 00:09:52,000 So now we can easily just solve for 217 00:09:52,000 --> 00:09:53,290 the molecular weight. 218 00:09:53,290 --> 00:09:57,210 And we'll find that molecular weight in this problem will be 219 00:09:57,210 --> 00:10:05,270 equal to 3.5 times 10 to the fifth grams per mole. 220 00:10:10,310 --> 00:10:11,870 There's another way to do this problem, actually. 221 00:10:11,870 --> 00:10:15,680 And what you can do is you can calculate n times 222 00:10:15,680 --> 00:10:17,860 the number of this-- 223 00:10:17,860 --> 00:10:20,170 this particular molecule-- 224 00:10:20,170 --> 00:10:24,980 and you can subtract off 3,090 orders. 225 00:10:24,980 --> 00:10:27,500 18 grams. 3,090 times eighteen. 226 00:10:27,500 --> 00:10:28,630 You subtract that off. 227 00:10:28,630 --> 00:10:32,940 Think about a little bit why it's 3,090 and not 3,091. 228 00:10:32,940 --> 00:10:35,160 So that's just maybe a brain teaser for you. 229 00:10:35,160 --> 00:10:37,320 That is the answer to part C. 230 00:10:37,320 --> 00:10:40,730 And now part D asks us if we're able to convert this 231 00:10:40,730 --> 00:10:41,920 into an elastomer. 232 00:10:41,920 --> 00:10:44,010 So this is where it's really required that you understand 233 00:10:44,010 --> 00:10:46,590 what an elastomer is and what that implies and what the 234 00:10:46,590 --> 00:10:47,820 structure looks like. 235 00:10:47,820 --> 00:10:49,640 So an elastomer-- 236 00:10:49,640 --> 00:10:52,240 and I sort of paraphrased this from the book and added my own 237 00:10:52,240 --> 00:10:56,260 thing to it-- is basically a randomly oriented amorphous 238 00:10:56,260 --> 00:10:58,960 polymer with some cross-linking such that you 239 00:10:58,960 --> 00:11:01,960 have the ability to move the chains but not slide them 240 00:11:01,960 --> 00:11:03,600 completely over each other. 241 00:11:03,600 --> 00:11:05,740 So in order to have an elastomer you must have 242 00:11:05,740 --> 00:11:06,680 cross-linking. 243 00:11:06,680 --> 00:11:07,970 In order to have cross-linking, 244 00:11:07,970 --> 00:11:09,090 what must you have? 245 00:11:09,090 --> 00:11:11,020 Well from lecture and from the book we know that to 246 00:11:11,020 --> 00:11:14,330 cross-link you must have double bonds in your carbons. 247 00:11:14,330 --> 00:11:15,940 So you must have 2 carbons-- 248 00:11:15,940 --> 00:11:17,020 going back over here-- 249 00:11:17,020 --> 00:11:18,550 which are double-bonded together. 250 00:11:18,550 --> 00:11:19,830 And why is that? 251 00:11:19,830 --> 00:11:22,200 Because what happens is those double bonds will be broken by 252 00:11:22,200 --> 00:11:26,120 some initiator element-- many times we'll have sulfur-- 253 00:11:26,120 --> 00:11:29,920 will come in, break the double bond and then you'll have 254 00:11:29,920 --> 00:11:30,910 links forming between things. 255 00:11:30,910 --> 00:11:33,250 So let me actually give you real-life example of this so 256 00:11:33,250 --> 00:11:35,070 you can see what I'm talking about. 257 00:11:35,070 --> 00:11:38,940 We're going to show disulfide bridges bounding and creating 258 00:11:38,940 --> 00:11:40,455 cross-links in a molecule. 259 00:11:55,320 --> 00:11:57,310 Something we're very familiar with is rubber. 260 00:11:57,310 --> 00:11:59,100 And we're also very familiar with car tires. 261 00:11:59,100 --> 00:12:02,530 The big difference between those two things is that 262 00:12:02,530 --> 00:12:05,470 rubber has vulcanized Which means we're actually putting 263 00:12:05,470 --> 00:12:08,520 sulphur into it to create a different type of polymer. 264 00:12:08,520 --> 00:12:12,080 We're looking at a cross-linked, much stronger 265 00:12:12,080 --> 00:12:14,060 rubber which we use on our car tires. 266 00:12:14,060 --> 00:12:16,550 So polyisoprene looks something like this. 267 00:12:56,490 --> 00:12:59,520 You should be familiar with this notation by now. 268 00:12:59,520 --> 00:13:05,100 We have this denotes it repeats. 269 00:13:05,100 --> 00:13:06,290 We have our double bonds, we have 270 00:13:06,290 --> 00:13:07,620 our carbons and hydrogens. 271 00:13:07,620 --> 00:13:09,770 This is polyisoprene. 272 00:13:09,770 --> 00:13:11,250 And this is what we're going to add sulphur 273 00:13:11,250 --> 00:13:12,430 to vulcanize rubber. 274 00:13:12,430 --> 00:13:14,750 We're going to create vulcanized rubber. 275 00:13:14,750 --> 00:13:19,300 sulphur generally actually looks, it's a ring of 8 276 00:13:19,300 --> 00:13:25,740 sulphur atoms. We'll then add it to rubber, 277 00:13:25,740 --> 00:13:27,230 maybe heat it up mixed. 278 00:13:27,230 --> 00:13:30,910 And you'll make what you're used to on car tires. 279 00:13:30,910 --> 00:13:33,040 So what's actually happening here is that you can see we 280 00:13:33,040 --> 00:13:34,930 have this double bond between carbons. 281 00:13:34,930 --> 00:13:37,970 The sulphur's going to come in, it's going to react with 282 00:13:37,970 --> 00:13:42,480 this double bond, and is actually going to pull it off. 283 00:13:42,480 --> 00:13:44,430 And it might form a chain like that. 284 00:13:44,430 --> 00:13:47,480 And then this sulfur will then connect to another one of 285 00:13:47,480 --> 00:13:49,820 these polyisoprene molecules that was 286 00:13:49,820 --> 00:13:51,390 originally double bonded. 287 00:13:51,390 --> 00:13:53,630 And I'm not going to draw the rest of it out. 288 00:13:53,630 --> 00:13:56,900 And it's going to react, kill this double bond, and it's 289 00:13:56,900 --> 00:13:57,870 going to connect it. 290 00:13:57,870 --> 00:14:00,100 So basically what's happened in this process of forming 291 00:14:00,100 --> 00:14:04,790 sulfide bridges is you've got this polymer, polyisoprene, 292 00:14:04,790 --> 00:14:08,260 connecting to this polymer, polyisoprene, and you have 293 00:14:08,260 --> 00:14:09,720 cross-linking going on. 294 00:14:09,720 --> 00:14:13,910 And that's what we would need to have in our case for the 295 00:14:13,910 --> 00:14:18,185 nylon 6 in order to get an elastomer, And because nylon 296 00:14:18,185 --> 00:14:21,840 6, because our mer unit, 6-aminohexanoic acid doesn't 297 00:14:21,840 --> 00:14:24,750 have double bonds, we can't break them. 298 00:14:24,750 --> 00:14:25,850 We can't network. 299 00:14:25,850 --> 00:14:27,380 We can't cross-link. 300 00:14:27,380 --> 00:14:29,460 And therefore we can't make an elastomer. 301 00:14:29,460 --> 00:14:31,010 So the answer to this problem-- 302 00:14:31,010 --> 00:14:33,440 in a very long-winded manner-- 303 00:14:33,440 --> 00:14:34,320 is no. 304 00:14:34,320 --> 00:14:38,060 You cannot create an elastomer given that mer unit.