Preparation of alkynes
Note that if a terminal alkyne is produced, the strong base will deprotonate the alkyne to form an alkynide anion. A proton source (like H2O) must be added to regenerate the terminal alkyne. Created by Jay.
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- Aren't phenols aromatic rings plus alcohol (OH) group? Why is he drawing arenes...or is the alcohol group implied? Makes it confusing for a chemistry beginner like me! Teachers please remember: what is implicit for you needs to be made explicit for us :)(14 votes)
- I think he was saying phenyl which is just a benzene attached to an R group. You're right about what a phenol is.(11 votes)
- I LOVE your videos...thanks SO much, you've saved me on numerous occasions. One thing, however, that frustrates me is when there's an NH3 there but no mention is made of it. I'm guessing it's a solvent but it always makes me nervous that I need to factor that in somehow so a word or to about its presence would be appreciated. Just my 2 cents and thanks again!(11 votes)
- You're right, NH3 is a solvent here. Note that when NaNH2 deprotonates the alkylhalide, it forms NH3, which just adds into the solvent rather than being a separate side product (if we used a different base or solvent).(9 votes)
- Why is it important to distinguish between geminal and vicinal dihalides? They seem to behave exactly the same in this video.(5 votes)
- Yes, they do react the same in this video, but they are different compounds, and they behave differently in other reactions. For example, geminal dihalides give aldehydes or ketones on hydrolysis, but vicinal dihalides give vicinal diols on hydrolysis.(14 votes)
- At1:19, why is the halide opposite the reacting hydrogen the one that is eliminated rather then than the halide on the same side of the bond?(4 votes)
- This is an E2 elimination reaction. The H and the Br must be antiperiplanar to each other. This enables the developing p orbitals to overlap and form the π bonds.(6 votes)
- what does deprotonate means..?(1 vote)
- The prefix de- means "away or "from".
The word proton refers to a "proton" or "hydrogen atom".
The suffix -ate tells us that this is a verb or action word.
Thu, to deprotonate something is to remove a proton or hydrogen atom from it..(6 votes)
- The reason it doesn't matter is because it is symmetrical? IF it were asymmetrical we would use markovics rule(2 votes)
- Markovnikov's Rule doesn't apply here - this rule is followed for E1 elimination reactions which proceed via a carbocation.(3 votes)
- At1:00I don't why there is NH3. Is it solvent?(2 votes)
- NH3 is the solvent. If you are talking about -NH2, that is from NaNH2 and is a very strong base.(3 votes)
- 6:36"double E2 elimination reaction" The halogenation video shows that the halogens add anti. How can there be an E2 reaction without anti-periplanar hydrogens?(2 votes)
- The compound can rotate about the central C-C bond to make the H and Br bonds anti-periplanar.
Although antiperiplanar elimination is preferred, syn-periplanar is also possible.(2 votes)
- At1:05, there should be two sets of non-bonded electrons on the nitrogen in the sodamide anion. As drawn, it is a neutral radical. (Maybe I'm just not seeing the fourth non-bonded electron.)(1 vote)
- I think there is. It's fainter than the others, but wasn't there before.(2 votes)
- At1:17the attack of the electrons must be to the upper halogen or it might be to the down halogen? does that makes any difference?(1 vote)
- Hello Hikiri2010,
The electrons must attack the upper halogen. This E2 reaction is concerted (meaning it happens in one step), so the upper halogen acts as the leaving group because it is trans from the hydrogen that is abstracted. Although the end result isn't affect by how whether the top or bottom halogen leaves, it is a very important concept and can have an important impact on the final stereo-chemistry in certain situations.
The reason the hydrogen and leaving group (in this case the halogen) have to be trans during an E2 elimination, is that the electrons from the C-H bond immediately form the C-C double bond with the sigma* (sigma star) orbital. This sigma* orbital is directly opposite from the C-X bond, and when electrons flow into the sigma* orbital they push out the electrons in the C-X bond. This is what causes the halogen to leave as X-. In order for this to all happen in one step, the electrons have to be properly aligned with this sigma* orbital.(2 votes)
Let's look at two ways to prepare alkynes from alkyl halides. So here I have an alkyl halide. So this is a dihalide, and my two halogens are attached to one carbon. We call this a geminal dihalide. So this is going to be a geminal dihalide reacting with a very strong base, sodium amide. So this is going to give us an E2 elimination reaction. So we're going to get an E2 elimination reaction, and this E2 elimination reaction is actually going to occur twice. And we're going to end up with an alkyne as our final product. So let's take a look at the mechanism of our double E2 elimination of a geminal dihalide. So let's start with our dihalide over here. And this time we're going to put in all of our lone pairs of electrons on our halogen, like that. So let me go ahead and put all of those in there, and then I have two hydrogens on this carbon. OK. Sodium amide is a source of amide anions, which we saw in our previous video can function as a strong base. So a strong base means that a lone pair of electrons here on our nitrogen is going to take this proton. And these electrons, in here, are going to kick in to form a double bond at the same time these electrons kick off onto our halogen. So an E2 elimination mechanism. You can watch the previous videos on E2 elimination reactions for more details. So we're going to form ammonia as one of our products. And our other product is going to be carbon double-bonded to another carbon. And then we're going to still have our halogen down here. And over here, in the carbon on the right, we're still going to have a hydrogen, like that. So we're not quite to our alkyne yet. So we've done one E2 elimination reaction, and we're going to do one more. So we get another-- another amide anion comes along, and it's negatively charged. It's going to function as a base. It's going to take this proton this time. And these electrons are going to move in here to form our triple bond. And these electrons are going to kick off onto our halogen, like that. So that is going to finally form our alkyne here. So you always have to have your base in excess, if you're trying to do this. Let's look at a very similar reaction, a double E2 elimination. This time the halogens are not on the same carbon. So let's go ahead and draw the general reaction for this. We have two carbons right here, and we have two halogens right here. And then hydrogen, and then hydrogen. This time we have two halogens on adjacent carbons. So this is called vicinal dihalides. So let's go ahead and write that. So this is vicinal, and the one we did before was geminal. So a vicinal dihalide will react in a very similar way if you add a strong base like sodium amide and you use ammonia for your solvent. So you're going to form an alkyne once again. So you're going to get an alkyne. It's going to be via a double E2 elimination reaction again. Let's look at the mechanism. So let's start with our vicinal dihalide down here. So let's go ahead and put our halogens in there. Lone pairs of electrons on our halogens, like that. And then we have hydrogen, and we have hydrogen right here. So we have our amide anion, and once again, functions as a strong base. It's going to take a proton. So it's going to take this proton right here. These electrons are going to move in to form our double bond the same time these electrons kick off on to our halogen. So that's our first E2 elimination reaction. So let's just go ahead and write E2 here to remind us this is yet another E2 reaction. And let's go ahead and draw the product of that. So now we're going to have carbon double-bonded to another carbon. And then we're going to have a hydrogen right here. And then we're going to have our halogen up here, like that. And then we're going to have-- we need one more reaction to form our alkyne. We're going to get another E2 elimination reaction. So sodium amide-- another anion of sodium amide comes along. So let's go ahead and put in those lone pairs of electrons, like that. It's going to function as a base. Lone pair of electrons takes this proton. These electrons kick in here to form our triple bond at the same time our halogen leaves. And so we form our alkyne like that. So you can produce alkynes from either vicinal or geminal dihalides via a double E2 elimination reaction. Let's see how we could use this in a synthesis reaction. So let's go ahead and try to make something-- try to make an alkyne from an alkene. OK, so let's start with an alkene here. And I'll put some benzene rings on this guy here. So here's a benzene ring like that. Put in my lone pairs-- sorry, put in my bonds like that. And then I'm going to put a double bond right here. And then I'm going to put another benzene ring attached like that. So this is 1,2-Diphenylethylene. . And I'm going to react this alkene with bromine. And you could use a solvent like carbon tetrachloride or something like that. And we're reacting an alkene with a halogen. And we've seen this reaction before in the videos on reactions of alkenes. We're going to add two bromines across our double bond. So we're going to draw the product of this reaction. Our benzene rings aren't going to react as readily as our double bonds will. So let's go ahead and draw in our other benzene ring here, like that. And we know that we're going to add a bromine to either side of our double bond. So let's go ahead and add a bromine to either side of our double bond here. And we'll also have a hydrogen bonded to each one of these carbons. That hydrogen was originally there as well, over here on the left. And we form 1,2-Dibromo-1,2-diphenylethane here. And now we have a vicinal dihalide. So if we add a strong base to our vicinal dihalide we can prepare an alkyne from that. So if we add an excess of sodium amide in ammonia we know that we're going to get a double E2 elimination reaction. And those halogens are going to go away in our double E2 elimination reaction and form a triple bond. So we're going to form a triple bond. So those two carbons, the ones that form a triple bond, are these two carbons right here. So when you run through the mechanism, you're going to get an alkyne. And then, on either side of that alkyne, you're going to get a phenyl groups. So let's go ahead and draw in our benzene rings, like that. So it doesn't really matter how we draw our electrons, so we'll go ahead and do this. So this would be our product. So let's go ahead and put in those electrons. This will be diphenylacetylene. So you can synthesize alkynes from alkenes, or you could synthesize an alkyne from a dihalide. So this is one way to do it.