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Current time:0:00Total duration:7:25

Video transcript

let's look at two ways to prepare alkynes from alkyl halide so here I have an alkyl halide so this is a dye halide and my two halogens are attached to one carbon we call this a geminal dihalide so it's 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 e to 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 e 2 elimination of a geminal dihalide so let's start with our di halide 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 hydrogen's on this carbon okay 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 when 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 e 2 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 have still have our halogen down here and over here on 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 we've done one e 2 elimination reaction and we're going to do one more right so we get another another amide anion comes along alright 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 bass in excess if you're trying to do this let's look at a very similar reaction a double e to elimination this time the halogens are not on the same carbon so let's go ahead and draw the general reaction for for this we have a two carbons right here and we have two halogens right here and then hydrogen and then hydrogen so this time this time we have we have our halogens two halogens on on adjacent carbons so this is called vicinal die halide so let's go ahead and write that so this is vicinal right the one we did before was geminal so a vicinal die halide will react in a very similar way if you add a strong base like sodium amide right and use ammonia for your solvent all right 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 e to elimination reaction again let's look at the mechanism alright so let's start with our vicinal die halide down here so let's go ahead and put our halogens in there right lone pairs of electrons on our halogens like that all right and then we have hydrogen and we have hydrogen right here so we have our amide anion right 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 onto our halogen so that's our first e to elimination reaction all right so let's just go ahead and write e 2 here to remind us this is yet another e 2 reaction and let's go ahead and draw the product of that alright so now we're going to have carbon double bonded to another carbon and then we're going to have 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 right to form our alkyne we're going to get another e 2 e 2 elimination reaction so sodium amide another another anion of sodium amide comes along alright so let's go ahead and put in those lone pairs of electrons like that it's going to function as a base electrons takes this proton these electrons kick in here to form our triple bond same time our halogen leaves and so we form our alkyne like that so you can produce alkynes from either vicinal or geminal dihalide x' be a double e to elimination reaction let's see how we could use this in a synthesis reaction so let's go ahead and and and try to make something try to make an alkyne from an alkyne okay so let's start with an alkene here I'll put some some benzene rings on this guy here so here's a here's a benzene ring like that put in my lone pairs so I 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 a one two diphenyl ethylene and i'm going to react this alkene with bromine alright you could use you know use a solvent like a carbon tetrachloride or something like that and and we're reacting an alkene with a halogen and we've seen this reaction before in the videos on reactions of alkenes right we're going to add two bro means across our double bond alright so we're going to draw the product of this reaction alright our benzene rings aren't going to react as readily as our double bond 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're also have a hydrogen bonded to each one of these carbons right that hydrogen was originally there as well over here on the left and we form one to dye bromo one to diphenyl methane here and now we have a vicinal die halide right so if we add a strong base to our business a LED we can prepare an alkyne from that so if we add alright if we add an excess of sodium amide right so in excess of sodium amide in ammonia alright we know that we're going to get a double e two elimination reaction and those halogens are going to go away and our double e to elimination reaction and form a triple bond alright so we're going to form a triple bond all right so the 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 gonna get an alkyne knot on either side of that alkyne right you're going to get phenyl group so let's go ahead and draw in our benzene rings like that so does it really matter how we draw our electron so we'll go ahead and do this all right so this would be our product so let's go ahead and put in those electrons so B diphenyl acetylene so you can synthesize alkynes from alkenes or you could synthesize now kind from a dye halide so this is one way to do it