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Let's look at two ways[br]to prepare alkynes

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from alkyl halides.

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So here I have an alkyl halide.

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So this is a dihalide,[br]and my two halogens

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are attached to one carbon.

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We call this a geminal dihalide.

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So this is going to be a[br]geminal dihalide reacting

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with a very strong[br]base, sodium amide.

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So this is going to give us[br]an E2 elimination reaction.

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So we're going to get an[br]E2 elimination reaction,

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and this E2 elimination[br]reaction is actually

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going to occur twice.

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And we're going to end up with[br]an alkyne as our final product.

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So let's take a look at the[br]mechanism of our double E2

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elimination of a[br]geminal dihalide.

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So let's start with[br]our dihalide over here.

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And this time we're[br]going to put in all

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of our lone pairs of electrons[br]on our halogen, like that.

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So let me go ahead and[br]put all of those in there,

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and then I have two[br]hydrogens on this carbon.

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OK.

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Sodium amide is a source[br]of amide anions, which

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we saw in our previous video[br]can function as a strong base.

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So a strong base means that[br]a lone pair of electrons

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here on our nitrogen is[br]going to take this proton.

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And these electrons,[br]in here, are

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going to kick in to form a[br]double bond at the same time

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these electrons kick[br]off onto our halogen.

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So an E2 elimination mechanism.

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You can watch the[br]previous videos

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on E2 elimination[br]reactions for more details.

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So we're going to form ammonia[br]as one of our products.

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And our other product[br]is going to be

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carbon double-bonded[br]to another carbon.

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And then we're going to still[br]have our halogen down here.

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And over here, in the[br]carbon on the right,

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we're still going to have[br]a hydrogen, like that.

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So we're not quite[br]to our alkyne yet.

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So we've done one E2[br]elimination reaction,

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and we're going to do one more.

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So we get another-- another[br]amide anion comes along,

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and it's negatively charged.

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It's going to[br]function as a base.

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It's going to take[br]this proton this time.

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And these electrons[br]are going to move

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in here to form our triple bond.

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And these electrons are going[br]to kick off onto our halogen,

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like that.

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So that is going to finally[br]form our alkyne here.

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So you always have to[br]have your base in excess,

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if you're trying to do this.

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Let's look at a very similar[br]reaction, a double E2

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elimination.

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This time the halogens are[br]not on the same carbon.

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So let's go ahead and draw[br]the general reaction for this.

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We have two carbons[br]right here, and we

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have two halogens right here.

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And then hydrogen,[br]and then hydrogen.

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This time we have two[br]halogens on adjacent carbons.

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So this is called[br]vicinal dihalides.

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So let's go ahead[br]and write that.

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So this is vicinal, and the[br]one we did before was geminal.

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So a vicinal dihalide will[br]react in a very similar way

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if you add a strong[br]base like sodium amide

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and you use ammonia[br]for your solvent.

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So you're going to form[br]an alkyne once again.

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So you're going[br]to get an alkyne.

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It's going to be via a double[br]E2 elimination reaction again.

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Let's look at the mechanism.

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So let's start with our[br]vicinal dihalide down here.

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So let's go ahead and put[br]our halogens in there.

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Lone pairs of electrons on[br]our halogens, like that.

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And then we have hydrogen, and[br]we have hydrogen right here.

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So we have our amide[br]anion, and once again,

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functions as a strong base.

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It's going to take a proton.

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So it's going to take[br]this proton right here.

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These electrons are[br]going to move in

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to form our double bond the same[br]time these electrons kick off

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on to our halogen.

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So that's our first E2[br]elimination reaction.

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So let's just go ahead[br]and write E2 here

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to remind us this is[br]yet another E2 reaction.

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And let's go ahead and[br]draw the product of that.

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So now we're going[br]to have carbon

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double-bonded to another carbon.

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And then we're going to[br]have a hydrogen right here.

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And then we're going to have[br]our halogen up here, like that.

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And then we're going to[br]have-- we need one more

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reaction to form our alkyne.

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We're going to get another[br]E2 elimination reaction.

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So sodium amide-- another anion[br]of sodium amide comes along.

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So let's go ahead and put in[br]those lone pairs of electrons,

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like that.

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It's going to[br]function as a base.

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Lone pair of electrons[br]takes this proton.

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These electrons kick in[br]here to form our triple bond

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at the same time[br]our halogen leaves.

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And so we form our[br]alkyne like that.

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So you can produce alkynes[br]from either vicinal or geminal

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dihalides via a double[br]E2 elimination reaction.

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Let's see how we could use[br]this in a synthesis reaction.

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So let's go ahead and[br]try to make something--

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try to make an alkyne[br]from an alkene.

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OK, so let's start[br]with an alkene here.

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And I'll put some benzene[br]rings on this guy here.

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So here's a benzene[br]ring like that.

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Put in my lone pairs-- sorry,[br]put in my bonds like that.

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And then I'm going to put[br]a double bond right here.

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And then I'm going to put[br]another benzene ring attached

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like that.

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So this is[br]1,2-Diphenylethylene. .

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And I'm going to react[br]this alkene with bromine.

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And you could use a[br]solvent like carbon

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tetrachloride or[br]something like that.

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And we're reacting an[br]alkene with a halogen.

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And we've seen this[br]reaction before

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in the videos on[br]reactions of alkenes.

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We're going to add two bromines[br]across our double bond.

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So we're going to draw the[br]product of this reaction.

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Our benzene rings[br]aren't going to react

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as readily as our[br]double bonds will.

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So let's go ahead and draw in[br]our other benzene ring here,

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like that.

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And we know that[br]we're going to add

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a bromine to either[br]side of our double bond.

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So let's go ahead[br]and add a bromine

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to either side of[br]our double bond here.

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And we'll also have[br]a hydrogen bonded

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to each one of these carbons.

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That hydrogen was[br]originally there as well,

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over here on the left.

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And we form[br]1,2-Dibromo-1,2-diphenylethane

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here.

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And now we have a[br]vicinal dihalide.

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So if we add a strong base[br]to our vicinal dihalide

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we can prepare an[br]alkyne from that.

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So if we add an excess of[br]sodium amide in ammonia

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we know that we're going to[br]get a double E2 elimination

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reaction.

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And those halogens are going[br]to go away in our double E2

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elimination reaction[br]and form a triple bond.

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So we're going to[br]form a triple bond.

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So those two carbons, the[br]ones that form a triple bond,

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are these two[br]carbons right here.

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So when you run[br]through the mechanism,

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you're going to get an alkyne.

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And then, on either[br]side of that alkyne,

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you're going to get[br]a phenyl groups.

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So let's go ahead and draw in[br]our benzene rings, like that.

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So it doesn't really matter[br]how we draw our electrons,

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so we'll go ahead and do this.

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So this would be our product.

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So let's go ahead and[br]put in those electrons.

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This will be diphenylacetylene.

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So you can synthesize[br]alkynes from alkenes,

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or you could synthesize[br]an alkyne from a dihalide.

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So this is one way to do it.