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Current time:0:00Total duration:7:57

Video transcript

let's look at the hydration of alkynes so we're going to start with a terminal alkyne over here you can see there's a hydrogen on one side of our alkyne and the other side of our alkyne let's say it's an alkyl group bonded to this carbon on the right so we're going to add water sulfuric acid and mercury to sulfate to our alkyne we're going to hydrate it now we've seen the hydration reaction before we we did a hydration reaction with an alkene and we added an H+ and an O h- across our double bond and we're going to do the same thing here we're going to add an H plus an O h minus this time across our triple bond and the O h- always adds to the most substituted carbon so we saw that in our earlier video so the regiochemistry the regio chemistry shows markovnikov's rule so this is a I'll just write Mark's rule here so we spent a lot of time talking about markovnikov's rule in earlier videos which means that this Oh H over here on the right is going to add to the most substituted carbon which happens to be the carbon on the right since that's going to be bonded to an alkyl group over here on the right the h plus right add it over here to this carbon on the left so now there are two hydrogen's on the left so we added h plus and o h minus we added water essentially across our triple bond and we get this molecule as our intermediate this molecule has a specific name is called an enol so it's called an enol the en comes from the fact that there's a double bond present in this molecule and the O L comes in the fact that there's an alcohol right this oxo this enol is the intermediate of this reaction however the enol is not the most stable form of this molecule so it's actually going to rearrange and we're going to get a ketone for our product and I'll show you the mechanism for this rearrangement in a few minutes here so we get a methyl ketone from a terminal alkyne so this reaction is is best used when you're looking to make a methyl ketone and you're starting with with a triple bond here so let's look at a reaction so let's start with a terminal alkyne so I'm going to put a methyl group on this side of the carbon like that and I'm going to react terminal alkyne with water and sulfuric acid and with mercury to sulfate like that so if we think about what's going to happen right we're adding H+ no h- across our triple bond we're going to add the O h- to the most substituted carbon so that would be the carbon on the right so we have carbon double bond it to another carbon we're going to add the O H to the carbon on the right we're going to add the H+ to the carbon on the left so there's water added across your triple bond and then we have still another hydrogen on this carbon the left here and then the methyl group is still over here on the right so that's our enol intermediate let me just draw this molecule again so this molecule is equivalent to this molecule right here so I'm going to draw in in this form from now on so this is my in all form and my email is going to rearrange right this is not the most stable form of this molecule so let's let's see the mechanism of how our enol rearranges so let's go ahead and redraw our enol down here so let's get a little room like that so here's my enol put in my double bond put in my lone pairs of electrons like that so this enol is present in an acidic mixture right h2o and h2 so4 going to give me hydronium ions in solution h3o plus all right so there's going to be h3o plus floating around here in solution right so we have our hydronium ion looking like that so this is a this of course is capable of donating a proton and the the pi electrons are going to actually move out here and pick up a proton like that and then these electrons are going to kick off onto your oxygen so this is a an acid-base reaction as the first step so therefore we go draw our equilibrium arrows here like that and let's see what happens well we're going to have this is our skeleton right the O H didn't really do anything yet and we're going to add a proton on to the carbon on the left side of your double bond like that so let's go ahead and show what electrons did the moving right the electrons in your pie on whichever one of these bonds here is your PI bond all right those electrons are going to move and form a bond with this proton like this so we ended up taking away a bond from this carbon right here so we had a double bond right that it took a bond away from that carbon that's going to make that carbon positively charged so there's a plus 1 formal charge on that carbon and we can draw a resonance structure for this intermediate so let's go ahead and draw our resonance bracket and our resonance arrows here and what can we do with our with our lone pairs of electrons to share that positive charge right so we can take one of these lone pairs of electrons and we can move them in here to form a double bond between our carbon and our oxygen so let's go ahead and do that a double bond forms between our carbon and our oxygen like that now there's only one lone pair of electrons on this top oxygen and if we think about what happened to our formal charge it actually moves out here to this oxygen all right so this action now is a plus 1 formal charge this is a resonance structure let me go ahead and keep this keep this hydrogen in here like that so that's our resonance structure for this molecule on the left now oxygen does not like having a plus 1 formal charge either so is there any way for this oxygen to get rid of its plus 1 formal charge and of course there is right there's water floating around so water is going to act as a base all right so a lone pair of electrons on water is going to take this proton right just that proton leaving these two electrons behind on my oxygen like that so we have another acid-base reaction so let's go ahead and show this reaction being at equilibrium so I draw my equilibrium arrows like that all right and so now I have my oxygen alright I had one lone pair of electrons around it now has two lone pairs of electrons around it so so now we're done right we formed our ketone version right so this is acetone or our simplest our simplest ketone so we have the keto form of the product all right this is the keto form of the product in the form of a ketone and then we have our enol over here all right so the keto in the enol form are said to be tautomers of each other and they're in equilibrium with each other and they can go back and forth and in this case we're using acid to catalyze this transition so this is called this this is called acid catalyzed acid catalyzed tautomerization so tautomerization like that and we're going back and forth between the keto with the enol form so you'll also hear this called keto enol tautomerization a lot like that so the keto and the enol form are in equilibrium with you with each other now the equilibrium is actually going to favor the keto form so this is actually the more stable form this is more stable so that carbon double bonded to an oxygen for the keto form is more stable than the carbon double bonded to another carbon in the enol form and essentially you're changing one proton one PI bond in this equilibrium and and this is the acid catalyzed form of this mechanism so in the next video we'll see a very similar reaction where we do a base catalyzed tautomerization