Hydroboration-oxidation of alkynes

here's the hydroboration oxidation of alkyne reaction so we start with our alkyne and once again usually it's a terminal alkyne right so here's your hydrogen here and on the other side of our triple bond let's say there's some R group attached to this carbon so to our alkyne we're going to add in the first step borane bh3 in THF as our solvent and in the second step we're going to add a hydrogen peroxide and hydroxide anion O h- and we're going to going to get addition of water across our double bond so it's very similar to the last reaction the difference here is the regiochemistry write this this regiochemistry is actually anti markovnikov addition so this is anti Marx rule right so anti markovnikov meaning the O H is going to add to the least substituted carbon which of course must be the carbon over here on the left so we end up adding h2o water across our triple bond so now there's a double bond present and then there's an O H in our molecule so just like in the last video we form an enol here like that and the enol of course is not the most stable form we're going to get some rearrangement and we're going to end up with an aldehyde here so we can see we get an aldehyde functional group from a terminal alkyne undergoing this reaction now sometimes if you use borane in this first step over here sometimes sometimes borane is so reactive that it's actually going to add two molecules across your triple bonds as your triple bond consists of two pi bonds so sometimes you'll see a different a different molecule you sometimes you'll see a dialkyl borane so put two alkyl groups up with R to be H and if you make them really bulky this steric hindrance prevents the second addition of the boring because in the mechanism you want only one addition of the boring and the boring is going to add on to the same carbon that the O H does so check out the video on hydroboration oxidation of alkenes for much more detail of that first mechanism that we discussed so the net result of this reaction is to form and aldehyde from your terminal alkyne so let's look at a reaction here so let's let's make let's make our triple bond right here like that and we'll make it a terminal alkyne and we'll make this side a benzene ring so it'll make a rather large R group here so let me put in some electrons on my benzene rings so let's just go like that alright so if I react this alkyne alright with either a die alkyl boron or I'll just put boring here and you you can use a dial Kilbourne if you want to whatever your professor wants to use in th F and second step hydrogen peroxides and Oh H minus like that so the first thing you have to think about is okay what's going to happen I have I'm going to be adding water across my triple bond and I need to think about what side gets the hydrogen and what side gets the o H so once again I have two choices right I could I could add the O H on to the carbon on the right or I could add the o H to the carbon on the left and the regio chemistry here is anti markovnikov so i'm going to add that o h to the least substituted carbon which of course is the carbon on the right side so i'm going to end up adding the o h to the carbon on the right side so let's go ahead and and draw our our double bond in here first let's go ahead and draw our benzene ring so I'll go ahead and make it like this and that benzene ring is connected to a carbon right this carbon right here there's now a double bond presence like that and I added the O H to the least substituted carbon I added the o H to the carbon on the right like that and I added a hydrogen to the carbon on the left so that's adding my my H 2 O across my triple bond and then there was still a hydrogen on this carbon on the right like that so that's my that's my enol intermediate like that so let's let's take a look at what's going to happen to that to enol intermediate it's going to rearrange and on the last video we saw an acid catalyzed tautomerization in this video base is present alright so we have our hydroxide anions in solution so let's go ahead and redraw our enol and show the enol reacting with base so let's go ahead and draw the enol here again so I'll draw that enol on the right side so there's my benzene ring and I have my double bond in here like that and I know I have an OHA like that and put in my lone pairs of electrons and I'm going to react this with hydroxide anions right so hydroxide anions are floating around hydroxides negatively charged so hydroxide will function as a base all right so let me get in my lone pairs of electrons here on my oxygen and going to pick up that proton right and kick these electrons off onto this oxygen so let's go ahead and draw the results of that acid-base reaction all right so it's an equilibrium so let me go ahead and draw my equilibrium arrows like that so what's going to happen I have my benzene ring still and I have my double bond and let's see now I have my oxygen with three lone pairs of electrons around it like that so negatively charged so now now I can draw a resonance structure for this molecule all right so let's go ahead and put in our resonance brackets here like that so what can I do to move that to spread that formal charge out throughout the molecule well I could take a lone pair of electrons right here and I can move it in to here to form a double bond and now I give too many bonds to this carbon so the PI bond here is going to break and the two electrons here are going to move out onto this carbon like that so let's go ahead and draw the result of the movement of those electrons I still have my benzene ring like that and there's my carbon skeleton here now I have a carbon double bonded to an oxygen there used to be three lone pairs of electrons around this top oxygen one moved in to form the double bonds now there's only two lone pairs like that and the electrons that used to be in this pi bond have moved onto this carbon right here so this carbon right here is still bonded to another hydrogen so the two electrons moving out on to it are actually going to form a carb anion here so this is actually a negatively charged carb anion like that and water is now present right o h- fix of a proton to become h2o so now water is floating around like that so now water is actually going to function as an acid it's going to donate a proton and this lone pair of electrons you're going to take that proton so they take that proton and kick these electrons in here off onto the oxygen atom like that this is an acid-base reaction so this is going to be at equilibrium I'm going to go ahead and put in our other resonance bracket like that so what are we going to get and we still have our benzene ring untouched over here and let's see now we have our carbonyl like that all right and we just added on a proton to this carbon right here like that so now we are done and you can see this is an aldehyde right I didn't draw in the hydrogen over here so you can leave that out if you want to but if you want to draw on your hydrogen there's a hydrogen there there's also a hydrogen on this carbon right here I just didn't draw it in two to show some clarity in the mechanism so the end result is we form a terminal a terminal um which we form an aldehyde from a terminal alkyne and we showed this tautomer Talmer ization using base so let me just go ahead and write one more thing here this is an example of a base catalyzed tautomerization alright where we go from the enol form right - in this case an aldehyde as your product which is more stable because of the carbon oxygen double bond