okay here we go again so I'm going to pick up on this slideshow on with a discussion of when Polly alkylation is sometimes desirable so it turns out that if you Paulie alkylate in the main you can create a starting material which can be subjected to conditions which will give you what's called the Hofmann elimination product this is a really great reaction a great way to generate alkenes in particular the less substituted alkene so as Isis rule predicts orcid you know so jess is more in a suggestion and actually the states that when you can form a bunch of different alkanes the excuse me have a cough drop in my mouth get that taken care of the more highly substituted alkene is favored coordinate is a central but with Hofmann elimination you get the least substituted alkene and Addison beauty of the the Hofmann elimination so let's change slides here let's talk see what the Hofmann elimination is all about okay so we're going to use crumpled amine as the sample of main but this will work with essentially any amine whether its primary or secondary and we'll get into tertiary tertiary at work as well okay so we're gonna treat this primary mean propylene with excess methyl iodide okay this will force poly alkylation to where the nitrogen first reacts put on one methyl group and then a second one and then a third one so recall when we talked about poly alkylation let me go back and show you poly alkylation very quickly if we take ammonia or any other amine and treat it with an alkyl halide it will first make the put on one R group and then to our groups and then three or four with ammonia you can put on four different ones obviously if this were you know an alkyl amine with something other than and three hydrogen's I can our NH 2 you can only put on three alkyl groups but bottom line is you can exchange all of the H's in your amine with an alkyl group via poly alkylation so let's go back to the example we're talking about propylene we only have two ages we'll exchange both of them and we'll put on a third alkyl amine some alkyl group okay so if exchange the first two put on two methyls we would have an N n n dimethyl amino group here is still a lone pair bottom line is you replace all hydrogen's and a lone pair okay all your hydrogen's and the lone pair with methyl groups and get what's called an ammonium plant a quaternary ammonium ion is what they call these quanta nari ammonium salts the salt corresponds to the negative charge species there as well but the ion itself is called a quandary ammonium ion and okay yeah so that's where you that's the poly alkylation step where you add excess methyl iodide and you'll replace all of the hydrogen's and a lone pair with methyl group so in this case you get three methyl groups adding to the nitrogen okay at this point you add silver oxide and water she doesn't show water over this arrow but and she should a lot of authors of MCATs and dats maybe don't do the best job you know including the water but if you see silver oxide you'd be thinking you know what there's water there but what's going on is what's shown here in red okay silver oxide silver rocks at AG to O is actually two silver pluses and 102 - okay that's what this guy is this guy is that okay - silver pluses and 102 - all right this is actually the conjugate base of hydroxide alright if you were able to pull a proton off of hydroxide you make you'd make this guy so hydroxide is an uber weak acid okay so super weak acid therefore this going to be a goober or super strong base right alright so in presence of Oh - - will deprotonate the water and make 2h o minuses ok this piece here ok forms that Oh H - and what the o2 - pulls off a proton it also becomes Oh H - ok so this one comes from the O pulling off the proton it makes that Oh H bond there she blows ok and then this leaves that makes that so you make two hydroxides / silver oxide alright so the reason you use silver is that this one of these silver pluses will react with the I - and form silver iodide and crash out of solution as a solid silver salt you may have done this in in chem 111 or kim 112 or chem 105 or chem 106 the lab dissociates goes with those gen chem classes you remember that silver iodide is has a really small KSP right solubility product constant and so very little of it soluble in water so if you add silver plus 2 I - boom that fast and equipment fingers came boom FSK silver iodide crashes a solution and hydroxide takes the iodide - negative iodide a nice place this is basically an anionic exchange reaction for exchanging the I - with an O H - okay and takes the I - is place that puts this Oh H - in close proximity to a hydrogen on the beta carbon okay let's go back and consider how we're using an alpha and beta here for a means we consider the first carbon here to be the Alpha carbon and the next carbon the beta carbon this would be the gamma carbon and then how did it go Delta is it Alpha Beta Gamma Delta I think that's how the freak alphabet goes I usually don't go beyond beta or gamma so I don't know the whole Greek alphabet so that's a hole in my knowledge at any rate that's the Alpha carbon s beta carbon your hydroxide in this hydroxide salt now is in close proximity to the hydrogen on the beta position and upon heating these require Heat okay and I wouldn't test you on this I wouldn't you know give an example without heat especially to know that it doesn't go without heat and neither will an MCAT or a debt so point is that sometimes people will forget to put the heat or the arrow like an MCAT or dad it won't surprise me at all if they don't you know if they forget the heat they just so methyl iodide excess and simmer oxide and forget the heat and probably forget the heat and the water okay but anyway if you see silver oxide and an amino group and excess methyl iodide you know kind of kind of recognize this - where you know you know what they're they're fishing for to see if I if I understand or remember the Hofmann elimination and so any rate after heating or upon heating the hydroxide pulls off that beta hydrogen so Daisy electrons attack the proton that bond breaks the electrons here move in to form a double bond between the Alpha and the beta carbons okay that's what's happening here hydroxide pulls off a proton that bond breaks make a double bond between the alpha and beta carbon as that bond breaks and this lays okay the driving force for this to leave is that nitrogen doesn't like a plus charge on it and if that bond breaks a lone Parenthood she didn't brothel on the sacrum out of your textbook she and draw the lone pair but there is a lone pair on that nitrogen okay as and it comes from that bond breaking okay so there's your leaving group and you've generated a double bond between the alpha and beta carbons okay they call that the Hofmann elimination so just going to review this think about it again the silver oxide is actually two silver pluses and an o2 minus in the presence of water you get deprotonation to form two hydroxide anions and one of these silver plusses reacts with the iodide minus and forms silver iodide Rockside now is in close takes the place of the I - ok this is where I - used to be ok upon heating it pulls off the beta proton you get those PI electrons those electrons move uniform a pi bond that group leaves and you generate an alkene between the alpha and beta carbons ok cool all righty then what I want to show you here this is just kind of mentioning that the ammonium ion here it's going to be your leaving group must be an T to the hydrogen this is classic e2 elimination they know s that's Spanish for C isn't that so Bay they know s it says right here D two of the e2 mechanism for Hofmann elimination this is an e2 mechanism we go through an anti-periplanar arrangement elimination of the leaving group and yeah you make your carbon-carbon double bond all righty here's an example we're going to add exit excuse me excess methyl iodide silver oxide and water often times your MCATs or desk won't show the water she doesn't show up and there's implied waters there you heat and that will leave your hydroxide will pull off the hydrogen off the beta carbon and via this kind of mechanism right here pulling off a hydrogen off the beta carbon she didn't she didn't label these but that should be the alpha carbon right there that is the Alpha and that one there is the beta okay yeah there's another example make a double bond between the alpha and beta carbons blah blah blah okay all right easy sneezy right okay yeah easy sneezy as long as all you have is one kind of beta hydrogen okay the plot thickens there are plenty of substrates where you have different kinds of beta hydrogen's and that's where this gets a little more tricky all right so here's an example of a compound that has several different kinds of beta actually - we call this the alpha carbon which I've labeled right there we have beta 1 and beta 2 take a look at that and if you need to pause this and think about why you know these two are different beta-1 and beta-2 why I'm saying that we have hydrogen's on on two different beta carbons before okay and these earlier examples only had one beta hydrogen okay didn't label that one beta because this is identical to that one we draw a plane a plane of symmetry right here okay that hydrogen is identical to that one so we don't we don't have two different kinds of beta hydrogen's here okay same thing here but in this example we do have two different kinds and that's where this gets tricky okay all righty so once we add our methyl iodide we convert the NH 2 into an end with three methyl groups on and a plus charge right upon heating the hydroxide can pull off either beta 1 or beta 2 if we call pull off beta 2 we get this product okay all right if you take a notes and these are on your slides on learning sweet it'd be good idea to write down that this one came from taking off beta 2 and this one came from taking off the hydrogen of beta 1 we're pulling off a hydrogen at beta 1 to make this product and we're pulling off of beta 2 okay beta 2 to make that one plane off 8 a 1 to make that one okay alright in the Hofmann elimination take a note of this and I'm a circle distance star okay the major alkyne has the less substituted or at least substituted double bond in a Hofmann elimination okay so Hofmann elimination gives the least substituted alkene okay should I make you dizzy here okay just make you dizzy okay it makes the less substituted alkene Hofmann elimination gives the less substituted alkene as a major product that's exactly opposite from his ICF rule so the e2 elimination we saw in and kim 351 actually gives the most substituted alkene not the least substituted alkene okay so that's an important difference between Hofmann elimination and regular e to elimination Hofmann elimination gives the fill-in-the-blank is it they most or least substituted alkene fill-in-the-blank what is it it's the least substituted alkene is the major product okay and so this that I call it the major this is your major right here it's they die subsidy alkene it is the least substituted and this is your minor product it is a trisubstituted alkene if I find this spoke on that I think I did I'm not going to change the slides and you can say one again that you know this is least substitutes got a one two substituents so a is a major product here and B is your minor because has one two three okay it's trisubstituted so let's see the made the minor product and this was that die substituted it's the major the least substituted alkene is major products any Hofmann elimination okay perfect I think I've been beat that dead horses move on okay here's another example where we have two different kinds of beta hydrogen there's one up here and these two are identical so labeling them both as beta 2 so our least substituted alkene which is a dye substitute alkane right let's die substituted it's less substituted in a trisubstituted so this will be our major product okay so plot thickens let's take a look at question 25.3 see and I think this numbering is from the fifth edition I'm not sure maybe from the fourth edition so take a look and when our Edition you're using there should be a very similar problem okay so I've written on here that the products that would be formed from that substrate but I think the next slide actually shows the whole thought process let's sort of the next slide okay alrighty so we're going to take in a first step we're going to show how this guy becomes poly alkylated and is the starting material in problem 25.30 artsy so we treated with excess methyl iodide first would get alkylation of the nitrogen to end to out to make that guy a base pulls off the proton gives us a secondary amine the secondary amine reacts with a second equivalent of methyl iodide and nickel in a second is a tertiary amine we started out with the secondary main we make a tertiary amine there and when the lone pair attacks that methyl group kicks out the iodide now we have our quanta Neri ammonium ion okay take a look that and digest it this is what happens when you add excess methyl iodide to this particular starting amine which is a secondary amine to start with we end up with a quaternary ammonium ion okay the next step will involve adding silver oxide and water which has the effect of replacing the I minus with a hydroxide anion that's an anion exchange reaction go back and look at when we first talked about this to verify that's the case okay so now here's the process if we have two here I say label the alpha and beta carbons and then draw all possible products where beta protons have been removed and draw a double bond between the alpha and beta carbons okay so this is the alpha carbon here and here is an alpha carbon okay so we've got two alpha carbons and two betas here's a beta okay and there's a beta and so we're going to get elimination at both this position and at that position we're going to cut here and yeah so let's take a look at the the products we'll get okay so this shows the exchange of the silver oxide with the quad salt to where hydroxide takes ionize place then we heat we're going to show removal of beta one okay this guy leaves and we end up with that product okay so this is formed by deprotonated beta one looks like we've got three I forgot about yeah we do have three different beta hydrogen sorry about that and forgot about that third one okay so hydroxide will pull off beta 2 and that guy leaves to give this product where we have a double bond between the alpha and beta two carbons and then last but not least we have beta 3 we can deprotonate there and I'm leaving to please and we get this products there's three total different products we can make if we ignore the cysts or fourth one if we considered the sis but the trans is favored so we're going to ignore this insist for right now but in a question that were to ask County from products that you would form the correct answer for this would be a total of four in that you're going to get a little bit of cysts here as well okay but you're going to get product resulting from deprotonation at beta 1 beta 2 and a beta 3 and dimethyl the dimethylamine oh group will leave and so yeah you get if you ignore the sis product there's three products if you include the sis product or before so take a look at that to digest it what's going on okay so this one I don't have the answers written out yes or just gonna kind of work through them I'll tell you verbally and then in a few minutes I'm going to pull out my iPad and see if I can figure out how to get it to record I don't this is all new to me guys I've never done this before so I'm I said in an email to the to this class and spring of 2020 I am an old dog I'm learning new tricks so we'll see what I can figure out but anyway we'll talk about these verbally let's take a look at this one and we'll see what would happen if we treat it with excess methyl iodide and silver oxide these two things ought to key you and if you see those those two in combination you know do not pass code and I'll collect $200 you can start salivating on Pavlov's dogs all that good stuff you're gonna know that hey we're gonna do a Hofmann elimination okay so I'm gonna look at this thing I say I've got an alpha carbon and two beta carbons okay so I'm gonna pull off hydrogen's off it let's call this beta one okay so I could double bond right there so I'll have an alkene with a double bond right there or I can pull it off of here okay yeah or I can have a double bond at this position nitrogen will leave okay that's the whole point is we're going to eliminate the nitrogen we'll have two alkenes one right here which is a mono substituted alkene and that is a die substituted alkene will have either probably sis and trans a little bit of sis not much but mostly trans so there'd be a total of three products counting the assistant trans isomers and the major one would be the least substituted okay with a terminal alkyne right here okay so that's the beauty of the Hofmann elimination it gives the least substituted alkene product all righty then that's number one number two let's see look maybe that's number two and there's number three okay yes where these are numbered um again we'll look at how many alpha carbons we have we have an alpha carbon here and an alpha carbon there okay so if this is the alpha carbon on the left hand side we'll call this beta 1 we'll be able to eliminate right here or make ethylene okay that's one product or right here we have beta 2 which is the same thing as that beta could have a plane of symmetry so this beta hydrogen and that beta hydrogen are equivalent so we won't get two different products we'll just just get one for pulling off a beta 2 and we'll get this alkane okay which is propylene or 1 propene where this group leaves ok and we get that so our major product is our least substituted alkene and that will be ethyl ain right here yeah this one's a little more complicated this is our alpha carbon we have a beta 1 and beta 2 and a beta 3 ok and so let's take a look at beta 1 if we remove a hydrogen off of there we'll make a double bond okay no keen and that will pull it off beta 2 will make a double bond here and that will leave or we pull off a beta 3 or make a double bond and that will leave if I'm not mistaken if we pull it off of here we make a dye substitute alkene if we pull of here we make it triceps to alkyne if we pull it off of there we make a tetra substituted alkene so we die substituted is less hindered there's gonna be less substituted and therefore is favored and so that that will be your major product okay so that's that let's take a look now at this one which is kind of fun one and gonna be the last one I do for chapter 25 and encourage you to take a look at your homework you know work your exercises take a look the solutions key solutions manual and kind of work some of these on your own but shown here on the left is methamphetamine game so we we talked earlier about how to make heroin we talked about how to make some time aspirin minna-san and yeah so now I'm going to talk about how to make morphine okay now the street approach for making morphine is from this guy which is pseudoephedrine and I'm not going to teach you how to do that I think it is known I think it is on the internet I don't encourage you to do that but I think it is known but this guy he's very easy to make and actually deals with chemistry discussed in this chapter okay so how you make it is you get your hands on this stuff which by the way is controlled we actually had some legitimate research community to do years ago where we wanted to use this as a substrate and they made us fill out all these forms and cetera et cetera and verify that we're working for a university and we weren't going to use it for nefarious purposes and all that kind of stuff but never say you can't this is not you can't get your hands on this okay it's if you did it'd be super easy to make methamphetamine but that's how you do it you take this ketone you treat it with methylamine and this undergoes some cowboy chemistry to make an amine okay so here's a carbon nitrogen double bond and then you either reduce with h2 and palladium or sodium cyano borohydride okay this is called a reductive amination okay this process where you reduce an amine to an amine is called a reductive amination and that's just that easy okay we have an instrument in our department that you used for doing hydrogenation it's called a par hydrogenate ER and I was told when I first started here 25 years ago that our hydrogenated or at one point was stolen by a meth lab in in Phoenix and recovered from a meth lab in Phoenix by FBI agents who I was told there was gunfire exchange so it sound like kind of exciting a place to not be there while it was happening and that instrument has since died we no longer we do have it but it's not functional you know this is ancient I mean it was just five years it was old 25 years ago I've seen the instrument but anyway they were somehow they got a they got their hands on this and methylamine were able to do this with a hydrogenate er but that that baby right there that is methamphetamine just that simple take a ketone you had methylamine get to do some cowboy chemistry to make the in Maine then you reduce the carbon nitrogen double bond with either h2 in play diem or sodium Santo borohydride those are the two different ways I circled h2 because that's for how they used the poor hydrogenated I was talking about that was in the FBI sting operation yeah that one anyway though I think here i show soon ephedrine stuff that's found in lots of over-the-counter cold medications and so the street method for making methamphetamines is to take that stuff and treat it with the secret ingredient I really don't know how to do it but there are reagents that will take off that hydroxyl and this is almost methamphetamine methamphetamine is basically this guy right and I don't know if that I think this is a racemic mixture okay but anyway if you get those Street this secret ingredients they'll take that off and you you can make bet amine from pseudoephedrine I believe it's true now that you this is not just over-the-counter I mean it I mean you don't have to have a prescription but this is not out on counters because four years ago like 25 years ago when I first started here BYU you used to be able to go into stores and just pick up you know some cold medications decongestants with pseudoephedrine put them in your cart and go out and buy as many as you wanted and people did that makers of methamphetamines are going and buy a whole cart load of you know great big basket of pseudoephedrine and then because meth got so out of control and it's not really a laughing matter I mean it it does destroy numerous lives and it's it's actually tragedy that the lives that are destroyed by methamphetamine but the street way for me making it involved purchasing cold medications and then treating them with secret ingredients to get your hands on this Lewis antigenic drug and this is how you make it in a lab lab setting using reductive amination which is just covered in this chapter okay let's talk about this one real quick and I'll just do it verbally and then see if I can get my iPad to work but here's a phone question twenty five point five four how would you prepare three phenyl 1-propanol mean from each compound and while we're on that topic nomenclature is important and any chapter that has nomenclature there will be at least ten points in the chapter about the nomenclature and aspects covered there okay okay so here we have one two three carbons and we want to make a primary amine okay want to somehow swap out the bromine for an NH two I would just do a Gabriel synthesis okay treat this with a lemon go back and look at the Gabriel synthesis and then the family mode and base and then hydrolyze off the that limit portion with h3o plus and that will give you the primary mean here we're going from a two carbon precursor and we want to put on a third carbon treat that with sodium cyanide so cyanide will do an sn2 substitution knock off the bromine and attach a nitrile there so I will give you a three carbon nitrile have a ch2 ch2 and C triple bond n you then then treat that with lah we discuss that one of the earlier slides about how you reduce night trials to make a means lah will do that and convert this guy into that product here we have the three carbons already there CH 2 CH 2 CH 2 in a nitro this is just how do you introduce an no2 to make an NH 2 and there are four ways one way is lah okay another is 10 and HCL or iron and HCl and the third way is hydrogen with palladium okay here we've got enamored with one two three carbons in it means this piece of here all these are just phenyl rings c6h5 that's the shorthand for phenyl okay we've got our three carbons and an amide here that's the keyboard way of writing and Abid that co represents AC double bond o if we treat this with lah that will convert the carbon eel into a ch2 so that's how you would convert that one then last but not least we have one two three carbons here's our Cho if you remember Cho is I told you before don't choke on show right Cho is the aldehyde if we react that with ammonia it will make an amine and if you treat that amine with h2 and in palladium or sodium cyano borohydride so na bh3 see in one of those two will can promote a reductive amination and convert the aldehyde into a primary amine so that's a great question showed you five different ways to make a primary amine and some interesting chemistry okay again this would be the Gabriel synthesis this is use Sinai and followed by reduction here you just reduce no.2 with those four reducing agents I talked about there it is the amman with lah and here you do a what's called a reductive amination by treating with ammonia and h2 in play diem or sodium cyano borohydride and i think we're going to stop here I'm gonna see if I can figure out how to do some of these with my iPad if I don't end up posting any it means I didn't figure it out and we'll just have to do some in class next time we're teaching this on campus so oh here we go closeout