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Video transcript

this video we're going to see how to how to prepare alkyl halides from alcohols and so if we start with this alcohol over here on the left and we add socl2 which is called venial chloride and pyridine to it we're going to substitute a chlorine atom for the O H group and this mechanism occurs via an sn2 type mechanism which means that it's only going to work with primary or secondary alcohols and it's possible to get and you will get inversion of configuration if you have a chirality Center present in your final product so let's take a look at the mechanism and we'll start with our alcohol and so the the oxygen is going to have to leave somehow but by itself o H's is not the best leaving group and so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group and so if we draw the dot structure for thionyl chloride we would have the sulfur double bonded to an oxygen here and then the sulfur is also bonded to chlorine so go ahead and put in those lone pairs of electrons on the chlorines like this and if you count up your valence electrons turns out you need two more and those go on the sulfur it's okay for sulfur to violate the octet rule since it is in the third period now so a lone pair of electrons on oxygen is going to form a bond with our sulfur atom which would therefore kick these electrons in here off onto the top oxygen so if we go ahead and draw what we get from that first step of our mechanism now our oxygen is bonded to our sulfur the oxygen is also bonded to a hydrogen one lone pair of electrons form that new bond so 1/4 of electrons is let one pair of electrons is left behind which would give this oxygen a plus 1 formal charge connected to the sulfur this top oxygen here had two lone pairs of electrons picked up one more lone pair which gives it a negative 1 formal charge and this sulfur is still bonded to chlorines so we can go ahead and draw those in and we can go ahead and draw that lone pair of electrons on that sulfur like that and so in the next step of the mechanism we're going to reform we're going to we're going to reform the double bond between oxygen sulfur so these electrons are going to kick in here and these electrons would kick off onto that chlorine so when we draw the next the next the next intermediate here we would now have our oxygen still bonded to a hydrogen still with a +1 formal charge like that and now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen the sulfur is also bonded to one chlorine now so one of the chlorines left and we can go ahead and draw in that chlorine alright so one of the chlorines left here and it's a negatively charged chloride anion and then still there's a lone pair of electrons on our sulfur like that so at this part of the mechanism the pyridine comes along so if we go ahead and draw the dot structure for pyridine it's a base and so it looks like a benzene ring except we have a nitrogen here instead and there would be a lone pair of electrons on this nitrogen and so that lone pair of electrons is going to function as a bronsted-lowry base and take this proton here on the oxygen and that would kick these electrons back off onto this oxygen so let me go ahead and draw that let's go ahead and get some more room here all right so what would we get we would now have our carbon bonded to our oxygen or oxygen now has two lone pairs of electrons around it and we have our our sulfur and our chlorine and our lone pair of electrons on the sulfur and now we've made a better leaving group so this is a better leaving group and then the O H was in the beginning and if we think about an sn2 type mechanism now we know that the bond between carbon and oxygen is polarized strong oxygen being more electronegative it will be partially negative and this carbon here will be partially positive right so carbon will be partially positive and so now we can think about our sn2 type mechanism our nucleophile will be this chloride anion up here that we formed in the mechanism so that's going to be the nucleophile and it's going to attack our partially our partially positive carbon an sn2 type mechanism so as the chloride attacks this is also the stuff on the right is going to leave all right so the electrons in magenta are actually going to move in here and then these electrons are going to kick off onto that chlorine so when we when we draw the product all right we can go ahead and show the the chlorine has now added on to our carbon on the left and on the right well if you follow the movement of those electrons are going to form sulfur dioxide so so2 and also the chloride anion so CL CL - like that and so we've done it we've we've substituted our chlorine atom for the O H and formed an alkyl halide so this is just a better way of forming forming an alkyl chloride from an alcohol so if we if we look at if we look at an example alright we'll just take something like ethanol here and if we react ethanol with thionyl chloride socl2 we had some pyridine as our base we're going to replace the O H with our chlorine like that and so once again if we look at our our alcohol is a primary alcohol and so primary alcohols will work the best because there's decreased steric hindrance and we don't worry about stereochemistry since we don't have any chirality centers in our product let's look at a way to form an alkyl bromide so we just formed an alkyl chloride let's look at the general the general reaction for forming and an alkyl bromide here so I go ahead and have my my alcohol and I react that with phosphorus tribromide so pbr3 and i'm going to get the O H group is going to leave and I'm going to put a bromine in its place and once again this mechanism is an sn2 type mechanism so primary or secondary alcohols only and possible inversion of configuration for your products depending on whether chirality centers are present or not so another sn2 mechanism and again we need to use phosphorus tribromide because the O H group is not the best leaving group so when we look at the mechanism I only look at this mechanism here let's go ahead and show that lone pair of electrons better like that we have phosphorus tribromide so i'm going to go ahead and draw the dot structure so we we would have these bro means here with lone pairs of electrons and there's three of them so again put in those bro means and we still have two more valence electrons to account for and those would go on our phosphorus like that so the first step it's it's analogous to our to our previous mechanism right lone pair of electrons on oxygen are going to form a bond with phosphorus and that would kick these electrons off onto one of the bromine so I just chose that one doesn't really matter which one you choose and so when we show we show the result of that we would now have our oxygen that bonded to a phosphorous the oxygen is still bonded to a hydrogen there's still a lone pair of electrons left behind this oxygen which give this oxygen a plus one formal charge and the phosphorous is now bonded to only two bro means all right so we can show the phosphorus bonded to only two bro means here so I can go ahead and put those in and I can put in the lone pair of electrons on phosphorus as well and so we lost one of our bro means and and that formed a bromide anion all right so we go ahead and draw in our bromide anion here and once again we've made a better leaving group so all the stuff here on the right is a better leaving group than the O H and so we can think about our sn2 type mechanism all right we're war once again the the carbon this carbon right here is going to be electrophilic so it wants negatively charged electrons which you could get from the bromide anion so nucleophilic attack and then that would kick these electrons in here off onto your oxygen and we can go ahead and show our products all right so when we draw the product all we need to do now is show we have substituted right the bromine is now attached to the carbon like that and our other product would be right there be a hydrogen attached to an oxygen and the oxygen we attached to two phosphorus and the phosphorus is attached to two bro means like that so we've we formed our alkyl bromide so if we if we just show a quick example once again we'll start with ethanol if we wanted to convert ethanol into ethyl bromide all we would have to do is add phosphorus tribromide like that and of course we're going to replace the O H with a bromine and so that's that's one way to prepare alkyl halides from alcohols