in this video we're gonna focus on reactions associated with the grignard reagent so let's talk about how to make it let's say if we have one bromo butane and if we react it with magnesium metal magnesium will insert itself between the carbon and bromine atoms given you a grignard reagent now what you need to know about the grignard reagent is the carbon that is attached to the magnesium atom that carbon is a nucleophilic carbon it bears a partial negative charge magnesium has is positively charged and this bromine atom is negatively charged I like to think
of it this way as the carbon having a negative charge the magnesium as having a positive two charge and the bromide ion as having a negative charge either case you need to realize that this carbon is nucleophilic now let's work on an example from so here we have an aldehyde functional group and we're gonna react it with methyl magnesium bromide followed by h3o plus so what do you think the major product of this reaction will be so in the first step the nucleophilic grignard reagent will attack the carbonyl carbon causing the pi bonds break and
keep in mind this carbon has a partial negative charge and so it's attracted to the partially positive carbon atom of the carbonyl functional group and so we're gonna get an oxygen for negative charge and here is the methyl group that we've added to the aldehyde so now in the next step the oxygen with a negative charge is going to receive a proton from the hydronium ion so it's going to grab a hydrogen turn in into a secondary alcohol and so this is the major product for this reaction now let's work on another example so here
we have cyclopentanone and let's react it with will use the same actually let's use a different reagent this time methyl magnesium bromide so feel free to pause the video and work on this example so we could follow the same process so the grignard reagent will attack the carbonyl carbon from the back given us an alkoxide ion and here is the ethyl group that we've added to it now in the next step it's going to pick up a hydrogen and so this is gonna give us a tertiary alcohol this time and so this is the major
product for the reaction so anytime you mix a grenade reagent with a ketone you're going to get a tertiary alcohol so now let's move on to our next example so here we have normal benzene and in the first step we're going to react it with magnesium and then in the second step carbon dioxide and then in the third step h3o plus so feel free to pause the video and work on this problem so once we add magnesium to bromobenzene magnesium will insert itself between this carbon atom and this bromine atom so this is gonna give
us phenol magnesium bromide now in the next step it's going to react with carbon dioxide so this carbon is partially positive but the carbon attached to the magnesium atom that carbon is partially negative and so it's going to attack this carbon breaking one of its PI bonds and so now we have benzoate we also have the magnesium ion in a solution with the bromide ion so if you want to you could ride this like this you could attach the oxygen with a negative charge to MGB R if you want to but I'm not going to
worry about the magnesium and the bromine atom at this time so the last step is to react it with h3o plus and so the final product for this reaction is benzoic acid now let's move on to our next example so what happens if we mix an ester with a grignard reagent so we're gonna react it with ethyl magnesium bromide followed by h3o plus now when we act in a grignard reagent with an aldehyde or a ketone you can only add one our group but with esters and acid chlorides you can add to our groups so
in this example you'll see that we're gonna add two methyl groups to the ester so let's go ahead and begin so this is gonna be the first nucleophilic addition methyl magnesium bromide will attack the carbonyl carbon giving us a tetrahedral intermediate and so here's the ethyl group that we've added at this point that's the first R group now we do have a relatively decent leaving group compared to this group so we can kick out the methoxide ion and thus we're gonna get a ketone now this same grinev reagent can react with the ketone so it's
going to attack the carbonyl carbon given us an alkoxide ion so here is the second R group but I'm gonna write it out for this problem so this is ch2 ch3 and here we have another one so now the last step is to react this with h3o plus turning this into an alcohol so what we have is a tertiary alcohol and so this is the major product for the reaction as you can see the Grenada V agent reacts twice with the ester it adds to our groups to it now sometimes you might be given the
reactant and the product of a reaction and you need to determine the reagent that's needed to convert the reactant into the product which is what we have in this example so what do we need in order to make this compound so we'll turn in the aldehyde into a secondary alcohol and notice that these four carbons were already present and so what we added were these carbons so our grignard reagent will need to contain this group so basically on this carbon you need to add the grinev agent and so basically all I did was draw what
I saw right there and just add em gbr to that carbon atom so that's gonna be the first step and remember when converting the aldehyde into an alcohol you only need to add the grinev agent once so we only need one equivalent of this grinev agent and then in the second step we simply need to use h3o plus and so that's how you can determine the reagents that you need in order to go from this aldehyde into the secondary alcohol now let's work on another example so here we have an acid chloride and we wish
to convert it to a tertiary alcohol what reagents do we need so we already have these six carbons well we need to add our these two propyl groups and recall that when mixing a grineer agent with an acid chloride or an ester to our groups will be added and those two our groups are the same we need to add two propyl groups so in step one I need two equivalence of propyl magnesium bromide and then in step two just h3o plus that's all we need to do now the mechanism by which the grinev agent reacts
with an acid chloride is virtually the same when it reacts with an ester the only difference is instead of having an och3 group you have a CL group which is about a leaving group go ahead and try this problem let's say we have one bromo butane and in the first step we're gonna be acting with magnesium and then in a second step ethylene oxide and then h3o plus in the third step what is the major product of that reaction so let's take this one step at a time so let's take butyl bromide and let's react
it with magnesium so this is going to give us butyl magnesium bromide and then in the next step we're gonna react out with ethylene oxide and so the green reagent will attack one of these carbons because it's symmetrical doesn't really matter which side we attack so we're gonna add two carbons to our four carbon chain so we're gonna have a total of six carbons and so on the last carbon that carbon is the carbon that's gonna bear the oxygen with a negative charge and in the next step we're gonna react us with the hydronium ion
so in the end we are going to get a primary alcohol so we have a total of six carbons and so this is called one hexanol so that's how you can make primary alcohols from alkyl halides using the grinev reagent and so that's it for this video thanks for watching
