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Alcohol Reactions - HBr, PBr3, SOCl2


 

in this  artic we're going to go over reactions associated with alcohols so what's going to happen if we react in alcohol with hydrobromic acid the oh group will be replaced with a bromine atom so that's going to be the end result and so basically the alcohol is converted into an alkyl halide but now let's go over the mechanism for this reaction now we have a primary alcohol and so the mechanism will go through an sn2 reaction but the first step in this reaction will be protonation of the o h group the o h group is a bad leaving group but once you add a hydrogen to it it becomes a good leaving group so the oxygen is going to acquire a hydrogen turning into this species and so whenever the oxygen has a positive charge it becomes a better leaving group in the second step the bromide ion which is now in the solution will attack from the back in an sn2 reaction expel an h2o and so that's how we can get our alkyl halide so that's the mechanism for the conversion of a primary alcohol into an alkyl halide primary alcohols react very slowly with hydrochloric acid due to the fact that chloride is a weaker nucleophile than bromide however this reaction can be enhanced if we use zinc chloride which is a powerful lewis acid and so this is known as the lucas reagent so this oxygen is going to attack zinc chloride but I need to draw this better so i'm going to draw zinc chloride like this so the oxygen attacks sync and it's going to expel a chloride ion and so we're going to have an oxygen that's attached to a hydrogen and the zinc species so right now the oxygen has a positive charge which means it's a good leaving group so in the second step the chloride ion comes in it attacks the carbon kicks out the oxygen and so that's how we can get our alkyl halide so if you want to convert an alcohol into an alkyl chloride where you have a primary alcohol you want to use the lucas reagent to do so now let's work on some more examples let's say we have a tertiary alcohol and this time let's react it with hydro iodic acid what is the major product in this reaction the end result is that we're going to replace the o h group with the iodine atom and so that's going to be the end result the let's propose a mechanism so because we have a tertiary alcohol this reaction will proceed by the sm1 mechanism but the first step is protonation we need to convert the hydroxyl group into a good leaving group so now because it's an sm1 reaction the iodide ion is not going to come in and attack this carbon that's going to be an sn2 reaction those methyl groups will prevent access to this carbon so what's going to happen instead is the leaving group is going to leave giving us a tertiary carbocation intermediate and then at this point that's when the iodide ion can come in and combine with the carbocation giving us an alkyl halide and so this is the product of the reaction now let's go ahead and work on another example so let's react this alcohol this is a 2-methylcyclohexanol let's react it with hydrobromic acid so go ahead and predict the major product of the reaction and also show a mechanism as well so the first step is protonation as always anytime you react an alcohol with an acid so now we have a good leaving group so what's going to happen is the leaving group is going to leave and once it leaves we're going to have a secondary carbocation but notice that the secondary carbocation is next to a tertiary carbon and so when you see that a hydride shift will occur and so this is going to give us a more stable tertiary carbocation intermediate and then at this point the bromide ion will attack the carbocation and so this is going to be the final answer so we have a tertiary alkyl halide and that's it now there are other ways in which we can convert alcohols into alkyl halides one reagent that we could use is pbr3 and this works through an sn2 mechanism converting the oh group into or replacing them for bromine atom and so this produces an alkyl bromide another example is using socl2 which also works through an sn2 mechanism but this time the o h group is replaced with cl now let's go over the mechanism for those two reactions let's begin by drawing one butanol and then pbr3 looks like this this is phosphorus tri-bromide the phosphorus atom has a lone pair now phosphorus is partially positive the reason for that is bromine is more electronegative than phosphorus so bromine is partially negative now the oxygen in the alcohol also has a partial negative charge so therefore it's attracted to the partially positive phosphorus atom so oxygen is going to behave as a nucleophile attacking the phosphorus atom causing one of the bromine atoms to be expelled and so we're going to get an intermediate that looks like this now whenever oxygen has three bonds it's going to have one lone pair and a positive charge now if you recall protonated alcohols are highly acidic whenever you have an oxygen with three bonds and the hydrogen on it that hydrogen is going to  be highly acidic so what's going to happen next is an acid-base reaction so we're going to use the solvent pyridine to remove a hydrogen pyridine is a weak base so pyridine is going to abstract a proton putting these two electrons back on the oxygen so this is what we have right now now in the final step a bromide ion is going to attack this carbon expel in this group and so we're going to get one bromobutane as our product and then this will be a side product which we can leave it like this so that's the mechanism for the reaction of an alcohol with pbr3 this last step here is an sn2 step where we get inversion at the carbon atom now let's go over the mechanism of the other reaction so let's start with our primary alkyl halide and let's react it with thionyl chloride which looks like this it has a sulfur atom an oxygen two chlorine atoms and a lone pair now the oxygen and the chlorine atoms are more electronegative than sulfur so the sulfur atom has a partial positive charge just like the phosphorus atom now the oxygen is going to attack the sulfur causing this pi bond to break and so we're going to get this intermediate so now the oxygen has a positive charge and in the next step the oxygen is going to use one of its lone pairs to reform the pi bond expelling a chlorine atom so this is what we now have what do you think is going to happen next now typically this reaction is carried out in pyridine and pyridine is a weak base which looks like this and the purpose of pyridine in this example is to get rid of the hydrogen in the last step the chloride ion attacks from the back and then break in the carbon oxygen bond those electrons will be used to form a pi bond between the sulfur and the oxygen atom and this will expel the chloride ion and so we're going to get this product so this will give us our alkyl chloride we're also going to get sulfur dioxide which looks like this the sulfur has a lone pair on it so it causes the molecule to have a bent shape and we have this other chloride ion and also pyridine has a hydrogen on it and so that's how you can show the mechanism for the conversion of an alcohol into an alkyl chloride using socl2 in pyridine now we need to discuss the stereochemistry of these reactions so let's say we have this particular alcohol two butanol and let's react it with hydrobromic acid what's going to happen so you need to know that this will occur by means of an sn1 mechanism since we have a secondary alcohol if we had a primary alcohol then it would be an sn2 reaction but secondary and tertiary alcohols react with hbr by means of an sn1 mechanism and so we're going to get a racemic mixture of products so the bromine atom it could be on the dash or it can be on the wedge but the key is we can get both stereo isomers in this example now let's say if we started with the same alcohol but this time instead of using hbi we chose to use pbr3 what's going to happen now this reaction proceeds by an sn2 mechanism and so we're going to have an inversion at the configuration or at this uh chiral center so we're going to have inversion of configuration and the bromine atom is going to be on the wedge as opposed to on a dash so we only get one of the two stereoisomers in this case now if we use socl2 something similar is going to happen so if we use thionyl chloride it's going to go through an sn2 reaction and so we're going to have an inversion at the chiral center but we're going to get a chlorine atom instead of a bromine atom so you need to be familiar with the stereochemistry of these reactions another one you need to know is tscl in case you see it on that test now this one it works through retention but it converts the oh into a good leaving group and the product that you get is ots which i'm going to talk about soon so here we have an alcohol and let's put the oh group in the front attached to some r group and then we have this compound which is para toluene cephalochloride abbreviated tscl and so to draw the product of this reaction all you need to do is remove hydrochloric acid and simply connect these two groups together and as you can see you're going to get retention at the oxygen so the stereochemistry doesn't change now for those of you who need the mechanism here's how you can show it oh by the way so this is tscl at the top and this compound is known as r ots so we have an ots group attached to the r group

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  6. علي منصرمحسن ذعفان.

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  8. محمد توفيق عبد الصمد البغدادى 01125597769محافظه دمياط 01201049626مصر01018558757مصر01019382712مصر

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