In my orgo textbook, it says that that one of the most common ways of making alkyl halides is: SN1 rxn with tertiary alcohol

 

However, a halide is a much BETTER leaving group than an OH group...so why does this reaction work?

 

Thanks in advance

Because OH- is not the leaving group. In the case of conc HCl, H2O will be the leaving group. In the case of SOCl2, -O-S(O)Cl- is the leaving group, in the case of PCl5, -O-PCl4- is etc etc.

i don't think that its -O-S(O)Cl- is the leaving group in case of SOCl2 and same for PCl5 , have a look on mech.

 

http://www.cem.msu.edu/~reusch/VirtualText/Images2/alcolrx1.gif

thanks for the responses...

 

ok, good point H20 is the actual leaving group. But halides are better leaving groups than water too. Cl-, Br-, and I- are very stable ions because they are the conjugate bases of strong acids

 

i would think a few H20 could pop off and be replaced by Cl-, but the equilibrium of this would lie much in the other direction.

 

....so i still dont see why this reaction would readily occur

thanks for the responses...

 

ok, good point H20 is the actual leaving group. But halides are better leaving groups than water too. Cl-, Br-, and I- are very stable ions because they are the conjugate bases of strong acids

 

i would think a few H20 could pop off and be replaced by Cl-, but the equilibrium of this would lie much in the other direction.

 

....so i still dont see why this reaction would readily occur

 

Hello,

 

Because you are dealing with a tertiary center and therefore automatically confined to an SN1 mechanism... for either route going from the alcohol to the chloride or the chloride to the alcohol you will go through the same carbocation intermediate.

 

Thus to effectively convert the alcohol to the chloride it is necessary to have an excess of Cl- anion present compared to water molecules present (Le Chatelier's principle)... This overcomes the equilibrium issue that you are referring to..

 

When actually doing this reaction in the lab often times molecular sieves or anhydrous calcium chloride is added to remove the water thereby further driving the equilibrium towards the production of the desired product - the tertiary halide.

 

As far as the other examples with SOCl2 etc.. they go through SN2 mechanisms and will not work on a tertiary alcohol...

 

hope this helps. : )

i don't think that its -O-S(O)Cl- is the leaving group in case of SOCl2 and same for PCl5 , have a look on mech.

 

http://www.cem.msu.edu/~reusch/VirtualText/Images2/alcolrx1.gif

 

If you look at your post you'll see I'm right. -O-S(O)Cl- leaves and rearranges giving SO2 and Cl-, likewise -O-PCl4- leaves giving POCl3 and Cl-.

Keep in mind that conversion of tertiary alcohols is usually done in concentrated HX, so there is quite a bit of hydronium floating around. Consider also that the tertiary halide is insoluble in water, so it seperates out and removes itself from the reaction.

 

tertiary alcohol + HX <--> tertiary alkyl oxonium halide <--> carbocation + water + halide ion --> tertiary alkyl halide + water

Hello,

 

Because you are dealing with a tertiary center and therefore automatically confined to an SN1 mechanism... for either route going from the alcohol to the chloride or the chloride to the alcohol you will go through the same carbocation intermediate.

 

Thus to effectively convert the alcohol to the chloride it is necessary to have an excess of Cl- anion present compared to water molecules present (Le Chatelier's principle)... This overcomes the equilibrium issue that you are referring to..

 

When actually doing this reaction in the lab often times molecular sieves or anhydrous calcium chloride is added to remove the water thereby further driving the equilibrium towards the production of the desired product - the tertiary halide.

 

As far as the other examples with SOCl2 etc.. they go through SN2 mechanisms and will not work on a tertiary alcohol...

 

hope this helps. : )

 

 

yes that explained a lot

 

thanks again for the responses everyone