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Author: Subject: Alkyl Chloride preparation without anhydrous zinc chloride?
hippo
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[*] posted on 31-8-2017 at 09:52
Alkyl Chloride preparation without anhydrous zinc chloride?


Is the presence of zinc chloride absolutely necessary for forming R-Cl from R-OH and HCl? I have been meaning to prepare n-Butyl Chloride and all the preparations I've seen require anhydrous zinc chloride which I don't currently have access to.
Also, is it necessary for the ZnCl2 to be anhydrous? Can a hydrated form be used?
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hippo
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clearly_not_atara
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[*] posted on 31-8-2017 at 10:42


I really don't understand why you would need anhydrous ZnCl2 when you are going to drop it right into an aqueous solution. Seems pretty silly to me. I think the real goal is just to decrease the water content of the rxn mixture as much as possible. This can be accomplished by adding zinc metal to hydrochloric acid and then passing additional HCl gas into the solution. HCl suitable for this purpose can be produced by the thermal decomposition of hydrated zinc chloride, which reduces the precursor burden.

Or maybe you can just use zinc sulfate, which is easy to dry.
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hippo
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[*] posted on 31-8-2017 at 11:08


The Wikipedia article states that:

Quote:

Hydrochloric acid alone reacts poorly with primary alcohols and secondary alcohols, but a combination of HCl with ZnCl2 (known together as the "Lucas reagent") is effective for the preparation of alkyl chlorides.


Although it doesn't state what part it has to play, every preparation I have seen thus far has stated that you need ZnCl2 present for the HCl to react
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XeonTheMGPony
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[*] posted on 31-8-2017 at 11:27


Step one: Get galvanized pipe or nails
Step two: Add hydrochloric acid
Step 3: watch it fizz
Step 4: Isolate your fresh Zinc Chloride from any iron salts.
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hippo
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[*] posted on 31-8-2017 at 11:43


It needs to be anhydrous I think
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Sigmatropic
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[*] posted on 31-8-2017 at 12:32


Simply refluxing butanol with HCl or in aqueous HCl is not going to afford butyl chloride. The affinity of water, or butanol, for HCl is simply too low. This means the concentration of Cl- is nowhere near high enough to establish a favorable equilibrium between butyl alcohol and butyl chloride.

Certain Lewis acids are usually employed to aid this. Zn is typically used but I don't see why Fe or Al wouldn't work.

The chlorides of these materials do not necessarily have to be anhydrous as they are added to concentrated but still aqueous HCl. Instead their oxychlorides may be employed. These oxychlorides are usually poorly defined and are thus not encountered in scientific literature. These oxychlorides are, however, obtained when one dries the aqueous chloride salts to (almost) dryness.

So no it does not need to be anhydrous yet the synthesis will not work when it is omitted entirely.

Good luck and report your findings!
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hippo
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[*] posted on 31-8-2017 at 13:01


Thanks Sigmatropic!
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Melgar
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[*] posted on 2-9-2017 at 02:46


Butanol forms an azeotrope with water that's about 50/50, so you should just be able to boil a solution of ZnCl2, water, and butanol, until chlorobutane starts coming over with it. Then just redistill whatever you collect. The chlorobutane should be the lowest-boiling component by quite a lot.



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[*] posted on 2-9-2017 at 03:56


Quote: Originally posted by Melgar  
...The chlorobutane should be the lowest-boiling component by quite a lot.

Chlorobutane and water will definitely form an azeotrope. That will have a lower boiling point than either component.
It''s possible that a ternary azeotrope will form with an even lower BP.
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[*] posted on 2-9-2017 at 11:17


In the Vogel's Handbook of Practical Organic Chemistry (7 or 5ed., I don't remember) you can find a procedure for cyclohexyl chloride from cyclohexanol, HCl and CaCl2. The zinc is therfore not necessary for some alcohols, but 1-butanol may need it as it is primary. I think that zinc may increase the amount of Cl- in solution as ZnCl2 to move the reaction forward is insanely soluble in water: 432g per 100g of H2O.
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[*] posted on 2-9-2017 at 12:33


Well, if copper doesn't interfere, the U.S. penny might be a good source of Zinc.

Modern pennies are 97.5 % Zinc. Copper plated, they are.
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[*] posted on 2-9-2017 at 13:49


You can always melt the penny and remove then remove the copper casing using a gas burner or even a kitchen stove. The zinc will drip out of the casing if you scratch or rub the copper layer off the penny.
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[*] posted on 3-9-2017 at 13:41


For the exact mechanism for ZnCl2 helping the reaction, on the 11th page of this pdf, it shows ZnCl2 as attaching to the oxygen of the alcohol and helping it become a much better leaving group.

https://crab.rutgers.edu/~alroche/Ch11.pdf
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[*] posted on 3-9-2017 at 21:10


Quote: Originally posted by MeshPL  
In the Vogel's Handbook of Practical Organic Chemistry (7 or 5ed., I don't remember) you can find a procedure for cyclohexyl chloride from cyclohexanol, HCl and CaCl2. The zinc is therfore not necessary for some alcohols, but 1-butanol may need it as it is primary. I think that zinc may increase the amount of Cl- in solution as ZnCl2 to move the reaction forward is insanely soluble in water: 432g per 100g of H2O.

I've seen this reaction described in later sources as the 'exceptional behavior' of cyclohexanol as opposed to other secondary alcohols.

So I wouldn't count on it working for too many other secondary alcohols.

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[*] posted on 4-9-2017 at 21:56


Cyclohexanol can be converted into cyclohexyl chloride with concentrated HCl at room temperature with perfect yields (90%+). Usual secondary alcohol require distillation temperature, large excess of azeotropic HCl-H2O, and still this yields only 50% of theory.
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[*] posted on 4-9-2017 at 23:04


Quote: Originally posted by byko3y  
Cyclohexanol can be converted into cyclohexyl chloride with concentrated HCl at room temperature with perfect yields (90%+). Usual secondary alcohol require distillation temperature, large excess of azeotropic HCl-H2O, and still this yields only 50% of theory.


Thanks for the details.
Even with the low yield that's still an attractive sounding way to make isopropyl chloride given how cheap and easy to get the reagents are. (But I'd certainly use the zinc chloride for more expensive alcohols)
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OrganoLeptic
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[*] posted on 6-9-2017 at 13:40


I can confirm that isopropyl chloride can be prepared from isopropyl alcohol with the use of 31% HCl(aq) and anhydrous CaCl2.
I got a 50 - 56% yield.

You can check out the video of the procedure here:
https://www.youtube.com/watch?v=KyeDmo0yOeA

I haven't yet used the isopropyl chloride I produced. But the boiling point of the product is in good agreement with the literature.

I don't know if this procedure would work for a primary alcohol.

I would be interested to hear if anyone had any success using HCl(aq)/CaCl2 on primary alcohols.

I will probably try it out myself in the next couple of weeks using a low boiling primary alcohol (probably n-propanol).




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byko3y
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[*] posted on 6-9-2017 at 22:49


You will not succeed with HCl+CaCl on saturated-chain primary alcohols. High pressure (2-5 atm) and high temperature needed.
Benzyl alcohol reacts with slight heating; activated benzyl alcohols e.g. methoxybenzyl easily reacts at r.t.; activated aliphatic primary alcohols like ethylene glycol only slowly react at reflux temperature.
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[*] posted on 7-9-2017 at 13:41


Thank you byko3y. Do you know this from personal experience or from an article? I know it is somewhat common knowledge
that primary alcohols require more vigorous conditions. But that is part of the problem since it is difficult (at least for me)
to find any data on the issue. (+ people usually don't publish negative results)

My previous post does not reflect this but I also think that I will not succeed. That is why trying this out is of low priority.

The thing is, I was unable to find any data on anybody trying this with removing the product from the reaction mixture as it is formed.
I don't know if removing it by distillation even makes a difference. The solubility of the products (alkyl chlorides) is typically very low
and one could argue that the formation of a second layer is already removing the product from the reaction mixture and thus pushing the reaction forward.
If that is the case, removing the product by distillation shouldn't really make much of a difference.

As far as I know the CaCl2 has no other function than to increase the concentration of chloride ions in the reaction mixture.
(If you have any data to the contrary I would like to hear about it.)
By this logic I would be content to see that someone tried to run the reaction with concentrated hydrochloric acid and removing
the product from the reaction mixture by distillation and failed miserably.

Byko3y if it isn't to much trouble I would also like to see a reference for the high pressure/high temperature
and the one for the activated aliphatic primary alcohols. I haven't seen any references like that, but I would like to.
Thank you.




[Edited on 7-9-2017 by OrganoLeptic]




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