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Question about Hydrolysis on Halogenoalkanes

It may sound stupid but I'm utterly confused about the reactivity/solubility with Halogenoalkanes and water. In some instances Halogenoalkanes react just fine with aqueous solutions but in other instances they need to be ethanol solutions. E.g Production of alcohol by reacting haloalkanes with aqueous potassium hydroxide but using ethanol solution of potassium cyanide when converting bromomethane into propanenitrile. Is the usage of ethanol the exception rather than the rule?

I think I somewhat understand why ethanol is used in that reaction, because the neucleophile needs to be CN ion?
Bananas also have potassium
Original post by kobutor
It may sound stupid but I'm utterly confused about the reactivity/solubility with Halogenoalkanes and water. In some instances Halogenoalkanes react just fine with aqueous solutions but in other instances they need to be ethanol solutions. E.g Production of alcohol by reacting haloalkanes with aqueous potassium hydroxide but using ethanol solution of potassium cyanide when converting bromomethane into propanenitrile. Is the usage of ethanol the exception rather than the rule?

I think I somewhat understand why ethanol is used in that reaction, because the neucleophile needs to be CN ion?


Halogenoalkanes are not very soluble in water as they can't do hydrogen bonding.

With the reaction between halogenoalkanes and aqueous KOH to form an alcohol, an aqueous solution of KOH is required as it is alkaline in nature, so OH- ions are produced. This will lead to the nucleophilic substitution forming an alcohol.

In the second reaction with KCN, if water was also used the reaction follows NC- + H2O ----> OH- + HCN (this should be a reversible arrow but I'm not sure how to insert it). This reaction shows that the cyanide ion is alkaline in nature when dissolved in water, and as you said leads to the substitution of OH- instead of CN-.

Whereas, if you used an ethanolic solution of KOH, the ethanol is a weak acid, it is deprotonated by the KOH to form C2H5OK, which dissociates to C2H5O- which is more basic than OH- meaning an elimination reaction occurs instead.

Haloalkanes + aq KOH = alcohol (SN2 reaction) ie substitution
Haloalkanes + ethanolic KOH = alkene (E2 reaction) ie elimination
Haloalkanes + aq KCN = alcohol
Haloalkanes + ethanolic KCN= nitrile

And so I hope you can see how the reagents influence the chemical reaction, in this case certain reagents are used in the reaction because they achieve certain products that you have to know- and I'd assume that the other reactions/ products aren't needed for your course. So I wouldn't say there is an exception/ rule on when ethanol is required it just depends on what you want to make.

(I'm also not sure if you need to know about SN2/ E2 reactions yet for your course, I just think it leaves a good way to distinguish between what's happening when you use an aqueous solution or alcoholic one)

I hope this kind of makes sense and hasn't confused you further. Let me know if you have questions!

Thanks

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