[souro10]

Hi.

I have noticed many books , study materials and websites writing that R-X reacts with Aq. KOH to form R-OH and with Alcoholic KOH to form Alkenes.

My question is, when they say " treated with Aq. KOH " does it mean the alkyl halide was taken in water as solvent ? If yes, how is this method at all suitable for primary halides , which will prefer to undergo SN2 reactions , but OH- will get solvated due H-bonding with water molecules as far as I know.

So, for primary halides shouldn't the reaction be carried out in conc. KOH and some polar but aprotic solvent like DMSO, which will dissolve both the organic substrate and the inorganic reagent, and solvate only the cationic part of the inorganic reagent and leave the nucleophile almost intact? 

The second question is, what does " treated with Alcoholic KOH " , once again, mean? Was the reaction carried out in alcohol solvent?

And thirdly, why does Aq. KOH lead to substitution and alcoholic KOH leads to elimination ?

Thanks.

[adianadiadi]

Look at the products in the elimination and apply Le Chatelier's principle.

[souro10]

You only answered the third question. And I do not agree with your answer because, Le chatelier's principle is only
applicable to reactions under equilibrium control. I doubt how much under equilibrium control an elimination rection would be. With no acid catalysts present , and with the alkene and the KCl being in two different layers, the probability of this being a correct explanation is low.

[adianadiadi]

Elimination under basic conditions too is reversible.

[orgopete]

I have noticed many books , study materials and websites writing that R-X reacts with Aq. KOH to form R-OH and with Alcoholic KOH to form Alkenes.

While I can understand that you be tempted to make that generalization, I think it is not correct. Textbooks sometimes write things too literally for students. Those reactions and their descriptions are not side by side comparisons. A primary halide is not especially prone to elimination unless a base like t-butoxide is used. One would not want to use an alcohol as a co-solvent or ether formation is likely to accompany the alcohol.

If the halide is secondary or tertiary, substitution will be somewhere between difficult to impossible. In that case, an alcohol as co-solvent will not interfere and the product will be the elimination product.

These are two different halides, different reactions, and different products.

[souro10]

@orgopete: I completely agree with you. I think many textbooks use this kind of over simplification, and that's not really good for the learning process.

@adianadiadi- kindly elaborate (propose a mechanism ) how you can go back to the alkyl halide from the alkene. (reversible)