[AdiDex]

A question is given in my text book ..!!
Q. why alcoholic.KOH gives elimination reaction but aqueous  KOH gives substitution reaction with alkyl hallide ?


I many books it is given that
In Water , OH^- get solvated so strongly due to which it's basic strength decreases .!!

Where as in Alcohol, It is due to presence of OR- ion which is stronger base than OH^- ,  So it leads to Elimination ..!!
Are they  correct or not ??

Now i have Some Doubts ..!!

1. We can't take 100% water , Alkyl hallide can't get dissolved in it , KOH and alkyl hallide will get seperate out and will not react....they will react only at interface.....so why we called it aqueous KOH ??   it can't be 100% H₂O

2. I think question given in my text book is inappropriate .
It is not necessary that elimination product will always be a major product even in alcoholic KOH. am i right ??
 I can't do anything as if this question comes in my boards examination then i have to answer it. so i'm confused that what they want to ask from me ..!!

3. How solvation affects basicity  only ?? its Nucleophilcity should also get decreased ...!!!

Someone please answr ...!!

[OrgXemProf]

The question doesn't afford information about the alkyl halide. Surely, the structure of the substrate (i. e., primary, secondary or tertiary alkyl halide) and the nature of the leaving group (Cl, Br or I) are likely to influence the course of reaction (substitution vs. elimination; unimolecular vs. bimolecular mechanisms).

Consider reaction of t-butyl chloride with KOH in either water or ethanol. The rate determining step of SN1 substitution is the same as that for E1 elimination, so whatever happens after rate-determining formation of the corresponding tertiary carbocation (substitution vs. elimination) does not affect the rate of either reaction.

Consider that "aqueous KOH" refers to an aqueous solution that contains sufficient alcohol to dissolve the alkyl halide completely. This solution provides a better ionizing solvent than 100% alcohol and also more effectively solvates hydroxide ion (nucleophile in SN1 reaction, base in E1 reaction) than does 100% alcohol.

So, how does a better ionizing and more highly ion-solvating medium favor substitution over elimination?

Is it possible that this question has been taken out of context, i. e., does it follow a lengthy discussion of competing reaction mechanisms that might account for the generalizations that are implicit in the question?

[orgopete]

I couldn't figure out the specifics of this question either. I don't know if this is the literal question and referring to a specific alkyl halide or what.

It virtually seems to refer to different alkyl halides and may not be the actual verbatim question(?). KOH doesn't give substitution except for primary halides and primary halides give little elimination.

Some clarification seems in order here.

[AdiDex]

I got my answer ....my sir  clarified evrything...
So he explained sumthing like this...
For elimination, the reason is partly correct.
The reason is simple: Solvation of base is very weak in alcohol. THat means, the reactivity is more. Also, the size of the solvated base is large enough to show neucleophilic beahvior. It rather finds hydrogen that lies on the surface.
But in water, it's highly deactivated. So, its selectivity increases. Bond strength, donation ability etc. dominate.
And water is smaller sized solvent.
So, small size, strong solvation... ony that thing is possible which would have larger enthalpy involved. So, carbon~base bond.