IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

umang

Pls tell me where to prefer S or E ??

Its really confusing !!!

khinart

It basically depends upon
1)The speices undergoing reaction.(That is whether it is alycyclic or substituted.
2)Concentration
3)Steric factors.

I will explain it to U with an example.

khinart

Now on reaction of ethyl chloride with KOH(Dillute) substitution SN2 occurs.
CH3CH2CL + KOH-------->Ch3CH2OH
It is mainly because the compound is alycylclic,the concentration of the OH negative ions id plenty to attack the carbon atom and there is no steric hindrance for the backword attack.

In case the reagent used here was Alcoholic KOH then beta elimination wud occur and an alkene wud be formed.

Ch3Ch2Cl + KOH(alc.)--------------->ethene

This is because the concentration of Oh negative ions is less.

khinart

I think for further details U must refer the 12 NCERT book

harsha_27

in a rxn vessel there will always be a competition b/w substitution and elimination but due to some factors one dominates the other....such important factors are...
1.temperature
2.type of attacking species(deg.1 ,deg2, deg3....)
3.gibbs free enegy(which must always be -ve for rxn to be spontaneous)
=>if temperature is high elimination dominates substitution and eliminated product will be the major one due to entropy factors at normal temperatures substitution occurs
=>if attacking species is a bulky one then elimination predominates.....or if the site of attack is more crowded then also elimination favors...
=>overall gibbs free energy must be -ve for the rxn to proceed in that direction ...of course 1 and 2 are resulting from 3
rate IFF useful

umang

in some rxns , its written C2H5O- ion , while in others , its written C2H5O- / C2H5OH . How to decide in this case ??? (- means -ve charge)

shevchenko

i dont get u when u mean sometimes its written as c2h50h or c2h5o-.in solution c2h5oh will exist as c2h5o- and h+

umang

i am talking about two diff questions having these reagents used -
C2H5O- and C2H5O- / C2H5OH .

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