between the two rkn which one will react more rapidly by sn2 rkn in PROTIC SOLVENT
A) 1. 
+ CH3CH2O-= product + Cl - 2. + CH3CH2OH= product +HCl
B) 1. + CH3CH2O-= product + Cl - 2. + CH3CH2S-= product + Cl -
c) 1. + (Ph)3P = product + Cl- 2. + (Ph)3N= product + Cl-
3a.. in first one as base is stronger in case 1
b.first one perhaps..same reason
c..check question ..both same
1question corrected...
no doubt for (A) first one is stronger nucleophile than second one , but its given to to be in protic solvent, then the frist one will be more solvated than the second one which will increase its size and in turn reduce the reactivity.....
and almost the same reason for the (B)
correct me if i m wrong somewhere......
1Condition 1 :
If a reaction is carried out in a protic solvent, whose molecules have a hydrogen bonded to fluorine /oxygen/nitrogen, the larger atom is a better nucleophile in an SN2 reaction. In other words, the weaker base is the better nucleophile in a protic solvent.
So answers :
A. 2
B. 2
C. 1
Condition 2: Protic solvent with No hydrogen bonding
A. 1
B. 1
C. 2
39Indeed there cannot...such a solvent would be a polar aprotic solvent then(which can dissociate into ions but cannot donate hydrogen)...polar protic solvents are meant to donate hydrogen ions.
Considering that there are oxygen and nitrogen here present in substrates, there has to be H-bonding.
1exactly but the ans given in the book are completly opposite of what we are thinking, i don't where i m going wrong
23A- 1 , B - 2 , B - 2
in A , RO- is more nucleophilic than ROH
in B , RS- is more nucleophilic than RO-, bcz of high polarisibility of RS- and comparatively lesser solvation than that of RO-
in C , P(Ph)3 is more nucleophile again due to high polarisibility of P , or u can say that P has more tendency than N to give electrons
is my ans matching ?
23@ The , protic solvent is defined as the one in which H is directly attached to very high electronegative atoms like , F , O , N ..
so H bonding is ought to happen