ROH +SOCl2+Pyridine →products
for the above RxN
Koverall=10-5[ROH][Cl-]+4*10-6[ROH][SOCl2]+10-6[ROH]
and if the alcohol is optically active (the only chiral carbon atom is the one containing OH group) then find % RACEMISATION
[Cl-] =1 MOLAR
[SOCl2] =1 molar
1doesn't the value of the rate constant play a role???????
u've given equal importance to all no ?
1dude. got a doubt . sorry for my Q's but this is what I got when I searched .
Tha presence of pyridine . stops and the Sni reaction which takes please like Sn2.
to yahan , thoda balav aega kya ? aap ka method abhi tak mughe convince nahi hua. please wrtie it in detail . total detail . every step.
please
1here SN1 racemises 100%
SN2 gives inversion and
SNi gives retention and it is lesser in % than inversion so it will cause 200% racemisation so
% racemisation == 2* 4*10-6+10-6/10-5+4*10-6+10-6
== 60%
1as if single rxn only SN1 causes racemisation but if retention and inversion go hand in hand the lesser causes racemisation
1Sn1 and Sn2 only causes racemization .
by Sn2 - 10-5 and due to Sn1 - .5* 10-6
no?
1Oh sry . Sn2 predominates .given by th rate constnat only. 10-5. messed it up . haan.
then what?
woh 5*10 -6 kahan se aa gaya?
getting it wrong dude.
please give total detail . da.
1no u r getting it wrong srinath i am giving imp to rate cons
look % S will be 4*10-6+.5*10-6/sum of all
multi ply it with 2 and get the ans
1This reaction proceeds by Sn1 , Sn2 , and Sni right???
so the one which contributes to racemization is only the first and last term , thus % racemization is given by that by the total . no?
1look for the question let assume that given comp is R
for Sn1 50% R and 50% S
for Sn2 100% S
for Sni 100% R
there fore % R=30%
%S=70%
% racemisation =2multiplied by % of lesser
1dude , here's my approach , check out and tell my mistake.
% racemizationis
10-5 + (10 -6 / 2) by 10-5 + ( 4* 10-6 ) + 10-6 / 2
I took it so , coz . in Sn1 . 100% racemization .and in Sn2 , 50% as it's a carbocationic intermediate.
where's the mistake.???? please clarify. this is fianl step after cancelling [ROH] and putting [cl-] = [socl2] =1
this reduces to .21 / 29 . which gives. 72.4 %
1no ithink the ans is 60% i have to do it again its a long time i have posted it
1the reactions u got completely right but i dont know why then ur getting the wrong ans there must be a silli mistake i think try it again
or tell me i will post the solution