Can someone help me put these in order from fastest reaction to slowest. I tried, but i keep getting it wrong
sn2 reaction is favured by less steric hinderance and less substituted alkyl halides.now try to arrange again on this basis.i will help u get it right if u dont get it right this time on ur own
is it that ?
no.u still it wrong
can you explain it to me ? please?
just a minute
ok
shall i explain u the whole working principle o an sn2 reaction???
no, i understand that. Its just when i try to apply what i know to put those in order i get it wrong
the general order of reactivity of halides in an sn2 reaction is, methyl halide>primary halide>secondary halide>tertiary halide first check this criteria and then which molecule offers the least steric hinderance for attack of an nucleophile from backside
and label the compounds so i cud give the correct order for u
the correct order is 1>2>4>3
from fastest to slowest
yes
thats what i did the first time and it said its wrong ...i didnt put a pic up
i think its the right order
it says its wrong
i am sure about the fastest and slowest one.its the two in the middle which are bit confusing.fine then go with this order: 1>4>2>3 check it out
its right
why would it be 4, 2, 3?
great
3 is a tertiary halide so its the slowest one. 4 and 2 are secondary halides so we need to check the steric hinderance offered to the attacking nucleophile,and in this case 4 offered less hinderance
could u help with another question
ya.sure
I'm very unclean when it says to add the formal charge and idk if i should include the leaving group
may be u shud try the carbon atomfrom the opposite side as in an sn2 reaction an invertedproduct is formed
inverted product
u mean the carbon beside the carbon i put the arrow to?
r the lone electrons and the formal charges right?
i mean the carbon being attacked(from which br is attached)
u mean if i just have the row from the top, not the bottom
there shudnt be positive charge on oxygen when br is alone,one more electron shud be on it
check it out
so the br should have a negative charge and two lone electrons , the O should be neutral and have 2 pairs of lone electrons
yes.that shud do it
still wrong
can i see the pic
what about other 4 lone electrons on br
sorry not 4 but 6
like that
ya.just like that.great
got it right.....if i have anymore questions today or tomorrow can i ask you ? It just that we have not done a lesson and i'm doing it with only reading the book and the notes
ya.no problem.never hesitate to ask anything.i will try my best to help
thank you
hey can u help
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