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Organic 10 Online
OpenStudy (anonymous):

Can someone help me put these in order from fastest reaction to slowest. I tried, but i keep getting it wrong

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

is it that ?

OpenStudy (anonymous):

no.u still it wrong

OpenStudy (anonymous):

can you explain it to me ? please?

OpenStudy (anonymous):

just a minute

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

shall i explain u the whole working principle o an sn2 reaction???

OpenStudy (anonymous):

no, i understand that. Its just when i try to apply what i know to put those in order i get it wrong

OpenStudy (anonymous):

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

OpenStudy (anonymous):

and label the compounds so i cud give the correct order for u

OpenStudy (anonymous):

OpenStudy (anonymous):

the correct order is 1>2>4>3

OpenStudy (anonymous):

from fastest to slowest

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

thats what i did the first time and it said its wrong ...i didnt put a pic up

OpenStudy (anonymous):

i think its the right order

OpenStudy (anonymous):

it says its wrong

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

its right

OpenStudy (anonymous):

why would it be 4, 2, 3?

OpenStudy (anonymous):

great

OpenStudy (anonymous):

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

OpenStudy (anonymous):

could u help with another question

OpenStudy (anonymous):

ya.sure

OpenStudy (anonymous):

OpenStudy (anonymous):

I'm very unclean when it says to add the formal charge and idk if i should include the leaving group

OpenStudy (anonymous):

may be u shud try the carbon atomfrom the opposite side as in an sn2 reaction an invertedproduct is formed

OpenStudy (anonymous):

inverted product

OpenStudy (anonymous):

u mean the carbon beside the carbon i put the arrow to?

OpenStudy (anonymous):

r the lone electrons and the formal charges right?

OpenStudy (anonymous):

i mean the carbon being attacked(from which br is attached)

OpenStudy (anonymous):

u mean if i just have the row from the top, not the bottom

OpenStudy (anonymous):

there shudnt be positive charge on oxygen when br is alone,one more electron shud be on it

OpenStudy (anonymous):

check it out

OpenStudy (anonymous):

so the br should have a negative charge and two lone electrons , the O should be neutral and have 2 pairs of lone electrons

OpenStudy (anonymous):

yes.that shud do it

OpenStudy (anonymous):

still wrong

OpenStudy (anonymous):

can i see the pic

OpenStudy (anonymous):

OpenStudy (anonymous):

what about other 4 lone electrons on br

OpenStudy (anonymous):

sorry not 4 but 6

OpenStudy (anonymous):

OpenStudy (anonymous):

like that

OpenStudy (anonymous):

ya.just like that.great

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ya.no problem.never hesitate to ask anything.i will try my best to help

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

hey can u help

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