Sn2

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atlanta213

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Hey guys!

In Sn2 reaction, I know product is inverted while optical activity is retained. Then if reactant were R configuration, the product changed to S configuration?

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sorry to steal your thread, but let me ask a question.

i know racemic mixture that is resulted from sn1 reaction loses its optical activity by having equal mixture of + and - products which add up to 0. ( if that make sense, i am an international student.)

and now, + means optical activity to the right? in other words dextero or something, and - means activity to the left (levo.)

However, my question is if there are two products that are enantiomeric to each other, than that does not mean anything about the optical activity.

right?

someone please clarify this thing.
 
sorry to steal your thread, but let me ask a question.

i know racemic mixture that is resulted from sn1 reaction loses its optical activity by having equal mixture of + and - products which add up to 0. ( if that make sense, i am an international student.)

and now, + means optical activity to the right? in other words dextero or something, and - means activity to the left (levo.)

However, my question is if there are two products that are enantiomeric to each other, than that does not mean anything about the optical activity.

right?

someone please clarify this thing.

don't confuse yourself. you know about R and S, last post explained that
in sn2 R will change to S
in sn1 we are going to have mixture of R and S ( don't confuse with + and -) that means one of the compond will be + and another one will be - ( I am sure you know R and S is not related + and - and you can not predict R is + or negative. posetive and negative is experimental) because you have equal amount of R and S then half of the part will be dexo and half will be levo and total not change in polarized light.
recall: (it is not related to your question but it is good to know) meso is one compond rasmic is two componds.
hope it helped
good luck 👍
 
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