(Small printing mistake-it’s n tends to infinity) @Sanjay_S
In the exam answer was given as D. I thought it should be B. Most websites also say it’s B. Pl explain…
This orgamic one was on yesterdays mypat test too.
In Sn2 one inverted product is predominant but that doesnt make D the answer. Its a risky question.
Right Bro…mypat test
What did you mark the answer as??
I think it is B,becuase we know the nucleophile attacks the chiral carbon,The configuration is inverted with leaving group going away,and this inversion product should be of opposite optical rotation i.e., B.
I marked B
Guys please try the maths sum too(I didn’t know how to start only)
Trying in some time. Aits na?
Friend sent. Yes,i think so
Manan correct me if i’m wrong coz I struggle with stereochemistry at times… Sn1 mei both backward attack and attack from front end are favorable (which would be option B). Sn2 mei backward attack is favorable? Therefore single stereoisomer? And hence D. I mean I answered D without checking below but I might certainly be wrong.
Ya bro… But even if the word often is not there? Assuming that the Sn2 reaction proceeds towards 100% completion, wouldn’t that mean only a single stereoisomer is formed?
I am also confused , earlier i thought both stereoisomers are formed with inversion product as a significant major. I think u r right , it should be D .
Bhai not getting don’t know why! The math question
bro whats the answer?
Friend shared it…didn’t give the answer…I’ll update you if i get it Bro!!!
bro one thing since they’ve given ordered pair where sum is K^2 i highly doubt whether the statement is correct because infinite pairs will be possible
I also have doubt , are they considering k to be perfect root(rational), then only question is correct.
because its not necessary that k is rational ( condition not given )
please clarify this with your friend and let me know the condition for k
if it’s perfect square only then aaram seh hojayega