Question

(ii) On treatment with aqueous sodium hydroxide, disubstitued cyclohexane 1 readily undergoes a ring-closing reaction to form epoxide 3 whereas compound 2 does not. Using appropriate conformational drawings, provide a mechanism which accounts for this observation. Classify the ring-closing reaction according to Baldwin's rules. OH ОН co con 'Br Br 1 2 3 (6 marks)

Answer 1

OH Naoh > Br The treatement of sodium hydroxide, with di Substituted Cyclohexane undergo ring - closing reaction In this reaction , 2 step are required to form the product It is sNy reaction Proceed via Backside attack leading to inversion of configuration To the st step the oo is formed by de protonation of group Оң 0 Не i NaH OH BY Br intermediate. io tramolecular In next step there is SN, ortachim lectures Scanned with Ca-e-Searner CS

o Bay epoxide Cinversion in Configuration) In this step there is Back side attack of oe is occures, and epoxide. we get ST OH Naoh No epoxide Br react with In this reaction when NaoH reactant the intern ediate is formed ) 00 DO Caree Вт SN2 reaction In this there intramolecular is possible there possible Because attack no Back side here 50 , when both substituents are in same plane there is no formation of epoxide is there Scanned with CamScanner CS

* In the disubsitituted cyclohexane 1 it under ring closing reaction via intramolecular SN2 mechanism .

In this reaction the back side attack is possible , to the epoxide is formed easily.

* In the disubsitituted cyclohexane 2 , Both subsituent are in same plane .

In this reaction , when the intermediate is formed , there is no intramolecular SN2 reaction is possible. Because there is no back side attack possible here.

So the epoxide is not formed in the 2 compound.