williamson ether synthesis

Williamson involves SN2 attack of alkoxide on alkyl halide............so for better reaction we have to identify better leaving group.
CH₃I=super primary(as sky said)[3]
EtI=1°

so 2nd combination will give better yield.[1]

for this u need to know the mechanism :)

see ... RNa has R- and Na+ .

this R- goes and attack R'I from behind ... and I- leaves..

so wat we can conclude??

>> R should be a good nucleophile..

>> I should be attached to such an alkyl chain where it will
not 'leave' n go away easily (like 3°) ... [since -I is a very very good leaving group]

now in the present case,

Et- is a better nucleophile than Me- .

so, it should be: CH3I and EtONa..

@eureka...

u wrote : CH3I=3°

[7] [7] [7] [7]

typo i hope ??

both are primary.. but..

the thing is that:

EtI will produce a more stable carbocation [hyperconjugation!]

so we will use CH3I and EtONa ...

are yaar.........kya kar rahe ho....bhavnao ko samjho........[3]

kiski bhavna?

aur kaunsi bhavna?