- Messages
- 37
- Reaction score
- 97
- Points
- 28
q 39
74.00g of butan-2-ol → 44.64 g of butanone
Moles of butan -2 -ol =1
moles of butanone= Mass/mr
=44.64/72= 0.62
now in theory 1 mole of butan 2 ol produces 1 mole of butanone.
but we got 0.62 moles of butanone
hence 62 percent yield
Use the same method for the rest of the reactions and you will gt 62 percent yield for all of them.
q 40
(CH3)3CBr + NaOH → (CH3)3COH + NaBr
(CH3)3CBr is a tertiary haloalkane .
It has three R groups attached to the Carbon bonded to the halogen
Only Tertiary haloalkanes undergo SN1 mechanism for nucleophillic substituition.
This is because tertiary haloalkanes have electron donationg methyl groups attached they can form an intermediate and stable carbocation.
The graph has two humps so this means that an intermediate is formed before the reaction proceeds.
2 and 3 are not tertiary haloalkanes so an intermediate is not formed
Thanks Maria.