Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
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Answer:
pOH=9.9
Explanation:
pH=-log[H+]= -log[0.0000877]
=4.06
pOH+ pH=14
pOH=14-4.06= 9.91
Na has a +1 charge, and O has a -1 charge.
