00:01
Benzile bromide c6h5 ch2br reacts rapidly with methanol to afford benzile -ethyre c6h5 ch2och3.
00:18
Draw a stepwise mechanism for the reaction and explain why this primary alkali halide reacts rapidly with a weak nucleophile under condition.
00:33
That favor an s -n -1 mechanism is the first part of the question and second part is would you expect the para -substituted benzylic halite that is ch3o c6 h4 ch2 br and no2 c6 h4 ch2 br to each be more or less reactive than benzyl bromide in this reaction.
01:09
Explain your reasoning.
01:12
So actually benzene bromide is primary halide.
01:18
It reacts rapidly with a weak nucleophile by sn1 mechanism because here benzene carbo -catine is formed which is more stable due to resonance.
01:35
So let us see how it is formed.
01:41
So step by step first we are making the ether starting from benzene bromide.
01:50
So here benzene bromide is taken benzene with ch2 br.
02:15
This is losing br and by that carbocatin is formed in the first step.
02:31
So by the loss of this br minus we are getting the structure that is carbocatine which is more stable due to resonance.
02:51
Double bond will remain here and ch2 will be positive at this carbonate.
03:03
This is the carbocation which is stable due to resonance.
03:06
How the resulting structures are formed that we can see this electron pair is given coming over here and by that positive charge appears at the pencil tree so in the next structure positive charge will be here double bond redemines same and this should be double bond and this should be double bolt and here here ch2.
03:52
In the next structure, next resonating structure, this double bond is shifted here and the positive charge must be here.
04:10
So in the next resonating structure, while making the resonating structure, we have to move only one bond and the remaining structure is not distinct.
04:32
So this double bond is as such and one double bond is formed here and now one positive charge is appearing here.
04:44
So in the next structure, next resulting structure, this bond is shifting and forming the positive charge.
05:06
And by that the structure will be double bond with this part.
05:19
Same ch2 and positive charge here and now the double bond is here so because of all these resonating structures what happens that this benzile carbocata and though it is coming from the primary halide it is quite stable and that is why the reaction is taking place fast after this resonating structures we are coming to the next structure where positive charge is going back to carbon and the structure left is double bond one double bond here one double bond is coming here one carbon with the positive charge this is the carbocatin which is found in the first step and after this here the attack of methanol is taking place so how the methanol is coming over here suppose it is ch3 oxygen with loan pairs and hydrogen here this loan pair is going to attack the positive carbon and the structure formed is benzene ring with double bonds at this position.
07:37
The carbon with hydrogen 2 and one part is now attest to oxygen, one hydrogen and with hydrogen and the positive part, positive charge will be at this oxygen, which is given the electron pair to carbon positive...