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I am confusing in between these two br groups. Which is the best Leaving group I am not deciding help me please

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And answer is C

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closed as off-topic by matt_black, orthocresol, Wildcat, Klaus-Dieter Warzecha, bon Oct 12 '16 at 12:40

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    $\begingroup$ welcome to chemistry SE. Be sure you are not infringing on anyone's copyright? ... meta.chemistry.stackexchange.com/questions/3268/… ... also also unclear ... please read through the tour chemistry.stackexchange.com/tour $\endgroup$ – Agriculturist Oct 12 '16 at 11:46
  • $\begingroup$ No I am not Pasting Any copyright material.I respect DMCA ,This image is available on google.If any problem i will remove this picture right now.Thanks For the suggestion $\endgroup$ – Akash Goel Oct 12 '16 at 12:02
  • $\begingroup$ @AkashGoel Just a random fact: That an image is available via Google does not imply, that no copyright law has been violated. $\endgroup$ – pH13 - Yet another Philipp Oct 12 '16 at 12:17
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It isn't best leaving group but which is the most reactive centre.

You have a choice of two reactions between the cyanide anion and (i) an alkyl bromide or (ii) a benzyl bromide. The benzyl bromide is more reactive than the alkyl bromide which leaves you with C or D.

The problem with D is that the bromine atom has an inverted stereochemistry which doesn't make sense.

C has the correct regiochemistry (the position in the molecule) and the correct stereochemistry. As to the stereochemistry: the reaction appears to be SN2 meaning that the cyanide anion attacks the carbon on the opposite face of the bromine atom of the starting material.

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The bromine closer to the aromatic ring system is replaced because it is easier to replace.

The reason is obviously not sterics, but it is the stability of the transition state for the SN2 reaction.

The transition state is stabilised by the pi electron density from the aromatic system, and hence, that is the preferred product.

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The answer is C because the nucleophilic attack of the C - Br adjacent to the benzene ring produces a carbanion transition state that can be stabilized by the pi system in the benzene ring. The geometry is obvious as the cyanide ion must come in from the side opposite to the Br. Try it using a model kit.

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