What Is the Final Product of Bromination and Cyanation of Phenyl Ethane?

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SUMMARY

The final product of the bromination and cyanation of phenyl ethane is 1-cyano-1-phenyl ethane. The reaction begins with bromination using ultraviolet light, leading to the formation of a stable free radical on the phenyl ring, resulting in 1-bromo-1-phenyl ethane. Subsequently, bromine, acting as a good leaving group, is replaced by the cyanide ion (CN-). The discussion concludes that the answer key may contain a misprint, as the expected product is not accurately represented.

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Homework Statement
Find the product of the following reaction
Question pic in My attempt section
Relevant Equations
None
IMG-20200412-WA0036.jpeg


So what I did was-:
The first step is bromination using ultraviolet light which proceeds via formation of free radical.Here free radical on phenyl ring attached carbon will be more stable due to resonance and 1 bromo 1 phenyl ethane will be formed.
In the second step,Bromine will leave as it's a good leaving group and CN- will come in place of bromine.So,according to me the final product would be 1 cyano 1 phenyl ethane.But the answer was given as d).Help.
 
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I think there is misprint in answer key.In place of one CN CH3 shoukd be there.Am i right?
 
Physics lover said:
Homework Statement:: Find the product of the following reaction
Question pic in My attempt section
Relevant Equations:: None

View attachment 260538

So what I did was-:
The first step is bromination using ultraviolet light which proceeds via formation of free radical.1) Here free radical on phenyl ring attached carbon will be more stable due to resonance and 2) 1 bromo 1 phenyl ethane will be formed.
In the second step,Bromine will leave as it's a good leaving group and CN- will come in place of bromine.So,according to me the final product would be 1 cyano 1 phenyl ethane.But the 3) answer was given as d).Help.
1). the phenyl carbon or the carbon alpha to the aromatic ring produces the most stable radical intermediate.
2) correct.
3) you were correct and 4) (not “d”) is the wrong structure. None of the printed answers are correct unless you assume that 4) is a misprint.
 
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