Chemistry Compounds undergoing Friedel-Crafts alkylation

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The 2nd, 3rd, and 6th compounds can undergo Friedel-Crafts alkylation, while the 4th compound cannot. The 4th compound's NH2 group, when exposed to AlCl3, loses its lone pair and forms a highly deactivating NH3+ cation. In contrast, the 2nd compound features a CH3 group that is activating and does not donate electrons, allowing for successful alkylation. The presence of the weakly deactivating Br group in the 4th compound does not compensate for the NH2's deactivating effect. Understanding the behavior of substituents in the presence of Friedel-Crafts catalysts is crucial for predicting reactivity.
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Homework Statement
Which of the following compounds can undergo Friedel-Crafts alkylation?
Relevant Equations
Friedel-Crafts alkylation
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The answer is that the 2nd, 3rd and 6th compounds can undergo Friedel-Crafts alkylation.

I don't understand why the 4th compound cannot undergo a Friedel-Crafts alkylation reaction, but the 2nd compound can. It has an activating NH2 group and a weakly deactivating Br group, and it's similar to the 2nd compound which has an activating CH3 group and a weakly deactivating Cl group – what makes it different?

Thank you.
 
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What happens to the NH2 group in the presence of the Friedel-Crafts catalyst?
 
jolly_math said:
Homework Statement: Which of the following compounds can undergo Friedel-Crafts alkylation?
Relevant Equations: Friedel-Crafts alkylation

View attachment 343324
The answer is that the 2nd, 3rd and 6th compounds can undergo Friedel-Crafts alkylation.

I don't understand why the 4th compound cannot undergo a Friedel-Crafts alkylation reaction, but the 2nd compound can. It has an activating NH2 group and a weakly deactivating Br group, and it's similar to the 2nd compound which has an activating CH3 group and a weakly deactivating Cl group – what makes it different?

Thank you.
The thing is the 4th compound has a NH2 group which has a lone pair but in presence of AlCl3 a strong electrophile , the NH2 gives up its lone pair and forms NH3+ cation which is a -M group and highly deactivating, but in the 2nd compound CH3 have activating nature and cannot give any electrons so the chlorine gets removed and CH3 gets added i think
 
PhysicsEnjoyer31415 said:
The thing is the 4th compound has a NH2 group which has a lone pair but in presence of AlCl3 a strong electrophile , the NH2 gives up its lone pair and forms NH3+ cation which is a -M group and highly deactivating, but in the 2nd compound CH3 have activating nature and cannot give any electrons so the chlorine gets removed and CH3 gets added i think
Im sorry but to clarify , not the chlorine on the compound
 
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