
Answer the following question:
The most stable carbonium ion is:
A) ${C_2}^ + {H_5}$
B) ${(C{H_3})_3}{C^ + }$
C) ${({C_6}{H_5})_3}{C^ + }$
D) ${C_6}{H_5}{C^ + }{H_2}$
Answer
562.8k+ views
Hint:This is the resonance stabilized, which means the pi-electron of all the three benzene rings are delocalized. Also, the central carbon atom has the vacant p- orbital, and if we move that pi bonds, the aromaticity of the phenyl groups allow the ion to go for many possible structures.
Complete step by step answer:
The most stable carbonium ion is ${({C_6}{H_5})_3}{C^ + }$ [triphenyl methyl carbonium ion]. In this ion, the positive charge on the carbon atom of this ion is distributed uniformly over a number of structures. The pi-electrons of all the three benzene rings are delocalized and the central carbon atom has the vacant p- orbital. If we move that pi bond, the aromaticity of the phenyl groups allow the ion to go for many possible structures. This further allows the ion to have more potential for the delocalization of positive charge. So, we can say that it is resonance stabilized.
Therefore the ratio is one and the correct answer is option C.
Additional information: Resonance stabilization decreases the energy of the carbocation. In the resonance stabilization process, hydrogen gets added to the carbon in the double bond, which produces the delocalization of electrons. Always remember, resonance allows for delocalization and the overall energy of the molecule is decreased since its electrons occupy a greater volume. It is obvious that the molecules, which experience resonance, are more stable than those, which do not.
Note:
Comparing the ${(C{H_3})_3}{C^ + }$ [tert-butyl carbocation] with ${({C_6}{H_5})_3}{C^ + }$, [triphenyl methyl carbonium ion] we can see that the positive charge on the carbon atom of tert-butyl carbocation leaves not a single place to move and due to this becomes a nucleophile magnet. The only thing which defines its stability is that it is the tertiary carbocation and is stabilized by the hyper conjugation, which is weaker in terms of stability than resonance.
Complete step by step answer:
The most stable carbonium ion is ${({C_6}{H_5})_3}{C^ + }$ [triphenyl methyl carbonium ion]. In this ion, the positive charge on the carbon atom of this ion is distributed uniformly over a number of structures. The pi-electrons of all the three benzene rings are delocalized and the central carbon atom has the vacant p- orbital. If we move that pi bond, the aromaticity of the phenyl groups allow the ion to go for many possible structures. This further allows the ion to have more potential for the delocalization of positive charge. So, we can say that it is resonance stabilized.
Therefore the ratio is one and the correct answer is option C.
Additional information: Resonance stabilization decreases the energy of the carbocation. In the resonance stabilization process, hydrogen gets added to the carbon in the double bond, which produces the delocalization of electrons. Always remember, resonance allows for delocalization and the overall energy of the molecule is decreased since its electrons occupy a greater volume. It is obvious that the molecules, which experience resonance, are more stable than those, which do not.
Note:
Comparing the ${(C{H_3})_3}{C^ + }$ [tert-butyl carbocation] with ${({C_6}{H_5})_3}{C^ + }$, [triphenyl methyl carbonium ion] we can see that the positive charge on the carbon atom of tert-butyl carbocation leaves not a single place to move and due to this becomes a nucleophile magnet. The only thing which defines its stability is that it is the tertiary carbocation and is stabilized by the hyper conjugation, which is weaker in terms of stability than resonance.
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