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Carbocation rearrangement involving three membered rings
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Mark B Fischer
Carbocation rearrangement involving three membered rings
Disclaimer: This is a very simplistic view of the cyclopropyl methyl cation and its reaction given above with $\ce{Br-}$, and is completely wrong. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.
How will the carbocation rearrange?
The following carbocation:
will not rearrange. It is a cyclopropyl methyl cation. This is undoubtedly the best method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.
The ring will not expand to form cyclobutane. No alkyl shift occurs. The $\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).
PS: The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail here (remark by @ron: "The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane "banana bonds" and the adjacent cationic center").
These two links (1) and (2) are for further reading regarding exceptional stability of cyclopropyl methyl cation.
Disclaimer: This is a very simplistic view of the cyclopropyl methyl cation and its reaction given above with $\ce{Br-}$, and is completely wrong. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.
How will the carbocation rearrange?
The following carbocation:
will not rearrange. It is a cyclopropyl methyl cation. This is undoubtedly the best method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.
The ring will not expand to form cyclobutane. No alkyl shift occurs. The $\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).
PS: The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail here (remark by @ron: "The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane "banana bonds" and the adjacent cationic center").
These two links (1) and (2) are for further reading regarding exceptional stability of cyclopropyl methyl cation.
Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.
It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.
In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.
Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]
Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.
It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.
In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.
Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]
Disclaimer: This is a very simplistic view of the cyclopropyl methyl cation and its reaction given above with $\ce{Br-}$, and is completely wrong. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.
The following carbocation:
will not rearrange. It is a cyclopropyl methyl cation. This is undoubtedly the best method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.
The ring will not expand to form cyclobutane. No alkyl shift occurs. The $\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).
PS: The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail here (remark by @ron: "The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane "banana bonds" and the adjacent cationic center").
These two links (1) and (2) are for further reading regarding exceptional stability of cyclopropyl methyl cation.
Disclaimer: This is a very simplistic view of the cyclopropyl methyl cation and its reaction given above with $\ce{Br-}$, and is completely wrong. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.
The following carbocation:
will not rearrange. It is a cyclopropyl methyl cation. This is undoubtedly the best method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.
The ring will not expand to form cyclobutane. No alkyl shift occurs. The $\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).
PS: The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail here (remark by @ron: "The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane "banana bonds" and the adjacent cationic center").
These two links (1) and (2) are for further reading regarding exceptional stability of cyclopropyl methyl cation.
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Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.
It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.
In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.
Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]
Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.
It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.
In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.
Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]
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