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maverick280857
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HI
This problem follows from a discussion in Morrison and Boyd (6th ed) Page 251.
"Our second example involves 1-chloro-2-propanol. Although technically a secondary alcohol, it reacts with hydrogen halides "abnormally" slowly, and at about the rate of a primary alcohol. This time we are dealing, not with a steric effect but with a polar effect. The rate of an [tex]S_{N1}[/tex] reaction, we have seen, depends upon the stability of the carbocation being formed. Let us compare, then, the 1-chloro-2-propyl cation with a simple secondary cation, the isopropyl caton, say. Electronegative chlorine has an electron-withdrawing inductive effect. As we have seen this intensifies the positive charge on the electron deficient carbon and makes the carbocation less stable. This same electron withdrawl destabilizes the incipient cation in the transition state, raises [tex]E_{act}[/tex], and slows down the reaction." (Quoted from the book)
Now if you draw the 1-chloro-2-propyl cation, you will see that there's a +ve charge on a carbon adjacent to the carbon bonded to the chlorine group (which in this state is indeed exerting a -I effect). But if I now perform a hydride shift from the adjacent carbon (the one bonded to chlorine) to this electron deficient carbon, I will get the 1-chloro-1-propyl cation, which I claim to be more stable than the precussor because now the lone pair of the chlorine can donate its electrons to the electron deficient carbon atom to give a new resonance form which will be substantially more stable than either cation we started out with. So this way, the reaction with 1-chloro-2-propanol should be faster than that with say isopropyl chloride.
One might argue that the reactivity depends on the initially formed (non-rearranged) cation in which case my arguments will fail, but a hydride shift is still possible and to me, nothing seems to preclude it. (I am not against experimental evidence anyway, but my reasoning follows from my organic chemistry lessons which tends to suggest just this).
Cheers
Vivek
This problem follows from a discussion in Morrison and Boyd (6th ed) Page 251.
"Our second example involves 1-chloro-2-propanol. Although technically a secondary alcohol, it reacts with hydrogen halides "abnormally" slowly, and at about the rate of a primary alcohol. This time we are dealing, not with a steric effect but with a polar effect. The rate of an [tex]S_{N1}[/tex] reaction, we have seen, depends upon the stability of the carbocation being formed. Let us compare, then, the 1-chloro-2-propyl cation with a simple secondary cation, the isopropyl caton, say. Electronegative chlorine has an electron-withdrawing inductive effect. As we have seen this intensifies the positive charge on the electron deficient carbon and makes the carbocation less stable. This same electron withdrawl destabilizes the incipient cation in the transition state, raises [tex]E_{act}[/tex], and slows down the reaction." (Quoted from the book)
Now if you draw the 1-chloro-2-propyl cation, you will see that there's a +ve charge on a carbon adjacent to the carbon bonded to the chlorine group (which in this state is indeed exerting a -I effect). But if I now perform a hydride shift from the adjacent carbon (the one bonded to chlorine) to this electron deficient carbon, I will get the 1-chloro-1-propyl cation, which I claim to be more stable than the precussor because now the lone pair of the chlorine can donate its electrons to the electron deficient carbon atom to give a new resonance form which will be substantially more stable than either cation we started out with. So this way, the reaction with 1-chloro-2-propanol should be faster than that with say isopropyl chloride.
One might argue that the reactivity depends on the initially formed (non-rearranged) cation in which case my arguments will fail, but a hydride shift is still possible and to me, nothing seems to preclude it. (I am not against experimental evidence anyway, but my reasoning follows from my organic chemistry lessons which tends to suggest just this).
Cheers
Vivek