In Halogen–metal exchange topic from Organic chemistry by Clayden and others There are few reaction like Ph-Br+Bu-Li -> Ph-Li+ Bu-Br. Why Ph-Br and LiBr are not produced? If so, the products would not be much stable?
Dear Mahmudul Hasan and Surabhi Gupta this reaction depends very much on the reaction conditions. When it is carried out in inert solvents at low temperature (as described in my previous answer), then it is kinetically controlled, because the phenyllithium precipitates. At room temperature, the thermodynamically favored (more stable) products n-butylbenzene and lithium bromide will form
Dear Mahmudul Hasan a key to understand this halogen-metal exchange is that this reaction is normally carried out in aliphatic hydrocarbon solvent such as n-pentane at low temperatures (e.g. –78 °C). Under these conditions the phenyllithium precipitates and is thus constantly removed from the equilibrium.
I think, at the same time butylbromide is also forming which is more stable than PhLi. After getting the more stability, the nucleophilic attach of butylbromide on Ph-ring (aromatic ring) doesn't supports or can say the rate of reaction will be very less to complete the substitution.
Dear Mahmudul Hasan and Surabhi Gupta this reaction depends very much on the reaction conditions. When it is carried out in inert solvents at low temperature (as described in my previous answer), then it is kinetically controlled, because the phenyllithium precipitates. At room temperature, the thermodynamically favored (more stable) products n-butylbenzene and lithium bromide will form
Dear Prof. Edelmann, Greetings!! It's very nice to hear the answer of this question from you in very simple as well as systematic manner. Thanks!!
I think also, that here we have a kinetic and thermodynamic controls of the reaction.
Dear Mahmudul Hasan i think this reaction depends on stability factor (on halogen-metal exchange) and the halogen metal exchange always goes through stability factor. as compare to butyl lithium, phenyl lithium is more stable and less reactive(pka 43) than butyl lithium. and I appreciate as prof. Frank T. Edelmann said reaction is always dependent upon the different parameters.
Phenyllithium and Bromobutane are produced by the reaction between Butyllithium and Bromobenzene , this is because of well-known metal-halogen exchange reaction . Bromine is better leaving group displaced by phenyl group from bromobutane.
Halogen-lithium exchange is not always done in solvents like pentane where one of the components could be insoluble. In synthetic organic chemistry, much more common is THF, which is capable of solubilizing both lithium species.
In general, it is an equilibrium reaction, and usually quite a fast one, driven by the difference in pKa: 43 for phenyl vs ~50-51 for normal alkyl and 53 for tert-butyl (that's why it is tert-BuLi that is used for generating n-alkyllithium species from the corresponding iodides).
You can call it "stability" of the lithium species if you wish, but this term is too general and in this case, in my opinion, misleading.
Dear Mahmudul Hasan and Vladimir Khlebnikov the use of inert solvents such as n-pentane is particularly advisable when it comes to isolate the pure phenyllithium derivative as a solid. The method can also be employed for ring-substituted phenyllithium derivatives, e.g. p-anisyllithium from p-iodoanisol. THF can be used when the phenyllithium species just needs to be generated and used in situ for further reactions.
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