Respuesta :

This is an electrophilic addition of a hydrogen halide to an alkene.

The pi bond is attacked by the electrophilic H⁺ from the H⁺Cl⁻, resulting in the loss of the double bond and the formation of a tertiary carbocation on C2. A less stable carbocation intermediate with the positive charge on C3 is also formed.

The nucleophilic Cl⁻ then attacks the carbocations, forming the haloalkane products. The major product will be 2-chloro-2-methybutane, where the Cl is bonded to the most substituted C (Markovnikov's rule). The minor product, if you have to worry about it, will be 2-chloro-3-methylbutane. The major product has no stereoisomers to worry about as it's achiral.

The attached image shows the scheme of this reaction. Only the major product is shown.  

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