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Pent-1-ene reacts with hydrogen bromide to produce two different haloalkane products. Identify the two products and use the mechanism of reaction to explain which of the two compounds is formed as the major product.

User Unexist
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Pent-1-ene reacts with hydrogen bromide as follows:


\begin{gathered} CH_3-CH_2-CH_2-CH=CH_2+HBr\text{ }\rightarrow \\ CH_3-CH_2-CH_2-CHBr-CH_3\text{ +} \\ CH_3-CH_2-CH_2-CH_2-CH_2Br \end{gathered}

The two products fromed are:

2-bromopentane and 1-bromopentane respectively.

The major product that is formed in the reaction is 2-bromopentane and this can be explained because the carbocation that is created during the reaction is more stable than the one that is form during the reaction that produces 1-bromopentane.

The reaction that formes 2-bromopentane follows Markovnikov's Rule. This rule sais:

When a compound HX (in this case HBr) is added to an unsymmetrical alkane (in this case pent-1-ene), the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.

The mechanism of the reaction that forms 2-bromopentane is the following, and it is called an electrophilic addition.

Step 1: Pent-1-ene is protonated and it gives rise to the more stable secondary carbocation.


CH_3-CH_2-CH_2-CH^+^{}_{}_{}-CH_3+Br^-

Step 2: The bromide ion reacts with the carbocation forming 2-bromopentane.


CH_3-CH_2-CH_2-CHBr-CH_3

The other carbocation that can be formed is the primary carbocation:


CH_3-CH_2-CH_2-CH_2-CH^+_2+Br^-

That later reacts with the bromide ion to form 1-bromopentane. However, as the secondary carbocation is more stable that the primary carbocation the major product of the reaction is 2-bromopentane.

User GodsBoss
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