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The only alkene which gives primary alkyl halides on hydrohalogenation is:
A.${C_2}{H_4}$
B.${C_3}{H_6}$
C.${C_4}{H_8}$
D.${C_5}{H_{10}}$

Answer
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Hint:Hydrohalogenation is the process of addition of a hydrogen halide across a double bond. The reaction can occur in both symmetrical and unsymmetrical alkenes and results in the breaking of the $\pi $ bond of the alkene. The addition takes place according to Markovnikov's rule.

Complete answer:
Alkenes are more reactive as compared to alkynes and undergo electrophilic addition reactions readily due to the $\pi $electron density of the double bond. An incoming electrophile attacks the carbon atom breaking the double bond.
The attacking electrophile breaks the bond and always attaches to the carbon atom at the end of the chain in case of terminal alkenes. As the bond breaks, a vacancy is created and a carbocation is formed. The carbocation is then attacked by a nucleophile, a compound without a degree of saturation is formed.
In hydrohalogenation reaction, the hydrogen halide breaks into a hydrogen cation and halide anion by the heterolytic cleavage. The hydrogen cation attaches to the carbon chain forming a carbocation. The carbocation formed then attacks the halide anion and an alkyl halide is formed.
This addition reaction takes place according to Markovnikov's rule. The hydrogen cation preferably attacks the terminal carbon of the alkene so that the carbocation formed is more substituted and hence stable. The incoming halide ion attaches to the inner carbon and thus a secondary halide is formed.
However, since ethene is a symmetrical alkene and has only two carbon atoms, the addition of hydrogen halide will give a primary halide as a product.
${C_2}{H_4}\xrightarrow[{}]{{HBr}}C{H_3} - C{H_2} - Br$

Thus, the correct answer is A.

Note:
Hydrohalogenation can also yield primary alkyl halides if performed in the presence of peroxide. This effect is known as Anti- Markovnikov effect or the peroxide effect. The reaction proceeds through a free radical addition mechanism.