Question

Addition of $HI$ on the double bond of propene yields isopropyl iodide and not n-propyl iodide as the major product. This is because the addition proceeds through
A. A more stable carbonium ion
B. A more stable carbanion
C. A more stable free radical
D. None of the above being a concerted reaction

Answer 1

Hint: When hydrogen halide is added to the double bonds saturated alkane compounds are produced. It is an example of electrophilic addition reaction. The products that are formed by this process are mainly depending upon the stability of carbocations. The major product is formed from more stable carbocation intermediate.

Complete Step by Step Answer:
In this problem we have an alkene, propene reacts with hydrogen iodide an electrophilic addition reaction occurs and forms isopropyl iodide $2-iodopropane$as the major product. The overall reaction is followed by Markovnikov’s rule.

In the first step, the pi electron attacks the electrophile proton,${{H}^{+}}$ and the iodide atom leaves with a bonded electron pair. By this electrophilic addition, two intermediate carbocations are formed.

The stability of carbocation follows an order: Tertiary(${{3}^{O}}$)>Secondary(${{2}^{O}}$)>Primary(${{1}^{O}}$)
Here secondary carbocation is more stable than primary carbocation. Hence the major product is formed from ${{2}^{O}}$ carbocation.

In the next step nucleophile iodide ion (${{I}^{-}}$) attacks carbocation to form $2-iodopropane$ isopropyl iodide as the main product. Another form but as a minor product.

Therefore the addition of $HI$ the double bond of propene yields isopropyl iodide and not n-propyl iodide due to a more stable secondary carbenium or carbocation.
Thus, option (A) is correct.

Note: The overall regioselectivity will be changed if we use peroxide in the reaction mixture. Then anti-Markovnikov's rule is followed and n-propyl iodide is formed as the major product. A negative iodide ion adds to the carbon atom which is linked with a higher number of hydrogen atoms. This is known as the Peroxide effect.