The reaction of CH_3-CH=CH_2 (propene) with H | vedclass

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Question

(A) The addition of $HI$ to propene follows Markovnikov's rule. In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation

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The reaction proceeds through a more stable carbocation.

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(A) The addition of $HI$ to propene follows Markovnikov's rule. In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation intermediate. Two possible carbocations can be formed: $1$. $CH_3-CH^+-CH_3$ (secondary carbocation,more stable) $2$. $CH_3-CH_2-CH_2^+$ (primary carbocation,less stable) Since the secondary carbocation is more stable due to the inductive effect and hyperconjugation,the reaction proceeds through this intermediate. Finally,the nucleophile $I^-$ attacks the secondary carbocation to form $CH_3-CH(I)-CH_3$ (isopropyl iodide).

The reaction of $CH_3-CH=CH_2$ (propene) with $HI$ gives isopropyl iodide instead of $n$-propyl iodide because:

Similar Questions

$CH_3-CH=CH_2$ $\xrightarrow{Br_2, hv (\text{low conc.})} \dots$ $\xrightarrow{Mg, \text{Dry ether}} \dots$ $\xrightarrow{CH_3-CO-CH_3, NH_4Cl} \dots$ $\xrightarrow{H^{+}, \Delta} (X)$. The end product $(X)$ of the above reaction is:

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Statement-$I$ :- Ozonolysis reaction is used to locate the position of a double bond in an alkene.
Statement-$II$ :- When $but-1-ene$ and $but-2-ene$ are subjected to ozonolysis separately,both yield the same product.
In the light of the above statements,select the correct option.

The stereochemistry of the product formed in the following reaction is:

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