Difficult
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
- DNone of the above being a concerted reaction
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Similar Questions
The correct order of the addition reaction rates of halogen acids with ethylene is
What are $X$ and $Y$ in the following reaction?
$Pent-2-ene \xrightarrow[(ii) Zn / H_2O]{(i) O_3} X + Y$
$Pent-2-ene \xrightarrow[(ii) Zn / H_2O]{(i) O_3} X + Y$
What is the product formed when propene reacts with $HCl$ in the presence of peroxide?
Medium
View Solution$1, 2$-dibromocyclohexane on dehydrohalogenation gives
Medium
View SolutionIdentify the correct statements with respect to $cis/trans$-but$-2$-ene from the following:
$I$. $cis$-but$-2$-ene is more polar than $trans$-but$-2$-ene
$II$. Melting point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
$III$. Boiling point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
Correct answer is
$I$. $cis$-but$-2$-ene is more polar than $trans$-but$-2$-ene
$II$. Melting point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
$III$. Boiling point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
Correct answer is
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