Question

Which of the following compounds will react faster?

Neopentyl iodide or 1-iodo pentane with ${CH}_{3}OH$

• A. $neo$-pentyl iodide > $1$-iodo pentane
• B. $neo$-pentyl iodide < $1$-iodo pentane
• C. $neo$-pentyl iodide = $1$-iodo pentane
• D. None of these

Answer 1

The correct option is A $neo$-pentyl iodide > $1$-iodo pentane

$C{H}_{3}OH$ is a nucleophile and protic. The presence of iodide in neopentyl iodide makes the compound electronegative.

Iodide is more readily available for the reaction. The compound on reaction with methanol readily forms carbocations by the removal of iodide ion.

Hence the reactivity of neopentyl iodide towards methanol is faster.

Considering the 1-iodo pentane the iodine group is less hindered by the nucleophile.

Hence the correct option is A.