Arrange the compounds of each set in order of reactivity towards $S_N2$ displacement: $1-$Bromo$-3-$methylbutane, $2-$Bromo$-2-$methylbutane, $3-$Bromo$-2-$methylbutane.
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$\ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \\\ \text{CH}_3\text{CH}\text{CH}_2\text{CH}_2\text{Br}\\1-\text{Bromo}-3-\text{methylbutane}(1^\circ)$
$\ \ \ \ \ \ \ \ \ \text{Br}\\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \\\ \text{CH}_3\text{CCH}_2\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ | \\\ \ \ \ \ \ \ \ \text{CH}_3\\2-\text{Bromo}-2-\text{methylbutane}(3^\circ)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \\\ \text{CH}_3\text{CHCH}\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ | \\\ \ \ \ \ \ \ \ \text{CH}_3\\2-\text{Bromo}-3-\text{methylbutane}(2^\circ)$
Since steric hindrance in alkyl halides increases in the order of $1^\circ < 2^\circ < 3^\circ,$ the increasing order of reactivity towards $S_N2$ displacement is $3^\circ < 2^\circ < 1^\circ .$
Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards $S_N2$ displacement as:
$2-$Bromo$-2-$methylbutane $< 2-$Bromo$-3-$methylbutane $< 1-$Bromo$-3-$methylbutane $[2-$Bromo$-3-$methylbutane is incorrectly given in $\ce{NCERT}]$
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A hydrocarbon $\ce{C_5H_{10}}$ does not react with chlorine in dark but gives a single monochloro compound $\ce{C_5H_9Cl}$ in bright sunlight. Identify the hydrocarbon.