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Arrange the compounds of each set-in order of reactivity towards $S{N^2}$ displacement.
A.2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
B.1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
C.1-Bromobutane, 1-Bromo-2, 2-dimethylpropane, D.1-Bromo-2-methylbutane,1-Bromo-3-methylbutane

Last updated date: 19th Jun 2024
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Hint:To answer this question, you should recall the concept of $S{N^2}$ reactions. These reactions involve the nucleophilic substitution reaction of the leaving group with a nucleophile in a given organic compound.

Complete Step by step solution:
We know that in $S{N^2}$ reactions, steric factors determine the reactivity. More reactive alkyl halides have less steric hindrance. Hence, the decreasing order of the reactivity of alkyl halides is \[{1^o} > {2^o} > {3^o}.\] Applying this rule to the given options we get:
A.1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methyl butane
B.1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane
C.1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2, D.2-dimethylpropane.

Note:This reaction proceeds through a backside attack by the nucleophile on the substrate. The nucleophile approaches the given substrate at an angle of\[{180^o}\] to the carbon-leaving group bond. Now, the leaving group is pushed out of the transition state on the opposite side of the carbon-nucleophile bond, forming the required product. It is important to note that the product is formed with an inversion of the tetrahedral geometry at the atom in the centre. You should remember the important points regarding $S{N^2}$ reactions:
1.$S{N^2}$ reactions are bimolecular reactions in which there are simultaneous bond-making and bond-breaking steps.
2.$S{N^2}$ reactions do not proceed via an intermediate.
3.$S{N^2}$ reactions result in inverted stereochemistry at the reaction centre.
4.Steric effects are particularly important in $S{N^2}$ reactions.
5.Unhindered back of the substrate makes the formation of carbon-nucleophile bonds easy. Therefore, methyl and primary substrates undergo nucleophilic substitution easily.
6.Strong anionic nucleophiles speed up the rate of the reaction. More negative charge more the nucleophilicity and a strong nucleophile can easily form the carbon-nucleophile bond.
7.Polar aprotic solvents do not hinder the nucleophile, but polar solvents form hydrogen bonds with the nucleophile. A good solvent for this reaction is acetone.
8.Stability of the anion of the leaving group and the weak bond strength of the leaving groups bond with carbon help increase the rate.