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Question
Arrange the following compounds in increasing order of reactivity towards SN2 reaction.
2-Bromopentane, 1-Bromopentane, 2-Bromo-2-methylbutane
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Solution
SN2 reaction involves the formation of transition state. Therefore, reactivity towards SN2 depends upon the steric hindrance:
Due to the presence of 2 methyl group in 2-Bromo-2-methylbutane it is least reactive towards SN2 reactions, 2-bromopentane is moderately reactive while 1-bromopentane is most reactive.
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
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RELATED QUESTIONS
Arrange the compounds of the following set in order of reactivity towards SN2 displacement:
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\[\begin{array}{cc}
\ce{CH3}\phantom{..}\\
|\phantom{....}\\
\ce{CH3-CH-CH2Cl}
\end{array}\]
(iv)
\[\begin{array}{cc}
\phantom{..}\ce{H}\\
\phantom{..}|\\
\ce{CH3-C-Cl}\\
\phantom{..}|\\
\phantom{.....}\ce{C2H5}
\end{array}\]
Which of the following alkyl halides will undergo SN1 reaction most readily?
Read the passage given below and answer the following question:
Nucleophilic substitution reaction of haloalkane can be conducted according to both SN1 and SN2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent.
Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is, the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I− < Br− < Cl− < F− and the order of their leaving tendency should be I− > Br− > Cl− > F−. Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on SN1 mechanism. If the leaving group is not easy to leave, the reaction is based on SN2 a mechanism.
Influences of solvent polarity: In SN1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In SN2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (SN1) of tertiary chlorobutane in 25℃ water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (SN2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both SN1 and SN2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for SN2 reaction, while strong polar solvent is favorable for SN1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on SN1 mechanism in solvents with a strong polarity (for example, ethanol containing water).
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Complete the reaction with the main product formed:
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
\[\begin{array}{cc}
\ce{CH3CHCH2CH2Br}\\
|\phantom{.............}\\
\ce{CH3}\phantom{..........}\\
\end{array}\] or \[\begin{array}{cc}
\ce{CH3CH2CHCH2Br}\\
|\\
\phantom{...}\ce{CH3}\\
\end{array}\]
