CBSE 2023 · Region 5 · Set 2 · Q32 · 4 marks
Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both $\displaystyle \mathrm{S_N}1$ and $\displaystyle \mathrm{S_N}2$ mechanisms. $\displaystyle \mathrm{S_N}1$ is a two step reaction while $\displaystyle \mathrm{S_N}2$ is a single step reaction. For any haloalkane which mechanism is followed depends on factors such as structure of haloalkane, properties of leaving group, nucleophilic reagent and solvent. Influences of solvent polarity : In $\displaystyle \mathrm{S_N}1$ reaction, the polarity of the system increases from the reactant to the transition state, because a polar solvent has a greater effect on the transition state than the reactant, thereby reducing activation energy and accelerating the reaction. In $\displaystyle \mathrm{S_N}2$ 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 ($\displaystyle \mathrm{S_N}1$) of tertiary chlorobutane at $\displaystyle 25$ $\displaystyle ^\circ$C in water (dielectric constant $\displaystyle 79$) is $\displaystyle 300000$ times faster than in ethanol (dielectric constant $\displaystyle 24$). The reaction rate ($\displaystyle \mathrm{S_N}2$) of $\displaystyle 2$-Bromopropane and NaOH in ethanol containing $\displaystyle 40$% water is twice slower than in absolute ethanol. Hence the level of solvent polarity has influence on both $\displaystyle \mathrm{S_N}1$ and $\displaystyle \mathrm{S_N}2$ reaction, but with different results. Generally speaking weak polar solvent is favourable for $\displaystyle \mathrm{S_N}2$ reaction, while strong polar solvent is favourable for $\displaystyle \mathrm{S_N}1$. Generally speaking the substitution reaction of tertiary haloalkane is based on $\displaystyle \mathrm{S_N}1$ mechanism in solvents with a strong polarity (for example ethanol containing water). Answer the following questions :(a)Why racemisation occurs in $\displaystyle \mathrm{S_N}1$ ?(b)Why is ethanol less polar than water ?Which one of the following in each pair is more reactive towards $\displaystyle \mathrm{S_N}2$ reaction ?(i)$\displaystyle \mathrm{CH_3-CH_2-I}$ or $\displaystyle \mathrm{CH_3CH_2-Cl}$(ii)chlorocyclohexane ($\displaystyle C_6H_{11}-Cl$) or (chloromethyl)cyclohexane ($\displaystyle C_6H_{11}-CH_2-Cl$) [structure images]Arrange the following in the increasing order of their reactivity towards $\displaystyle \mathrm{S_N}1$ reactions :(i)$\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 1$-Bromopentane, $\displaystyle 2$-Bromopentane(ii)$\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 3$-methylbutane
Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both $\displaystyle \mathrm{S_N}1$ and $\displaystyle \mathrm{S_N}2$ mechanisms. $\displaystyle \mathrm{S_N}1$ is a two step reaction while $\displaystyle \mathrm{S_N}2$ is a single step reaction. For any haloalkane which mechanism is followed depends on factors such as structure of haloalkane, properties of leaving group, nucleophilic reagent and solvent. Influences of solvent polarity : In $\displaystyle \mathrm{S_N}1$ reaction, the polarity of the system increases from the reactant to the transition state, because a polar solvent has a greater effect on the transition state than the reactant, thereby reducing activation energy and accelerating the reaction. In $\displaystyle \mathrm{S_N}2$ 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 ($\displaystyle \mathrm{S_N}1$) of tertiary chlorobutane at $\displaystyle 25$ $\displaystyle ^\circ$C in water (dielectric constant $\displaystyle 79$) is $\displaystyle 300000$ times faster than in ethanol (dielectric constant $\displaystyle 24$). The reaction rate ($\displaystyle \mathrm{S_N}2$) of $\displaystyle 2$-Bromopropane and NaOH in ethanol containing $\displaystyle 40$% water is twice slower than in absolute ethanol. Hence the level of solvent polarity has influence on both $\displaystyle \mathrm{S_N}1$ and $\displaystyle \mathrm{S_N}2$ reaction, but with different results. Generally speaking weak polar solvent is favourable for $\displaystyle \mathrm{S_N}2$ reaction, while strong polar solvent is favourable for $\displaystyle \mathrm{S_N}1$. Generally speaking the substitution reaction of tertiary haloalkane is based on $\displaystyle \mathrm{S_N}1$ mechanism in solvents with a strong polarity (for example ethanol containing water). Answer the following questions :
(a)
Why racemisation occurs in $\displaystyle \mathrm{S_N}1$ ?
(b)
Why is ethanol less polar than water ?
Which one of the following in each pair is more reactive towards $\displaystyle \mathrm{S_N}2$ reaction ?
(i)
$\displaystyle \mathrm{CH_3-CH_2-I}$ or $\displaystyle \mathrm{CH_3CH_2-Cl}$
(ii)
chlorocyclohexane ($\displaystyle C_6H_{11}-Cl$) or (chloromethyl)cyclohexane ($\displaystyle C_6H_{11}-CH_2-Cl$) [structure images]
Arrange the following in the increasing order of their reactivity towards $\displaystyle \mathrm{S_N}1$ reactions :
(i)
$\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 1$-Bromopentane, $\displaystyle 2$-Bromopentane
(ii)
$\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 3$-methylbutane
Marking-scheme solution
(a) Because of the formation of planar carbocation \(\displaystyle / \mathrm{sp}^{2}\) hybridized carbocation.
(b) Due to the +I effect / electron-releasing nature of the ethyl group in ethanol.
(c) (i) \(\displaystyle \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}\)
(ii)
Haloalkanes and HaloarenesChemical Reactions of Haloalkanes and HaloarenesAnalysecase_studymedium
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.