CBSE 2023 · Region 5 · Set 1 · Q31 · 4 marks

Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ and $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ mechanisms. $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ is a two step reaction while $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ 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 $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ 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 $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ 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 $\displaystyle \left(\mathrm{S}_{\mathrm{N}} 1\right)$ of tertiary chlorobutane at $\displaystyle 25$ °C in water (dielectric constant $\displaystyle 79$) is $\displaystyle 300000$ times faster than in ethanol (dielectric constant $\displaystyle 24$). The reaction rate $\displaystyle \left(\mathrm{S}_{\mathrm{N}} 2\right)$ 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 $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ and $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ reaction, but with different results. Generally speaking weak polar solvent is favourable for $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ reaction, while strong polar solvent is favourable for $\displaystyle \mathrm{S}_{\mathrm{N}} 1$. Generally speaking the substitution reaction of tertiary haloalkane is based on $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ mechanism in solvents with a strong polarity (for example ethanol containing water). Answer the following questions:

(a)

Why racemisation occurs in $\displaystyle \mathrm{S}_{\mathrm{N}} 1$?

(b)

Why is ethanol less polar than water?

(c)

Which one of the following in each pair is more reactive towards $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ reaction?

(i)

$\displaystyle \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{I}$ or $\displaystyle \mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{Cl}$

(ii)

(c) Arrange the following in the increasing order of their reactivity towards $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ 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 $\displaystyle \mathbf{2} \boldsymbol{\times} \mathbf{1}$ 32. Rahul set-up an experiment to find resistance of aqueous $\displaystyle \mathrm{KC} l$ solution for different concentrations at $\displaystyle 298$ K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a. c. power in the audio frequency range $\displaystyle 550$ to $\displaystyle 5000$ cycles per second. Once the resistance was calculated from null point he also calculated the conductivity K and molar conductivity $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}$ and recorded his readings in tabular form.

S. No.	Conc. (M)	$\displaystyle \mathrm{k} \mathrm{S} \mathrm{cm}^{-1}$	$\displaystyle \mathbf{\Lambda}_{\mathbf{m}} \mathbf{S ~ c m}^{\mathbf{2}} \mathbf{~ m o l}^{\boldsymbol{-} \mathbf{1}}$
1.	$\displaystyle 1.00$	$\displaystyle 111.3 \times 10^{-3}$	$\displaystyle 111.3$
2.	$\displaystyle 0.10$	$\displaystyle 12.9 \times 10^{-3}$	$\displaystyle 129.0$
3.	$\displaystyle 0.01$	$\displaystyle 1.41 \times 10^{-3}$	$\displaystyle 141.0$

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.