Problem 186
Question
In \(\mathrm{S}_{\mathrm{N}}^{2}\) reactions, the correct order of reactivity for the following compounds: \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) is: \([2014]\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (c) \(\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)
Step-by-Step Solution
VerifiedKey Concepts
Steric Hindrance
Let's consider the compounds mentioned in our problem. Looking at the bulky groups attached to the carbon atom bonded to the chlorine:
- **Primary Alkyl Halides:** Compounds like \( \mathrm{CH}_{3}\ \mathrm{Cl} \) and \( \mathrm{CH}_{3}\ \mathrm{CH}_{2}\ \mathrm{Cl} \) have minimal steric hindrance since only one or no additional carbon is attached.
- **Secondary Alkyl Halides:** In \( (\mathrm{CH}_{3})_{2} \mathrm{CHCl} \) , the central carbon is bonded to two other carbon atoms adding significant steric hindrance.
- **Tertiary Alkyl Halides:** Complexes such as \( (\mathrm{CH}_{3})_{3} \mathrm{CCl} \) exhibit the highest steric hindrance with three carbon atoms crowding the reactive center, slowing the reaction the most.
Nucleophilic Substitution
A nucleophile is an electron-rich species that seeks out positive or electron-deficient areas in molecules to form bonds:
- In SN2 reactions, the nucleophile must directly attack the carbon atom bonded to the leaving group from the backside (opposite side of the leaving group).
- This backside attack leads to an inversion of configuration, sometimes known as the "Walden Inversion," reversing the stereochemistry of the molecule.
- The simultaneous bond-breaking and bond-forming make SN2 reactions a single-step mechanism without intermediates.
Alkyl Halides
The carbon-halogen bond is polarized because the halogen is more electronegative than carbon:
- This polarization creates a partial positive charge on the carbon, making it an ideal site for nucleophiles to attack in substitution reactions.
- Different types of alkyl halides (primary, secondary, tertiary) have different reactivity in SN2 reactions based on their structural steric hindrance.
- The leaving group ability, primarily dictated by the halogen, also influences reactivity. Better leaving groups aid faster reaction rates.