Arrange the following in order of increasing property as indicated :
(i) CH3F, CH3Cl, CH3Br, CH3I →reactivity towards nucleophilic substitution reaction.
(ii) (CH3)3CCl, (CH3)2CHCl, CH3CH2Cl →reactivity towards SN1 reaction.
(iii) CH3Cl, CH3CH2Br, (CH3)2CHBr →boiling point.
The correct answer and explanation is :
Answers:
- Reactivity towards nucleophilic substitution reaction (SN2) (Increasing Order):
CH₃F < CH₃Cl < CH₃Br < CH₃I - Reactivity towards SN1 reaction (Increasing Order):
CH₃CH₂Cl < (CH₃)₂CHCl < (CH₃)₃CCl - Boiling Point (Increasing Order):
CH₃Cl < CH₃CH₂Br < (CH₃)₂CHBr
Explanation:
(i) Reactivity towards Nucleophilic Substitution (SN2)
The SN2 mechanism depends on the leaving ability of the halide and steric hindrance. Among the given molecules (CH₃F, CH₃Cl, CH₃Br, CH₃I), the carbon-halogen bond strength decreases as we move down the group (F > Cl > Br > I). This means iodine is the best leaving group due to its larger size and weaker bond with carbon. Fluorine, on the other hand, forms the strongest bond, making CH₃F the least reactive in SN2 reactions.
Thus, the reactivity order is:
CH₃F < CH₃Cl < CH₃Br < CH₃I
(ii) Reactivity towards SN1 Reaction
The SN1 mechanism is favored by the stability of the carbocation intermediate. The order of carbocation stability is:
Tertiary (3°) > Secondary (2°) > Primary (1°)
Thus, the reactivity order is:
- CH₃CH₂Cl (1°) < (CH₃)₂CHCl (2°) < (CH₃)₃CCl (3°)
This follows the stability trend of their respective carbocations.
(iii) Boiling Point Order
Boiling points depend on molecular weight and intermolecular forces (dipole-dipole and van der Waals forces). As we move from CH₃Cl to CH₃CH₂Br to (CH₃)₂CHBr, the molecular weight increases, leading to stronger van der Waals forces. Thus, the boiling point increases as follows:
CH₃Cl < CH₃CH₂Br < (CH₃)₂CHBr