Question
Explain unimolecular nucleophilic substitution (SN1) reaction with mechanism.
✓
Answer
→ SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid etc.)
→ The reaction between tert- butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics,
→ The rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.
→ Rate = K[(CH3)3 C Br]
$
\left( CH _3\right)_3 CBr +{ }^{-} OH \longrightarrow\left( CH _3\right)_3 COH + Br{-}
$
2-Bromo-2-methylpropane 2-methylpropane-2-01
→ It occurs in two steps. In step I, the polarised C-Br bond undergoes slow cleavage to produce a carbocation and a bromide ion.
→ The carbocation formed is then attacked by nucleophile in step II to complete the substitution reaction.
→ Step I is the slowest and reversible. It involves the C-Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent.
→ Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.
→ Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3º alkyl halides undergo SN1 reaction very fast because of the high stability of 3º carbocations.
→ Reactivity order for SN1 reaction
→ 3º - halide 2º - halide > 1º - halide > CH3 - X
→ The reaction between tert- butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics,
→ The rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.
→ Rate = K[(CH3)3 C Br]
$
\left( CH _3\right)_3 CBr +{ }^{-} OH \longrightarrow\left( CH _3\right)_3 COH + Br{-}
$
2-Bromo-2-methylpropane 2-methylpropane-2-01
→ It occurs in two steps. In step I, the polarised C-Br bond undergoes slow cleavage to produce a carbocation and a bromide ion.
→ The carbocation formed is then attacked by nucleophile in step II to complete the substitution reaction.
→ Step I is the slowest and reversible. It involves the C-Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent.
→ Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.
→ Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3º alkyl halides undergo SN1 reaction very fast because of the high stability of 3º carbocations.
→ Reactivity order for SN1 reaction
→ 3º - halide 2º - halide > 1º - halide > CH3 - X
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