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Elimination reaction of 1-bromo-2-phenylcyclopentane
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Ole Henry Norback
Elimination reaction of 1-bromo-2-phenylcyclopentane
There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.
$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.
$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.
Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.
As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.
There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.
$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.
$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.
Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.
As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.
Well, if there is a carbocation formed, and there is a rearrangement, then, the charge will be shifted to the attaching point of the benzene and cyclopentane ring, as it gives maximum stability(tertiary and also resonance). Then the products will also change.More
The carbocation formed by releasing $\ce{Br^-}$ ion will be a secondary carbocation. So how can we say that it will not be a stable one, as the relative order of stability is tertiary>secondary>primary (ignoring resonance effects, if any, of the benzene ring)?More
@ShoubhikRajMaiti Assume only tertiary and benzylic (maybe allylic) carbocations to be stable enough to be formed. Everything else would immediately undergo Wagner-Meerwein rearrangements.More
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The product structure is wrong, the Br should be gone.
There's no carbokation here, this is a reaction with E2-mechanism, which works best if the abstracted hydrogen is at a dihedral angle of 180° to the leaving group.
The product structure is wrong, the Br should be gone.
There's no carbokation here, this is a reaction with E2-mechanism, which works best if the abstracted hydrogen is at a dihedral angle of 180° to the leaving group.
Then the choice of product between 3-Phenylcyclopentene and 1-Phenylcyclopentene would depend upon the diastereomer, right? I was thinking about an E1-mechanism here...More
There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.
$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.
$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.
Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.
As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.
There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.
$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.
$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.
Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.
As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.
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The product structure is wrong, the Br should be gone.
There's no carbokation here, this is a reaction with E2-mechanism, which works best if the abstracted hydrogen is at a dihedral angle of 180° to the leaving group.
The product structure is wrong, the Br should be gone.
There's no carbokation here, this is a reaction with E2-mechanism, which works best if the abstracted hydrogen is at a dihedral angle of 180° to the leaving group.
More
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