When 1,3-butadiene is treated with HBr in the absence of air, a reaction occurs by an ionic mechanism to form both the 1,2- and 1,4-addition products. However, the proportions in
which these two products are obtained are markedly affected by the temperature at which the reaction is carried out.
When the reaction is carried out at low temperature (–80°C), a mixture containing 80 percent of the 1,2-product and 20 percent of the 1,4-product is obtained, i.e., the major product is formed by 1,2-addition.
|
2 2 2 3 2 3
80 C
Br
CH CH — CH CH HBr CH CH CH CH BrCH CH CHCH
1, 3-Butadiene 3-Bromobut-1-ene(80%) 1-Bromobut-2-ene(20%) (1, 2-product) (1, 4-product)
== == + ổổổổ- ∞ ặ == — + ==
Mechanism: The reaction proceeds through the following two steps:
Step 1: In this step (which is the same whether the 1,2-product or the 1,4-product is being formed), the addition of H! takes place at C-1 (but not at C-2 which give a much less stable 1° carbocation) of 1,3-butadien because the resulting carbocation is a relatively more stable (resonance-stabilized) allylic cation. It is the rate-determining step of the reaction.
Step 2: This is the product-determining step. The allylic cation is, in fact, a resonance hybrid of a secondary allylic cation (I) and a primary allylic cation (II). Because a secondary allylic cation is more stable than a primary allylic cation, the contribution of I to the hybrid is greater than that of II. Because of this, in the allylic cation, the interior carbon becomes relatively more positive than the terminal carbon and, therefore, the activation energy for the formation of 1,2-product is less than that for the formation of 1,4-product.
As a consequence, the bromide ion attacks C-2 (the more positive carbon) more readily than C-4 (the less positive carbon) to give the 1,2-addtion product, 3-bromobut-1-ene, predominantly. Under these conditions, the reaction is said to be under kinetic (or rate) control because the product formed more rapidly predominates in the product mixture. The predominant product is called the kinetic product. The 1,4-addition product 1-bromobut- 2-ene, however, is relatively more stable than the 1,2-addition product 3-bromobut-1-ene and it is called the thermodynamic product.
Energy profile diagram: This behaviour of 1,3-butadiene and hydrogen bromide can be more fully understood if we examine the energy profile diagram for the reaction shown in the following figure.
The energy of activation leading to the 1,2-product is less than that leading to the 1,4- product. As a result, the 1,2-product is formed at a rate faster than the 1,4-product and predominates at lower temperature. Since an internal double bond is more stable than a terminal double bond because of hyperconjugative effect, therefore, the 1,4-product is thermodynamically more stable than the 1,2-product and so, its valley is placed at a lower level than that of the 1,2-product. The product formation is irreversible because lower temperature does not provide sufficient energy to lift either product out of its deep potential energy valley.
At higher temperature (40°C), the intermediate ions have sufficient energy to cross both the barriers with relative case. Sufficient energy is also available for ionization of the products. Therefore, both the reactions are now reversible. Since the 1,4-product is thermodynamically more stable, its activation energy for ionization is much greater than that of the 1,2-product and consequently, the 1,4-product ionizes more slowly than the 1,2-
product. Under these conditions, the 1,2-product, which is still formed rapidly, rearranges to the more stable 1,4-product through the allyl cation because the overall change is energetically favourable. So, when an equilibrium is established between the products, the more stable 1,4-product predominates in the product mixture.
Under these conditions, the reaction is said to be under thermodynamic (or equilibrium) control because by equilibration the more stable product predominates. The predominant product is called thermodynamic (or equilibrium) product.
1. Which out of 2-hexene and 3-hexene is expected to be more suitable for using as the substrate for synthesizing 3-bromohexane(CH3CH2CHBrCH2 CH2CH3)? Give your reasoning.
Solution Being an unsymmetrical alkene 2-hexene (CH3CH==CHCH2CH2CH3) reacts with HBr to form a mixture of 2-bromohexane (50%) and 3-bromohexane (~50%) and 3-bromohexane (~50%) and this is because the reaction proceeds through the formation of two carbocations. However, the symmetrical alkene 3-hexene (CH3CH2CH==CHCH2CH3) reacts with HBr to form 3-bromohexane exclusively because the reaction proceeds through the formation of a single carbocation. Therefore, 3-hexene is more suitable as the substrate for synthesizing 3-bromohexane.
2. The addition of HBr to which of the following alkenes is more regioselective and why?
==
CH2 CH3
Methylenecyclohexane 1-Methylcyclohexene
Solution When H! adds to methylenecyclohexane, one of the carbocations that could be formed is tertiary and the other is primary. A tertiary carbocation is very much stable than a primary carbocation, which, in fact, is too unstable to be formed. Thus, 1-bromo-1- methylcyclohexane will be only product of this reaction.
On the other hand, when H! adds to 1-methylcyclohexene, one carbocation that could be formed is tertiary and the other is secondary. Since a tertiary carbocation and a secondary carbocation are closer in stability, therefore, both 1-bromo-1-methylcyclohexane and 1-bromo-2-methylcyclohexane will be formed as the major product and the minor product, respectively.
Hence, the addition of HBr to methylenecyclohexane is relatively more regioselective.
3. The addition of HCl to 3,3-dimethylbut-1-ene leads to the formation of an unexpected product, 2-chloro-2,3-dimethybutane, in somewhat greater yield than 3-chloro-2,2-dimethylbutane, the expected Markownikoff product.
== ổổổặ ==
3 3 2 HCl 3 3 3 3 2 3 2
(CH ) C — CH CH (CH ) CCHCl CH (CH ) CClCH (CH ) 3,3-Dimethylbut-1-ene 3-Chloro-2,2-dimethylbu tane 2-Chloro-2,3-dimethylbutane
(minor product) (major product)
Explain this observation.
Solution The reaction starts like a normal addition of HCl, that is, by protonation of the double bond to yield the more stable carbocation with the greater number of alkyl groups at the electron-deficient carbon. Reaction of this carbocation with Cl* occurs as expected to yield the expected Markownikoff product as the minor product.
This intermediate secondary (2∞) carbocation undergoes rearrangement (the methyl group moves with its pair of bonding electrons from the carbon adjacent to the positive carbon) to give a relatively more stable tertiary carbocation. The major product is formed by the attack of Cl* on the new carbocation.
4. When HCl is allowed to react with 3-methylcyclohexene, 1-chloro-3- methylcyclohexane and 1-chloro-1-methylcyclohexane are obtained.
Draw a mechanism to explain this result.
Solution The addition of H! to 3-methylcyclohexene forms two different secondary carbocations. However, one of them undergoes rearrangement by a hydride shift to yield a more stable tertiary carbocation. Nucleophilic attack of Cl* on the secondary and the tertiary carbocations produces 1-chloro-3-methylcyclohexane and 1-chloro-1- methylcyclohexane, respectively.
5. Explain the mechanism of the following reaction:
Solution When H! adds to a-pinene, a stable 3∞ carbocation is formed predominantly.
But this 3∞ carbocation readily rearranges to a less stable 2∞ carbocation which is normally not favourable. In fact, the rearrangement expands a strained four-membered ring to a much less-strained five-membered ring, and this relief of strain provides a driving force for the rearrangement. Nucleophilic attack of Cl* on the 2∞ carbocation forms bornyl chloride.
6. Methylenecyclobutane react with hydrogen chloride to give chlorometh- ylcyclobutane (the anti-Markownikoff product), while methylenecyclo- hexane reacts with hydrogen chloride to yield 1-chloro-1-methylcyclo- hexane (the Markownikoff product). Account for these observations.
Solution When H! adds to methylenecyclobutane, one of the carbocations that could be formed is primary (I) and the other is tertiary (II). The 3∞ carbocation II suffers from considerable angle strain (120∞ – 90∞ = 30∞), whereas the 1∞ carbocation has small angle strain (109.5∞ – 90∞ = 19.5∞). Because of greater angle strain, the carbocation II, although a 3∞ one, is less stable than the 1∞ carbocation I and the addition of HCl proceeds through the carbocation I to give anti-Markownikoff product chloromethylcyclobutane.
Since a primary carbocation is too unstable to be formed, the anti-Markownikoff product is probably obtained through a concerted addition process.
When H! adds to methylenecyclohexane, one of the carbocations that could be formed is tertiary (III) and the other is primary (IV). The carbocation III suffers from a very little angle strain. Then, from electronic point of view the 3∞ cabocation III is relatively more stable than the 1∞ carbocation IV and therefore, the reaction proceeds through the carbocation III to give the Markownikoff product 1-chloro-1-methylcyclohexane.
7. Predict the major product in each of the following reactions and give your reasoning.
(a) CH2== ==C O+HCl ổổặ (b) CH C3 ∫∫CH+2HBrổổặ
(c) (d)
Solution
(a) Addition of H! to ketene yields a resonance-stabilized acylium ion instead of a primary carbocation which is further destabilized by electron-with drawing inductive effect (–I) of the –CHO group.
(b) The addition of H! to the two sp carbons of propyne may produce one secondary vinylic cation (I) and one primary vinylic cation (II). Since the secondary vinylic cation is relatively more stable (stabilized by the +I and hyperconjugative effect of the methyl group) than the primary vinylic cation, therefore, the reaction actually proceeds through the secondary vinylic cation which subsequently reacts with Br* to yield 3-bromopropene.
2-Bromopropene adds another molecule of HBr to form 2,2-dibromopropane as follows:
Addition of H! to the C-1 carbon of 2-bromopropene leads to the formation of the carbocation III in which bromine withdraws electrons by its –I effect and donates electron by its +R effect. Since –I 〉 + R, bromine is net electron-withdrawing and it destabilizes the carbocation. But the +I and hyperconjugative stabilizing effect of the two methyl groups is large enough to compensate this destabilization and in fact, it is more stable than the 1∞ carbocation IV (that could be formed by the addition of H! to C-2) which is further destabilized by the –I effect of bromine. The reaction thus proceeds through the carbocation III to give 2,2-dibromopropane.
(c) When H! adds to styrene (PhCH == CH2), one of the carbocation that could be formed is secondary benzylic and the other is primary. The secondary benzylic carbocation is much more stable (stabilized by resonance) than the primary carbocation and for this reason, the reaction proceeds through the secondary benzylic carbocation to give PhCHBrCH3 nearly exclusively.
(d) Since one mole of HCl is allowed to react with this unsymmetrical diene, it adds preferentially to the more reactive or more electron-rich C-1—C-2 double bond to give 5-methylhex-1-ene (CH2==CHCH2CH2CClMe2) as the major product through the formation of a stable 3° carbocation. HCl also adds to the C-5—C-6 double bond to give 5-chloro-2-methylhex-1-ene as the minor product through the formation of a less stable 2° carbocation.
8. Explain the following observation:
CH3