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Hot answers tagged c-c-addition

14

For the hydroboration of isobutene there is a steric aspect: And though sterics may play some part, it doesn't explain near 100% regioselectivity. More important in determining regiochemistry, I would say, are the electronics of the transition state: Boron is electrophilic, and as such will draw the electron density of the alkene π-bond towards itself. ...

11

From ChemGuide: Using bromine water as a test for alkenes If you shake an alkene with bromine water (or bubble a gaseous alkene through bromine water), the solution becomes colourless. Alkenes decolourise bromine water. The chemistry of the test This is complicated by the fact that the major product isn't 1,2-dibromoethane. The water ...

10

With bromination, the first step is the formation of an adduct complex, which can be explained that the π-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ...

10

The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test. However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture....

10

Key The reaction is run in heptane, a non-polar solvent. A non-polar solvent will not stabilize or support ions as well as a polar solvent. Background When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed. In the case of chlorine, it is not always clear whether a cyclic halonium ion (the 3-membered structure on the ...

9

Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws: This approach removes the alkene and there is no obvious way to get it back (there is a way) More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases,...

8

Markovnikov's rule is used to predict how (for example) HX, where X is a halide, adds to an unsymmetrical olefin. As the diagram below shows, Markovnikov addition follows the rule that the "H" will add to the least substituted carbon while the "X" will add to the most substituted carbon. This is due to the fact that these reactions proceed by protonaton of ...

8

Yes, I believe the rearrangement can happen. However, the product from the rearrangement is probably not the major one. The closest reaction to what you have drawn that I could find on SciFinder is this: The paper is mentioned below in the reference. In the paper, the researchers performed similar reactions of alkenes with $\ce{I-Cl}$ and in cases where ...

7

The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition. The final step, the oxidation by hydrogen peroxide, is ...

7

Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is ...

7

My instinct is that there are probably too many variables here for a definitive and accurate answer to be given strictly from first principles. When comparing molecules in which there is only one key difference, especially when that difference leads to steric and electronic effects which are essentially additive, it's fairly easy to draw accurate conclusions....

6

This looks like a typical Kharasch (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material: Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence ...

6

TL;DR - Sometimes you need to do the experiment. Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations: At C2 (Bromo) $$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$ The two issues to consider are resonance and induction. Resonance You could draw a resonance structure to show that the ...

6

The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors. Carbonate 1 is a racemate. No products ...

5

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate ...

5

There's a very simple explanation for this. Your reasoning regarding $\ce{Br-}$ being a better nucleophile is absolutely correct. The reason as to why methanol attacks instead of $\ce{Br-}$ is because it is the $\mathsf{\color{red}{solvent}}$. The solvent concentration is simply too high. Only one $\ce{Br-}$ is generated. More or less the answer is this ...

5

One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.

5

In electrophilic aromatic substitution the activation energy is very high ( as it should be, since the aryl carbocation transition state is very unstable) The role of $\ce{FeBr_3}$ or any lewis acid in general is not exactly promoting the electrophilic aromatic substitution as much as making a good electrophile. $\ce{AlCl_3}$ plays the same role much better* ...

5

Friedel–Crafts reactions proceed through the alkylation or acylation of an aromatic ring using an alkyl halide. Considering the alkyl halide compound of the form $$\ce{R-X}$$ where $\ce{R}$ is the alkyl group and $\ce{X}$ is the halide (whatever it may be), I assume the question is asking what compound, of the previously noted form, could be added to an ...

4

Entropy would favour a random orientation of each monomer unit when building up the chain, however whenever you have two different substituents at the ethylene unit, then electrostatics and sterical hindrance will make one side be preferred to make the connection to the growing chain end. Polymers that do grow via chain growth therefore have a regular ...

4

Grignards unless modified with Cu(I) salts add 1,2 not 1,4 so the EtMgBr will add to the ketone to give the t-alcohol and leaving the double bond intact. Consider then the HCl step - if it is strong enough to protonate the double bond leading to Cl addition then it will protonate the t-alcohol in preference leading to elimination or t-Cl formation. So the ...

4

As you already know, the alkene has to approach the catalyst surface in the first step. For optimum adsorption, the molecule must be able to orient itself parallel to the surface of the metal. Less number of substituents promote this. The activated hydrogen is in the form of $\ce{Pt-H}$ or $\ce{Pd-H}$ bonds on the surface of metal particles, and ...

4

You're close. You are right in thinking that the first step involves protonation of the double bond. But which end of the double bond will you protonate? If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation. However if you protonate the other end of the double bond you wind up with a carbocation that has the ...

4

Ben is right that your product is simple dimer of substrate. However, there is much easier approach to conduct dimersisation of butene - it happens easily if you use strongly acidic catalyst, like conc. sulfuric acid or ion-exchange resin, such reaction is used for example in production of isooctane. It's similar to cationic polymerization of alkenes, so ...

4

You misunderstood what hyperconjugation is. It occurs when the electrons in a $\ce{C-H}$ $\sigma$-bond are shared with an adjacent atom's empty or partially filled $\mathrm{p}$-orbital, giving the species an extended molecular orbital and thereby stabilizing it. This means that hyperconjugation doesn't really come into play until bromine is added across one ...

4

Here is a potential solution using old-school chemistry, though with several steps. Hydroborate the alkyne and work up with $\ce{H2O2}$ to give the aldehyde. Alkylate the aldehyde enolate with $\ce{RI}$, either directly (LDA or LiHMDS) or via enamine/silyl enol ether Reduce aldehyde to alcohol ($\ce{NaBH4/EtOH}$) Form triflate/mesylate and eliminate (DBU/...

4

Without further details, I would say that both products are possible. The reaction that you have given here is an Iodolactonization reaction. You have correctly identified that the tert-butyl group would be removed and that the oxygen atom would prefer to form a 6-member ring. In terms of stereochemistry, however, it is tough to make a decision, as: The ...

4

TL;DR: The final product will depend on what solvent you are using, what temperature the reaction is being operated and the concentration of the reactants. Long answer From the abstract of this paper1: The $\ce{AlCl3}$ and $\ce{SnCl4}$- catalyzed Friedel Craft alkylation (methylation, ethylation, isopropylation and tert-butylation) of naphthalenes with ...

3

I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater. A Kinetic Perspective The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the ...

3

There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE. To start, ...

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