10

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. ...


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

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

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

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 ...


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 ...


7

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 ...


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....


7

The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : G. van der Rest et al. Eur. Mass ...


6

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 ...


6

Markovnikov's rule is usually stated in terms of subsitution on the alkene, for example: In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms). This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{...


5

Hydrogenation of the alkene introduces a stereo centre, thus the products are:


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

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

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 $\ce{PCl5}$.


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

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

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 ...


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/...


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

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 ...


2

You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another: the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion: The bromonium would only be able to form after a rotation ...


2

In example A, you are right that the cyclohexane ring is more stable than the cycloheptane ring. There are two choices for rearrangement in this case: 1) Ring expansion to form the cycloheptane ring 2) 1,2-hydride shift to yield the tertiary alkyl bromide The second product will be more stable and thus be the major product In example B, you are right, ...


1

1,6-additions are indeed possible (1) but since you have competing 1,4-addition and 1,2-addition it becomes more difficult. To push it to the 1,6-addition bulky organometallic catalysts are used. Reaction pathways are also determined by the electrostatic interactions, and these diminish the further away you get from the carbonyl. (1) https://onlinelibrary....


1

The nucleophilic substitution with another equivalent of the primary bromide will never be favourable as the nucleophilicity of the enolate generated is pretty low (the charge is delocalized!) Even if E1cb was a possible route, you would end up with an extremely reactive vinyl aldehyde, which would likely participate in a Michael addition with an enolate in ...


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