What do alkenes react with

Although diborane itself does not react easily with alkene double bonds, H. Brown Purdue, Nobel Prize discovered that the solvated monomer adds rapidly under mild conditions.

Boron and hydrogen have rather similar electronegativities, with hydrogen being slightly greater, so it is not likely there is significant dipolar character to the B-H bond. Since boron is electron deficient it does not have a valence shell electron octet the reagent itself is a Lewis acid and can bond to the pi-electrons of a double bond by displacement of the ether moiety from the solvated monomer.

As shown in the following equation, this bonding might generate a dipolar intermediate consisting of a negatively-charged boron and a carbocation. Such a species would not be stable and would rearrange to a neutral product by the shift of a hydride to the carbocation center.

Indeed, this hydride shift is believed to occur concurrently with the initial bonding to boron, as shown by the transition state drawn below the equation, so the discrete intermediate shown in the equation is not actually formed. Nevertheless, the carbocation stability rule cited above remains a useful way to predict the products from hydroboration reactions.

You may correct the top equation by clicking the button on its right. Note that this addition is unique among those we have discussed, in that it is a single-step process. Also, all three hydrogens in borane are potentially reactive, so that the alkyl borane product from the first addition may serve as the hydroboration reagent for two additional alkene molecules.

To examine models of B 2 H 6. As illustrated in the drawing on the right, the pi-bond fixes the carbon-carbon double bond in a planar configuration, and does not permit free rotation about the double bond itself. We see then that addition reactions to this function might occur in three different ways, depending on the relative orientation of the atoms or groups that add to the carbons of the double bond: i they may bond from the same side, ii they may bond from opposite sides, or iii they may bond randomly from both sides.

The first two possibilities are examples of stereoselectivity , the first being termed syn-addition , and the second anti-addition. Since initial electrophilic attack on the double bond may occur equally well from either side, it is in the second step or stage of the reaction bonding of the nucleophile that stereoselectivity may be imposed.

If the two-step mechanism described above is correct, and if the carbocation intermediate is sufficiently long-lived to freely-rotate about the sigma-bond component of the original double bond, we would expect to find random or non-stereoselective addition in the products. On the other hand, if the intermediate is short-lived and factors such as steric hindrance or neighboring group interactions favor one side in the second step, then stereoselectivity in product formation is likely.

The interesting differences in stereoselectivity noted here provide further insight into the mechanisms of these addition reactions. The selectivity is often anti, but reports of syn selectivity and non-selectivity are not uncommon. Such reactions are most prone to rearrangement when this is favored by the alkene structure. The halogens chlorine and bromine add rapidly to a wide variety of alkenes without inducing the kinds of structural rearrangements noted for strong acids first example below.

The stereoselectivity of these additions is strongly anti , as shown in many of the following examples. An important principle should be restated at this time. The alkenes shown here are all achiral, but the addition products have chiral centers, and in many cases may exist as enantiomeric stereoisomers.

In the absence of chiral catalysts or reagents, reactions of this kind will always give racemic mixtures if the products are enantiomeric.

On the other hand, if two chiral centers are formed in the addition the reaction will be diastereomer selective. This is clearly shown by the addition of bromine to the isomeric 2-butenes. Anti-addition to cisbutene gives the racemic product, whereas anti-addition to the trans-isomer gives the meso-diastereomer.

We can account both for the high stereoselectivity and the lack of rearrangement in these reactions by proposing a stabilizing interaction between the developing carbocation center and the electron rich halogen atom on the adjacent carbon. This interaction, which is depicted for bromine in the following equation, delocalizes the positive charge on the intermediate and blocks halide ion attack from the syn-location.

The stabilization provided by this halogen-carbocation bonding makes rearrangement unlikely, and in a few cases three-membered cyclic halonium cations have been isolated and identified as true intermediates. A resonance description of such a bromonium ion intermediate is shown below.

The positive charge is delocalized over all the atoms of the ring, but should be concentrated at the more substituted carbon carbocation stability , and this is the site to which the nucleophile will bond.

The stereoselectivity described here is in large part due to a stereoelectronic effect. This aspect of addition reactions may be explored by clicking here.

Because they proceed by way of polar ion-pair intermediates, chlorine and bromine addition reactions are faster in polar solvents than in non-polar solvents, such as hexane or carbon tetrachloride. However, in order to prevent solvent nucleophiles from competing with the halide anion, these non-polar solvents are often selected for these reactions. Such reactions are sensitive to pH and other factors, so when these products are desired it is necessary to modify the addition reagent.

By adding AgOH, the concentration of HOCl can be greatly increased, and the chlorohydrin addition product obtained from alkenes. The more widely used HOBr reagent, hypobromous acid, is commonly made by hydrolysis of N-bromoacetamide, as shown below. Both HOCl and HOBr additions occur in an anti fashion, and with the regioselectivity predicted by this mechanism OH bonds to the more substituted carbon of the alkene.

Vicinal halohydrins provide an alternative route for the epoxidation of alkenes over that of reaction with peracids. As illustrated in the following diagram, a base induced intramolecular substitution reaction forms a three-membered cyclic ether called an epoxide.

Both the halohydrin formation and halide displacement reactions are stereospecific, so stereoisomerism in the alkene will be reflected in the epoxide product i. A general procedure for forming these useful compounds will be discussed in the next section. This page is the property of William Reusch. Comments, questions and errors should be sent to whreusch msu.

Palladium gets electroplated on the carbon surface a few atomic layers making it a cheap alternative to the very expensive platinum group metal. While catalytic hydrogenation of alkenes is a powerful reaction, it is a bit too powerful. This reaction will reduce all double or triple bonds to single bonds in your molecule!

So, you should be very careful when using this method in your synthesis. Also, the reaction is a syn addition that adds both hydrogens to the same face of the molecule.

This is an important consideration as it may influence your synthetic decisions. The only tricky point in this reaction might be the use of mixed halogens.

Also, generally only chlorine and bromine are used in this reaction. Interestingly, in this reaction you have a competition of two nucleophiles: bromide anion that you form after the initial attack on the alkenes and water. Since negatively charged bromide is a much better nucleophile than water, you need to play the statistics game here and use a large excess of water. This way, the chances of bromide acting as a nucleophile are statistically small. Thus, water is going to be the attacking species opening the bromonium ion.

An important thing to remember about this reaction is regioselectivity. Many reactions of alkenes have a certain stereochemistry and regiochemistry associated with them. Markovnikov regioselectivity means that the electrophile the halogen here going to end up on the less substituted carbon of the double bond, while the hydroxyl -OH will be on the most substituted carbon. Mechanistically, the reaction is similar to the previous two examples.

Like the other two, it also gives anti-addition stereospecificity, and gives you a Markovnikov product. A really cool thing about this reaction is that it allows you to form 5- or 6-membered cyclic ethers when the -OH is a part of the original alkene:. Hydrohalogenation is regioselective and gives the Markovnikov product. And where you have carbocations, you have potential troubles! Can you draw the mechanism for this reaction to explain the structure of the final product?

In this reaction you end up adding water to your alkene. You would typically see something like sulfuric acid H 2 SO 4 as a catalyst in this reaction. Make sure to never use HCl or HBr as a catalyst as those will just do the hydrohalogenation reaction!!! Since this reaction also forms a carbocation as an intermediate, beware of possible rearrangements:. Remember, that if a carbocation can rearrange to give you a more stable carbocation, it will! So always check your reactions for any possible shifts or you might miss some points on the test.

This is very useful reaction although you get to work with very poisonous mercury compounds. It gives the same product alcohol as the hydration of alkenes. And no carbocation means no rearrangements! This reaction is very similar to the regular oxymercuration. It seems easy for you to just memorize the rule or just memorize the fact that 2-bromomethylpropane is the product for above reaction, without understanding why.

However, you will notice soon that your memorization will be overwhelmed and mixed up with many more reactions coming up. The proper way to study organic reactions is to learn and understand the mechanism, unify the principles of reactions based on mechanism.

The mastery of the contents will much easier and a lot more fun in this way, rather than trying to memorize tons of reactions. For the addition of HBr to 3-methylbutene, two products were observed. Show the reaction mechanism to explain the formation of both products.

Here we will learn that the hydrobromide, HBr, can also add to alkene in a way that gives anti-Markovnikov product. The anti-Markovnikov product are obtained through different mechanism, that is the radical mechanism. To initiate radical mechanism, peroxide must be involved in order to generate the radical in the initiation step of the mechanism. The peroxide therefore acts as radical initiator by generating radicals, and the addition is called radical addition.

The detailed radical addition mechanism of the above addition of HBr to 2-methylpropene is given here. The initiation involves two steps for the radical addition mechanism. It shown clearly in the propagation steps that the order of the addition is reversed in radical addition comparing to that of electrophilic addition.

Specifically, the bromine radical Br is added to the double bond first followed by the abstraction of hydrogen atom H , therefore the anti-Markovnikov product is produced as a result.

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