Strong acid addition to alkynes is very similar to strong acid addition to alkenes. The regional chemistry continuesRegrá de Markownikow, but the stereochemistry is generally different. Addition to alkenes is generally non-stereospecific, while alkynes are generally subjectAntiAdditive.
Strong acids generally protonate alkenes on their less substituted carbon, thus developing a positive charge on the more substituted carbon of the reacting double bond. This intermediate carbocation can rearrange or simply incorporate a nucleophile to form the expected addition product. The situation is different with alkynes. A similar protonation of the \(\pi\) bond of an alkyne would produce a very unstable vinyl cation, so in most cases this does not happen. A vinyl cation can have only one alkyl group adjacent to the vacant p orbital to stabilize its positive charge, therefore the stability of the vinyl cation is similar to that of the monosubstituted primary alkyl cation (see below). As a result, a strong acid can form a \(\pi\) complex with the \(\pi\) alkyne bond, but proton transfer to the alkyne is extremely difficult. However, the strong acid polarizes the \(\pi\) bond by developing a positive charge on both carbons of the \(\pi\) bond. The most substituted carbon retains most of the positive charge and becomes the target for any nucleophiles present in solution. The nucleophile binds to the more substituted carbon, allowing the acid to protonate the less substituted carbon of the \(\pi\) bond.
emNucleophile Additionreactions, a nucleophile attacks an electron-poor species. Let us now consider electrophilic addition reactions, specifically of alkynes. As you will see, the reaction mechanisms are very similar to the addition reactions of electrophilic alkenes. The triple bonds of alkynes are easily attacked by electrophiles due to their high electron density, but are less reactive than alkenes due to the compact C-C electron cloud. less substituted carbon. Here we will review the following reactions listed below to learn the mechanisms behind these electrophilic additions of alkynes: (1) Addition of HX toAlkene, (2) halogenation ofAlkynesand (3) hydration of alkynes.
Adding an electrophile to an alkene or alkyne follows the same steps outlined below.
- Start with a reactant (either an alkene or an alkyne) that has \(\pi\) electrons. A pair of electrons moves from the \(\pi\) bond to the electrophilic proton to form a new covalent bond. An intermediate carbocation forms on the more stable carbon. As shown in the diagram above, the intermediate consists of hydrogen covalently bonded to the carbon and a positive charge on the other carbon.
- Addition of the halide ion to the positively charged intermediate forms the second covalent bond, giving you the haloalkane product (as seen on the left side of the diagram above) and a haloalkene product (as seen on the right side of the diagram above). The halide ion acts as a nucleophile that is attracted to the positively charged carbon because it can donate or share electrons. So if the nucleophilic halide (represented as X-) attacks, it targets the positively charged carbon leading to this second covalent bond, as shown above.
Note: The way to determine where the addition of the proton and halide occurs is based on Markovnikov's rule, which states that the proton will add to the carbon with the most hydrogen atoms and the halogen will prefer the most substituted carbon.
Now let's look at our first reaction, the addition of hydrogen halide.
Reaction 1: Addition of hydrogen halide to an alkyne
Summary:Order of reactivity ofhydrogen halides: HI > HB r > HCl > HF.
Follow the Markovnikov rule:
- Hydrogen adds carbon with the most hydrogens, halogen adds carbon with the fewest hydrogens.
- Protination occurs at the more stable carbocation. When HX is added, haloalkenes are formed.
- With the addition of excess HX you getAntiAddition to a geminal dihaloalkane.
Addition of an HX to an internal alkyne
As shown in Figure 1, the \(\pi\) electrons attack the hydrogen of HBr and since it is a symmetric molecule it doesn't matter which carbon it is added to, but in an asymmetric molecule it binds the hydrogen covalently bonded to carbon more hydrogen. Once the hydrogen is covalently bonded to one of the carbons, it receives an intermediate carbocation (not shown but looks the same as shown in Figure 1) on the other carbon. Again, this is a symmetric molecule, and if it were asymmetric, which carbon would have the positive charge?
The final step is to add bromine, which is a good nucleophile because it can donate or share electrons. Thus, the bromine attacks the intermediate carbocation and places it on the highly substituted carbon. As a result, you will get 2-bromobutene from your 2-butyne reagent as shown below.
Now what if you have too much HBr?
Encore due to excess HX available ? produces a geminal dihaloalkane
Here, the electrophilic addition proceeds with the same steps used to obtain the product from the addition of an HX to an internal alkyne. The \(\pi\) electrons attacked the hydrogen and added it to the carbon on the left (shown in blue). Why was hydrogen added to the left carbon and bromine to the right?
It now has its intermediate carbocation followed by bromine attack on the right carbon, resulting in a haloalkane product.
Adding HX to Alcino Terminal
- Here is an addition of HBr to an asymmetric molecule.
- First, try to understand how the reactant got into the product, and then look at the mechanism.
The electrons \(\pi\) attack the hydrogen, represented by the arrows pushing the electrons, and the bromine acquires a negative charge. The intermediate carbocation forms a positive charge on the left carbon after adding hydrogen to the carbon with more hydrogen substituents.
Bromine, which is negatively charged, attacks the positively charged carbocation and forms the final product with the nucleophile on the more substituted carbon.
Addition present due to excess HBr
Reaction: Halogenation of alkynes
- Stereoselectivity: anti-addition
- The reaction proceeds via the cyclic halonium ion.
In addition toBrother2
- The addition of Br2to an alkyne is analogous to adding Br2to an auk.
- once brother2approaches the nucleophilic alkyne, becomes polarized.
- The \(\pi\) electrons of the triple bond can now attack the polarized bromine, form a C-Br bond and displace the bromine ion.
- You now get an intermediate electrophilic carbocation that immediately reacts with the bromine ion to give you the dibromo product.
First you see that Br is polarized2are attacked by electrons \(\pi\). After forming the C-Br bond, the other bromine is released as the bromine ion. The intermediate here is a bromonium ion, which is electrophilic and will react with the bromine ion to give you the dibromo product.
Reaction: Hydration of alkynes
Summary:Upon addition of water, alkynes can hydrate to enols, which spontaneously tautomerize to ketones. The reaction is catalyzed by mercury ions. Follow Markovnikov's rule: terminal alkynes produce methyl ketones
- The first step is an acid/base reaction in which the \(\pi\) electrons of the triple bond act as a Lewis base and attack the proton, protinating the carbon with the most hydrogen substituents.
- The second step is the attack of the nucleophilic water molecule on the electrophilic carbocation, creating an oxonium ion.
- It is then deprotonated by a base, creating an alcohol called an enol, which then tautomerizes into a ketone.
- Tautomerism is a simultaneous change of protons and double bonds from the enol form to the keto isomer form as shown in Figure 7 above.
Now let's look at some hydration reactions.
Hydration of the terminal alkyne generates methyl ketones.
As shown in Figure 7, the \(\pi\) electrons attack a proton and form a carbocation, which is then attacked by the nucleophilic water molecules. After deprotinization we generate an enol which then tautomerizes to the ketone form shown.
As you can see here, the \(\pi\) electrons of the triple bond attack the proton, which forms a covalent bond to carbon with more hydrogen substituents. Once the hydrogen is bonded, you have a carbocation that is being attacked by the water molecule. You now have a positive charge on the oxygen, which causes a base to come in and deproteinize the molecule. Once deprotonated you have an enol that tautomerizes.
Tautomerism occurs here when the proton is attacked by the electrons of the \(\pi\) double bond, creating a covalent bond between carbon and hydrogen on the least substituted carbon. The electrons from the oxygen eventually move to the carbon, forming a double bond with the carbon and giving it a positive charge, which is then attacked by the base. The base deprotonates the oxygen, which at equilibrium leads to the most stable end product, a ketone.
- Volhardt. Schore, Organic Chemistry Structure and Function, 5. Auflage, New York: W.H. Freeman und Company, 2007.
- Solomons, T.W. Graham and Craig B. Fryhle. Eighth edition of organic chemistry. Maryland: John Wiley and Sons Incorporated, 2008.
1) Reaction: addition of hydrogen halide to an alkyne
1a)What product would result from the reaction of 1-pentyne with hydrogen bromide?
1b)Take your answer from 1a and add excess hydrogen bromide. what would the product be
1c)What is the mechanism of this reaction?
2) Reaction: Addition of X2for alkynes
2a)What product would result from the reaction of 1-pentyne with Br?2?
2b)Take your answer from 2a and add excess Br2. what would the product be
2c)What is the mechanism of this reaction?
3) Reaction: hydration of alkynes
3a)What would be the product if 1-pentyne were reacted with mercury(I) sulfate, water and sulfuric acid?
4)What is the product when 3-methylbutyne reacts with HCl? Include in your answer:
a) the reaction mechanism
b) a clear explanation of why hydrogen and chlorine combine where they combine
5)a) Draw the enol form of the ketone below.
b) Draw the keto form of the enol below.
- Professor Hilton Weiss (bardic college)
- Aleksandra Milmann