How can alkane be produced by Grignard reagent

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What evidence is there for this mechanism?


The reaction order of the substitution can be precisely determined by kinetic measurements. Since the leaving group is directly substituted, we expect only one transition state, and the rate of the reaction must depend on the concentration of both partners, because the chances of the reactants meeting one another increase with increasing concentration of one or both partners.

This reaction of the second order corresponds to the following rate equation:

A reaction that obeys this law is called bimolecular. Indeed, many nucleophilic substitution reactions exhibit a second order law of time.

A concerted substitution response profile is as follows:


Influences the SN2-reaction the spatial arrangement of the substituents at the reaction center?

To answer this question, we need to trace the stereochemical fate of a chiral halogen:

e.g. (S.) -2-bromobutane and hydroxide:

So the reaction went under Inversion of stereochemistry. This is only possible with one Rear attack :

OtherN2 reactions give us the same result (INVERSION).

With this we can also explain the interesting results of Paul Walden (Chapter 8.3). Therefore, such a sterochemical result (inversion of the configuration during a substitution) is also called Walden inversion designated.

The designation S.N2 is now understandable: S.ubstitution Nucleophilic2. order

8.4.2 Influence of nucleophile, leaving group and steric factors on the SN2 reaction

Now let us examine different nucleophiles, and their relative nucleophilic strength, theirs Nucleophilicity, consider. The higher the nucleophilicity, the faster S will runN2 reactions. Nucleophilicity depends on a number of factors: the charge, the basicity, the solvent, the polarizability and the substituents.

Of two nucleophiles with the same reactive atoms, the one with one is negative charge the stronger nucleophile. The reaction must be faster, the more electron-rich (negative) the attacking species is. e.g .:

  HO:- > H2O     RO:- > RAW    RS:- > RSH    H2N:- > NH3

Also the Basicity of O and N (1st row) seems to correlate with nucleophilicity: the more basic species is obviously also the more reactive nucleophile. It is also important that the nucleophilicity is also increasing within a group in the periodic table down to (Polarizability):

NH2-   > OH-   > NH3   > H2O
Basicity and nucleophilicity increasing
R.3P: > R.3N: MeS- > MeO- I.- > Br- > Cl-

Relative reaction rates of different nucleophiles with iodomethane
Nu:- + Me-I ----> Me-Nu + I-

Nucleophile Relative speed Nucleophile Relative speed
CH3OH1 CH3SCH3347 000
NO3-~32 N3-603 000
F.-500 Br-617 000
SO42-3 160 CH3O-1 950 000
CH3COO-20 000 CN-5 010 000
Cl-23 500 (CH3CH2)3As7 940 000
CH3CH2SCH2CH3219 000 I.-26 300 000
NH3316 000 HS-100 000 000

The rate of nucleophilic substitution is also determined by the structure of the Leaving group influenced. The leaving group in a nucleophilic substitution is often negatively charged. Hence, the relative ease with which it is substituted becomes apparent to her Leaving ability, correlate with their ability to stabilize a negative charge:

Leaving ability

I.-   > Br-   > Cl-   >> F.-

Whether halides are good or bad leaving groups should depend on their base strength: the weaker basic X- is (the stronger the conjugate acid HX), the better X- functions as a leaving group, HF is the weakest acid, HI is the strongest. Other good leaving groups are:

The -OH group is a poor leaving group (H.2O is a weaker acid). Because of this, simple alcohols do not enter into nucleophilic substitution reactions. But the OH group becomes a good leaving group when it is converted into a sulfonate:

This can be achieved even more easily by protonation. By protonating the lone pair of electrons on the oxygen atom, an oxonium ion (H.3O+ is a good acid). The bad leaving group OH becomes a good leaving group, namely water:

For this reason, one obtains by reacting various alcohols with concentrated hydrohalic acids the corresponding haloalkanes in good yields. Likewise, the RO group in ethers can use ROR (Chapter 9.8) very strong acids are protonated and thus form a good leaving group.

Sterically hindered haloalkanes react only very slowly:

The reason for this can be found by looking at how the nucleophile approaches the electrophile:

The successive introduction of alkyl groups thus has a cumulative effect on the steric hindrance of substitution. So the following reactivity series applies:

A 1-halogen-2,2-dimethylpropane is also almost completely inaccessible to attack from the rear. The structure of the alkyl part of haloalkanes can therefore have a pronounced effect on nucleophilic attack.

8.4.3 The SN1 reaction

Nucleophilic substitution reactions on sp3Centers can also expire according to an alternative mechanism. In the SN1 reaction, the reaction takes place gradually. A typical example would be a reaction of a tertiary alcohol in aqueous or alcoholic solution in the presence of a strong acid, or the reaction of a tertiary haloalkane in aqueous or alcoholic (strongly polar) Solution:

Reactivity series:

Nu:- + R-X ----> R-Nu + X:-
R = 3O > 2O > 1O >> methyl
applies to SN1 reactions (see on the other hand pN2 reversed)


The change in energy during this course of the reaction can be shown in an energy profile diagram:


Such a reaction is called a unimolecular nucleophilic substitution, abbreviated S.N1. This means that kinetic studies of the reaction of 2-bromo-2-methylpropane with water show that the rate of substitution only depends on the concentration of the educt used:

Speed ​​= k [R-X]   (Nu: is not involved in the GBS)

The first-order kinetics observed suggest a unimolecular mechanism in which only the haloalkane participates in the rate-limiting step.


A carbenium ion is planar and therefore achir al. If you use an optically active tertiary haloalkane, you should racemic S.N1 products because the nucleophile can attack the carbenium ion from both sides:

i.e. the reaction does not take place stereospecifically.

8.4.4 The influence of the substrate structure, the strength of the nucleophile, the leaving group, and the solvent on the SN1 response.

We have already seen (Chapter 4.2) that the stability of carbenium ions in the series 1O < 2O < 3O increases, i.e. the stability of carbenium ions increases with increasing alkyl substitution:

It is therefore to be expected that in substitution reactions the degree of substitution of the reactive carbon atom essentially determines whether haloalkanes (and similar derivatives) with nucleophiles preferentially according to S.N1 or after pN2 respond:

For this reason, 3O-Halogenalkanes according to the SN1 mechanism implemented, 1O Haloalkanes according to S.N2.

The special stability of the 3O Carbenium ions are the reason that 3O Haloalkanes (and similar derivatives) according to S.N1 react: they dissociate into ions relatively easily.

We have already seen that 2O Haloalkanes and others 2O Alkyl derivatives SNReceive 2 reactions. Can 2OCompounds due to the increased stability of the corresponding carbenium ions (compared to methyl or 1ODerivatives) SN1 reactions received? Under certain conditions: yes.

For example :

The heterolytic cleavage of a C-X bond in the rate-limiting step (GBS) the SN1 reaction proceeds through a strongly polar transition state. Due to the strongly polar transition state, the SN1 reaction with increasing Solvent polarity more quickly.

The SN2 reaction will increase with Nucleophilicity of the attacking species accelerated. At the GBS of an SNHowever, the nucleophile is not involved in 1 reaction, so it should not have any influence on the reaction rate. But it can Product distribution essentially determine.

Since the leaving group in GBS leaves the molecule, this practices Leaving ability has a major impact on the speed of reaction. Therefore, 3O Iodoalkanes hydrolyzed faster than 3O Bromoalkanes, these in turn faster than 3O Chloroalkanes.

Substitution reactions on sp2Centers?

Halogen atoms on sp2- Centers are opposite SN1- and SN2- reactions completely unreactive. This means that nucleophilic substitution reactions with vinyl and aryl halogen are normally not observed.

8.5 Elimination reactions

In addition to substitution reactions, haloalkanes can also react with nucleophiles in another way, because often good nucleophiles are also good ones Bases are: they can Eliminations received:

Elimination reactions are very useful for the preparation of alkenes. Haloalkanes can therefore enter into substitution and elimination reactions. Which is preferred is often a question of the reaction conditions. From a mechanistic point of view, substitution and elimination reactions are similar. Again, there are two mechanisms for elimination reactions that are common:

8.5.1 Bimolecular Elimination: E2

In the E2-Reaction takes place a concerted anti- Elimination of HX instead.


The evidence for this mechanism comes from kinetic and stereochemical Investigations.

e.g .:

Speed ​​= k [RX] [Base] 2nd order

Kinetic studies of the E2 reaction suggest a bimolecular transition state. It is also characterized in that a proton is preferred in anti-Position to the exiting halide is released, and that the following processes take place simultaneously: bond formation between base and proton; Rehybridization of sp3 to sp2 and formation of the double bond through the resulting p orbitals; Breaking the bond between the carbon atom and the leaving group.

How can we prove that this is the leaving group anti-continuous proton is given preferentially to the base? This knowledge can be gained through stereochemical investigations, e.g. let's look at them again first anti-Elinimation to:

anti-periplanar geometry:

The SP3sigma orbitals in the source material must overlap and become p orbitals. The overlap must already begin in the transition state, and that can only take place if the orbitals are in one plane (or periplanar) to one another.

These considerations have important stereochemical consequences if we start from certain chiral molecules because they change the configuration of the alkene (E. or Z-alkene), e.g .:

Both spatial arrangements can be
by simply rotating around the central C-C bond
merge into each other.
Fischer projection antiperiplanar

Which configuration (E.- or Z-) is for the alkene, which by an E2 elimination at (1S.,2S.) -1,2-dibromo-1,2-diphenylethane is formed, to be expected?

First we need to draw this molecule so that the H and Br groups that are lost in the elimination are anti-periplanar to each other:

The product is (Z) -1-bromo-1,2-diphenylethylene.

8.5.2 Unimolecular Elimination: E1

The elimination, like the nucleophilic substitution, can also take place as a unimolecular or bimolecular reaction, E1 or E2.

The E1 reaction proceeds - just like in the SN1 reaction - via a carbenium ion as an intermediate. Instead of the carbenium ion combining with a nucleophile, it splits off a proton and thus produces an alkene:

e.g. starting from a tert-Alkyl bromide:

This process is called E1because he's one unimolecular kinetics follows:

Speed ​​= k [R-Br]

(Base is not involved in the GBS)

8.5.3 E1 and E2 reactions: competition?

The same factors that make an SN1-opposite an SNPromote 2 reaction, also promote E1 versus E2 eliminations, i.e. tert-Alkyl groups in the starting material, because they stabilize the carbocation intermediate, and a good ion-solvating solvent (polar solvent), e.g .:

Weakly basic reagents (water, alcohol) usually cause the elimination to proceed according to E1 if it can proceed via carbenium ions (i.e. at tertCenters).

Strong baseslike OH- or RO- however, both carry with tert-Haloalkanes as well as with 1O or 2O-Haloalkanes to an E2 elimination.

8.5.4 pN1 and E1 processes occur as competing reactions

The first steps of the SN1 and E1 processes are identical. When a carbenium ion is formed, two competing processes can occur. Either the nucleophilic attack on the positively charged carbon, or the loss of a proton, creating an alkene. In fact, products from both processes are often observed, e.g .:

tert-Butyl bromide is unstable in methanol because Br- can easily be split off, with a relatively stable 3O-Carbocation arises. However, the cation continues to react quickly, either with the loss of a proton or as a nucleophile due to attack by the solvent.

Another example :

Elimination often takes place more quickly at elevated temperatures.

8.6 Biological substitution reactions

Certain reactions in biological systems are also nucleophilic substitution reactions on sp3 Centers. One of the most common examples is that Methylation - The transfer of a methyl group from an electrophile to a nucleophile. In the laboratory, methyl iodide would be used for this purpose. S-adenosylmethionine (SAM) is often used in nature:

The sulfur of the S-adenosylmethionine has a positive charge, and is therefore a good leaving group in an S.N2 response. For example, in the brain, adrenaline is built up from noradrenaline, with a methyl group in an SN2 reaction from SAM to noradrenaline (enzyme-catalyzed) is transferred:

A second example would be one of the many methyltransferases that methylate bases in DNA to modify restriction sites:

8.7 Grignard connections

In this section we will learn how to make a reagent with a nucleophilic carbon atom (R.-) through a new class of compounds, the organometallic reagentsCan describe. These reagents contain a metal atom, usually lithium or magnesium, bound to a carbon atom of an organic molecule. They are characterized by strong basicity and nucleophilicity and play an extremely important role in organic syntheses.

Organomagnesium compounds, RMgX, are named after their discoverer V. Grignard also Grignard connections. The C-metal bond is covalent, but strongly polarized by the electropositive metal:

Grignard compounds result from a reaction between magnesium metal and a haloalkane, suspended in an inert aprotic solvent (non-protonic) such as diethyl ether or tetrahydrofuran (THF), e.g .:

The chemical behavior of such compounds corresponds to that of a negatively charged carbon atom, a Carbanions. Such compounds have strong nucleophilic and basic properties and therefore react quickly with electrophiles, e.g .:

1) You respond vigorous with acids such as HCl, H2O or ROH (alcohols) forming alkanes and MgOHX:

They must therefore be produced in a non-protonic solvent (e.g. an ether!) With the exclusion of moisture.

2) But much more useful is the exploitation of the nucleophilicity of the alkyl group attached to the metal for the preparation of alcohols by the reaction of Grignard reagents with Carbonyl compounds or cyclic ethers strained by nucleophilic ring opening (Epoxies). We will cover these topics in the next chapters to treat.