Alkyl fluorides generally give Hoffman’s elimination product. This is because, the carbon–fluorine bond in fluorides is the weakest among all the halogens and hence, cleaved easily. Due to this, steric effects are observed to be more pronounced than electronic effects. Hence, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product
Alkyl fluorides generally give Hoffman’s elimination product.
Alkyl fluorides generally give Hoffman’s elimination product. This is because:
The carbon–halogen bond in fluorides is the weakest among all the halogens and hence, cleaved easily. Due to this, steric effects are observed to be more pronounced than electronic effects.
This is because, the carbon–halogen bond in fluorides is the weakest among all the halogens and hence, cleaved easily
(1) The carbon–halogen bond in fluorides is the weakest among all the halogens and hence, cleaved easily.
(2) When an alkyl fluoride acts as a nucleophile, it donates its electron pair to the carbonyl carbon atom. This results in formation of a covalent bond between the carbonyl carbon and fluoride ion.
(3) Carbon–fluorine bond is stronger than carbon–chlorine bond because of higher electronegativity of fluorine compared to chlorine.
Due to this, steric effects are observed to be more pronounced than electronic effects.
The halogen atom in an alkyl fluoride, like any other halide, can be regarded as a Lewis acid. In the presence of a Lewis base, such as ammonia or pyridine, it is possible for both steric and electronic effects to be observed. This is because the alkyl group may not be able to reach all the way around the halogen atom because of its bulkiness, but still has some effect on its bonding interactions with other groups.
Hence, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product.
Therefore, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product.
This is because:
The carbon-fluorine bond is very weak, so it breaks easily and the electrons move to form a new bond between the carbon and oxygen atom.
Because the C-F bond is weaker than C-O bonds, this means that it would be easier for both electrons to break from their respective atoms at some point during this process.
In summary, E2 reactions typically give Hoffman’s product when an alkyl fluoride is used as the electrophile. This is because the alkyl fluoride is able to donate its fluorine atom to form a carbocation intermediate (see below).
Hence, it can be concluded that the carbon–fluorine bond is the weakest among all the halogens and hence, cleaved easily. In addition to this, the steric effects are more pronounced than electronic effects which leads to an E2 mechanism for Hoffman elimination product formation.
Hofmann’s elimination (also known as the E2 reaction) is a method of organic synthesis that involves the elimination of hydrogen halides from alkyl halides. The reaction is named after the German chemist Ernst Otto Fischer, who first described it in 1883. It can be used to synthesize esters from alcohols and carboxylic acids, as well as other products:
The carbocation formed by the E1cB pathway is stabilized by the fluorine substituent
The carbocation formed by the E1cB pathway is stabilized by the fluorine substituent. The strong electron withdrawing effect of fluorine helps to stabilize the cation, since it has a much greater electronegativity than hydrogen. Fluorine also has a large electron density that can be used to delocalize electrons to make bonds more stable, something that’s often seen with leaving groups such as water or bromide ions. The high electronegativity of fluorine gives it an unusually strong bond with carbon atoms in molecules like alkanes and alkyl chlorides, making those molecules more acidic than similar compounds with other halogens like chlorine or bromine
E2 gives the product with an antiperiplanar hydrogen beta to the leaving group.
The stereochemistry of the product is determined by the beta hydrogen. The beta hydrogen is antiperiplanar to the leaving group, so it will add anti to form a chair as shown below.
Hofmann’s eliminations are observed for alkyl fluorides.
In the Hofmann’s elimination, the hydrogen is replaced by a fluorine.
This reaction is an E2-like reaction in which a fluorine acts as an electrophile to attack an alkyl halide. As you can see from this mechanism, it does not require any additional acid or base to be added to aid in deprotonation of the carbocation intermediate.
In conclusion, Hofmann’s eliminations are observed for alkyl fluorides. This is because the carbocation formed by E1cB pathway is stabilized by the fluorine substituent. The hydrogen beta to the leaving group of this species is antiperiplanar to it, which gives rise to E2 mechanism as a result.
In this chapter you will consider what would happen if the nucleophilic water molecule in an S N 2 reaction were replaced by a tertiary carbon-based nucleophile. In the first reaction of this chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2. There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book. The Hofmann product (2) was obtained almost exclusively under these conditions. It is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism. A subsequent reaction of the same product in 3 with Me2NH followed by methanol gave 4 without rearrangement. If we compare this figure with Fig 1 1 (p 6), we can see that the conformation of substrate has not changed much on loss of fluoride and that product 4 is formed with retention configuration in this case
In this chapter you will consider what would happen if the nucleophilic water molecule in an S N 2 reaction were replaced by a tertiary carbon-based nucleophile.
Imagine you were a student in an organic chemistry course, and the professor asked you to consider what would happen if the nucleophilic water molecule in a S N 2 reaction were replaced by a tertiary carbon-based nucleophile. Specifically, imagine that this hypothetical reaction involved an alkene geometrically similar to butane, with two alkyl groups bonded to its terminal carbon atoms. The following diagram shows such a molecule:
The alkoxide ion, CH3CH2CH2O–, is obviously a good nucleophile for reactions with primary and secondary alkyl halides, but what about tertiary halides?
The alkoxide ion, CH3CH2CH2O–, is obviously a good nucleophile for reactions with primary and secondary alkyl halides, but what about tertiary halides?
The answer lies in the fact that tertiary alkyl groups are more electron withdrawing than primary or secondary ones. You might be able to guess that this means they’re harder for your nucleophilic reagent to get close to, which makes it harder for the reaction to occur. In order for an E1 mechanism (especially) to work, the transition state has to have a lot of sigma bonding between carbon and oxygen atoms; this means there needs to be a lot of σ donation from oxygen toward carbon (or vice versa), not just from carbon toward oxygen. And if you recall from our discussion above regarding carbocations in terms of stability when compared with other carbocations like methylene carbocation or vinyl cation, you should know that even though these ions are electronically similar (both having two sp³ hybridized bonds), they differ in how much resonance stabilization they receive—methylene gets much more resonance stabilization than vinyl does because it has an extra p-orbital available; so too do tertiary carbocations get more resonance stabilization than secondary ones because their extra π bond makes them less bent and therefore less reactive than the latter type
The first reaction of the chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2.
The first reaction of the chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2. The mechanism of this reaction has been discussed in our textbook [1]. It is known that the alkoxide ion is a good nucleophile and can attack at either C2 or C3 with retention of configuration. The process may be described as follows:
There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book.
There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book.
In the case of the alkyl fluoride, you can see that since the leaving group is not a primary or secondary carbon centre, but rather a tertiary carbon centre, it must be displaced by an incoming nucleophile (a hydrogen ion). The end result is elimination product formation.
The Hofmann product (2) was obtained almost exclusively under these conditions.
The Hofmann elimination is an E1cB reaction in which a tertiary halide undergoes nucleophilic substitution with a tertiary alcohol to form an alkoxide ion, which then eliminates HF. The mechanism for this reaction begins with a bimolecular nucleophilic substitution (Sb) of the halide by the alcohol. The alkoxide ion is then formed from protonation of the intermediate and then eliminated as HF, which results in cleavage of both bonds to give the final product.
It is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism.
At lower temperatures, it is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism.
A subsequent reaction of the same product in 3 with Me2NH followed by methanol gave 4 without rearrangement.
The reaction of 5 with Me2NH gives the same product as that obtained from 4, namely 6. This indicates that the intermediate formed in the Hofmann elimination is an enolate anion which can be trapped by a nucleophile without rearrangement to give a different compound.
If we compare this figure with Fig. 1.1 (p. 6), we can see that the conformation of the substrate has not changed much on loss of fluoride and that the product, 4, is formed with retention of configuration in this case.
Elimination reactions are governed by sterics rather than electronic effects
The Hoffman elimination reaction is a good example of this phenomenon. In the bulk, alkyl fluorides generally give the product with excellent regioselectivity. But when a small amount of solvent is added, rapid dissolution of the fluoride occurs and all products can be formed equally well. This is because solvation affects steric effects more than electronic effects, thereby shifting from high-valent to low-valent alkyl intermediate structures (see Figure 1).
In conclusion, we can say that the Hofmann elimination product is generally formed. The mechanism for this is concerted and involves a nucleophilic attack on the tertiary carbon centre by water (or methanol). This occurs via an intermediate alkoxide ion.
The Hoffman elimination is an organic reaction in which the hydrogen atoms of an alkyl halide are replaced with hydroxyl groups to produce a primary alcohol. It is named after Hermann Ludwig Ferdinand von Hoffmann who first reported the reaction in 1881. The mechanism involves two steps:
Alkyl Fluorides Generally Give Hoffman’s Elimination Product. This Is Because
Alkyl fluorides generally give Hoffman’s elimination product. This is because the leaving group of alkyl fluorides is fluoride ion, which, as a strong base, can be very nucleophilic. The mechanism for the reaction involves a sequence of two SN2 processes:
First step: SN2 nucleophilic attack by fluoride ion to form an allylic carbocation intermediate.
Second step: In this step, there is an electrophilic attack by an H+ on carbon atom 5 to generate a pentacoordinated phenolate anion and a tetrahedral carbon-fluorine bond (this process is known as S1).
The Leaving Group Of Alkyl Fluorides Is Fluoride Ion, Which, As A Strong Base, Can Be Very Nucleophilic.
The fluoride ion is a strong base, which means that it can be very nucleophilic. This makes it a good leaving group (a group that is easily replaced by another molecule).
The Mechanism For The Reaction Involves A Sequence Of Two Sn2 Processes.
The mechanism for the reaction involves a sequence of two Sn2 processes. The first step is the alkylation of the carbon atom, which forms an alkoxide ion when protonated by water. In this step, an alkyl group is transferred from BH3 to formate and generate a carbocation intermediate (a carbon-centered species with a positive charge). In the second step, an alcohol molecule attacks this carbocation at its most acidic carbon (as shown here), forming another carbocation intermediate with added oxygen in its place (see below).
In The First One, The Hydrogen Atoms Of An Sp3 Carbon Atom Is Replaced By An Alkyl Group (R) To Produce A Primary Alkyl Halide.
In the first one, the hydrogen atoms of an sp3 carbon atom are replaced by an alkyl group (R) to produce a primary alkyl halide.
In A Second Step Another Alcohol Molecule Replaces The Other Hydrogen Atom On The Carbon Atom Giving Hoffman’s Elimination Product.
Hoffman’s elimination is a two-step process. The first step is the addition of an alkyl group to the carbon atom, forming a carbocation intermediate. In the second step, another alcohol molecule replaces the hydrogen atom on that same carbon atom, resulting in Hoffman’s elimination product.
Why do alkyl fluorides generally give the hoffman elimination product?
Why do alkyl fluorides generally give the Hoffman elimination product?
The leaving group is fluoride ion. The mechanism for this reaction involves a sequence of two Sn2 processes. In the first one, hydrogen atoms of an sp3 carbon atom are replaced by an alkyl group (R) in order to produce a primary alkyl halide:
This article was written to give you an understanding of the mechanism behind this reaction. We hope that it has been helpful and look forward to reading your comments below!
Answers ( 4 )
Alkyl fluorides generally give Hoffman’s elimination product. This is because, the carbon–fluorine bond in fluorides is the weakest among all the halogens and hence, cleaved easily. Due to this, steric effects are observed to be more pronounced than electronic effects. Hence, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product
Alkyl fluorides generally give Hoffman’s elimination product.
Alkyl fluorides generally give Hoffman’s elimination product. This is because:
This is because, the carbon–halogen bond in fluorides is the weakest among all the halogens and hence, cleaved easily
(1) The carbon–halogen bond in fluorides is the weakest among all the halogens and hence, cleaved easily.
(2) When an alkyl fluoride acts as a nucleophile, it donates its electron pair to the carbonyl carbon atom. This results in formation of a covalent bond between the carbonyl carbon and fluoride ion.
(3) Carbon–fluorine bond is stronger than carbon–chlorine bond because of higher electronegativity of fluorine compared to chlorine.
Due to this, steric effects are observed to be more pronounced than electronic effects.
The halogen atom in an alkyl fluoride, like any other halide, can be regarded as a Lewis acid. In the presence of a Lewis base, such as ammonia or pyridine, it is possible for both steric and electronic effects to be observed. This is because the alkyl group may not be able to reach all the way around the halogen atom because of its bulkiness, but still has some effect on its bonding interactions with other groups.
Hence, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product.
Therefore, fluorides generally give E2 reaction (concerted mechanism) giving Hoffman’s product.
This is because:
In summary, E2 reactions typically give Hoffman’s product when an alkyl fluoride is used as the electrophile. This is because the alkyl fluoride is able to donate its fluorine atom to form a carbocation intermediate (see below).
Hence, it can be concluded that the carbon–fluorine bond is the weakest among all the halogens and hence, cleaved easily. In addition to this, the steric effects are more pronounced than electronic effects which leads to an E2 mechanism for Hoffman elimination product formation.
Hofmann’s elimination (also known as the E2 reaction) is a method of organic synthesis that involves the elimination of hydrogen halides from alkyl halides. The reaction is named after the German chemist Ernst Otto Fischer, who first described it in 1883. It can be used to synthesize esters from alcohols and carboxylic acids, as well as other products:
The carbocation formed by the E1cB pathway is stabilized by the fluorine substituent
The carbocation formed by the E1cB pathway is stabilized by the fluorine substituent. The strong electron withdrawing effect of fluorine helps to stabilize the cation, since it has a much greater electronegativity than hydrogen. Fluorine also has a large electron density that can be used to delocalize electrons to make bonds more stable, something that’s often seen with leaving groups such as water or bromide ions. The high electronegativity of fluorine gives it an unusually strong bond with carbon atoms in molecules like alkanes and alkyl chlorides, making those molecules more acidic than similar compounds with other halogens like chlorine or bromine
E2 gives the product with an antiperiplanar hydrogen beta to the leaving group.
The stereochemistry of the product is determined by the beta hydrogen. The beta hydrogen is antiperiplanar to the leaving group, so it will add anti to form a chair as shown below.
Hofmann’s eliminations are observed for alkyl fluorides.
In the Hofmann’s elimination, the hydrogen is replaced by a fluorine.
This reaction is an E2-like reaction in which a fluorine acts as an electrophile to attack an alkyl halide. As you can see from this mechanism, it does not require any additional acid or base to be added to aid in deprotonation of the carbocation intermediate.
In conclusion, Hofmann’s eliminations are observed for alkyl fluorides. This is because the carbocation formed by E1cB pathway is stabilized by the fluorine substituent. The hydrogen beta to the leaving group of this species is antiperiplanar to it, which gives rise to E2 mechanism as a result.
In this chapter you will consider what would happen if the nucleophilic water molecule in an S N 2 reaction were replaced by a tertiary carbon-based nucleophile. In the first reaction of this chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2. There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book. The Hofmann product (2) was obtained almost exclusively under these conditions. It is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism. A subsequent reaction of the same product in 3 with Me2NH followed by methanol gave 4 without rearrangement. If we compare this figure with Fig 1 1 (p 6), we can see that the conformation of substrate has not changed much on loss of fluoride and that product 4 is formed with retention configuration in this case
In this chapter you will consider what would happen if the nucleophilic water molecule in an S N 2 reaction were replaced by a tertiary carbon-based nucleophile.
Imagine you were a student in an organic chemistry course, and the professor asked you to consider what would happen if the nucleophilic water molecule in a S N 2 reaction were replaced by a tertiary carbon-based nucleophile. Specifically, imagine that this hypothetical reaction involved an alkene geometrically similar to butane, with two alkyl groups bonded to its terminal carbon atoms. The following diagram shows such a molecule:
The alkoxide ion, CH3CH2CH2O–, is obviously a good nucleophile for reactions with primary and secondary alkyl halides, but what about tertiary halides?
The alkoxide ion, CH3CH2CH2O–, is obviously a good nucleophile for reactions with primary and secondary alkyl halides, but what about tertiary halides?
The answer lies in the fact that tertiary alkyl groups are more electron withdrawing than primary or secondary ones. You might be able to guess that this means they’re harder for your nucleophilic reagent to get close to, which makes it harder for the reaction to occur. In order for an E1 mechanism (especially) to work, the transition state has to have a lot of sigma bonding between carbon and oxygen atoms; this means there needs to be a lot of σ donation from oxygen toward carbon (or vice versa), not just from carbon toward oxygen. And if you recall from our discussion above regarding carbocations in terms of stability when compared with other carbocations like methylene carbocation or vinyl cation, you should know that even though these ions are electronically similar (both having two sp³ hybridized bonds), they differ in how much resonance stabilization they receive—methylene gets much more resonance stabilization than vinyl does because it has an extra p-orbital available; so too do tertiary carbocations get more resonance stabilization than secondary ones because their extra π bond makes them less bent and therefore less reactive than the latter type
The first reaction of the chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2.
The first reaction of the chapter, from 1980, was a non-concerted E1cB elimination of the tertiary alcohol from 1 to 2. The mechanism of this reaction has been discussed in our textbook [1]. It is known that the alkoxide ion is a good nucleophile and can attack at either C2 or C3 with retention of configuration. The process may be described as follows:
There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book.
There are many examples of nucleophilic displacement reactions on tertiary carbon centres in this book.
In the case of the alkyl fluoride, you can see that since the leaving group is not a primary or secondary carbon centre, but rather a tertiary carbon centre, it must be displaced by an incoming nucleophile (a hydrogen ion). The end result is elimination product formation.
The Hofmann product (2) was obtained almost exclusively under these conditions.
The Hofmann elimination is an E1cB reaction in which a tertiary halide undergoes nucleophilic substitution with a tertiary alcohol to form an alkoxide ion, which then eliminates HF. The mechanism for this reaction begins with a bimolecular nucleophilic substitution (Sb) of the halide by the alcohol. The alkoxide ion is then formed from protonation of the intermediate and then eliminated as HF, which results in cleavage of both bonds to give the final product.
It is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism.
At lower temperatures, it is likely that there is a change in the mechanism at lower temperatures and that reactions occur by the concerted E1cB mechanism.
A subsequent reaction of the same product in 3 with Me2NH followed by methanol gave 4 without rearrangement.
The reaction of 5 with Me2NH gives the same product as that obtained from 4, namely 6. This indicates that the intermediate formed in the Hofmann elimination is an enolate anion which can be trapped by a nucleophile without rearrangement to give a different compound.
If we compare this figure with Fig. 1.1 (p. 6), we can see that the conformation of the substrate has not changed much on loss of fluoride and that the product, 4, is formed with retention of configuration in this case.
Elimination reactions are governed by sterics rather than electronic effects
The Hoffman elimination reaction is a good example of this phenomenon. In the bulk, alkyl fluorides generally give the product with excellent regioselectivity. But when a small amount of solvent is added, rapid dissolution of the fluoride occurs and all products can be formed equally well. This is because solvation affects steric effects more than electronic effects, thereby shifting from high-valent to low-valent alkyl intermediate structures (see Figure 1).
In conclusion, we can say that the Hofmann elimination product is generally formed. The mechanism for this is concerted and involves a nucleophilic attack on the tertiary carbon centre by water (or methanol). This occurs via an intermediate alkoxide ion.
The Hoffman elimination is an organic reaction in which the hydrogen atoms of an alkyl halide are replaced with hydroxyl groups to produce a primary alcohol. It is named after Hermann Ludwig Ferdinand von Hoffmann who first reported the reaction in 1881. The mechanism involves two steps:
Alkyl Fluorides Generally Give Hoffman’s Elimination Product. This Is Because
Alkyl fluorides generally give Hoffman’s elimination product. This is because the leaving group of alkyl fluorides is fluoride ion, which, as a strong base, can be very nucleophilic. The mechanism for the reaction involves a sequence of two SN2 processes:
First step: SN2 nucleophilic attack by fluoride ion to form an allylic carbocation intermediate.
Second step: In this step, there is an electrophilic attack by an H+ on carbon atom 5 to generate a pentacoordinated phenolate anion and a tetrahedral carbon-fluorine bond (this process is known as S1).
The Leaving Group Of Alkyl Fluorides Is Fluoride Ion, Which, As A Strong Base, Can Be Very Nucleophilic.
The fluoride ion is a strong base, which means that it can be very nucleophilic. This makes it a good leaving group (a group that is easily replaced by another molecule).
The Mechanism For The Reaction Involves A Sequence Of Two Sn2 Processes.
The mechanism for the reaction involves a sequence of two Sn2 processes. The first step is the alkylation of the carbon atom, which forms an alkoxide ion when protonated by water. In this step, an alkyl group is transferred from BH3 to formate and generate a carbocation intermediate (a carbon-centered species with a positive charge). In the second step, an alcohol molecule attacks this carbocation at its most acidic carbon (as shown here), forming another carbocation intermediate with added oxygen in its place (see below).
In The First One, The Hydrogen Atoms Of An Sp3 Carbon Atom Is Replaced By An Alkyl Group (R) To Produce A Primary Alkyl Halide.
In the first one, the hydrogen atoms of an sp3 carbon atom are replaced by an alkyl group (R) to produce a primary alkyl halide.
In A Second Step Another Alcohol Molecule Replaces The Other Hydrogen Atom On The Carbon Atom Giving Hoffman’s Elimination Product.
Hoffman’s elimination is a two-step process. The first step is the addition of an alkyl group to the carbon atom, forming a carbocation intermediate. In the second step, another alcohol molecule replaces the hydrogen atom on that same carbon atom, resulting in Hoffman’s elimination product.
Why do alkyl fluorides generally give the hoffman elimination product?
Why do alkyl fluorides generally give the Hoffman elimination product?
The leaving group is fluoride ion. The mechanism for this reaction involves a sequence of two Sn2 processes. In the first one, hydrogen atoms of an sp3 carbon atom are replaced by an alkyl group (R) in order to produce a primary alkyl halide:
This article was written to give you an understanding of the mechanism behind this reaction. We hope that it has been helpful and look forward to reading your comments below!