Its basic character. .Consequently, ammonia is a base and a nucleophile.Nitrogen is less electronegative than oxygen, so ammonia is a much stronger base than water, and also a much better nucleophile.Additionally, amino acids, which are organic derivatives of ammonia, are also tetrahedrally hybridized and are comparable to ammonia in terms of basicity and nucleophilicity.The carboxylic acid provides complete neutralization (protonation) of amines.

Often, the acidity of amines is discussed in relation to the acidity of the corresponding conjugate acids. .A corresponding concept, pKb, has already been developed to measure the acidity of Bronsted acids, and we will see that these two quantities are very closely related to each other.Here are the corresponding acid dissociation temperatures for ammonia in dilute aqueous solution, and a secondary, tertiary, and primary amine:


In terms of pKa, ammonia has the lowest value.This means that ammonia is the weakest of the four bases.

You should also note that the alkylstabilizing effect is purely inductive!

In the dimethylammonium ion, the second alkyl group has only a very slight influence, whereas the third group (in the trimethylammonium ion) causes an increase in acidity (decrease in basicity) relative to the dimethylammonium ion.Ionized trimethylammonium is less acidic than ammonia, however.

There is no difference between primary and secondary amines in terms of basicity.

Third alkyl groups also inhibit solvation sterically.In the presence of three alkyl groups, the solvent's ability to stabilize the corresponding ammonium ion is weakened dramatically, causing a reversal of the alkyl group's tendency to decrease acidity and increase basicity.

In water, pKa + pKb = 14 in respect of the relationship between pKb and pKa.

The following is an explanation of pKbis:


Ammonia, methylamine, dimethylamine, and trimethylamine have pKbs of 4.74, 3.34, 3.27,.4.19.In relation to their pKb values, the strongest bases have the least positive values, as was the case for acidity when comparing pKa values.

AMINES ARE THE MOST BASIC CLASS OF ORGANIC COMPONENTS. AMINES ARE DISTINCTIVE FROM ORGANIC COMPONENTS WITH OTHER FUNCTIONALITIES.These are one of the only organic compounds that are substantially basic in aqueous solution and that can be protonated completely by diluted solutions of strong acids.Assuming the R groups are small enough, the protonated alkyl ammonium ions will form salts, which is water soluble (I assume).This means that amines can be separated from other compounds such as halides, ethers, alcohols, and ketone (not to mention alkanes, alkenes and alkynes, of course), using a simple extraction method.

Soluting a mixture of organic compounds in an organic solvent such as ether with dilute aqueous acid results in an exothermic reaction.

As a result, the amine is protonated and released as an ammonium salt into the aqueous solution, while other functionality, such as ketones, remains in the organic process.

.As a result of drying and evaporation, amine is obtained.

Primary and secondary amines, like ammonia, have protic hydrogens, so they are acidic (unlike tertiary amines, which lack acidic hydrogen).Ammonia has a pKa value of 38, and is a very weak acid.Secondary amines and primary amines both have pKa's that are very similar in magnitude.Thus, such amines are considerably more basic (pKb about 4) than they are acidic (pKa 38), so that their aqueous solutions are relatively strongly alkaline.

Because amines are weakly acidic, their conjugate bases - RNH- or R2NH- are very strong bases!

.Since the fourth group is an electron pair, all four groups are different from one another.

It has been found that when chiral amines are prepared, they are very rapidly inverted by anumbrella-like procedure to make their corresponding enantiomer, resulting in speedy racemization.Certain amines, for which this process is particularly difficult, are readily prepared and stable as a single enantiomer.

Added to a tetraalkylammonium ion with a fourth R group, this inversion cannot happen and the quaternary ammoniumions can thus be chiral and stable.


Given that amines are relatively basic functional groups, they are also likely to be rather nucleophilic.

As amines are far more basic than oxygenated functional groups such as alcohols and ethers or ketones, they are also predicted to be much more nucleophilic as well.As a result, they can be used as nucleophiles in SN2 reactions with alkyl halides.The same holds true for ammonia, the inorganic parent of organic amines.

A reaction with an appropriate (methyl, primary, or secondary)alkyl halide in an SN2 reaction to prepare primary amines does work, but requires an excess of ammonia, because the primary amine is also reactive toward the alkyl halide.A secondary amine is produced, and even further reaction with alkyl halides results in tertiary amine.As a result, a mixture of primary, secondary, and tertiary amines will be formed unless excessive amounts of ammonia are used.

If ammonia is present in large excess over the alkylhalide, then the alkylhalide has much more ammonia than amine to react with.


Note that the initially formed product has alkylammonium cation, which cannot act as a nucleophile (no unshared electron pair) as it would not be able to react with alkyl bromide in order to give a dialkylamine.


Another way to form amines -- specifically and exclusively primary amines -- is to utilize another nitrogen nucleophile that is readily available, the azide anion.An alkyl halide can be used to create anorganic azide, which is then reduced to the primary amine using lithium aluminum hydride.We will not discuss the specific mechanism of this last reaction.


.Once formed, the organic azide cannot react with the alkyl halide.Consequently, we don't need to use the excess of the nucleophile to get only the primary amine.

.With lithiumaluminum hydride, these nitriles can also be reduced to primary amines.Primary amines have an extra carbon atom compared to alkylhalides.


As with nitriles, amides are also reducible.



An explanation of.

The nitro functionality (which is another nitrogen-containing functionality which can be easily introduced onto an aromatic ring, as you are aware) can easily be installed on an aromatic ring, followed by the reduction to the amino function.

It is essentially phenylamine (aniline), the simplest aromatic amine.It is easily synthesized as shown in the following reactions.


It is basic.

While it is less basic than normal alkylamines, replacing an alkyl group with an aryl or phenyl group will greatly diminish the basicity.

In this case, anilinium (the conjugate acid of aniline) has a pKa of 4.6 while methylamine has one of 10.7.In this case, the anilinium ion is a million times stronger acid than methylaminium. In consequence, aniline is a million times weaker base than methylamine!



In general, stabilization of the reactant side of the equation tends to reduce acidity(since the hydronium ion is on the right hand side of the equation), but stabilization of the product side tends to increase acidity.

It has been found that aniline is rather strongly stabilized by resonance, but anilinium is not.These resonance structures for aniline can be seen in the figure below, where it is seen that the ring is electron rich, with partial negative charge (carbanion character) at the ortho and para positions, while the nitrogen tends to be electrondeficient (partial positive charge).


This resonance or delocalization is possible because the unshared electrons on nitrogen are in conjugation with (capable of directly overlapping with) the 2p AOon the directly attached ring carbon.Electrons in this region are then delocalized to the positions indicated around the ring.The orbital image displays the following overlap:


A conjugation like this is possible only when the orbital external to the ring is benzylic (i.e., on an atom directly attached to the ring).

Aniline thus undergoes resonance stabilization, but methylamine does not, since the p-type orbital cannot overlap with the p-type orbital of this reactant.Thus, aniline is thermodynamically more stable than methylamine or anyalkylamine, and therefore less likely to be protonated (weak base).

Lastly, theanilinium ion (the conjugate acid of aniline) lacks this conjugated system because nitrogen is positively charged (highly electron deficient) and it therefore cannot contribute any electrons to the ring.Since it doesn't have any vacant orbitals available to accept electrons from the ring (violating the octet rule), it cannot accept electrons from the ring.Recall that the resonance structure below on the right is not a valid resonance structure.


The Pyridine.A pyridine is an aromatic amine, but in a different way and in a different sense than an aniline.It is essentially benzene with one of the CH groups of benzene replaced with a N atom.

As shown below, the unshared electron pair in pyridine is perpendicular to the pi system consisting of overlapping pz AO's, which means that it is not part of the aromatic system and is independent of it.

.In this regard, pyridine is less proton susceptible than typical aliphatic amines such as piperidine.Pyridine has a pKa of 5.25.

In addition, the pyridinium ion (the conjugate acid) remains aromatic because, when it is joined to a proton, it is in the trigonal plane, so there are no electrons removed from the pi electron system, or 6 pi electron system, just like benzene.



To begin with, theamide ions are even stronger basic than ahydroxide anion, so they would be terrible groups to leave.

In your opinion, can acids be used to generate better leaving groups than alcohols?

Since an acidic solution does not have a good base to abstract the beta protons, water will have to be the base in such a solution.There's a concern that ammonia is not an effective leaving group when water with a weak base is the best base available. It is not!

What are the solutions to this apparent dilemma?. .Because the amide ion is such a terrible leaving group, it cannot remain neutral unless it is converted to the ammonium form.

As amines are pretty decent nucleophiles, as well as bases, he can react with alkyl halides via SN2 displacement, as shown below.


First, primary amine is converted to secondary amine, then those secondary amines are converted to tertiary amine which then reacts with a third molecule of methyl iodide to produce quaternary ammonium salt.There are no more acidic protons at this point, so a base can be used to complete the E2 reaction.

The first step in doing this is to react the quaternary ammonium iodide with silver oxide, which in turn produces insoluble silver iodide (often called aquaternary ammonium hydroxide).Our departure group is reasonable at this point, which gives us a solid base.When trimethylamine is heated up to approximately 80 degrees, it usually gets eliminated.


Hoffmann Elimination Transition State.

This reaction, like all E2 reactions, is concerted.

We would therefore like to develop a transition state model for the reaction, which would allow us to rationalize and/or predict things like selectivity (especially regioselectivity).

As usual, we use canonical structures for the reactant and the product, and then an X structure for the non-reactant or non-product.



When we developed the TS model for the elimination of HX from an organic halide by a base such as hydroxide ion, we used only the reactant and product-like structures, so the TS only exhibited an alkene character.Alkyl halides are mainly characterized by their alkene character.

.It is crucial to include the complete chemical treatmentwhich reveals the carbanion character in this instance (eliminations that require an amine leaving group) since it is the carbanion character that is now prevailing over the alkene character, leading to a drastic change in the regiochemical selectivity.

The elimination of alkyl halides tended to favor the more highly substituted alkene, which is more stable.We called this Saytzeff regiochemistry.

.Hoffmannregiochemistry is a form of this regiochemistry.

The elimination of quaternary ammonium salts, for example, favors 1-butenes over 2-butenes despite the alkene character of the TSwhich would tend to favor the former.


Q These are the TS models of the two competing TSs:


As a result, the intermediate leading to 1-butene is favored, since it exhibits primary carbanion character, while the intermediate leading to 2-butene exhibits secondary carbanion character.

Carbunions are destabilized by alkyl groups (remember that alkyl groups donate electrons so they stabilize positive charge, while destabilizing negative charge. Compounds containing methyl are more stable than primary, secondary, and tertiary.

There is only one carbanion character in the TS leading to the 1-butene, and one for either of the 2-butenes.

It is approximately 90:10 for 1-to-2-butenes.

As a final question, we ask why there is a drastic change between the relative amounts of carbanion and alkene character in the two types of elimination (alkyl halides andalkylammmonium salts).In our drawing above, the last canonical structure for the TS is the one with the alkene character.In this structure, the leaving group contributes more energy to the structure (in this case it's left).The better the leaving group, the greater the alkene content of the structure.In contrast to alkylammonium ioneeximinations, chloride (or bromide) ions are much better leaving groups than trimethylamine, so alkyl halide eliminations tend to be more alkene-like.

qIt is intriguing that the fluoride ion, the worst of the halogen leaving groups, tends to have regiochemical preferences more similar to those of the alkylammonium ions, that is, favoring the less substituted and less stable alkene.


nitrous acid is readily able to convert all primary amines into diazonium salts. .A major cause of this high level of reactivity is the extremely thermodynamically stable state of dinitrogen, making it an outstanding leaving group.


It should be noted that the positive charge is delocalized, and is shared by both nitrogens.Although the positive charge on nitrogen is not very favorable (electronegative atom), resonance stabilization makes this ion stable enough to form.

The diazonium ions, on the other hand, readily decompose through an SN1 mechanism, with dinitrogen as the leaving group.

The very best of all leaving groups may be nitrogen because the molecule is so thermodynamically stable.When these diazonium ions are formed at ice bath temperatures, they will quickly lose nitrogen, forming a carbocation that will then react with available nucleophiles (e.g., solvent or chloride ion).


If R is an aryl group, such as phenyl, the diazonium ion is relatively stable at about zero degrees centigrade, but when heated it decomposes rapidly as it approaches room temperature.This allows the diazonium ions to react with substances after their generation.

It is thought that aryldiazonium ions are more stable than alkyldiazonium ions because the Ar-N bond is partially double in the resonance structure below. This is an additional small contribution over the main two resonance structures previously discussed.As a result, the positive charge is delocalized onto the orthoand para positions of the benzene ring due to the overlap between the pi systems of the N-N pi bond and benzene rings.


Anazo compounds

.Because of their resonance stabilization, they are relatively mild (not extremely reactive, but very selective) electrophiles.

Electrophilic aromatic substitution reactions are one of their main uses. .The react with amines that have the powerful electron-donating ability.The reaction of N,N-dimethylaniline with aryl diazonium ions can be seen in the following example:


A generic name for this product is azo compound.

qAzo compounds are densely colored compounds often used in textile dyes.

Additionally, aryldiazonium ions can be employed for converting amino groups in anilines or aniline derivatives to other functionality, such as halides or nitriles.As part of the process, a salt containing the desired nucleophile is added to a cold aqueous solution containing a diazonium ion, and the temperature is allowed to rise.By this process, the diazonium ion decomposes into a thearyl carbocation, which reacts then with a nucleophile.


A chloro, bromo, iodo, or nitrile function can be produced (or reduced to hydrogen using an appropriate reducing agent).

arylhalides cannot be substituted by either SN1 or SN2, since the double bond in the aryl halogen atom is too strong to break easily.On the other hand, evenaryl systems can undergo an SN1 substitution reaction when the potent leaving group is dinitrogen.