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Why is Iodination of Benzene Difficult?

Here we will try to understand 'why Iodination of Benzene is difficult?' by looking at the reaction mechanism, nature of the reactants and also try come up with few steps to overcome the challenge.

For a faster electrophilic aromatic substitution reaction, any one of the reagent Benzene or the halogen in the starting material has to be slightly more reactive than the other. For satisfying this condition, electron donating groups attached to the phenyl ring making it more nucleophilic are preferred over unsubstituted Benzene. Also, the electrophilicity of the halogen is increased by using a Lewis acid catalyst thereby making it more reactive. These changes in turn help to achieve transition state faster and stabilize it better.

electrophilic aromatic iodination


For the halogens, the electronegativity and electrophilicity decrease from F to I in the periodic table. Fluorine is most electrophilic, and Iodine is least. Therefore, Fluorination is highly reactive, and Iodination is highly unreactive for electrophilic aromatic substitution reactions.

The exothermic rates of aromatic halogenation also decrease from Fluorine to Iodine. Fluorination reaction being highly exothermic and explosive, the reaction cannot be controlled resulting in polyfluorinated products. For Iodination, the reaction is endothermic with 12kJ/mol of energy absorbed. Therefore, it cannot be done using the conventional method using Lewis acid catalyst and requires strong oxidizing agents.


reactivity halogens iodination of benzene

This is because, I2 adds to the Benzene reversibly generating HI. HI being a strong reducing agent regenerates I2 from aryl iodide giving back the aromatic hydrocarbon. However, in the presence of oxidizing agents such as HIO3, HI is converted back to Iodine thereby increasing the concentration of Iodine in the reaction mixture. According to Le-Chatelier’s principle, if the concentration of one of the reagents is increased then the equilibrium shifts in the forward direction to give aryl iodide as the desired product.

Another way to obtain aryl iodide is to remove HI as soon as it is formed in the reaction mixture, by forming salts. For example, when the Mercuric oxide is used, it converts Hydrogen Iodide to Mercuric Iodide that is then discarded.

Iodination benzene methods

As we know that for faster reactions, the nucleophilicity of the ring should be increased by adding electron donating groups or the electrophilicity of the electrophile that is halogen should be increased but unsubstituted Benzene is weakly nucleophilic. Also the Iodine is non-reactive and non-polar but its electrophilicity can be increased by forming a positive ion. These ions are more reactive than the complex of halogen with Lewis acid catalyst and react instantaneously with Benzene to give aryl iodide.  

Increase electrophilicity halogen

Oxidizing agents, when used, oxidize I2 to iodonium ion I+. The earliest reported reaction was using HNO3 or H2SO4 in the presence of Acetic acid and Iodine to give iodonium ion and was called Tronov-Novikov reaction.

Iodine can be easily oxidized using, nitric acid. HNO3 in the presence of an acid and Iodine generates Iodine cation that reacts with Benzene giving Iodobenzene, Nitrogen dioxide, and water. As HNO3 is consumed in the reaction, hence it is a reagent and not a catalyst.

Iodination of benzene HNO3 nitric acid iodine

A similar mechanism is seen for Cupric Chloride and Hydrogen peroxide for generation of Iodine cation.

iodination benzene CuCl

Few examples of other oxidizing agents are Oleum, Cupric chloride, Silver sulfate, Sodium periodate, Periodic acid, Sodium hypochlorite, Iodic acid (HIO3), Nitric acid and Sulfuric acid, Sulfur trioxide, and Hydrogen peroxide. Sometimes Lewis acid catalyst is used along with the oxidizing agents.

The best reagent for Iodination is Iodine monochloride ICl, an interhalogen compound. Chlorine is more electronegative and pulls the electron cloud towards itself giving Iodine a (delta) positive charge. The Benzene ring can pick up this quickly than Iodine from the non-polar I2 bond thereby giving aryl iodide as the desired product.

iodination benzene iodine monochloride ICl

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Nature of Carbon-Halogen Bond

In this section, we will learn about the nature of Carbon-Halogen Bond and the trends observed in Electronegativity, Bond Polarity, Bond Length, Bond Enthalpy and the Dipole Moment. 

Nature of Carbon-Halogen Bond

Halogens on forming a covalent bond with Carbon pulls the electrons in the bond towards itself. This property is known as the electronegativity. The unequal sharing of the electron cloud results in electron-deficient Carbon to get a partial positive charge denoted as (delta)+ and the electron-rich halogen atom to get a partial negative charge shown as (delta)-. This charge separation results in making Carbon-halogen bond, a polar covalent bond.


polar covalent bond electronegativity difference

Electronegativity value changes with the atomic size and the atomic number. It increases if we go from left to right and from bottom to top of the periodic table making Fluorine the most electronegative element. Linus Pauling assigned it a value of 4 and calculated the electronegativity values of other elements relative to Fluorine. This scale is called Pauling scale of electronegativity.

In general, a bond of Carbon with more a electronegative element from group 15-17 is considered polar with Carbon carrying (delta)+ charge. However, when carbon forms a bond with metal atoms, its polarity is reversed as Carbon is now more electronegative and thereby carries a (delta)- charge.

Electronegativity difference to predict nature of bond

Electronegativity difference is a great tool to predict the nature of bond whether it is polar, non-polar or ionic. If it is less than 0.5, then it is a non-polar covalent bond. Example, Carbon-Hydrogen or Hydrogen-Hydrogen. If electronegativity difference is in the range of 0.5-2, the bond is a polar covalent bond. A carbon-halogen bond is a perfect example so is the hydrogen-fluorine bond. For values above 2, the bond is an ionic bond. For example, Silver Fluoride, Potassium Fluoride, Sodium Bromide and Sodium Chloride. Therefore, due to decreasing electronegativity of the halogens, as we go down the group 17 of the periodic table, a Carbon-Fluorine Bond is found to be the most polar followed by Carbon-Chlorine, Carbon-Bromine and least is for Carbon-Iodine.

Nature of Carbon-Iodine bond

nature of carbon halogen bond

If you take a look at the electronegativity values of Carbon and Iodine; for both, it is 2.5. That means, there is not much difference in the electronegativity, and hence, it would be treated as a non-polar bond. However, polarity not only depends on the electronegativity difference but also on the polarizability of the atoms involved in a covalent bond.

'Polarizability is the ability of the electrons in the atom to distort or shift due to external influences.' It could be an external electric field or presence of polar molecules nearby. Iodine due to its position in the periodic table has large size due to the involvement of more shells and therefore, the higher number of electrons. The large size makes it difficult for the attractive force of the nucleus to pull the outside electrons towards itself thereby decreasing the nuclear charge. These outside electrons are therefore available for distortion or dispersion depending on the nature of the surrounding medium.

Let us take the example of Methyl Iodide which is a Carbon-Iodine bond. Even though there is no electronegativity difference between them, it has a polarizable atom, Iodine. So in a chemical reaction, due to the presence of the other polar molecules in the reaction medium, the polarity is induced in Carbon-Iodine bond. Iodine being a polarizable atom would pull the electron clouds towards itself carrying a partial negative charge and carbon, partial positive charge thereby making Carbon-Iodine a polar bond.

Two things we learned here, that electronegativity and polarizability of halogen influence the polarity of carbon-halogen bond and also, electronegativity and bond polarity decrease if we go down in a group from Fluorine to Chlorine, Bromine, and Iodine. 

Bond length

Another change that happens is in the bond length. As new shells are added to the group, and size of the halogen increases; and on forming a bond with Carbon, its bond length would also increase. As Fluorine is the smallest element and Iodine, the largest, Carbon-Fluorine Bond would be smallest and Carbon-Iodine bond the longest. 

Bond Enthalpy of Carbon-Halogen Bond

Due to the increasing bond length between the Carbon-Halogen, the energy required to break such a bond would be least. This energy is called bond enthalpy and it minimum for Carbon-Iodine bond and maximum for, smallest Carbon-Fluorine Bond. As bond enthalpy is high for Fluoroalkanes, it is highly unreactive, and Iodoalkanes are very reactive.

Dipole moment

Another physical property affected by the bond polarity is the dipole moment. Dipole moment calculates the total molecular polarity by taking into consideration the electronegativity difference leading to separation of charges and the distance between the two charges (µ= q xd). As electronegativity and the bond polarity decreases from Fluorine to Iodine, the dipole moment is expected to decrease. However, the dipole moment of Chloromethane is greater than that of Fluoromethane because the bond length of the Carbon-Fluorine is shortest even though it is the most polar.

Attempt this quiz to check your understanding of this concept.

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Preparation of Amines

In this section, we will learn about the various methods for the preparation of amines and tips to remember those using mnemonics.

A way to remember all the preparation methods is by memorizing this statement,

synthesis of amines

‘In search of amines, two brothers, Schmidt-Hofmann-Gabriel-Hofmann, wanted to reduce ANI-ANO. ’ in which, ANI stands for Amide, Nitrile, and Imine and ANO for Azide, Nitro and Oxime.

The names of the two brothers are for four name reaction for the preparation of amines, namely:

1) Schmidt Reaction

2) Hofmann degradation of primary amides

3) Gabriel Phthalimide reaction

4) Hofmann Ammonolysis

From the statement, we learn that the two brothers want to Reduce ANI-ANO. It means that all the methods are for functional group reduction specifically, nitrogen-containing functional groups.

The amide, in general, is represented as carbonyl attached to N, nitrile also known as cyano is written as C triple bond N, imine is a functional group containing Carbon-nitrogen double bond. Azide functional group is shown as N3; Nitro is shown as NO2, and oxime is similar to imine in having carbon-nitrogen double bond with nitrogen attached to hydroxyl group shown as C=N-OH.

The Most commonly used reducing agents for amine preparation are H2 gas in Raney Ni and LAH.

The first A of the ANI-ANO stands for Amide, and the reaction is the reduction of the amide functional group to an amine.

What we observe is that primary, secondary and tertiary amides are reduced to the respective amines using LAH in solvent diethyl ether.  In the second step, water is added to end the reaction and to isolate the product.

synthesis of amines from amides

From the general reaction scheme, we can see that the change is from CO of the amide to the CH2 of the amine. This change is brought about by reducing agent LAH that is the best reducing agent for the reduction of carbonyls such as amides, esters, aldehydes, ketones  etc. It reduces amide by donating hydride, i.e., H-. that is an anion of hydrogen.

All the four hydrides of LAH can be used in the reduction, and the method can be best explained by showing you the example of a reduction of primary alkyl amide to the primary alkyl amine.

The hydride of the LAH adds to the carbonyl of the amide to give an intermediate imine, with one hydrogen coming from LAH. This imine is highly reactive due to the positive charge on electronegative nitrogen. It then undergoes further addition of the hydrogen from LAH to give primary alkyl amine as the final product. This method can be used for the synthesis of alkyl amine and aryl amine from alkyl or aryl amides.

Let’s look at a few examples;

When acetamide is treated with LAH, it gives ethyl amine.

Similarly, secondary alkyl amide, N-methyl acetamide undergoes reduction to give ethyl methyl amine.

The third example is of tertiary alkyl amide, N,N-Dimethylacetamide that gets reduced to form ethyl dimethyl amine.

Benzamide is the case of aromatic amide that gives benzylamine on treatment with LAH.

Test your knowledge by attempting this quiz.

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Electrophilic Aromatic Substitution

In this section, we will look into nature, mechanism and the reactivity of electrophilic aromatic substitution reaction in which the electrophile is a halogen.

Benzene is an aromatic hydrocarbon with pie bonds similar to an alkene. However, in case of an alkene, halogen adds instantaneously across the double bond by the anti-addition method and the reaction is exothermic. The halogen addition across the double bond is not observed in the case of Benzene under standard reaction conditions.

If an analogous comparison is done between halogen addition across alkene and Benzene, it is observed that for Benzene, the reaction is thermodynamically unfavorable and endothermic due to the decrease in entropy and loss of the ring aromaticity.

why does benzene undergo substitution reactions

Benzene is an aromatic hydrocarbon but behaves differently from alkane hydrocarbons. It does not undergo free radical halogenation reactions like alkanes in the presence of a halogen source and UV light.

Take the example of Methane; it undergoes free-radical substitution reactions that give a mixture of mono and polyhalogenated products.


Benzene under similar reaction condition gives a polychlorinated hydrocarbon; Benzene hexachloride (BHC) also known as 1,2,3,4,5,6- hexachlorocyclohexane as the product by losing its aromaticity entirely. Such kind of polyhalogenation is observed only for highly reactive halogen Chlorine and not for Bromine.

If an alkyl group is attached to Benzene, then the alkyl group behaves like an alkane to give mono and polyhalogenated product but Benzene ring is unaffected in this case.

To summarize, we can say reactions that favor aromaticity of the Benzene ring is preferred over-reactions that destroy it. Also, halogen addition to the ring does not occur by free-radical substitution reactions or by electrophilic addition reactions. It occurs by electrophilic substitution reaction on treatment with halogen and a Lewis acid catalyst in which a proton is substituted with the halogen. This suggests that only an electrophile can attack the ring and atoms can only the attack the alkyl substituent attached to the phenyl ring.

As the electrophiles, in this case, are the halogens, and we know that the electronegativity and electrophilicity decrease from F to I in the periodic table. Fluorine is most electrophilic, and Iodine is least. Therefore, fluorination is highly reactive, and iodination is highly unreactive for electrophilic aromatic substitution reactions.

The exothermic rates of aromatic halogenation also decrease from Fluorine to Iodine. That means Fluorination is highly exothermic and can be explosive. Such reaction is hard to control at the mono fluorinated stage and usually gives polyfluorinated products.

For Iodination, the reaction is endothermic and cannot be done using conventional method using Lewis acid catalysts and requires other oxidizing agents.

The other method to synthesize 1- fluorobenzene is by diazotization method starting from Aniline. The diazonium salt formed is treated with Fluoroboric acid giving the desired product.

For Chloro substituted Benzene, the reaction is done using Chlorine gas and Lewis acid catalysts such as AlCl3 or FeCl3. Similarly, for a Bromo substituted product, Bromine and Lewis acid catalyst such as AlBr3 or FeBr3 is used.

Let us look at the general reaction mechanism and discuss the changes happening using an energy profile diagram for more understanding;


electrophilic aromatic substitution mechanism

In the first step, the pie electrons in the ring accept electrophile and forming a carbocation intermediate. This intermediate formed has sp3 carbon and is not aromatic.

As the bond formed with the electrophile is a sigma bond, the complex is called a sigma complex. As aromaticity of the ring is lost in this step, it is a slow step and, therefore, the rate-determining step of the reaction. It is also endothermic in nature with almost 36 kcal of energy consumed. The intermediate carbocation is then stabilized by resonance lowering the energy; with the positive charges delocalized at the ortho and the para positions.

In the next step, the base tries to abstract the proton resulting in an energy rise and finally, due to proton abstraction, the aromaticity is regained with significant loss in energy. In the end, a stable product with less energy than the starting material is formed.

Why is the substitution of a proton with a halogen favorable reaction? This is because, as compared to proton, halogen with its electron-donating resonance effect and electron-withdrawing inductive effect makes the product more stable than the starting material. The halogens, tries to stabilize the carbocation arenium ion at the transition state by its stronger –I effect than weaker +R effect. So, any group that stabilizes the carbocation will favor the transition state. Carbocation can be stabilized by making the ring more nucleophilic that is by increasing the electron density of the ring by introducing electron donating groups. For e.g. groups such as –CH3, -OCH3, -NH2, -NR2

A look at Bromination reaction mechanism will help us understand more. As the halogens are homonuclear diatomic molecules and electrophilicity decreases down the group, its reactivity towards Benzene ring is low. The reactivity of the halogens is increased by making it more positive using Lewis acid catalysts.

Lewis acid catalysts make it more electrophilic than diatomic Br2 but less than when it exists as bromonium ion. Lewis acid such as Fe or Al has an empty orbital that attracts the lone pair of electrons from one of the bromine in the bond making the terminal Bromine quite electrophilic. This terminal halogen is picked by the Benzene ring for the reaction, and the intermediate formed is stabilized by resonance. The central electrophilic Bromine cannot be picked as it is already trivalent and cannot expand its valency to four.

The Lewis acid complex FeBr4 then acts as a base by picking up the proton. In the end, the Lewis acid catalyst is regenerated for taking part in the reaction again along with Bromobenzene and HBr as the final products. The overall reaction is exothermic releasing 45kJ/mol of energy.

In short for faster reactions, groups that increase nucleophilicity and electrophilicity favor the reaction by stabilizing the transition state.

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Why Anilines and Phenols do not undergo Friedel Craft Acylation Reaction?

Friedel Craft acylation reaction is a type of electrophilic aromatic substitution reaction wherein the hydrogen of the aromatic ring is substituted with the acyl group (R-CO-), and a new Carbon-Carbon bond.  The acylation reaction requires a Lewis acid catalyst such as AlCl3 and an acid chloride (R-CO-Cl). The byproduct formed is HCl.

Friedel Craft carbon carbon bond formation reaction

Under the Friedel Craft Reaction condition, the aniline (Lewis base) binds to the electrophile AlCl3 (Lewis acid) to give a coordination complex (salt).

Acid + Base = Salt

The positive charge on the nitrogen is electron withdrawing, and it pulls the electron density of the ring by negative inductive effect (-I effect). It therefore, deactivates the ring for further acylation reactions. The complex precipitates out of the reaction mixture and the acylation reaction is not observed. 

friedel craft acylation of aniline

A similar case is seen for mono and disubstituted anilines.

Friedel craft reaction substituted anilines

In the case of Phenols, the acylation takes place at the oxygen (O-acylation) and not at the carbon of the ring (C-acylation) to give a phenolic ester. 

Friedel craft reaction phenol

Nevertheless, a method does exist to convert the O-acylated product to the C-acylated product. The O-acylated product after isolation is treated with an excess of AlCl3, a reaction known as   Fries rearrangement to obtain ortho and para hydroxy acetophenone. The ortho and the para isomer can be later separated by chromatographic techniques.

fries rearrangement phenol

The Phenol does not form a coordination complex with AlClsimilar to the aniline as the lone pair on the OH is tightly held and is less basic than the -NH2 group.

But if the oxygen of the phenol is substituted, (example, Anisole) it then prefers to undergoes acylation reaction at the carbon of the benzene ring (C-acylation). The para isomer would be the major product.

anisole friedel craft reaction

In summary, Phenols do not undergo FC acylation reaction because of O-acylation and in the case of Anilines, it forms a salt with AlCl3 and precipitates out of the reaction mixture. 

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Suggested Reading


Definition- Annealing

The process of heating a solid metal or glass to a specific high temperature and gradual cooling is called annealing. The physical property(strength, elasticity and crystalline property) of a solid  can be altered by this method.

what is annealing definition


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