Haloalkanes and Haloarenes Class 12 NCERT notes

Haloalkanes And Haloarenes

HALOALKANES AND HALOARENES Class 12

When hydrogen atoms or atoms of alkanes are replaced by a corresponding number of halogen atoms, the compounds are called halogen derivatives of alkanes. They are classified according to the number of halogen atoms that replace hydrogen atoms in the alkane. Monohalogen derivatives: They contain only one halogen atom.

E.g. CH3Cl Methyl chloride

CH3CH(Br)CH3 2-bromopropane

Monohalogen derivatives of alkane are called alkyl halides. Dihalogen alkanes contain two halogen atoms. Trihalogen alkanes contain three halogen atoms.

Monohaloalkanes

The general formula is RX where R is an alkyl group and X is a halogen.

Haloalkanes and Haloarenes Class 12 NCERT notes

Nomenclature of Haloalkanes and Haloarenes

Common system:

‘Alkyl halides’ are the monohalogen derivatives of alkanes. These are named by naming the alkyl group attached to halogen and adding the name of the halide. E.g. Methyl halide, Isobutyl halide. The name of the alkyl group and halide are written as two separate words. The prefixes used to distinguish alkanes like n-, iso-, sec-, tert, etc. are also written.

IUPAC system:

Rules for naming haloalkanes that have branches in carbon chains:

The monohalogen derivatives of alkanes are called haloalkanes. The name of haloalkanes is written by prefixing the word ‘halo’ (bromo or chloro or iodo or fluoro) to the name of the alkane corresponding to the longest continuous carbon chain holding the halogen atom. E.g. Bromoethane E.g. Trichloromethane

(a) The longest continuous chain containing the carbon attached to the halogen group is selected as the parent alkane (principal chain or parent chain). While naming alkanes, all the rules that apply to alkane names should be followed.

(b) The carbon atoms are numbered in such a way that the halogen carrying carbon atom gets the lowest number.

(c) The position of the halogen atom and other substituents are indicated by numbers 1,2,3….etc. E.g. 1-Iodo-2-methylpropane

Dihalo derivatives

(a) When two halogen atoms are attached to the same Carbon-atom, these are called geminal dihalides. Alkylidene dihalides or alkylidene dihalides are also names used for such compounds. E.g. ethylidine dichloride

(b) When two halogen atoms are attached to adjacent Carbon-atoms, they are called vicinal dihalides. As they are prepared from alkenes, they are named as the dihalide of the alkene from which they are prepared. E.g. ethylene dichloride

Polyhalo derivatives:

Polyhalo derivatives are compounds with multiple halogen atoms. These have important application in the agricultural industry. Fully halogenated hydrocarbons are also called perhalohydrocarbons under a common system.

Nomenclature of aryl halides:

Aryl halides are termed Haloarenes in IUPAC systems. ‘Halo’ (bromo or chloro or iodo or fluoro) is prefixed before the name of the aromatic hydrocarbon. In the case of disubstituted compounds, the relative positions are indicated by (1,2), (1,3) or (1,4). Ortho, meta and para are also used to indicate the positions. E.g. Chlorobenzene, Bromobenzene.

PHYSICAL PROPERTIES OF ALKYL HALIDES

(a) Boiling point:

The below chart shows the boiling point of some simple haloalkanes.

Haloalkanes and Haloarenes Class 12 NCERT notes

Notice that three of these have b.ps’ below room temperature (taken as being about 20° C). These will be gaseous at room temperature. All the other you are likely to come across are liquids.

Haloalkanes and Haloarenes Class 12 NCERT notes

(b) Boiling point of some isomers:

The example shows that the boiling point fall as the isomers goes from a primary to a secondary to a tertiary haloalkane.

To put it simply, this is the result of the fall in the effectiveness of the dispersion forces. The temporary dipoles are greatest for the longest molecule. The attractions will also be stronger if the molecules can lie closely together. The tertiary haloalkane is very short and fat, and won’t have much close contact with its neighbours.

(c) Solubility of haloalkanes

(i) Solubility in water: The haloalkanes are very slightly soluble in water. In order to dissolve haloalkane in water, you have to break attractions between the haloalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Energy is released when new attractions are set up between the haloalkane and the water molecules. These will only be dispersion forces and dipole-dipole interactions. These aren’t as strong as original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules.

(ii) Solubility in organic solvents: Haloalkanes tend to dissolve in organic solvents because the new intermolecular attractions have the same strength as the ones being broken in the separate haloalkane and solvent.

CHEMICAL REACTIVITY OF HALOALKANES

The importance of bond strengths:

The pattern in strengths of the four carbon-halogen bonds are:

Haloalkanes and Haloarenes Class 12 NCERT notes

Bond strength falls as you go from C-F to C-I (C-F being the strongest)

In order for anything to react with the haloalkanes, the carbon-halogen bond has got to be broken. As that gets easier when you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are non-reactive and thus, not considered.

The influence of bond polarity: Out of the four halogens, fluorine is the most electronegative and iodine the least. This means that the electron pair in the C-F bond will be dragged most towards the halogen end. Let’s look at the methyl halides as a simple example:

Haloalkanes and Haloarenes Class 12 NCERT notes

One of the important set of reactions of haloalkanes is substitute reactions, which involves replacing the halogen by something else. These reactions involve:

(a) The carbon-halogen bond breaking to give positive and negative ions. The ion with the positively charged carbon atom then reacts with something either fully or slightly negatively charged. Or,

(b) Something either fully or negatively charged attracted to the slightly positive carbon atom and pushing off the halogen atom.

The thing that governs the reactivity is the strength of the bonds which have to be broken. It is difficult to break a C-F bond, but easy to break a C-I one.

CHEMICAL REACTIONS OF HALOALKANES

(a) Nucleophilic substitution reaction

(i) Nucleophilic substitution in primary haloalkanes

Nucleophiles:

A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge.

Haloalkanes and Haloarenes Class 12 NCERT notes

Nucleophiles are either fully negative ions, or have a strong –ve charge. Common nucleophiles are hydroxide ions, cyanide ions, water and ammonia. Notice that each of these contains at least one lone pair of electrons either on an atom carrying a full negative charge or on a very electronegative atom carrying a substantial-charge.

The nucleophilic substitution reaction –an SN2 reaction:

We’ll discuss this mechanism by using an ion as a nucleophile because it’s slightly easier. The water and ammonia mechanisms involve an extra step which you can read about on the pages describing those particular mechanisms. We’ll take bromoethane as a typical primary haloalkane. The bromoethane has a polar bond between the carbon and the bromine. We’ll look at its reaction with a general purpose nucleophilic ion which we’ll call Nu. This will have at least one lone pair of electrons. Nu could, for example, be OH or CN. The lone pair on the Nu ion will be strongly attracted to the +carbon and will move towards it and begin making a co-ordinate (dative covalent) bond. In the process, the electrons in the C-Br bond will be pushed even closer towards the bromine, making it increasingly negative.

Haloalkanes and Haloarenes Class 12 NCERT notes

The movement goes on until the –Nu is firmly attached to the carbon, and the bromine has been expelled as a Br ion.

Note:

We haven’t shown all the lone pairs on the bromine here. These other lone pairs aren’t involved in the reaction, and including them simply clutters the diagram to no purpose.

Things to notice:

The Nu- ion approaches the carbon from the far side of the bromine atom. The large bromine atom hinders attack from the side nearest to it and, being –ve would repel the incoming Nu anyway. This attack from the back is important if you need to understand why tertiary haloalkanes have a different mechanism. There is obviously a point in which the Nu- is half attached to the carbon, and the C-Br bond is half way to being broken. This is called a transition state. It isn’t an intermediate. You can’t isolate it – even for a short time. It’s just the mid-point of a smooth attack by one group and the departure of another.

Technically, this is known as an SN2 reaction. S stands for substitution, N for nucleophilic, and the 2 refers to the initial stage of the reaction that involves two species –the bromoethane and the Nu- ion.

Mechanism:

The step-wise mechanism needs to be drawn as shown with very clear details as it gives one a picture of the molecule’s arrangement in space.

Haloalkanes and Haloarenes Class 12 NCERT notes

Notice that the molecule has been inverted during the reaction-rather like an umbrella being blown inside-out.

(ii) Nucleophilic substitution in tertiary haloalkanes

Remember that a tertiary haloalkane has three alkyl groups attached to the carbon with the halogen on it. These alkyl groups can be the same or different. Consider a simple one, (CH3)3CBr – 2 – bromo-2-methylpropane.

The nucleophilic substitution reaction-an SN1 reaction

Why is a different mechanism necessary?

You will remember that when a nucleophile attacks a primary haloalkane, it approaches the +ve C from the side away from the halogen atom. With a tertiary haloalkane, this is impossible. The back of the molecule is completely cluttered with CH3 groups.

Haloalkanes and Haloarenes Class 12 NCERT notes

The alternative mechanism:

The reaction happens in two stages. In the first, a small proportion of the haloalkane ionizes to give a carbocation and a bromide ion.

This reaction is possible because tertiary carbocations are relatively stable compared to secondary or primary ones. Even so, the reaction is slow. However, once the carbocation is formed, it will react immediately when it comes into contact with a nucleophile like Nu. The lone pair on the nucleophile is strongly attracted towards the +ve C and moves towards it to create a new bond.

The speed of the reaction is governed by the ionization of haloalkane. Because this initial slow step only involves one species, the mechanism is described as SN1 -substitution, nucleophilic, one species taking part in the initial slow step.

Why don’t primary haloalkanes use the SN1 mechanism?

If a primary haloalkane uses this mechanism, the first step would be, for example:

Haloalkanes and Haloarenes Class 12 NCERT notes

A primary carbocation would be formed, and this is much more energetically unstable than the tertiary one formed from tertiary haloalkanes-and therefore, much more difficult to produce. This instability brings in a very high activation energy for the reaction involving a primary haloalkane. The activation energy is much less if it undergoes an SN2 reaction.

(iii) Nucleophilic substitution in secondary haloalkanes

There isn’t anything new in this. Secondary haloalkanes will use both mechanisms-some molecules will react using the SN2 mechanism and other, the SN1. The SN2 mechanism is possible because the back of the molecule isn’t completely cluttered by alkyl groups and so, the approaching nucleophile can still reach the carbon atom. The SN1 mechanism is possible because the secondary carbocation formed in the slow step is more stable than a primary one. It isn’t as stable as a tertiary one though, and so the SN1 route isn’t as effective as it is with tertiary haloalkanes.

(iv) Factors affecting nucleophilic substitution reactions:

Steric Nature of the Substrate.

Steric accessibility of the electrophilic center in the substrate is probably the most important factor that determines if a nucleophilic substitution will follow an SN1 or an SN2 mechanism.

Examples of SN2 (sterically accessible) substrates

Haloalkanes and Haloarenes Class 12 NCERT notes

Examples of SN1(sterically hindered) substrates

Haloalkanes and Haloarenes Class 12 NCERT notes

Nature of the nucleophile:

Both SN1 and SN2 reactions prefer small nucleophiles. Large nucleophiles have more difficulty accessing the electrophilic center in the substrate. They also have an increased tendency to act as Bronsted bases, seeking acidic protons rather than electrophilic centers. This is due to the lower activation energy of acid-base reactions compared to nucleophilic substitutions.

Haloalkanes and Haloarenes Class 12 NCERT notes

Small, strong nucleophiles that favor SN2 reactions are shown below. Most of them have a localized negative charge. It is also better if they are weak bases, such as bromide and iodide ions, but they can be strong bases such as hydroxide and alkoxide ions (conjugate bases of alcohols).

Weak, small nucleophiles that favor SN1 reactions are shown below. Notice that several of them are the conjugate acids of strong nucleophiles. They are also typically neutral, but some have a delocalized negative charge.

Haloalkanes and Haloarenes Class 12 NCERT notes

Large nucleophiles, especially if they are strong, have a tendency to act as Bronsted bases rather than as nucleophiles. They should be avoided if a nucleophilic reaction is desired.

Solvent used:

It has already been mentioned that SN2 mechanisms are favored by low to moderate polarity solvents such as acetone and N, N-dimethylformamide(DMF). SN1 mechanisms are favored by moderate to high polarity solvents such as water and alcohols. In SN1 reactions, quite frequently, the solvent also doubles as the nucleophile. Water and alcohols are prime examples of this practice.

Haloalkanes and Haloarenes Class 12 NCERT notes

 Leaving group:

The nature of the leaving group has more of an effect on the reaction rate (faster or slower) than it does on whether the reaction will follow an SN1 or SN2 mechanism. The most important thing to remember in this regard is that good leaving groups are weak bases.

o Except for fluorine, all halogens are good leaving groups
o Groups that leave as resonance stabilized ions are also weak bases and therefore, good leaving groups.
o Water is a good leaving group frequently used to prepare alkyl chlorides and bromides from alcohols.

The OH group in alcohols is not a good leaving group because it leaves as a hydroxide ion, which is a strong base. However, if the hydroxyl group is protonated first with a strong acid, it can leave as a water molecule, which is a good leaving group.

(b) Elimination reactions:

We have seen that alkyl halides may react with basic nucleophiles such as NaOH via substitution reactions.

Haloalkanes and Haloarenes Class 12 NCERT notes

When a 2° or 3° alkyl halide is treated with a strong base such as NaOH, dehydrohalogenation occurs producing an alkene-an elimination (E2) reaction.

Haloalkanes and Haloarenes Class 12 NCERT notes

There are 2 kinds of elimination reactions, E1 elimination reaction and E2 elimination reaction.
E2 = Elimination, Bimolecular (2nd order). Rate = K [RX] [Nu:]

E2 reactions occur when a 2ο or 3o alkyl halide is treated with a strong base such as OH, OR,  NH2, H, etc.

Haloalkanes and Haloarenes Class 12 NCERT notes

The Nu: removes an H+ from a β-carbon, the halogen leaves forming an alkene

All strong bases, like OH, are good nucleophiles. In 2° and 3º alkyl halides, the α-carbon in the alkyl halide is hindered. In such cases, a strong base will ‘abstract’ (remove) a hydrogen ion (H+) from a β-carbon, before it hits the α-carbon. Thus, strong bases cause elimination (E2) in 2° and 3 alkyl halides and cause substitution (SN2) in unhindered methyl and 1° alkyl halides.

In E2 reactions, the Base to H σ bond formation, the C to H σ bond breaking, the C to C π bond formation, and the C to Br σ bond breaking all occur simultaneously. There are no intermediate forms of the carbocation. Reactions in which several steps occur simultaneously are called ‘concerted’ reactions.

(i) Zaitsev’s Rule:

Recall that, in the elimination of HX from alkenes, the more highly substituted (more stable) alkene product predominates.

Haloalkanes and Haloarenes Class 12 NCERT notes

E2 reactions, do not always follow Zaitsev’s rule. E2 eliminations occur with anti-periplanar geometry, i.e. periplanar means that all 4 reacting atoms H, C, C, & X all lie in the same plane. Anti means that H and X (the eliminated atoms) are on opposite sides of the molecules.

 Look at the mechanism again and note the opposite side and same plane orientation of the
mechanism:

When E2 reactions occur in open chain alkyl halides, the Zaitsev product is actually the major product. Single bonds can rotate to the proper alignment to allow the antiperiplanar elimination.

Haloalkanes and Haloarenes Class 12 NCERT notes

In cyclic structures, however, single bonds cannot rotate, in regards to the stereochemistry. See the following example.

E.g. Trans-1-chloro-2-methylcyclopentane undergoes E2 elimination with NaOH. Draw and name the major product.

Haloalkanes and Haloarenes Class 12 NCERT notes

(ii) Substitution vs Elimination:

As with E2 reactions, E1 reactions also produce the more highly substituted alkene (Zaitsev’s rule). However, unlike E2 reactions where no C+ is produced, C+ arrangements can occur in E1 reactions. E.g. t-butyl chloride + H2O  (in EtOH) at 65ºC.

Haloalkanes and Haloarenes Class 12 NCERT notes

In most unimolecular reactions, SN1 is preferred to E1, especially at low temperatures. If the E1 product is desired, it is better to use a strong base and force the E2 reaction.  Note that increasing the strength of the nucleophile favors SN1 over E1. Can you postulate an explanation? Mixtures of products are usually obtained.

(iii) E1CB elimination:

In any E1CB reaction, a base first removes a proton from the α carbon of the substrate to give an intermediate carbanion (a species with a negatively charged carbon). This carbanion then loses the leaving group (: L) to form alkene products (s). The E1CB mechanism usually occurs with strong bases and with substrates where groups directly attached to the carbanion center can stabilize that center’s negative charge.

Haloalkanes and Haloarenes Class 12 NCERT notes

(iv) Elimination of X-X:

Alkenes also form from the loss of both X’s of a 1,2-dihaloalkane.

Haloalkanes and Haloarenes Class 12 NCERT notes

These dehalogenation reactions do not involve bases. They use metals such as Mg or Zn that react with the halogens (Cl, Br, and/or I) to form metal salts such as MgX2 or ZnX2. Their mechanisms probably involve the formation of intermediate organometallic compounds on the metal surface that then eliminate as +Mg ‒ X or +Zn ‒ X and X

Haloalkanes and Haloarenes Class 12 NCERT notes

Reaction with organometallic compounds:

(a) Grignard reagent:

(i) The reaction of RX with Mg in the ether of THF
(ii) The product is RMgX – an organometallic compound (alkyl-metal bond) and Carbanions (CH3Mg+) are very strong.

Haloalkanes and Haloarenes Class 12 NCERT notes

(iii) Alkylithium (RLi) forms from RBr and Li metal
(iv) RLi reacts with copper iodide to give lithium dialkylcopper (Gilman reagents)
(v) Lithium dialkylcopper reagents react with alkyl halides to give alkanes

Haloalkanes and Haloarenes Class 12 NCERT notes

(b) Miscellaneous reactions:

Haloalkanes and Haloarenes Class 12 NCERT notes

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