AS Chemistry - Halogenoalkanes

Oct 25, 2019

[latexpage]

Isomerism and nomenclature

A halogenoalkane is calssed as primary, secondary or tertiary, depending on how many alkyl groups are attached to the carbon atom joined to the halogen atom:

image-20190420120908224.png

Reactions of halogenoalkanes

Nucleophilic substitution reactions

Halogens are more electronegative than carbon, and so the $\ce{C-X}$ bond is strongly polarized. During the reactions of halogenoalkanes, this bond breaks heterolytically to give the halide ion:

image-20190420121525319.png

The carbocation formed is then vulnerable for nucleophiles – lewis bases – to attack:

Often the nucleophile will attack the partially positive carbon before the halide ion has left:

image-20190420122446255.png

The dominant reaction of halogenoalkanes is therefore nucleophilic substitution. The halide ion is replaced by a nucleophile, and the chirality of the molecule reverses. Whether the halide ion is ‘squeezed off’ by the nucleophile or it leaves by itself, is determined on the structure of the halogenoalkane (primary, secondary, tertiary).

If the halogenoalkane is primary, the latter will happen: The nucleophile attacks the halogenoalkane in whole, squeezing off the halide ion.

If the halogenoalkane is tertiary, the repulsive force of the R groups attached to the carbon hinderes the interaction of nucleophile with the molecule. This is called steric hindrance. Therfore, the halide ion is more likley to leave by itself, and then the nucleophile will attack the carbocation.

Therefore, the nucleophilic substitution reaction can be separated into two categories:

The $S_N1$ reaction

The hydrolysis of 2-bromo-2-methylpropane is a first-order reaction.

image-20190420123839811.png


$$
\text{rate}=k_1[\text{RBr}]\quad\quad\quad(\text{where RBr is }\ce{(CH3)3Br})
$$
This means that the rate doubles if we double [RBr], but if we double [$\ce{OH-}​$], the rate does not change at all. The rate depends only on the concentration of the bromoalkane, it is independent on the hydroxide concentration. Hydroxide ions cannot therefore be involved in the rate-determining step – that is, the step in the overall reaction that is the slowest; the one that limits the overall rate of reaction.

image-20190420124544540.png

The first step involves the heterolysis of the $\ce{C-Br}$, forming the carbocation and a bromide ion. This is the slow step in the reaction, and the hydroxde ion does not take part in it.

The sequence of events is called the $S_N1$ mechanism.

This is the reaction profile of the $S_N1$ hydrolysis of 2-bromo-2-methylpropane:

image-20190420125325043.png

The $S_N2$ reaction

The hydrolysis of bromomethane is a second-order reaction:
$$
\ce{CH3-Br + OH- -> CH3-OH + Br-}\
\text{rate}=k_2[\text{RBr}][\ce{OH-}]
$$
Doubling either [RBr] or [$\ce{OH-}​$] will double the rate of this reaction. We can therefore deduce that both bromomethane and hydroxide are involved in the rate-determining step.

image-20190420130555267.png

The reaction is a conitinuous one-step process. The complex shown in square brackets is not an intermediate (unlike the carbocation in the $S_N1$ reaction) but a transition state. It is a half-way stage in the reaction, where the $\ce{C-Br}$ bond is still getting longer and the $\ce{O-C}$ bond is getting shorter.

The sequence of events is called the $S_N2​$ mechanism.

The energy profile of this $S_N2$ reaction is:

image-20190420131545338.png

How the relative rates of $S_N1$ and $S_N2$ reactions depend on structure

Due to the inductive effect, the stability of carbocations increases in the order primary < secondary < tertiary. As the carbocation becomes more stable, the activation energy for the reaction leading to it also decreases.

We therefore expect that the rate of the $S_N1$ reaction will increase in the order primary < secondary < tertiary.

Also, because alkyl groups are much larger than hydrogen atoms, the more substituted the halogenoalkane is, the more crowded the molecules in the transition state of the $S_N2$ reaction is. So the rate of $S_N2$ reaction orders oppositely: tertiary < secondary < primary.

Overall, we can deduce that the predominant reation of primary halogenoalkanes is the $S_N2$ reaction, and the predominant reaction of tertiary halogenoalkanes is the $S_N1$ reaction, while the rate of both $S_N2$ and $S_N1$ reaction on secondary halogenoalkanes is basically the same.

image-20190420132846085.png
1. Hydrolysis

Some halogenoalkanes are reactive enough to be hydrolyzed to form alcohols just by heating with water:
$$
\ce{C(CH3)3Cl + H2O -> C(CH3)3OH + HCl}
$$
The hydrolysis is quicker if it is carried out in hot aqueous sodium hydroxide:
$$
\ce{CH3CH2Br + NaOH (aq) ->[\text{boil under reflux}] CH3CH2OH + NaBr (aq)}
$$
Iodoalkanes are seen to be the most reactive halogenoalkanes, and chloroalkanes are the least reactive.

2 With ammonia

The ammonia molecule contains a lone pair of electrons. The lower electronegativity of nitrogen compared with oxygen makes ammonia a stronger nucleophile than water. The reaction is as follows:

image-20190513141730172.png

The solution for this reaction is ethanol.

The product is called ethylamine, which like ammonia, the nitrogen atom possesses a lone pair. But the ethyl group, as an electron donating group, makes the molecule even more nucleophilic. So if an excess of bromoethane is used, further reactions can occur:
$$
\ce{CH3CH2NH2 + CH3CH2Br -> (CH3CH2)2NH + HBr}\
\ce{(CH3CH2)2NH + CH3CH2Br -> (CH3CH2)3N + HBr}
$$
Finally, even the triethylamine possesses a lone pair of electrons on its nitrogen atom. It can react as follows:
$$
\ce{(CH3CH2)3N + CH3CH2Br -> (CH3CH2)4N+ + Br-}
$$
The $\ce{(CH3CH2)4N+}$cation is called the tetraethylammonium

To avoid these further reactions, an excess of ammonia is used.

3 With cyanide ions

When a halogenoalkane is heated under reflux with a solution of sodium (or potassium) cyanide in ethanol, a nucleophilic substitution reaction occurs and the halogen is replaced by the cyano group:
$$
\ce{CH3CH2Br + Na+CN- -> CH3CH2CN + Na+Br-}
$$
The product is called propanenitrile. The name indicates that it has three carbon atoms. So we must take in the account of the carbon atom on the cyano group when naming nitriles. This reaction increases the carbon chain by 1, so it may be useful in organic syntheses. The propanenitrile can be hydrolyzed to carboxylic acids by heating under reflux with dilute sulfuric acid:
$$
\ce{CH3CH2CN + 2H2O + H+ ->[\text{dilute }\ce{H2SO4 (aq)}][\text{heat under reflux}] CH3CH2CO2H + NH4+}
$$
Nitriles can also be reduced to amines by a hydrogen with a nickel or platinum catalyst:
$$
\ce{CH3CH2CN + 2H2 ->[\ce{Ni}\text{ or }\ce{Pt}\text{ cat.}]CH3CH2CH2NH2}
$$

Elimination reactions of halogenoalkanes

Nucleophiles are electron-pair donors, donating to a positve or partially positive carbon atom. Bases are also electron-pair donors, donating a pair of electrons to a positive hydrogen atom.

The hydroxide ion can act as a nucleophile, donating one of its lone pair on the oxygen atom to the partially positive charged carbon in bromomethane, or it can act as a base, donating one of its lone pair on the oxygen atom to a hydrogen atom in a hydrogen chloride molecule.

The hydrogen atom attached to the carbon are slightly partially positive. They are therefore very slightly acidic. When a halogenoalkane is reacted with hydroxide ions under certain conditions, an acid-base reaction can occur at the same time as the carbon-halogen bond is broken:

image-20190422195021589.png

An alkene has been formed by the elimination of hydrogen bromide from the bromoalkane. The same reagent (hydroxide ion) can therefore carry out two different types of reaction (substitution or elmination) when reacted with the same bromoalkane. But the proportion in which each reaction undergoes is determined by the following factors:

  • The nature of the bromoalkane (primary, secondary or tertiary)
  • The strength and physical size of the base
  • The solvent used for the reaction

In general, nucleophilic substitution and elimination reactions are favored by:

ConditionsNucleophilic substututionElimination
Nature of bromoalkanePrimaryTertiary
BasesSmall, weakBig bulky, strong
Solvent polarityPolarLess polar
TemperatureLowHigh

The most commonly used reagent is potassium hydroxide dissolved in ethanol and heated under reflux.

A common ‘big and bulky’ base is $\ce{(CH3)3C-O- K+}$

Chlorofluorocarbons in the ozone layer

The chlorofluorocarbons jabe almost ideal properties for use as aerosol propellants and refrigerant heat-transfer fluids.

  • They are chemically and biologically inert
  • They can be easily liquified by pressure a little above atmospheric pressure

Although fairly expensive to produce (compared to ther refrigerants like ammonia), they rapidly replaced other fluids that had been used.

Recently, there are concerns that CFCs are breaking the ozone layer in the atmosphere that blocks most of the ultraviolet rays coming from the sun.

Although the carbon-fluorine bond is quite strong, the carbon-chlorine bond is weak enough that ultraviolet rays can break them homolytically, forming two radicals:
$$
\ce{CF3Cl ->[\text{ultraviolet light}].CF3 + Cl.}
$$

Some stratospheric chemistry

  1. Production of oxygen atoms: $\ce{O2->[\text{UV light at 250nm wavelength}]2O}$

Once oxygen atoms have formed, they can react with oxygen molecules to produce ozone, which by absoption of ultraviolet light decomposes to re-form oxygen atoms and oxygen molecules.

  1. Natural ozone formation: $\ce{O + O2 -> O3}$
  2. Natural ozone depletion: $\ce{O3 -> O2 + O}$

This absorption needed to naturally deplete ozone is what prevents the UV light from reaching us.

A stady rate is set up in which the rate of production of ozone equals the rate of its breakdown. Chlorine atoms disrupt the balance of this steady state by acting as a homogeneous catalyst for the destruction of ozone:

  1. $\ce{Cl. + O3 -> ClO. + O2}$
  2. $\ce{ClO. + O -> Cl. + O2}$

Reaction 5, in which chlorine atoms are regenerated, involves the destruction of the xoygen atoms needed to make more ozone by the natural formation reaction (2). In this way, chlorine atoms have a doubly depleting effect on ozone concentration in the atmosphere.

Preparing halogenoalkanes

Radical substitution

As mentioned in Chapter 13, halogenoalkanes can be prepared using radical substitution.

Shared formula:
$$
\text{Alkene + Halogen}\rightarrow \text{Halogenoalkane + Hydrogen halide}
$$

Electrophilic addition of alkenes

As mentioned in Chapter 14, halogenoalkanes can be prepared using electrophilic addition of either halogen diatomic molecules or hydrogen halides to alkenes.

Reminder: The regioselectivity of electrophilic addition of hydrogen halides to alkenes depends on the Markovnikov’s rule.

Nucleophilic substitution of alcohols

Yes. We will mention these in detail in Chapter 16.

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