Halogenoalkanes (7) Elimination reactions

Reactions of the Halogenoalkanes

How does competition between substitution and elimination reactions in haloalkanes work out?

What we find is this:

A) In ethanol solvent where OH^— acts as a base rather than a nucleophile then the corresponding alkene forms by elimination of HBr like this:

CH₃CH₂CH₂CH₂Br     +    OH^—       =      CH₃CH₂CH=CH₂    +   H₂O   +   Br^—

1-bromobutane                    base               but-1-ene

B) But if the solvent is changes for water then the substitution reaction dominates as the hydroxide ion acts as a nucleophile in the more polar solvent:

CH₃CH₂CH₂CH₂Br   +          OH^—   =          CH₃CH₂CH₂CH₂OH   +   Br^—

1-bromobutane                                            butan-1-ol

The important point to remember is that both substitution and elimination mechanisms occur together.

The use of a solvent favours one over the other but does not mean that one works exclusively instead of the other. 

So ethanolic potassium hydroxide allows the elimination mechanism to predominate

Aqueous potassium hydroxide favours the substitution mechanism in the haloalkane.

I have discussed the substitution mechanism here

The elimination mechanism is a concerted mechanism, one step following on the next…

First, the base abstracts a proton attached to the carbon atom attached to the C—Br bond.

Second, the two electrons remaining from the C—H bond form a double carbon-carbon bond

Third, these two electrons repel the electrons in the C—Br bond (rather than the C—CH₃ bond) because the bromine atom forms a good leaving group as a bromide ion (Br^—). 

You ought also to realise that breaking a C—H or a C—CH3 bond costs much more energy than breaking a C—Br bond.

The result is an alkene see below:

In this example, 2-bromo-2-methyl propane is converted into methylpropene.

You should be able to see why this is called an elimination reaction.

HBr is removed (eliminated) from the haloalkane molecule  (It is product and should be in the equation).

The experiment can be carried out in the laboratory by use of a simple test tube apparatus collecting the gaseous alkene over water.

You should be able to draw the apparatus used as it is essentially the same kit as that used below in cracking liquid paraffin.

You should of course be able accurately to add labels to your diagram

You should be able to give the equation for the reaction of 2-bromobutane and ethanolic KOH in which two different alkenes are formed. 

You should be able to name and draw the displayed formulae of these two alkene products.

Halogenoalkanes (6) Nucleophilic substitution in tertiary haloalkanes

#3 Nucleophilic substitution in halogenoalkanes

In a previous post I discussed the basics of this reaction between haloalkanes and aqueous alkali. 

Here is the general reaction scheme:

We made some criticisms of this image from the web.

The curly arrows are pretty random and sloppily drawn.

We’d like to see them start on a lone pair of the nucleophile and end on a specific positive centre.

We’d also like to see them start on the R—X bond and end as they do on the halide lone pair. 

These generic mechanisms do not indicate anything about the rate at which these reactions proceed. 

Let’s look at another representation of this hydrolysis reaction mechanism from the web:

Now this is the hydrolysis of 2-bromo-2-methyl propane: a tertiary haloalkane

Note how the arrows are drawn, where they start and where they end but we don’t like to see the arrow from the hydroxide ion coming from the negative charge rather it should come from a lone electron pair drawn on the OH. 

Let’s consider what is happening here.

We have a two step reaction in which step one involves the formation of a tertiary carbocation.

The formation of this carbocation is possible because the three electron pushing methyl groups stabilise (delocalise) the positive charge that remains on the ion after the bromide ion has left. 

The use of a polar solvent such as silver nitrate solution (e.g. 0.02M AgNO3) facilitates the formation of the carbocation. 

Not only is it a polar solvent the Ag+ ion is able to form strong bonds with halide ions according to ROC Norman (p130)

There is no intermediate transition state as in the hydrolysis of a primary haloalkane.

The second step, which is faster than the first, involves the attack of the nucleophile (OH^–) on the positive carbocation.

This reaction is always going to be much quicker than the first step because the second step involves the combination of oppositely charged ions. 

The slowest or rate determining step of the reaction involves just the haloalkane and not the hydroxide ion i.e. it is monomolecular. 

We can summarise the reaction mechanism as follows: substitution (S), nucleophilic(N) and monomolecular(1)

That’s why it is called an Sn1 reaction!!

The stereo chemistry of this reaction is important too but in a different way to the hydrolysis of a primary haloalkane.

The hydroxide ion can attack from either side of the carbocation.

Let’s look at this more closely:

See how the nucleophile can attack the carbocation from either side.

The attack of the nucleophile leads to two distinctively different molecules if R¹, R² and R³ are different.

You should try re-writing this equation using actual structures for the R groups to see the effect here.

The point is this: the two different products are stereoisomers of each other.

Their carbon atom is a chiral centre.

They possess one of the structural features that gives rise to optical isomerism in organic chemistry that is they have four different functional groups attached to the chiral carbon atom. 

Solutions of each product will rotate the plane of plane polarised light. 

However, because there is a 50:50 chance of the formation of either isomer the resulting product mixture is a non-optically active substance: it is a racemic mixture. 

One molecule of the D- form cancels out the rotation effect of a molecule of the L- form. 

In my next post, I’ll discuss competing elimination reaction in tertiary halogenoalkane hydrolysis. 

Halogenoalkanes (5) Nucleophilic substitution in primary haloalkanes

#2 Nucleophilic substitution in primary halogenoalkanes.

In a previous post I discussed the basics of this reaction between haloalkanes and aqueous alkali. 

Here is the general reaction scheme:

We can make some criticisms of this image from the web:

The curly arrows are pretty random and sloppily drawn.

We’d like to see them start on a lone pair of the nucleophile and end on a specific positive centre.

We’d also like to see them start on the R—X bond and end as they do on the halide lone pair. 

These generic mechanisms do not indicate anything about the rate at which these reactions proceed. 

Let’s look at another representation of this reaction mechanism from the web:

Now this is the hydrolysis of chloromethane: a primary haloalkane

Note how the arrows are drawn, where they start and where they end: brilliant, copy this!! 

This reaction proceeds in one flowing movement.

The only thing missing is the intermediate transition state.

But this mechanism does show the way in which the geometry of the primary haloalkane ‘flips over’.

There will be more about the consequences of this ‘flip over’ in a later post.

The reaction involves both the haloalkane and the hydroxide ion in the slowest or rate determining step i.e. it is bimolecular. 

The reaction is substitution (S), nucleophilic(N) and bimolecular(2)

That’s why it is called an Sn2 reaction!!

Of course the orientation of the attacking group has to be right too. 

The stereo chemistry of this reaction is important.

The hydroxide ion attacks from the opposite side to the leaving group the chloride (halide) ion. 

So what about the intermediate transition state?

Let’s look at another image from the web (from chemguide) to show this transition state:

The thing here is to see that the transition state is negatively charged

We write the transition state in square brackets.

I prefer to see curly arrows in red, unless it is on an exam paper, there’s a long historic reason for me saying that that I shall talk about on this blog sometime. 

The other striking geometry is that the transition state has five covalent bonds and the central structure of it is planar.

What is the CH₃-C-H bond angle in the transition state?

Two of the bonds in the transition state are partial, one forming as one breaks, hence the use of dotted lines to show this feature. 

The transition state then is a temporary structure with a half life of around 10^—12 seconds.

Primary haloalkanes hydrolyse by this mechanism because there are insufficient methyl groups on the primary carbon to stabilise a carbocation as an intermediate structure. 

In my next post, I’ll discuss hydrolysis of tertiary halogenoalkanes