ASIGNACIÓN DEL VALOR RELATIVO A LAS TAREAS DE LOS USUARIOS. USUARIOS

Elimination reactions involving halogenoalkanes Elimination from 2-bromopropane . . .

The formation of an alkene (propene) from 2-bromopropane.

The elimination reaction involving 2-bromopropane and hydroxide ions

2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.

Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask.

The mechanism

In elimination reactions, the hydroxide ion acts as a base - removing a hydrogen as a hydrogen ion from the carbon atom next door to the one holding the bromine.

The resulting re-arrangement of the electrons expels the bromine as a bromide ion and produces propene.

Elimination from unsymmetrical halogenoalkanes . . .

2-bromobutane is an unsymmetric halogenoalkane in the sense that it has a CH3 group one side of the C-Br bond and a CH2CH3 group the other.

You have to be careful with compounds like this because of the possibility of more than one elimination product depending on where the hydrogen is removed from.

The basic facts and mechanisms for these reactions are exactly the same as with simple halogenoalkanes like 2-bromopropane. This page only deals with the extra problems created by the possibility of more than one elimination product.

Background to the mechanism

You will remember that elimination happens when a hydroxide ion (from, for example, sodium hydroxide) acts as a base and removes a hydrogen as a hydrogen ion from the

halogenoalkane.

For example, in the simple case of elimination from 2-bromopropane:

The hydroxide ion removes a hydrogen from one of the carbon atoms next door to the carbon-bromine bond, and the various electron shifts then lead to the formation of the alkene - in this case, propene.

With an unsymmetric halogenoalkane like 2-bromobutane, there are several hydrogens which might possibly get removed. You need to think about each of these possibilities.

Where does the hydrogen get removed from?

The hydrogen has to be removed from a carbon atom adjacent to the carbon-bromine bond. If an OH^- ion hit one of the hydrogens on the right-hand CH3 group in the 2-bromobutane (as we've drawn it), there's nowhere for the reaction to go.

To make room for the electron pair to form a double bond between the carbons, you would have to expel a hydrogen from the CH2 group as a hydride ion, H^-. That is energetically much too difficult, and so this reaction doesn't happen.

That still leaves the possibility of removing a hydrogen either from the left-hand CH3 or from the CH2

group.

If it was removed from the CH3 group:

The product is but-1-ene, CH2=CHCH2CH3. If it was removed from the CH2 group:

This time the product is but-2-ene, CH3CH=CHCH3.

In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.

Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an

explanation of the two ways of naming these two compounds. Which isomer gets formed is just a matter of chance.

Geometric isomerism: Isomerism is where you can draw more than one arrangement of the atoms for a given molecular formula.

Geometric isomerism is a special case of this involving molecules which have restricted rotation around one of the bonds - in this case, a carbon-carbon double bond. The C=C bond could only rotate if enough energy is put in to break the pi bond. Effectively, except at high temperatures, the C=C bond is "locked".

In the case of but-2-ene, the two CH3 groups will either both be locked on one side of the C=C (to give the cis or (Z) isomer), or on opposite sides (to give the trans or (E) one).

Beware! It is easy to miss geometric isomers in an exam. Always draw alkenes with the correct 120° bond angles around the C=C bond as shown in the diagrams for the cis and trans isomers above. If you take a short cut and write but-2-ene as CH3CH=CHCH3, you will almost certainly miss the fact that cis and trans forms are possible.

The overall result

Elimination from 2-bromobutane leads to a mixture containing:

 but-1-ene

 cis-but-2-ene (also known as (Z)-but-2-ene)

 trans-but-2-ene (also known as (E)-but-2-ene)

Elimination v. substitution . . .

The reaction between a halogenoalkane and hydroxide ions can lead to either an elimination reaction or nucleophilic substitution.

The reactions

Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution.

Nucleophilic substitution

The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction.

In the example, 2-bromopropane is converted into propan-2-ol.

Elimination

Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.

The 2-bromopropane has reacted to give an alkene - propene.

Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.

What decides whether you get substitution or elimination?

The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.

For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions.

The solvent

The proportion of water to ethanol in the solvent matters.

 Water encourages substitution.

 Ethanol encourages elimination.

The temperature

Higher temperatures encourage elimination.

Concentration of the sodium or potassium hydroxide solution Higher concentrations favour elimination.

In summary

For a given halogenoalkane, to favour elimination rather than substitution, use:

 heat

 a concentrated solution of sodium or potassium hydroxide

 pure ethanol as the solvent

The role of the hydroxide ions

The role of the hydroxide ion in a substitution reaction

In the substitution reaction between a halogenoalkane and OH^- ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon.

This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place.

The role of the hydroxide ion in an elimination reaction

Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH^- ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br^-.

The dehydration of alcohols The dehydration of ethanol . . .

Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C. Concentrated phosphoric(V) acid, H3PO4, can be used instead.

The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H2SO4" over the top of the arrow.

The mechanism - the full version

We are going to discuss the mechanism using sulphuric acid. Afterwards, we'll describe how you can use a simplified version which will work for any acid, including phosphoric(V) acid.

In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated.

The protonated ethanol loses a water molecule to give a carbocation (a carbonium ion).

Finally, a hydrogensulphate ion (from the sulphuric acid) pulls off a hydrogen ion from the carbocation.

The mechanism - a simplified version

People normally quote a simplified version of this mechanism.

Instead of showing the full structure of the sulphuric acid, you write it as if it were simply a hydrogen ion, H⁺. That leaves the full mechanism:

An advantage of this (apart from the fact that it doesn't require you to draw the structure of sulphuric acid) is that it can be used for any acid catalyst without changing it at all. For example, if you use this version, you wouldn't need to worry about the structure of phosphoric(V) acid.

Note: Although most people probably write the mechanism in this form, it is actually quite misleading because it suggests the possibility of a free hydrogen ion in a chemical system. A free hydrogen ion is a raw proton and this is always attached to something else during a chemical reaction.

The dehydration of more complicated alcohols . . .

You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated.

Butan-2-ol is just an example to illustrate the problems

The basic facts and mechanisms for these reactions are exactly the same as with a simple alcohol like ethanol. This page only deals with the extra problems created by the possibility of more than one dehydration product.

Background

To make the diagrams less cluttered, we'll use the simplified version of the mechanism showing gain and loss of H⁺.

Remember that the mechanism takes place in three stages:

 The alcohol is protonated by the acid catalyst.

 The protonated alcohol loses a water molecule to give a carbocation (carbonium ion).

 The carbocation formed loses a hydrogen ion and forms a double bond.

So, in the case of the dehydration of a simple alcohol like ethanol:

The dehydration of butan-2-ol The first two stages

There is nothing new at all in these stages.

In the first stage, the alcohol is protonated by picking up a hydrogen ion from the sulphuric acid.

In the second stage, the positive ion then sheds a water molecule and produces a carbocation.

The complication arises in the next step. When the carbocation loses a hydrogen ion, where is it going to come from?

Where does the hydrogen get removed from?

So that a double bond can form, it will have to come from one of the carbons next door to the one with the positive charge.

If a hydrogen ion is lost from the CH3 group

But-1-ene is formed.

If a hydrogen ion is lost from the CH2 group

This time the product is but-2-ene, CH3CH=CHCH3.

In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.

Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene.

Which isomer gets formed is just a matter of chance.

The overall result

Dehydration of butan-2-ol leads to a mixture containing:

 but-1-ene

 cis-but-2-ene (also known as (Z)-but-2-ene)

 trans-but-2-ene (also known as (E)-but-2-ene)

In document LA ESTRUCTURA CONCEPTUAL DEL REGISTRO BIBLIOGRÁFICO: ANÁLISIS DE LA FUNCIONALIDAD DE LAS “REGLAS DE CATALOGACIÓN” ESPAÑOLAS Y DEL FORMATO IBERMARC BIBLIOGRÁFICO (página 132-139)