NUCLEOPHILIC ADDITION / ELIMINATION IN THE REACTIONS OF ACYL CHLORIDES

NUCLEOPHILIC ADDITION / ELIMINATION IN THE REACTIONS OF ACYL CHLORIDES Background What are acyl chlorides? Acyl chlorides (also known as acid chlorides) are one of a number of types of compounds known as "acid derivatives". This is ethanoic acid: If you remove the -OH group and replace it by a -Cl, you have produced an acyl chloride. This molecule is known as ethanoyl chloride and for the rest of this topic will be taken as typical of acyl chlorides in general. Acyl chlorides are extremely reactive. They are open to attack by nucleophiles - with the overall result being a replacement of the chlorine by something else. Nucleophiles A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else. Nucleophiles are either fully negative ions, or else have a strongly - charge somewhere on a molecule. The nucleophiles that we shall be looking at all depend on lone pairs on either an oxygen atom or a nitrogen atom. Nucleophiles based on oxygen atoms We shall be talking about water and alcohols, taking ethanol as a typical alcohol. Notice how similar these two molecules are around the oxygen atom. That's what turns out to be important. Nucleophiles based on nitrogen atoms We shall be considering ammonia and primary amines, taking ethylamine as a typical primary amine. A primary amine contains the -NH2 group attached to either an alkyl group (as it is here) or a benzene ring. As far as these reactions are concerned, the nature of any hydrocarbon attached to the nitrogen makes no difference. The nitrogen atom is the important bit. Again, notice how similar these two molecules are around the nitrogen atom - and also how similar they are to the previous ones containing oxygen. Both types of molecule have an active lone pair of electrons attached to one of the most electronegative elements. All of these molecules also have at least one hydrogen atom attached to the oxygen or nitrogen. Why are acyl chlorides attacked by nucleophiles? The carbon atom in the -COCl group has both an oxygen atom and a chlorine atom attached to it. Both of these are very electronegative. They both pull electrons towards themselves, leaving the carbon atom quite positively charged.
	Note:  If you aren't sure about electronegativity and bond polarity follow this link before you go on. Use the BACK button on your browser to return to this page.
The overall reaction We are going to generalise this for the moment by writing the reacting molecule as "Nu-H". Nu is the bit of the molecule which contains the nucleophilic oxygen or nitrogen atom. The attached hydrogen turns out to be essential to the reaction. The general equation for the reaction is: In each case, the net effect is that you replace the -Cl by -Nu, and hydrogen chloride is formed as well. Since the initial attack is by a nucleophile, and the overall result is substitution, it would seem reasonable to describe the reaction as nucleophilic substitution. However, the reaction happens in two distinct stages. The first involves an addition reaction, which is followed by an elimination reaction where HCl is produced. So the mechanism is also known as nucleophilic addition / elimination.
	Note:  You will find both terms in use - and to confuse the issue still further, these are also examples of condensation reactions. A condensation reaction is one in which two molecules join together to make a bigger one, and in the process shed a little molecule. The only Exam Board to require these mechanisms (AQA) calls them addition / elimination reactions.
The general mechanism The addition stage of the mechanism As the lone pair on the nucleophile approaches the fairly positive carbon in the acyl chloride, it moves to form a bond with it. In the process, the two electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, leaving that oxygen negatively charged.
	Note:  If you aren't happy about the use of curly arrows in mechanisms, follow this link before you go on. Use the BACK button on your browser to return to this page.
Notice the positive charge that forms on the nucleophile. Just accept this for the moment. It's much easier to explain why that charge must be there if you have a real example in front of you. This is fully explained in the pages on the reactions involving water, ammonia and so on. The elimination stage of the mechanism This happens in two steps. In the first step, the carbon-oxygen double bond reforms. To make room for it, the electrons in the carbon-chlorine bond are repelled until they are entirely on the chlorine atom - forming a chloride ion. Finally, the chloride ion plucks the hydrogen off the original nucleophile. It removes it as a hydrogen ion, leaving the pair of electrons behind on the oxygen or nitrogen atom in that nucleophile. That cancels the positive charge.

THE REDUCTION OF ALDEHYDES AND KETONES

THE REDUCTION OF ALDEHYDES AND KETONES This page gives you the facts and mechanisms for the reduction of carbonyl compounds (specifically aldehydes and ketones) using sodium tetrahydridoborate (sodium borohydride) as the reducing agent. Only one UK A level Exam Board (AQA) is likely to ask for these mechanisms, and they are happy with a simplified version of what is quite a complex mechanism. Because of that simplification, these reactions are dealt with entirely on this page - without the "talk through" page that you will find for other mechanisms on this site. The reduction of aldehydes and ketones by sodium tetrahydridoborate The facts Sodium tetrahydridoborate (previously known as sodium borohydride) has the formula NaBH4, and contains the BH4- ion. That ion acts as the reducing agent. There are several quite different ways of carrying out this reaction. Two possible variants (there are several others!) are: The reaction is carried out in solution in water to which some sodium hydroxide has been added to make it alkaline. The reaction produces an intermediate which is converted into the final product by addition of a dilute acid like sulphuric acid. The reaction is carried out in solution in an alcohol like methanol, ethanol or propan-2-ol. This produces an intermediate which can be converted into the final product by boiling it with water. In each case, reduction essentially involves the addition of a hydrogen atom to each end of the carbon-oxygen double bond to form an alcohol. Reduction of aldehydes and ketones lead to two different sorts of alcohol. The reduction of an aldehyde For example, with ethanal you get ethanol: Notice that this is a simplified equation - perfectly acceptable to examiners. The H in square brackets means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol. A primary alcohol is one which only has one alkyl group attached to the carbon with the -OH group on it. They all contain the grouping -CH2OH.

THE NUCLEOPHILIC ADDITION OF HYDROGEN CYANIDE TO ALDEHYDES AND KETONES

THE NUCLEOPHILIC ADDITION OF HYDROGEN CYANIDE TO ALDEHYDES AND KETONES This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic addition reactions between carbonyl compounds (specifically aldehydes and ketones) and hydrogen cyanide, HCN. If you want the mechanisms explained to you in detail, there is a link at the bottom of the page. Aldehydes and ketones behave identically in their reaction with hydrogen cyanide, and so will be considered together - although equations and mechanisms will be given for both types of compounds for the sake of completeness. The reaction of aldehydes and ketones with hydrogen cyanide The facts Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile: With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
	Note:  When you are naming these compounds, don't forget that the longest carbon chain has to include the carbon in the -CN group. In both of the above examples, the longest carbon chain is 3 carbons - hence the "prop" in both names. The carbon with the nitrogen attached is always counted as the number 1 carbon in the chain.
The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.
	Note:  If the reaction is done using hydrogen cyanide itself, a little sodium hydroxide solution is added to produce some cyanide ions from the weakly acidic HCN. Again the pH of the solution is adjusted to around pH 5 - in other words, the sodium hydroxide is not added to excess. The rate of the reaction falls if the pH is any higher. Whichever set of reagents you use, the reaction contains the same mixture of hydrogen cyanide and cyanide ions.
The mechanisms These are examples of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the cyanide ion acting as a nucleophile.
	Nucleophile:  A species (molecule or ion) which attacks a positive site in something else. Nucleophiles are either fully negative ions or contain a fairly negative region somewhere in a molecule. All nucleophiles have at least one active lone pair of electrons. When you write mechanisms for reactions involving nucleophiles, you must show that lone pair.
The mechanism for the addition of HCN to propanone In the first stage, there is a nucleophilic attack by the cyanide ion on the slightly positive carbon atom. The negative ion formed then picks up a hydrogen ion from somewhere - for example, from a hydrogen cyanide molecule. The hydrogen ion could also come from the water or the H3O+ ions present in the slightly acidic solution. You don't need to remember all of these. One equation is perfectly adequate.
	Note:  The product molecule here has been drawn differently from the one in the equation further up this page. It has been rotated through 90°. There is no reason why you can't do that if it makes the appearance of the mechanism easier to follow.
The mechanism for the addition of HCN to ethanal As before, the reaction starts with a nucleophilic attack by the cyanide ion on the slightly positive carbon atom. It is completed by the addition of a hydrogen ion from, for example, a hydrogen cyanide molecule.
	Note:  Again, the product molecule looks different from the one in the equation further up this page. The central carbon atom still has the same four groups attached, but to make the mechanism easier to follow, they are simply arranged differently. That's not a problem - we're still talking about the same substance.
Optical isomerism in 2-hydroxypropanenitrile When 2-hydroxypropanenitrile is made in this last mechanism, it occurs as a racemic mixture - a 50/50 mixture of two optical isomers. It is possible that you might be exected to explain how this arises.
	Note:  You almost certainly won't be able to tell whether or not you need to know this from the syllabus. You need to refer to recent exam papers and mark schemes. If you haven't already got these, you can obtain them from your Exam Board via links on the syllabuses page.
Optical isomerism occurs in compounds which have four different groups attached to a single carbon atom. In this case, the product molecule contains a CH3, a CN, an H and an OH all attached to the central carbon atom. The reason for the formation of equal amounts of two isomers lies in the way the ethanal gets attacked. Ethanal is a planar molecule, and attack by a cyanide ion will either be from above the plane of the molecule, or from below. There is an equal chance of either happening. Attack from one side will lead to one of the two isomers, and attack from the other side will lead to the other.
	Note:  This is probably as much as you need to know for exam purposes, but a full explanation of this is given on the "talk through" page. Follow the link below.
All aldehydes will form a racemic mixture in this way. Unsymmetrical ketones will as well. (A ketone can be unsymmetrical in the sense that there is a different alkyl group either side of the carbonyl group.) What matters is that the product molecule must have four different groups attached to the carbon which was originally part of the carbon-oxygen double bond.