Pause for thought

1. Use of curly arrows

When teaching and describing organic mechanisms involving the use of curly arrows it is much clearer to use 3-D diagrams rather than 2-D diagrams. A curly arrow represents the movement of a pair of electrons from where they start to where they end up. If you only use a 2-D diagram this can cause considerable confusion. Consider the substitution of the bromine atom in bromoethane by the hydroxide nucleophile. Firstly the curly arrow should start from a lone pair on the oxygen atom of the hydroxide ion not from the hydrogen atom - but where should the arrowhead go? Some books show it going to the carbon atom but the pair of electrons ends up forming the bond between the carbon atom and the oxygen atom of the -OH group so the arrowhead should really be going there. In 2-D this is exactly where the electron pair between the carbon atom and the bromine atom is already located. So the diagram will look like this:

Although 'correct' in 2-D there are two big problems. The first is that the existing electron pair forming the C-Br bond will repel the incoming electrons and the second is that the angle between the C^….OH and C^....Br bonds in the transition state looks to be 90^o or even less. In reality the hydroxide nucleophile will approach the δ+ carbon atom of the polar C—Br bond from the opposite side. This means that the bond angle between the C^....OH and the C^....Br bonds in the transition state will be 180^o. This can easily be shown using a 3-D representation and avoids all the problems caused by a 2-D diagram.

Full details about using curly arrows correctly (and the common errors that students make) can be found on the page 'Curly arrows' in Areas of difficulty. It is worth talking about the work of Paul Walden (1863 - 1957), a Latvian chemist, who provided strong evidence for the S_N2 mechanism using secondary halogenoalkanes. If an optically active halogenoalkane is used and the nucleophile does 'attack' from the other side then the product should rotate the plane of plane-polarized light in the opposite direction, which is indeed the case. This is known as Walden inversion and is a good way to bring some TOK into the lesson and also cover this part of the syllabus.

2. How important are electron deficient carbon atoms?

Chemistry is not always as simple and obvious as it would appear to be. When describing the mechanism for nucleophilic substitution it is usual to state that the nucleophile, which contains a non-bonding pair of electrons, is attracted to a carbon atom deficient in electrons in some way. This electron deficient carbon atom may be an actual positive ion as it is in the case of S_N1 where the first step is the breaking of the carbon-halogen bond to form an intermediate carbocation. Alternatively it may be the carbon atom in the halogenoalkane which is electron deficient due to the more electronegative halogen atom and hence attracts the nucleophile as in the S_N2 mechanism.

This is normally shown by using δ⁺ and δ^– with the nucleophile being attracted to the δ⁺ on the carbon atom.

It would seem logical that the stronger the polarity of the carbon-halogen bond the more the nucleophile will be attracted to the δ⁺ carbon atom. The strength of the polarity will depend on the electronegativity of the halogen compared to the carbon atom. By far the most polar bond is the carbon to fluorine bond.

Difference in electronegativity values for the carbon-halogen bond

On this logic alone one would expect nucleophiles to react most readily with fluoroalkanes and least readily with iodoalkanes. In fact exactly the reverse is the case. Fluoroalkanes are generally so unreactive that one of their main uses is as fire extinguishers and they are not even included in this sub-topic. From their own experimental work students will be able to see that 1-iodobutane, for example, reacts much faster with nucleophiles than 1-brombutane which in turn reacts much faster than 1-chlorobutane. This suggests that the polarity of the carbon atom is relatively insignificant and that some other factor must be dictating the ease of the reaction.

Even though the 'Application and skills' section uses the words 'Explanation of how the rate depends on the identity of the
halogen (i.e. the leaving group)', there is no mention in the 'Guidance' as to exactly what explanation is required. The syllabus and some books talk about the halogen being a 'leaving group' and it is the ease with which this group leaves that is the important factor, not the polarity of the bond. The ease with which the halogen leaves is, of course related to the strength of the carbon-halogen bond. Iodo- compounds are the most reactive because the carbon-iodine bond is very weak compared to the other carbon-halogen bonds. The fact that the carbon to fluorine bond is so strong accounts for the inertness of fluorocarbons even though they have high polarity.

One other interesting aside concerns the first 'Understandings' statement. It is easy to explain why the hydroxide ion is a better nucleophile than water but for reactions following the S_N1 pathway the concentration of the nucleophile does not appear in the rate equation. It could be asked how important is the nature of the nucleophile in the substitution of tertiary halogenoalkanes?