Guest Post on SN1/SN2/E1/E2 (2): Three And A Half Steps To Any Substitution Or Elimination Reaction

by James

in Organic Chemistry 1

The second in a series of guest posts on SN1/SN2/E1/E2 by Chemistry Blog author and “instructor of organic chemistry at a small Midwestern liberal arts college” @azmanam  . For the first post, click here. 


3 ½ Steps To Any Substitution Or Elimination Reaction. Step (½): What is the nature of the leaving group? by @azmanam

If you haven’t looked through James’ fantastic posts over the four possible mechanisms (SN1, SN2, E1, E2), be sure to go back and read them before today’s installment.

Our first (half) step is to assess the nature of the leaving group. This is a relatively quick step and ensures we can even perform one of these reactions. For substitution/elimination reactions, the leaving group must satisfy two conditions. It must be a ‘good’ leaving group, and it must be attached to an sp3 hybridized carbon atom.

What does it mean to be a ‘good’ leaving group? Remember from our first post that it all has to do with the electrons. If we can rationalize these reactions in terms of electrons and electron density and stability, we can predict how these reactions will play out.

So we need to think about what happens to the leaving group during the reaction and what factors will make this better or worse. These reactions are of a nucleophile/base with an electrophile bearing a leaving group to perform a substitution or elimination reaction – but in either case the leaving group is expelled along with a pair of electrons.

This is important: when the leaving group leaves, it gains a new lone pair of electrons and an excess of electron density. For a leaving group to be a ‘good’ leaving group, it would have to accept this new glut of electron density with ease. A ‘bad’ leaving group would have to become significantly unstable with this new electron density. A ‘good’ leaving group is able to stabilize this excess electron density, and a ‘bad’ leaving group is not.

Let’s take a look at the following potential leaving groups. Let’s look at the ‘good’ leaving groups versus the ‘bad’ leaving groups and see how electron density stability helps explain the ranking of these leaving groups.

 1-adamlg

When the leaving group leaves, it leaves with the pair of electrons that used to be in the bond. How the leaving group handles that sudden build up in electron density determines whether the leaving group will be a ‘good’ or ‘bad’ leaving group. Remember that it’s all about electron density – which is like a hyperactive child. If that electron density can be spread out over a larger volume of our leaving group, it will be a more stable and better leaving group than if the electron density cannot be spread out and is forced to be concentrated in a very small volume. 

2-adamlg

Side note: there is another phenomenon that works by this same thought process: base strength. When an acid gets deprotonated and gets turned into its conjugate base, it accepts a pair of electrons and extra electron density. The more stable the electrons, and the more stable the base, the weaker the base will be.

This leads to a convenient mnemonic for substitution and elimination reactions: the best leaving groups are the weakest bases. Cl, Br, MsO (OSO2CH3) are all weak bases and fantastic leaving groups.

The other criterion a leaving group has to meet is the nature of the carbon atom to which it is attached. The carbon atom must be sp3 hybridized. For our purposes in introductory organic chemistry, substitution and elimination will not occur at sp2– or sp-hybridized  carbon atoms.

adam-2

So our first (half) step to determining which substitution or elimination reaction will occur is to do a quick check to make sure the leaving group is a good leaving group (a weak base) and that the leaving group is attached to an sp3-hybridized carbon atom. If even one of these criteria is not met, no substitution or elimination reaction will happen. If both of these criteria are met, it doesn’t help us decide which of the four reactions occur, but we can move on to the next three steps to figure out which of the four reactions is the one that will take place. The first big step we’ll take is to assess the nature of the nucleophile, and that will be the topic of the next post.

Next post: 3½ Steps To Any Substitution Or Elimination Reaction, Part 1: The Nucleophile

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{ 5 comments… read them below or add one }

Jon

For one of these reactions to occur, the leaving group does NOT need to be attached to an sp3 hybridized carbon. For example, [1-bromoethenyl]benzene and sodium tert-butoxide will form phenylacetylene.

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James

This is true – the one exception is elimination to form alkynes.

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E

errata?

The more stable the electrons, and the more stable the base, the weaker the ACID will be.

Reply

James

Not a typo. Anything you do to stabilize the conjugate base will mean that its conjugate acid is a stronger acid.
“The stronger the acid the weaker the conjugate base”.
“The weaker the acid the stronger the conjugate base”.

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Prashant

Thanks a lot sir. That was so simply explained.

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