Introduction to Rearrangement Reactions

by James

in Organic Chemistry 1, Organic Reactions

The previous four posts on acid-base, substitution, addition, and elimination covered the 4 main reactions in organic chemistry I. Now it’s time to go beyond those mainstays to introduce a few of the less common (but still important) reactions you learn in organic chemistry 1. They will be rearrangements, radical substitution, and cleavage (oxidative cleavage).

Let’s look at rearrangements in this post. As with everything in this series, the point is not to understand why just yet, but to be able to see from the diagrams what bonds are broken and formed. You need to understand how to read line diagrams. But other than that no further skills are required. The point here is to be able to follow the plot – to see what is happening. A later series of posts will go into more detail as to why things happen, but it takes time to build up that knowledge.

Rearrangement reactions are really interesting. They can accompany many of the reactions we’ve previously covered such as substitution, addition, and elimination reactions. In fact, if you don’t look closely, sometimes you can miss the fact that a rearrangement reaction has occurred. Let’s look at a substitution reaction first.

On the top is a “typical” substitution reaction: we’re taking an alkyl halide and adding water. The C-Br bond is broken and a C-OH bond is formed. If you look at the table on the right you’ll see this follows the typical pattern of substitution reactions.

However if we change one thing about this alkyl halide – move the bromine to C-3 instead of C-2 – now when we run this reaction we see a different product emerge. It is also a substitution reaction (we’re replacing Br with OH) but it’s on a different carbon. That’s because if you look closely, you can see there are actually 3 bonds broken and 3 bonds formed. The C2-H bond broke and the C3-H bond formed.

Very strange!

This represents a rearrangement reaction – when you see a group “move” from one carbon to another. Let’s look at another example.

Here we have an addition reaction. On top, nothing special – as with all additions, we break a C-C double bond (π bond )and form two new single bonds to the adjoining carbons (H and Cl). But look at the bottom example. If we use that alkene instead, we find that the Cl ends up on C3, not C-2. Again, examining the bonds broken/formed, we see that there’s an extra pair of events: the C3-H bond was broken and the C-2H bond was formed. In other words, the hydrogen “migrated” from one carbon to another. Weird!

Finally, let’s look at an elimination reaction. If you take an alcohol like the one below and add an acid (like H2SO4, pictured) and help the reaction along with some heat, you break the C1-OH and C2-H bonds, and form a new double bond between C1-C2. This is, in other words, a typical elimination reaction.

But if you take a slightly modified alcohol like the bottom example (with an extra methyl group on C1) and try the same reaction, something strange happens again. Analyzing the bonds broken and formed,  there’s an “extra” bond being broken and an “extra” bond being formed here. If you look closely you can see that one of the methyl groups on C1 (we’ll call it C8) moved over to C2.

So what can we conclude about rearrangement reactions?

1. they can accompany many of the reactions we’re already familiar with, such as substitution, addition, and elimination reactions.

2. They involve the “movement” of an atom (H in the top two examples, C in the third) from one carbon to another.

What other insight might we glean from these examples? Here’s two questions.

1. look up, if you don’t know already, what “primary, secondary and tertiary alcohols and alkyl halides are.

2. In the substitution reactions and the elimination reactions, classify every alcohol (or alkyl halide) according to whether it is primary, secondary or tertiary. Notice any difference between the “normal” cases and the “rearrangement” cases?

For more detail on rearrangement reactions, start here: Rearrangement Reactions (1) – Hydride Shifts

Next Post: Introduction To Free-Radical Substitution Reactions

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{ 6 comments… read them below or add one }

AzaPrins October 17, 2011 at 11:10 pm

Thanks for the easy-to-understand tutorial!

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Temperature Conversion October 18, 2011 at 10:35 am

The Rearangement reaction are very tough and now i clear for this tutorial……thanks………..

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james October 19, 2011 at 3:09 am

Thanks. Glad you found it helpful.

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Diana October 19, 2011 at 4:41 am

thanks so much for the hammond postulate explanation. I was wondering which aspects of a reaction I should be looking at when i have a relatively hindered secondary alkyl halide and wondering if it goes into SN1 or E1. We keep on coming up to this conclusion. It seems to me the relative strength of the nucleophile and solvent seem to matter. For SN1 there seems to be a need for a polar solvent to stabilize the carbocation formed. In E1 it doesn’t seem to matter. Any words of perspective?

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james October 23, 2011 at 2:34 am

For these types of problems I’d look at these variables in order of importance: 1. primary/sec/tert 2. nucleophile (strong/weak) 3. solvent 4. temp

Hindered secondary alkyl halide. First thing I’d look at is strength of nucleophile. Is it a good base/nucleophile (read: does it bear a negative charge)? If so, E2 and SN2 are more likely (E2 probably more than SN2 if hindered). If the nucleophile is weak (e.g. water, alcohols – anything neutral) you’ll be comparing E1 vs. SN1.

Next after strength of nucleophile I’d rank solvent. For determining SN2/E2 (strong nucleophile/base) polar protic solvent (e.g. ethanol) generally signifies elimination. Polar aprotic signifies substitution.

For SN1/E1 (weak nucleophile/base) the solvent will generally be polar protic. Heat favors elimination over substitution.
Hope this helps!

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sanele August 7, 2012 at 11:55 am

thank you guys, the notes really helped. I’m going to ace that paper tomorrow!!!!!

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