Organic Chemistry: Competing Reactions, Part 4 Quick Guide

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Alright, so we are finally to the last installment of the Sn1,Sn2, E1, E2 guide bonanza. From here on in I will refer to the substitutions and eliminations as fundamental reactions collectively. I really wanted to get this entry out before people had finals, hopefully it worked out for you as well.  So, let’s get the last entry about fundamental reactions out of the way.

Competing reactions

Eventually the average student will be able to discriminate between the basics of the fundamental reactions. The harder part of the course usually involves being able to discriminate with a fair degree of certainty when a reaction will undergo a  Sn1, Sn2, E1 or E2 process. At times the reaction may favor only one mechanism, or it may favor all of them– competing reactions are often why you’ll spend those extra hours in Organic lab purifying the product. Sometimes the reaction is impossible, therefore the answer to the question stem may be the dreaded ‘no reaction’. Interestingly, it takes a remarkable amount of courage and confidence to write ‘no reaction’ on an exam.

Organic Chemistry will be the first course where you are truly inundated with details, i.e. the first time you had to try to drink from a fire hydrant so they say. They key is to work a lot of problems, and keep the details straight.  Working a lot of problems will make you familiar with they manner that I will demonstrate to make competing reactions less troublesome.

First, let’s take all the information we already know about fundamental reactions and place it into a form that’s easy on the eyes:

 

Reactions

 Substrate

Sn1

Sn2 E1 E2
Me-X

X

1

X

X

2

X

X

X

X

3

X

X

X

Table 1 

Some will say Sn1 can also reaction with a primary alkyl halide, but in practice this would quite a monumental feat because primary carbocations are extremely unstable under normal conditions. So, I left it out to avoid confusion.

 

The table listed above is just an amalgamation of the previous tables I’ve listed so far in the previous guides. From this chart we can state several axioms:

  • Me-X alkyl halides are restricted to Sn2 only.
  • Primary alkyl halides can undergo all fundamental reactions except E1 and Sn1.
  • Secondary alkyl halides can undergo all fundamental reactions.
  • Tertiary alkyl halides can undergo all the fundamental reactions except  Sn2

In other words, if you get a problem and it’s a Me-X, it’s a slam-dunk, it has to be an Sn2 or NR if the proper reagents aren’t there. Whereas, secondary alkyl halides a multitude of products can be formed all depending on the conditions. So, let’s pretend you had a pesky secondary alkyl halide, such as 2-chloro-3-methylbutane. If we glance at table 1 we are quickly reminded that it can undergo all of the fundamental reactions, given the right conditions.

Selecting by Reagents:

If we put 2-chloro-3-methylbutane with a strong nucleophile we should unequivocally expect an Sn2 and/or E2 reaction. If we put 2-chloro-3-methylbutane with a nucleophile with a solvent (or other conditions) favoring favoring the formation of the carbocation intermediate we should expect to see only Sn1 and/or E1. To summarize what I just said:

Similar Conditions:-Weak Nu: or base-Polarizing solvent- Rate liming step: carbocation intermediate formation Similar Conditions:-Strong Nu: or base-Any solvent works, but aprotic solvents best-Rate limiting step:  transition state formation

Uni-molecular

Bi-molecular

Substrates

Sn1

E1

Sn2

E2

Me-X

X

1

X

X

2

X

X

X

X

3

X

X

X

table 2, Nu: = nucleophile

Now, if we go back to our 2-chloro-3-methylbutane, we’ve already figured out that if given the same conditions we  can assert that the reactions will compete: Sn1 versus E1 or Sn2 versus E2. That is, given the same conditions you should expect either a unimolecular reaction competition (E1 versus Sn1) or a bimolecular competition (E2 versus Sn2).

Sn1 versus E1 Discriminating with Conditions/Reagents:

Similar Conditions:-Weak Nu: or base-polarizing solvent- Rate liming step: carbocation intermediate formation

Uni-molecular

Substrates

Sn1

E1

Me-X
1
2

X

X

3

X

X

For all intents and purposes, for the first year of Organic Chemistry, it’s best in my opinion to just keep this as a rule of thumb:

When it’s a uni-molecular reactions (Sn1 and E1) expect a mixture products of both Sn1 (watch for rearrangements, and racemic mixtures) and E1.

In practice, selectivity between Sn1 and E1  is poor, as a result  lots of products may form, with the most thermodynamic-ally stable product(s) being more abundant.  In lab expect to use a lot of fancy lab techniques to isolate the actual product.

Adding heat to the reactants creates a carbocation either way, making the reaction susceptible to both Sn1 and E1. However, as heat is continually put into the reaction the alcohol (substitution) product will tend to dehydrate into the elimination product(s) (alcohols -> alkenes), for this reason in general higher and prolonged temperature biases towards elimination products. Therefore if you started off with 2-chloro-3-methylbutane, reacted it with water and heat, you’d form a mixture of alcohols and alkenes. But, if you left the lab and let the solution boil too long uncontrollably you’d come back only have mostly mostly alkenes. Cold temperatures would shut down both pathways.

Sn2 versus E2: Discrimination with Conditions/Reagents

Similar Conditions:-Strong Nu: or base-any solvent works, but aprotic solvents best-Rate limiting step:  transition state formation

Bi-molecular

Substrates

Sn2

E2

Me-X

X

1

X

X

2

X

X

3

X

When it’s a bi-molecular reaction (Sn2 and E2) expect a mixture of products for secondary and primary alkyl halides. However, know that E2 and Sn2 do not work with Me-X and tertiary alkyl halides respectively.

Discriminating between Sn2 and E2 is typically more clear-cut, therefore it’s an extremely testable point. If Sn2 and E2 are competing cold favors Sn2 and heat favors eliminations. You can select for E2 products by using a poor nucleophile but moderate/strong base.

Last Thoughts

If you are just starting Organic Chemistry it’s probably hard to imagine how these four reactions could possibly be important. However, as you gain more experience in synthesizing compounds you’ll soon find the trend to be more obvious. You’ll start the course by looking at alkanes, then perhaps looking at free radical reactions with halogens to create alkyl halides. Pretty much once you’ve loaded your alkane with a halide the sky’s the limit of what you can form given the proper reagents and lab equipment (with proper usage).

Remember, Biochemistry is just Organic chemistry with fancy nomenclature, the better you know Organic the more intuitive everything ‘chemistry’ will seem to be — though in Biochemistry you’re more interested in the results of the reaction than the physics of the mechanism itself.

As always, you can always find me on twitter: https://twitter.com/masterofsleep

Alright folks, back to writing tips about how to apply to medical school Good luck all you cool people with finals! I heart goes out to you as a former premed.

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