Substitution Reactions

By James Ashenhurst

Introduction to Nucleophilic Substitution Reactions

Last updated: December 2nd, 2022 |

Following up on the 4 most important patters of reactions in Org 1, and introduction to acid-base reactions, here’s the second major pattern. It’s called nucleophilic substitution. I won’t go into nucleophiles and electrophiles in this post. The point is actually not to understand what’s going on here!

Don’t understand the reactions below. Just focus on being able to see one thing: what bonds are formed and what bonds are broken. 

Just that one thing. Later on we’ll go into more detail about why things happen. But not today.

Here are three examples of nucleophilic substitution reactions.

three examples of nucleophilic substitution reactions bond forms and breaks at the same carbon

What’s the pattern? In each case, we’re breaking a bond at carbon, and forming a new bond at carbon. (and yes, salts form too… but this is organic chemistry, so we’re carbo-centric.)

This is an extremely common pattern for reactions and you will see it over and over again in Org 1 and Org 2.

Now let’s set the stage for later discussion. Here’s some interesting results that experiments tell us. We don’t get this information by thinking about what happens and predicting – we have to interrogate nature in order to get her to give up her secrets.

Interesting observation #1: when we measure the reaction rate.  In some substitution reactions the rate is proportional to  the concentration of two different species (e.g. the alcohol and HBr). In other cases, it only depends on the concentration of one species (i.e. the alcohol). Interesting!

 Interesting observation #2: In certain cases (like the reaction below) the reaction rate depends on the type of alkyl halide. Primary alkyl halides are slow, but tertiary alkyl halides are fast.

type of alkyl halide can affect the rate primary versus tertiary hydrolysis with h2o tertiary much faster than primary

Interesting observation #3: In other cases, the reaction rate depends on the type of alkyl halide, but it is the primary alkyl halide that is fast. And the tertiary alkyl halide does not even give substitution products… you’ll notice the double bond there. This  is actually an elimination reaction.

tertiary and primary alkyl halide naoh gives different results sn2 versus elimination

Interesting observation #4: Finally, there is a property that *some* molecules have of rotating plane polarized light. It’s been known since the late 19th century. In some substitution reactions, molecules that are “optically active” retain their optical activity in the reaction… but in others, this “optical activity” disappears.

optical activity of alkyl halide can remain or be removed in some substitution reactions

What’s going on? How can we come up with a hypothesis for why and how these reactions work?

Next Post In The Series on Key Reactions: Introduction To Addition Reactions


Get Started Learning About Substitution Reactions, in the Series: Walkthrough Of Substitution Reactions


Comment section

10 thoughts on “Introduction to Nucleophilic Substitution Reactions

  1. 1) In example #3, should the products be 2-methyl propan-2-ol and NaBr instead of HBr (in the reaction that doesn’t take place, assuming it did).
    2)In #4, why isn’t the second compound optically active? Isn’t C3 acting as a chiral centre for the compound?

    1. a) Because the SN2 reaction does not operate on tertiary alkyl halides. b) C3 is a chiral centre, but the molecule loses optical activity (becomes racemic) due to the fact that it proceeds through an SN1 mechanism (containing a carbocation)

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