## Contents

II. Nucleophilic Substitution of 2-Bromobutane with NaOH

II.1. How Can This Reaction Take Place?

II.1.2. The Less Scientific Way

II.2. Why is Br^{–} the Leaving Group?

## I. Introduction

In Acid-Base V, we saw an example of an organic reaction that was obviously an acid-base reaction. Unfortunately, a lot of organic reactions aren’t as obvious as the deprotonation of alkyne. I reckon that most of them don’t immediately look like acid-base reactions, but in fact they really are.

So what are the acid-base reactions that are not obvious? They are reactions that involve leaving groups! Substitution, addition, elimination.. Basically almost all of them!

The first time students are introduced to the concept of leaving group is when they learn about the nucleophilic substitution.

A wise man once said:

He implies that the conjugate acid of a good leaving group has to have a small p*K*_{a} value. In other words, the p*K*_{aH} of the leaving group has to be small. The smaller the value, the better it is as a leaving group.

Let’s have a look at one example in this post, to prove the theory above. We will see more complex examples in the next post.

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## II. Nucleophilic Substitution of 2-Bromobutane with NaOH

For this particular example, let’s not worry about things like what mechanism is involved, stereochemical consequences, etc. Our main concern here is to look at the possible leaving groups in a molecule, judge which one is the best and explain why.

The nucleophilic substitution of 2-bromobutane with NaOH yields 2-butanol. The species that is actually doing the job is the OH^{–} anion, so we don’t really have to include Na^{+} in the equation. This is similar to what we previously saw in the deprotonation of acetylene with sodium amide in Acid-Base V section II.2.

So here is the complete chemical equation for this reaction:

The main question is:

** Why does this reaction go the way it does?**

This can be answered by asking two other questions:

** 1. How can this reaction take place?**

** 2. Why does Br ^{–} become the leaving group?**

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### II.1. How Can This Reaction Take Place?

There are two ways to look at this, one is the scientific way, while the other one is the less scientific way.

The reason I want to show you the less scientific way is because that was the way I learnt about this. I couldn’t really wrap my head around the scientific way at first, but then I came to realise the logic after I considered the less scientific way.

Both methods will give the *K*_{eq} value of this equilibrium, from which we can determine why this reaction take place.

**NOTICE**: I wrongly quoted the p*K*_{aH} value of OH^{–} as –1.7, when it should’ve been 15.7. This does not change the conclusion though, only different *K*_{eq} values. I am currently busy doing other stuffs, so I haven’t got the chance to fix this anytime soon. I will do it as soon as I can, and when I’m done with it, I will remove this notice. Sorry for the inconvenience.

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**II.1.1. The Scientific Way**

First, we’ll have to assign the role of each species involved. We know that a nucleophile is a base and that in nucleophilic substitution, a leaving group is also a base. Because the leaving group is one of the products, it is a conjugate base. Therefore, both nucleophile and leaving group have their own p*K*_{aH}.

**Note 1**:

I’m sure everyone has been taught that in a reaction, a nucleophile is different from a base, depending on the role. Nucleophiles attack electrophiles, whilst bases deprotonate acidic hydrogens.

What I want to show here is a reaction, the species involved in the reaction, and their acid-base properties. The role of the hydroxide anion is still a nucleophile, but in terms of acid-base, it is a base. Likewise, the role of bromide anion is a leaving group, but in terms of acid-base, it is a conjugate base.

**Note 2**:

Some people may argue that the p*K*_{a} values of –1.7 and –9 are of H_{2}O and HBr, not of 2-butanol and 2-bromobutane. How can we use those values for completely different compounds?

**Answer**: Because we kick out Br^{–}, then the p*K*_{a} that matters is the p*K*_{a} of the conjugate acid of Br^{–}, which by definition is HBr, not 2-bromobutane. And here, since the role of Br^{–} is as a conjugate base, then HBr is the acid.

Similar answer can be given for OH^{–}: because OH^{–} is the base, then the p*K*_{a} that matters is the p*K*_{a} of its conjugate acid, which by definition is H_{2}O.

So, now we have determined that:

- In the reactant side, OH
^{–}is the base - In the product side, Br
^{–}is the conjugate base

Therefore, since a base has to react with an acid, we can also determine that:

- In the reactant side, 2-bromobutane is the acid
- In the product side, 2-butanol is the conjugate acid

Now that we have identified the acid-base properties of each species, let’s put the p*K*_{a} values only on the reactant side.

Now, remember the steps we performed in Acid-Base V? If you want, you can do it yourself, try to come up with the *K*_{eq} value by yourself using those steps and compare your result with the one shown here.

To make this post shorter, I will use the formula we came up with at the end of Acid-Base V.

Judging from the magnitude of *K*_{eq}, we can say that this reaction can happen.

Additionally, we can also come up with another formula. We know we’re dealing with nucleophiles and leaving groups, and we also know that leaving groups are conjugate bases of acids. Also, both nucleophile and leaving group have their own p*K*_{aH} values. Therefore:

**p K_{aH} of leaving group = pK_{aH} of conjugate base = pK_{a} of acid**

So for the substitution of 2-bromobutane with NaOH:

**p K_{aH} of nucleophile **= p

*K*

_{aH}of base =

**p**

p= p

*K*_{aH}of OH^{–}p

*K*_{aH}of leaving group*K*

_{a}of acid = p

*K*

_{aH}of conjugate base =

**p**

*K*_{aH}of Br^{–}From this, we can derive the equation below, where we don’t need to identify which species is the acid. We can come up with *K*_{eq}, simply by knowing the p*K*_{aH} of the nucleophile and leaving group. Nucleophile is the base, and leaving group is the conjugate base. Therefore:

We can apply it to the reaction to prove that this formula is correct:

Let’s calculate *K*_{eq} now:

There you go! Hope this is clear enough for you! If not, have a look at the less scientific way in the next section below. If you can understand the explanation above then good, go ahead and continue to Section II.2..

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**II.1.2. The Less Scientific Way**

If you can’t really follow what’s going on in the previous section, then here I’d like to introduce a trick that I invented myself. To the best of my knowledge, I’ve never seen this method in any textbooks and I’ve never seen any other people doing this. I do know for certain that if you write this in an exam paper, your professor will definitely question you about it. That being said, this is a good trick for you to look at some reactions that are not really obvious acid-base reactions.

Have a look at steps 1 and 2 below:

The first step is just the rewritten full chemical equation. In 2-bromobutane and 2-butanol, we know that both Br and OH are partially negative. So in 2-bromobutane, we can separate Br^{–} from the butane skeleton; while in 2-butanol, we can separate OH^{–} from the butane skeleton. Because we separated negatively charged species, both butane skeletons become positively charged in step 2.

**Note**: the separation does not really happen in real life. We only separate them for the reason that I will explain in the following paragraph.

So in step 2, we have postively-charged butane skeletons separated from their negative counterpart. Now, let’s substitute the positively-charged butane skeletons with a positively-charged species we’re more familiar with: H^{+}.

Can you see where we’re going? In step 3 now we have substituted the positively-charged butane skeletons with H^{+}. Of course, H^{+} and Br^{–} becomes HBr, whilst H^{+} and OH^{–} becomes H_{2}O. Hence, we end up with step 4: a usual acid-base reaction between HBr and OH^{–}, that I’m sure everyone is familiar with.

Overall, we have transformed the nucleophilic substitution of 2-bromobutane with NaOH into the reaction of HBr with NaOH!

We can now predict the *K*_{eq} of this reaction. I think it is quite obvious that this reaction definitely can and will take place. However, let’s just do the calculation anyway just to show you.

First, let’s determine which species is the acid/base and conjugate acid/base. I don’t think you’ll have any problem doing that by now. And let’s assign their p*K*_{a} and p*K*_{aH} values.

**Note**: You can probably see more clearly here why we use the p*K*_{a} values for H_{2}O and HBr. However, I’d like to remind you once again that this method is not the scientific way and that you definitely should not write this in your exam paper. You can use this method as a way for you to think from a different perspective.

So, the *K*_{eq} of this reaction is:

There you go. Can you see how the not-so-obvious acid-base reactions such as the nucleophilic substitutions are actually one? This is what I came up with during my attempt to understand these things, and I hope it somehow can help you too.

Again, anyone with comments, suggestions, or even complaints about this method, feel free to do so! I’d love to hear what you think about this.

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### II.2. Why is Br^{–} the Leaving Group?

Now let’s have a look at why Br^{–} is the leaving group. I have mentioned at the beginning of this post that good leaving groups are weak bases, therefore possess small p*K*_{aH} value. The smaller the p*K*_{aH} value of a base, the better it is as a leaving group.

The carbon bonded to the Br^{–} group is also bonded to three other groups: H^{–}, CH_{3}^{–}, and C_{2}H_{5}^{–}. Let’s pretend for now that we don’t know that Br^{–} is the leaving group. So, we must give all four of these groups equal chance to be one. We do four theoretical reactions in which each of them becomes a leaving group, and then by calculating the Keq of each reaction, we analyse which one has the largest *K*_{eq}. Therefore determining which species has the best chance to hold the title of a good leaving group. From there, it will be crystal clear which group is the best leaving group and why.

First, let’s start with what we have done in the previous section: **Br ^{–} as the leaving group**.

We have calculated the *K*_{eq} of this reaction in the previous section and found out that the *K*_{eq} is approximately 2 × 10^{7}. I’ll put the calculation again here.

Now let’s consider **H ^{–} as the leaving group**.

Using the formula we came up with earlier, we are able to calculate the *K*_{eq} of this reaction too.

**The remaining leaving groups are CH _{3}^{–} and C_{2}H_{5}^{–}**. Since both of them are anion of alkanes, both of them have the same p

*K*

_{aH}value, which is 50. So let’s just combine the

*K*

_{eq}calculation of both reactions here.

Again, using the formula we came up with earlier, we are able to calculate the *K*_{eq} of both of these reactions.

**Let’s now compile all four reactions and their respective K_{eq}s.**

No |
Nu |
pK_{aH} Nu |
LG |
pK_{aH} LG |
K_{eq} |

1 | OH^{–} |
–1.7 | Br^{–} |
–9 | 2 × 10^{7} |

2 | OH^{–} |
–1.7 | H^{–} |
35 | 2 × 10^{–37} |

3 | OH^{–} |
–1.7 | CH_{3}^{–} |
50 |
2 × 10^{–52} |

4 | OH^{–} |
–1.7 | C_{2}H_{5}^{–} |
50 | 2 × 10^{–52} |

**Note**: Nu stands for nucleophile and LG stands for leaving group.

With this example, we have proved that smaller p*K*_{aH} of leaving group gives larger *K*_{eq}. Larger *K*_{eq} means a reaction can take place. This proves that good leaving groups are weak bases (small p*K*_{aH}), and explains why Br^{–} is the leaving group.

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## III. Conclusion

In this post we have proved:

**The best leaving group has the smallest p***K*_{aH}

- The best leaving group is therefore a weak base

We can therefore predict which species will act as a leaving group by comparing the p*K*_{aH}s of potential leaving groups and choosing the one with the smallest value. This is much easier and simpler than having to ‘deconstruct’ them and calculate their respective *K*_{eq}s like what we’ve done in this post.

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Page last updated: 02ii17

RP

**Other Posts in Acid, Base, and p K_{a} Series**