Animated Solution for Chemistry - Basic Concepts in Chemistry: Consider the above reaction where $6.1\text{ g}$ of benzoic acid is used to get $7.8\text{ g}$ of m-bromo benzoic acid. The percentage yield of the product is ......... . (Round off to the nearest integer). [Given : Atomic masses : $\text{C}=12.0\text{ u},\text{H}=1.0\text{ u},\text{O}=16.0\text{ u},\text{Br}=80.0\text{ u}$]

Paper vs. Practice

When we write chemical equations on a whiteboard or in a notebook, they represent an idealized universe. In this perfect world, every single molecule of reactant collides with perfect geometry and sufficient energy to magically transform into the desired product. There are no spills, no side reactions, and no stubborn unreacted starting materials.

However, the moment you step into a real organic chemistry laboratory, this illusion shatters. The laboratory is a chaotic, messy place governed by thermodynamics and kinetics. Reactions rarely go to $100\%$ completion. Side products form because molecules don't always behave exactly as we want them to. Furthermore, the very act of isolating and purifying the product—through techniques like crystallization, extraction, or chromatography—inevitably leads to some physical loss of the material.

This fundamental disconnect between the ideal and the real is why the concept of percentage yield is so critical in chemistry. The amount of product we actually isolate at the end of an experiment—the actual yield—is almost always less than the maximum amount we calculated on paper—the theoretical yield. In this problem, we are going to explore exactly this concept by analyzing the bromination of benzoic acid. We are given a specific starting mass and an actual product mass, and our mission is to determine just how efficient this chemical transformation truly was.

The Chemistry of Bromination

Before we dive into the numbers, let's take a moment to appreciate the chemistry happening in the flask. We are starting with benzoic acid, an aromatic compound where a carboxylic acid group ($-\text{COOH}$) is attached directly to a benzene ring. We are reacting this with molecular bromine (${\text{Br}}_2$) in the presence of iron(III) bromide (${\text{FeBr}}_3$).

This is a classic example of an Electrophilic Aromatic Substitution (EAS) reaction. The benzene ring is electron-rich, but the bromine molecule is non-polar and not electrophilic enough to attack the ring on its own. This is where the ${\text{FeBr}}_3$ catalyst steps in. It acts as a Lewis acid, coordinating with the bromine molecule and polarizing the $\text{Br}-\text{Br}$ bond, effectively creating a highly reactive, positively charged bromine ion (${\text{Br}}^{+}$).

Now, where does this electrophile attack the ring? The carboxylic acid group is strongly electron-withdrawing due to the electronegative oxygen atoms. It pulls electron density away from the ring, making it less reactive than benzene itself. More importantly, it is a meta-directing group. It deactivates the ortho and para positions more strongly than the meta position. Therefore, the incoming bromine electrophile is directed to the meta position, resulting in the formation of m-bromobenzoic acid.

Molar Mass

To analyze this reaction quantitatively, we must bridge the gap between the macroscopic world (the grams we measure on a balance) and the microscopic world (the individual molecules reacting). The bridge we use is the molar mass.

Let's calculate the molar mass for our starting material, benzoic acid. Its molecular formula is ${\text{C}}_6{\text{H}}_5\text{COOH}$, which can be condensed to ${\text{C}}_7{\text{H}}_6{\text{O}}_2$.

Next, we calculate the molar mass for our product, m-bromobenzoic acid. In this molecule, one of the hydrogen atoms on the benzene ring has been replaced by a heavy bromine atom. Its formula is ${\text{C}}_7{\text{H}}_5{\text{BrO}}_2$.

Notice how significantly the mass increases just by swapping one tiny hydrogen atom for a massive bromine atom!

Calculating the Theoretical Yield

Now that we have our molar masses, let's step into the ideal world and calculate how much product we should have obtained if everything went perfectly.

We started the experiment with $6.1\text{ g}$ of benzoic acid. To find out how many molecules this represents, we convert this mass into moles:

We have exactly $0.05\text{ moles}$ of our starting material. Now, we look at the balanced chemical equation. The stoichiometry is beautifully simple and straightforward: $1\text{ mole}$ of benzoic acid reacts to yield exactly $1\text{ mole}$ of m-bromobenzoic acid.

Because the ratio is $1:1$, our theoretical yield in moles is identical to our starting moles:

To find out what this theoretical amount looks like on a weighing scale, we convert these moles back into grams using the molar mass of the product:

If our reaction was $100\%$ efficient, and we didn't lose a single molecule during the entire process, we would be holding exactly $10.05\text{ g}$ of pure m-bromobenzoic acid.

Actual Yield and Reaction Efficiency

However, the problem explicitly states that we only obtained $7.8\text{ g}$ of the product. This value is our actual yield. It is the harsh reality of the laboratory.

Why did we lose over two grams of potential product? There are many possible culprits. Perhaps the reaction reached an equilibrium state before all the benzoic acid was consumed. Perhaps some of the benzoic acid underwent a different, unintended side reaction. Or, most likely, a significant amount of the product was lost during the workup and purification steps—maybe it stayed dissolved in the solvent, or got stuck to the filter paper during recrystallization.

To quantify the efficiency of our reaction, we calculate the percentage yield. This is simply the ratio of what we actually got to what we theoretically could have gotten, expressed as a percentage. It tells us how close we came to perfection.

Let's plug in our numbers:

Our reaction was approximately $77.6\%$ efficient. The question asks us to round off our final answer to the nearest integer. Looking at the first decimal place, which is a $6$, we apply standard rounding rules and round the number up.

Thus, the final percentage yield of our reaction is $78\%$. This is a very respectable yield for an organic synthesis in a real-world laboratory setting!