Consider the reaction sequence from P to Q shown below. The overall yield of the major product Q fro

Q: Why do we use a basic medium for the coupling reaction with $\beta$-naphthol?

In a basic medium, $\beta$-naphthol is converted into the naphthoxide ion. The negatively charged oxygen is a much stronger electron-donating group than the neutral hydroxyl group, which highly activates the aromatic ring toward electrophilic attack by the weak diazonium ion.

Q: Why is the temperature kept at $0-5^\circ\text{C}$ during diazotization?

Benzene diazonium chloride is highly unstable and can easily decompose into phenol and nitrogen gas at higher temperatures. Keeping the reaction ice-cold ensures the diazonium salt remains stable enough to undergo the subsequent coupling reaction.

Q: How did we get the molecular formula $\text{C}_{16}\text{H}_{12}\text{N}_2\text{O}$ for product Q?

The product is formed by joining a phenyl group ($\text{C}_6\text{H}_5$) and a naphthol group ($\text{C}_{10}\text{H}_6\text{OH}$) via an azo bridge ($-\text{N}=\text{N}-$). Adding these up: $6+10=16$ carbons, $5+6+1=12$ hydrogens, $2$ nitrogens, and $1$ oxygen.

Q: What is the purpose of acetic acid ($\text{CH}_3\text{CO}_2\text{H}$) in the final step?

The coupling reaction takes place in a basic medium. Adding acetic acid at the end neutralizes the excess base, ensuring the final azo dye product is obtained in its neutral, stable form rather than as a sodium salt.

Q: I forgot to multiply by the 75% yield and got 24.8 g. Why is this wrong?

In real-world chemistry, reactions rarely go to 100% completion due to side reactions or losses during purification. The problem explicitly states the overall yield is 75%, meaning you only obtain 75% of the theoretical maximum amount. You must multiply your theoretical moles by 0.75.

Question

Consider the reaction sequence from $\mathbf{P}$ to $\mathbf{Q}$ shown below. The overall yield of the major product $\mathbf{Q}$ from $\mathbf{P}$ is $75\%$. What is the amount in grams of $\mathbf{Q}$ obtained from $9.3	ext{ mL}$ of $\mathbf{P}$ ?

(Use density of $\mathbf{P} = 1.00	ext{ g mL}^{-1}$, Molar mass of $	ext{C} = 12.0$, $	ext{H} = 1.0$, $	ext{O} = 16.0$ and $	ext{N} = 14.0	ext{ g mol}^{-1}$)

Sigma Percentile · Accepted Answer

### Decoding the Reaction Sequence

Imagine you are a molecular architect, and your job is to build a vibrant, complex dye from a simple starting material. This problem tests exactly that skill, blending the elegance of organic synthesis with the rigorous accounting of physical chemistry.

We start with compound $\mathbf{P}$, which is given as aniline ($	ext{C}_6	ext{H}_5	ext{NH}_2$). The first step is a classic transformation: **diazotization**.

When aniline is treated with sodium nitrite ($	ext{NaNO}_2$) and hydrochloric acid ($	ext{HCl}$) at ice-cold temperatures ($0-5^\circ	ext{C}$), it is converted into benzene diazonium chloride.

$$ 	ext{C}_6	ext{H}_5	ext{NH}_2 \xrightarrow{	ext{NaNO}_2, 	ext{HCl}, 0-5^\circ	ext{C}} 	ext{C}_6	ext{H}_5	ext{N}_2^+	ext{Cl}^- $$

This diazonium salt is highly reactive and serves as a fantastic electrophile for our next step.

### The Art of Azo Coupling

Now, we introduce $\beta$-naphthol in a basic medium ($	ext{NaOH}$). This is where the magic happens.

In a basic environment, $\beta$-naphthol loses a proton to become a naphthoxide ion. The negatively charged oxygen atom is a powerful electron-donating group, pushing electron density into the aromatic ring and making it highly nucleophilic.

The weak diazonium electrophile attacks the electron-rich alpha position (C-1) of the naphthoxide ring. This electrophilic aromatic substitution is known as an **azo coupling reaction**.

The result is our major product $\mathbf{Q}$, an intensely colored azo dye known as 1-phenylazo-2-naphthol. The final addition of acetic acid simply neutralizes the reaction mixture.

### The Physical Chemistry Accounting

With the organic chemistry sorted, let's put on our accountant hats. We need to find out exactly how much of product $\mathbf{Q}$ we made.

First, let's determine how much aniline we started with. We are given a volume of $9.3	ext{ mL}$ and a density of $1.00	ext{ g/mL}$.

$$ 	ext{Mass of } \mathbf{P} = 9.3	ext{ mL} 	imes 1.00	ext{ g/mL} = 9.3	ext{ g} $$

The molar mass of aniline ($	ext{C}_6	ext{H}_7	ext{N}$) is $93	ext{ g/mol}$. Dividing the mass by the molar mass gives us the initial moles.

$$ 	ext{Moles of } \mathbf{P} = \frac{9.3	ext{ g}}{93	ext{ g/mol}} = 0.1	ext{ mol} $$

### The Catch: Percentage Yield

According to the balanced chemical equation, one mole of aniline should produce one mole of the azo dye. So, theoretically, we should obtain $0.1	ext{ mol}$ of product $\mathbf{Q}$.

However, the problem throws a curveball: the overall yield is only **$75\%$**.

In the real world, reactions are rarely perfect. To find the actual moles produced, we must multiply our theoretical yield by $0.75$.

$$ 	ext{Actual Moles of } \mathbf{Q} = 0.1	ext{ mol} 	imes 0.75 = 0.075	ext{ mol} $$

### The Final Calculation

To find the final mass in grams, we need the molar mass of our complex product $\mathbf{Q}$.

Let's count the atoms in 1-phenylazo-2-naphthol. It consists of a phenyl ring ($	ext{C}_6	ext{H}_5$), an azo bridge ($	ext{N}_2$), and a naphthol ring ($	ext{C}_{10}	ext{H}_6	ext{OH}$).

Combining these gives the molecular formula **$	ext{C}_{16}	ext{H}_{12}	ext{N}_2	ext{O}$**.

Now, we carefully calculate its molar mass using the given atomic weights:

$$ M_{\mathbf{Q}} = 16(12) + 12(1) + 2(14) + 16 = 248	ext{ g/mol} $$

Finally, we multiply the actual moles of $\mathbf{Q}$ by its molar mass to find the total mass obtained.

$$ 	ext{Mass of } \mathbf{Q} = 0.075	ext{ mol} 	imes 248	ext{ g/mol} = \mathbf{18.6	ext{ g}} $$

And there we have it! A beautiful synthesis of organic theory and physical calculation leading us to the final answer of $18.6	ext{ g}$.

Enter Numerical Value:

Visualized Solution

$Diazotization of Aniline$

$Azo Coupling Reaction$

$Calculating Moles of P$

$Applying Percentage Yield$

$Molar Mass of Product Q$

$Final Mass of Q$

$Why Basic Medium?$

The Sigma Insight: Amines

Decoding the Reaction Sequence

The Art of Azo Coupling

The Physical Chemistry Accounting

The Catch

The Final Calculation

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