In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is__

Q: Why does the haloform reaction occur here?

Acetophenone contains a methyl ketone group ($\text{-CO-CH}_3$), which is the specific structural requirement for the haloform reaction to take place.

Q: What happens to the missing carbon in the Hoffmann bromamide degradation?

The carbonyl carbon of the amide is lost as a carbonate ion ($\text{CO}_3^{2-}$), effectively stepping down the carbon chain by one.

Q: Why does aniline undergo tribromination instead of mono-bromination?

The $-\text{NH}_2$ group is a strongly activating ortho/para director. Without protection (like forming an acetanilide first), it activates the ring so much that all available ortho and para positions are rapidly brominated.

Q: How do we calculate the yield for multiple steps?

You must multiply the starting moles by the fractional yield of each step sequentially. For example: $10 \times 0.6 \times 0.5 \times 0.5 \times 1.0 = 1.5$ moles.

Q: Is acetic acid a catalyst in the last step?

Acetic acid acts as a polar solvent that facilitates the polarization of the bromine molecule, making the electrophilic attack on the highly activated aromatic ring easier.

Question

In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is_______.
(Atomic weight in $	ext{g mol}^{-1}$: $	ext{H} = 1, 	ext{C} = 12, 	ext{N} = 14, 	ext{O} = 16, 	ext{Br} = 80$. The yield (\%) corresponding to the product in each step is given in the parenthesis)

Sigma Percentile · Accepted Answer

This problem is a beautiful journey through a multi-step organic synthesis. It tests not only your knowledge of classic name reactions but also your ability to meticulously track stoichiometry and percentage yields across sequential transformations. Let's break down the chemistry step by step.

### The Synthetic Journey Begins

We start with $10 	ext{ mol}$ of **Acetophenone**, a classic aromatic ketone. The first set of reagents is $	ext{NaOBr}$ followed by acidic hydrolysis ($	ext{H}_3	ext{O}^+$). This should immediately ring a bell: it's the **Haloform Reaction**.

The methyl ketone group is highly susceptible to oxidation by hypohalites. The methyl group is cleaved off as bromoform ($	ext{CHBr}_3$), leaving behind a carboxylate salt, which upon acidification yields **Benzoic Acid** (Product A).

Since the yield for this step is $60\%$, we calculate the moles of A formed:
$$n_A = 10 	ext{ mol} 	imes 0.60 = 6 	ext{ mol}$$

### Stepping Down the Chain

Next, Benzoic Acid is treated with ammonia ($	ext{NH}_3$) and heated. Initially, an acid-base reaction occurs to form ammonium benzoate. Upon heating, a molecule of water is eliminated, converting the salt into an amide. Thus, Product B is **Benzamide**.

The yield for this transformation is $50\%$. Applying this to our available moles:
$$n_B = 6 	ext{ mol} 	imes 0.50 = 3 	ext{ mol}$$

Now comes a pivotal carbon-chain-shortening step. Benzamide is reacted with bromine and potassium hydroxide ($	ext{Br}_2/	ext{KOH}$). This is the famous **Hoffmann Bromamide Degradation**. The carbonyl carbon is expelled as a carbonate ion, and the nitrogen attaches directly to the aromatic ring, yielding a primary amine. Product C is **Aniline**.

With another $50\%$ yield, the moles of Aniline formed are:
$$n_C = 3 	ext{ mol} 	imes 0.50 = 1.5 	ext{ mol}$$

### Electrophilic Aromatic Substitution

In the final synthetic step, Aniline is treated with $3$ equivalents of bromine in acetic acid. The $-	ext{NH}_2$ group is a strongly activating, ortho/para-directing group. It pumps electron density into the benzene ring via resonance, making the ring highly nucleophilic.

Because the ring is so activated, it undergoes rapid polyhalogenation. Bromine atoms substitute at all available ortho and para positions, resulting in **2,4,6-Tribromoaniline** (Product D). The yield here is a perfect $100\%$, so we retain our $1.5 	ext{ mol}$ of product.

### The Final Calculation

To find the final mass in grams, we must determine the molar mass of 2,4,6-Tribromoaniline. Its molecular formula is $	ext{C}_6	ext{H}_4	ext{NBr}_3$.

Let's sum the atomic weights:
$$M = (6 	imes 12) + (4 	imes 1) + 14 + (3 	imes 80)$$
$$M = 72 + 4 + 14 + 240 = 330 	ext{ g/mol}$$

Finally, we multiply the number of moles by the molar mass to get the total mass of Product D:
$$	ext{Mass} = 1.5 	ext{ mol} 	imes 330 	ext{ g/mol} = 495 	ext{ g}$$

The final amount of D formed is exactly **$495 	ext{ g}$**.

Enter Numerical Value:

Visualized Solution

$n_{initial} = 10 mol$

$Haloform Reaction$

$Amide Formation$

$Hoffmann Bromamide Degradation$

$Electrophilic Aromatic Substitution$

$Molar Mass of D$

$Final Mass Calculation$

$Synthetic Strategy Insight$

The Sigma Insight: Amines

The Synthetic Journey Begins

Stepping Down the Chain

Electrophilic Aromatic Substitution

The Final Calculation

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