The reaction sequence given below is carried out with 16 moles of X. The yield of the major product

Q: Why does the first step halve the number of moles?

The sodium-mediated coupling joins two molecules of the aryl bromide together to form a single biphenyl molecule. Thus, 16 moles of monomer yield 8 moles of dimer.

Q: What exactly happens during the Cannizzaro step?

Because the intermediate P has two aldehyde groups and no alpha-hydrogens, heating it with a strong base causes an intramolecular redox reaction. One aldehyde is oxidized to a carboxylic acid, and the other is reduced to an alcohol.

Q: Why doesn't the soda-lime affect the alcohol group?

Soda-lime decarboxylation is highly specific to carboxylic acids. It removes the `-COOH` group as carbon dioxide, leaving the primary alcohol (`-CH_2OH`) completely intact.

Q: Where does the extra compound R come from in the final step?

The reaction sequence implies that the product R from step 3 is used as a reagent in step 5. The 1 mole of T formed in step 4 reacts with the available R to form the final ether.

Q: How do I quickly calculate the molar mass of such a large molecule?

Break it down into its components. The final molecule S consists of two biphenyl units, two `-CH_2-` linkers, and one oxygen atom. Counting the atoms systematically gives $ \text{C}_{26}\text{H}_{22}\text{O} $, which sums to 350 g/mol.

Question

The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is_______.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80

Sigma Percentile · Accepted Answer

This problem is an absolute masterpiece of multi-step organic synthesis. It tests not only your knowledge of diverse organic reactions but also your ability to meticulously track stoichiometry and percentage yields across a long sequence. Let's embark on this chemical journey step by step.

### Analyzing the Starting Material

We begin with compound **X**, which is 2-(2-bromophenyl)-1,3-dioxolane. Notice the two key features here: an aryl bromide and an acetal group. The acetal is essentially a masked aldehyde, protecting it from unwanted side reactions early in the sequence. We are given exactly $16 	ext{ moles}$ of this starting material.

### Step 1: Wurtz-Fittig Coupling and Deprotection

The first set of reagents is sodium in dry ether, followed by acid hydrolysis ($H_3O^+$). The sodium triggers a coupling reaction, joining two molecules of the aryl bromide together to form a biphenyl derivative.

Immediately after, the acid swoops in and unmasks the acetal groups, reverting them back to aldehydes. Because two molecules of **X** combine to form one molecule of **P** (biphenyl-2,2'-dicarbaldehyde), our initial $16 	ext{ moles}$ are halved. With a $100\%$ yield, we obtain exactly $8 	ext{ moles}$ of **P**.

### Step 2: The Intramolecular Cannizzaro Reaction

Compound **P** possesses two aldehyde groups and lacks alpha-hydrogens. When heated with a strong base like $NaOH$, it undergoes a beautiful disproportionation known as the intramolecular Cannizzaro reaction.

One aldehyde group is oxidized to a carboxylic acid ($-COOH$), while the other is simultaneously reduced to a primary alcohol ($-CH_2OH$). This yields compound **Q**. However, the yield for this step is only $50\%$. Therefore, our $8 	ext{ moles}$ of **P** produce just $4 	ext{ moles}$ of **Q**.

### Step 3: Decarboxylation

Next, we treat **Q** with soda-lime ($NaOH/CaO$) and heat it. This is a classic setup for decarboxylation. The carboxylic acid group is completely ripped off as carbon dioxide gas, leaving the primary alcohol untouched.

This transforms **Q** into compound **R** (biphenyl-2-ylmethanol). With another $50\%$ yield, our $4 	ext{ moles}$ of **Q** are reduced to $2 	ext{ moles}$ of **R**.

### Step 4: Bromination of the Alcohol

Moving forward, we react **R** with phosphorus tribromide ($PBr_3$). This reagent has one specific job: it replaces the primary alcohol group with a bromine atom via nucleophilic substitution.

Compound **R** is thus converted into compound **T** (2-(bromomethyl)biphenyl). Factoring in the $50\%$ yield, our $2 	ext{ moles}$ of **R** give us exactly $1 	ext{ mole}$ of **T**.

### Step 5: Williamson Ether Synthesis

Here is where the sequence culminates. We take our newly formed compound **T** and react it with some of the compound **R** we made earlier, in the presence of sodium hydride ($NaH$).

The $NaH$ deprotonates the alcohol in **R**, creating a strong alkoxide nucleophile. This alkoxide then attacks the bromomethyl group of **T** in a classic Williamson ether synthesis, forming our final massive ether, compound **S**.

Since **T** is our limiting reagent at $1 	ext{ mole}$, and the yield is $50\%$, we end up with $0.5 	ext{ moles}$ of **S**.

### Final Calculation

To find the final mass, we must calculate the molar mass of **S**. The chemical formula for this large ether is $C_{26}H_{22}O$.

$$ M_S = 26(12) + 22(1) + 16 = 350 	ext{ g/mol} $$

Multiplying this molar mass by our final molar quantity gives us the answer:

$$ W_S = 0.5 	ext{ moles} 	imes 350 	ext{ g/mol} = \mathbf{175 	ext{ g}} $$

This sequence is a brilliant reminder to always keep a strict eye on stoichiometry and percentage yields at every single step of a synthesis!

Enter Numerical Value:

Visualized Solution

$n_{X} = 16 mol$

$X Na, H_{3} O^{+} P$

$P NaOH, Δ Q$

$Q NaOH/CaO R$

$R PBr_{3} T$

$T + R NaH S$

$W_{S} = n_{S} \times M_{S}$

$Conclusion$

The Sigma Insight: Carbonyl Compounds

Analyzing the Starting Material

Step 1

Step 2

Step 3

Step 4

Step 5

Final Calculation

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