The stoichiometric reaction of 516 g of dimethyldichlorosilane with water results in a tetrameric cy

Q: Why does the reaction form a cyclic product instead of a long linear polymer?

While linear polymers are common, under specific controlled conditions (like certain solvents or concentrations), the ends of a growing short chain can react with each other faster than they react with new monomers, leading to cyclic structures. The problem explicitly dictates that a tetrameric cyclic product is formed, so we follow that constraint.

Q: How do I quickly figure out the formula of the tetramer?

When dimethyldichlorosilane hydrolyzes, it loses two $\text{Cl}$ atoms and gains one $\text{O}$ atom per silicon atom during condensation (since water is eliminated). The repeating monomer unit is therefore $(\text{CH}_3)_2\text{SiO}$. A 'tetramer' simply means you multiply this entire unit by 4, giving $[(\text{CH}_3)_2\text{SiO}]_4$.

Q: Why did we divide 516 by 129?

To perform stoichiometric calculations, we must work in moles, not grams. $129 \text{ g/mol}$ is the molar mass of the reactant, dimethyldichlorosilane. Dividing the given mass ($516 \text{ g}$) by the molar mass gives us the number of moles ($4 \text{ moles}$).

Q: Can I calculate the mass directly using a unitary method without finding moles?

Yes, you can! You know that $4 \times 129 \text{ g}$ of reactant theoretically yields $296 \text{ g}$ of product. So, $516 \text{ g}$ (which happens to be exactly $4 \times 129$) will yield $296 \text{ g}$. Then apply the $75\%$ yield. However, converting to moles first is generally safer and less prone to arithmetic errors.

Q: What would happen if we used trimethylchlorosilane instead?

Trimethylchlorosilane, $(\text{CH}_3)_3\text{SiCl}$, only has one chlorine atom. Upon hydrolysis, it forms $(\text{CH}_3)_3\text{SiOH}$. Because it only has one reactive hydroxyl group, it can only condense with one other molecule to form a dimer: $(\text{CH}_3)_3\text{Si-O-Si}(\text{CH}_3)_3$. It cannot form chains or rings, and is often used as a 'chain terminator' to control polymer length.

Question

The stoichiometric reaction of $516	ext{ g}$ of dimethyldichlorosilane with water results in a tetrameric cyclic product $\mathbf{X}$ in $75\%$ yield. The weight (in g) of $\mathbf{X}$ obtained is_______.
[Use, molar mass ($	ext{g mol}^{-1}$): $	ext{H} = 1, 	ext{C} = 12, 	ext{O} = 16, 	ext{Si} = 28, 	ext{Cl} = 35.5$]

Sigma Percentile · Accepted Answer

### The Chemistry of Silicones

Welcome to a fascinating journey into the world of inorganic polymers! Silicones are incredibly versatile materials, and their synthesis is a classic topic in JEE Advanced Chemistry. In this problem, we are tasked with finding the actual yield of a cyclic silicone product formed from the hydrolysis of dimethyldichlorosilane.

Let's break down the chemistry step-by-step. We start with our reactant, dimethyldichlorosilane, which has the chemical formula $(	ext{CH}_3)_2	ext{SiCl}_2$. When this molecule is exposed to water, it undergoes a rapid hydrolysis reaction. The highly electronegative chlorine atoms are excellent leaving groups, and they are replaced by hydroxyl ($	ext{-OH}$) groups from the water molecules.

This initial step produces an intermediate compound known as a silanediol, specifically dimethylsilanediol, $(	ext{CH}_3)_2	ext{Si(OH)}_2$, along with hydrochloric acid ($	ext{HCl}$) as a byproduct.

### The Condensation Phase

Now, silanediols are notoriously unstable and highly reactive. They do not simply remain in solution; instead, they undergo a process called **condensation polymerization**. In a condensation reaction, molecules join together while losing a small molecule—in this case, water.

Depending on the exact reaction conditions (like temperature, concentration, and solvent), these silanediols can link up to form long, linear polymer chains, or they can bite their own tails to form cyclic structures.

The problem explicitly states that a **tetrameric cyclic product** $\mathbf{X}$ is formed. This means exactly four molecules of dimethylsilanediol condense together, releasing four molecules of water, to form an eight-membered ring consisting of alternating silicon and oxygen atoms.

The overall stoichiometric equation for this transformation is:

$$4(	ext{CH}_3)_2	ext{SiCl}_2 + 4	ext{H}_2	ext{O} ightarrow [(	ext{CH}_3)_2	ext{SiO}]_4 + 8	ext{HCl}$$

### Crunching the Numbers

With the chemistry understood, we can now move on to the stoichiometry. First, we need to determine how many moles of our starting material we have.

Let's calculate the molar mass of dimethyldichlorosilane, $(	ext{CH}_3)_2	ext{SiCl}_2$:

$$M = 2(12) + 6(1) + 28 + 2(35.5) = 129 	ext{ g/mol}$$

We are given $516 	ext{ g}$ of the reactant. To find the number of moles ($n$), we divide the given mass by the molar mass:

$$n = \frac{516}{129} = 4 	ext{ moles}$$

Looking back at our balanced overall equation, we see a beautiful stoichiometric relationship: exactly $4 	ext{ moles}$ of dimethyldichlorosilane are required to produce $1 	ext{ mole}$ of the tetrameric cyclic product $\mathbf{X}$. Therefore, our reaction will theoretically yield exactly $1 	ext{ mole}$ of $\mathbf{X}$.

### Calculating the Final Yield

To find the theoretical mass of the product, we need the molar mass of $\mathbf{X}$, which has the formula $[(	ext{CH}_3)_2	ext{SiO}]_4$.

The mass of a single repeating unit, $(	ext{CH}_3)_2	ext{SiO}$, is:

$$M_{	ext{unit}} = 2(12) + 6(1) + 28 + 16 = 74 	ext{ g/mol}$$

Since it is a tetramer, we multiply this by four:

$$M_X = 4 	imes 74 = 296 	ext{ g/mol}$$

So, our theoretical yield is $296 	ext{ g}$. However, chemical reactions in the real world are rarely perfectly efficient. The problem states that the reaction has a **$75\%$ yield**.

To find the actual weight of $\mathbf{X}$ obtained, we simply take $75\%$ of our theoretical yield:

$$W_{	ext{actual}} = 296 	imes \frac{75}{100} = 296 	imes \frac{3}{4} = 222 	ext{ g}$$

And there we have it! The final weight of the cyclic product obtained is **$222 	ext{ g}$**.

Enter Numerical Value:

Visualized Solution

$(CH_{3})_{2} SiCl_{2} + 2 H_{2} O \to (CH_{3})_{2} Si(OH)_{2} + 2 HCl$

$4 (CH_{3})_{2} Si(OH)_{2} \to [(CH_{3})_{2} SiO]_{4} + 4 H_{2} O$

$M = 2 (12) + 6 (1) + 28 + 2 (35.5) = 129 g/mol$

$n = \frac{516}{129} = 4 moles$

$M_{X} = 4 \times 74 = 296 g/mol$

$W_{actual} = 296 \times \frac{75}{100} = 222 g$

$Chain Termination vs Polymerization$

The Sigma Insight: Group 14 Elements

The Chemistry of Silicones

The Condensation Phase

Crunching the Numbers

Calculating the Final Yield

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