Animated Solution for Chemistry - Organic Chemistry: If the reaction sequence given below is carried out with 15 moles of acetylene, the amount of the product D formed (in g) is ______. The yields of A, B, C and D are given in parentheses. [Given : Atomic mass of H = 1, C = 12, O = 16, Cl = 35]

Welcome to a beautiful journey through organic synthesis! Today, we are not just solving a problem; we are stepping into the shoes of a chemical engineer. We have a multi-step reaction sequence, and our mission is to track the exact amount of product formed at every single stage, paying close attention to the percentage yields.

Let's start with our raw material: $15$ moles of acetylene ($HC\equiv CH$).

Our first step is a classic transformation. When acetylene gas is passed through a red-hot iron tube, it undergoes cyclic polymerization. Three molecules of acetylene stitch together to form one beautiful, stable benzene ring ($C_6{H}_6$).

The stoichiometry tells us that $3$ moles of acetylene yield $1$ mole of benzene. Therefore, our $15$ moles of acetylene should theoretically give us $\frac{15}{3}=5$ moles of benzene.

However, the real world is rarely perfect. The problem states that the yield for this step is only $80\%$. So, we must adjust our expectations. The actual amount of benzene (Product A) formed is $5\times 0.80=4$ moles.

Now, we take our $4$ moles of benzene and react it with isopropyl chloride in the presence of anhydrous $AlC{l}_3$. This is the famous Friedel-Crafts alkylation. The $AlC{l}_3$ acts as a Lewis acid catalyst, generating an isopropyl carbocation that electrophilically attacks the benzene ring.

The result is isopropylbenzene, commonly known in the industry as cumene. Theoretically, $4$ moles of benzene should yield $4$ moles of cumene. But wait, the yield here drops to $50\%$.

Applying this yield, the actual amount of cumene (Product B) we obtain is $4\times 0.50=2$ moles.

Next up is one of the most important industrial reactions: the Cumene process. We subject our cumene to aerial oxidation to form cumene hydroperoxide, followed by acid hydrolysis ($H_3{O}^{+}$). This elegant sequence cleaves the molecule to produce two highly valuable chemicals: phenol and acetone.

Our primary focus is phenol (Product C). From our $2$ moles of cumene, we theoretically expect $2$ moles of phenol. Once again, the yield is $50\%$.

This means we successfully isolate $2\times 0.50=1$ mole of phenol.

We have finally reached the last step. We react our $1$ mole of phenol with acetyl chloride ($C{H}_{3}COCl$) in the presence of pyridine. Pyridine acts as a base to mop up the $HCl$ byproduct, driving the reaction forward.

This is an acetylation reaction, converting the phenol into an ester: phenyl acetate (Product D). The fantastic news here is that the yield is a perfect $100\%$! Therefore, our $1$ mole of phenol converts entirely into $1$ mole of phenyl acetate.

The question asks for the amount of product D in grams. To find this, we need the molar mass of phenyl acetate. Its molecular formula is $C_8{H}_8{O}_2$.

Let's calculate the molar mass: $M=(8\times 12)+(8\times 1)+(2\times 16)=96+8+32=136\text{ g/mol}$.

Since we have exactly $1$ mole of phenyl acetate, the final mass is simply $1\text{ mole}\times 136\text{ g/mol}=136\text{ g}$.

And there we have it! By carefully tracking the moles and applying the yields step-by-step, we've successfully navigated this synthetic pathway.