Animated Solution for Chemistry - Organic Chemistry: The number of -CH2- (methylene) groups in the product formed from the following reaction sequence is ________.

Welcome, future engineers and doctors! Today, we are going to embark on a fascinating molecular journey. This isn't just a sequence of reactions; it's a story of transformation, where a simple straight-chain alkene is chopped, oxidized, stitched back together, and finally curled into one of the most elegant structures in organic chemistry: the benzene ring.

Before we unleash our chemical reagents, we must first understand what we are working with. Look at the zig-zag structure provided in the question. It might look like a simple line drawing, but every vertex and every line tells a story.

If we count the carbon atoms from either end, we find an eight-carbon chain. Right in the middle, between carbon-4 and carbon-5, sits a double bond. Because the molecule is perfectly symmetric, it doesn't matter which side we start counting from. This starting material is oct-4-ene. Identifying the starting material correctly is the crucial first step; a single miscounted carbon here would derail the entire sequence!

Ozonolysis

Our first set of reagents is ozone (${\text{O}}_3$) followed by zinc dust and water (${\text{Zn/H}}_2\text{O}$). This is the classic reductive ozonolysis.

Imagine ozone as a pair of molecular scissors. It doesn't just break the pi bond; it completely cleaves the carbon-carbon sigma bond as well. The double bond is sliced right down the middle. Because we are using zinc, the reaction is reductive, meaning it stops at the aldehyde stage and doesn't over-oxidize.

Each half of our cleaved oct-4-ene receives an oxygen atom. Since the original molecule was symmetric, we get two identical molecules of a four-carbon aldehyde. This molecule is butanal (${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{CHO}$).

Next, we introduce potassium permanganate (${\text{KMnO}}_4$). If ozone was a pair of scissors, ${\text{KMnO}}_4$ is a sledgehammer of oxidation. It is a incredibly strong oxidizing agent.

Aldehydes are highly susceptible to oxidation. The hydrogen atom attached directly to the carbonyl carbon is a prime target. ${\text{KMnO}}_4$ swiftly replaces this hydrogen with a hydroxyl ($-\text{OH}$) group.

This elegant substitution transforms our butanal into a carboxylic acid. Specifically, we now have two moles of butanoic acid (${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{COOH}$).

Now we reach step three: treating our butanoic acid with sodium hydroxide ($\text{NaOH}$) and subjecting it to electrolysis. This is the famous Kolbe electrolysis, a brilliant method for synthesizing symmetric alkanes.

First, the sodium hydroxide neutralizes the butanoic acid to form sodium butanoate. When an electric current is passed through the aqueous solution, the carboxylate ions migrate to the anode. Here, they lose an electron and undergo decarboxylation—meaning they release carbon dioxide (${\text{CO}}_2$) gas.

What's left behind is a highly reactive propyl radical (${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2^{\bullet }$). Radicals hate being alone. Two of these propyl radicals will quickly find each other and couple together. Three carbons plus three carbons gives us a straight six-carbon chain. We have successfully synthesized n-hexane (${\text{C}}_6{\text{H}}_{14}$)!

The Final Transformation

We are in the home stretch. Our n-hexane is subjected to chromium oxide (${\text{Cr}}_2{\text{O}}_3$) at a blistering temperature of $770\text{ K}$ and a crushing pressure of $20\text{ atm}$.

These are the textbook conditions for aromatization (or dehydrocyclization). Alkanes with six or more carbons, when subjected to these extreme conditions over a catalyst, undergo a remarkable transformation. The straight hexane chain curls up, loses four molecules of hydrogen gas (${\text{H}}_2$), and forms a highly stable, conjugated aromatic ring.

Our n-hexane has been transformed into the king of aromatic compounds: benzene (${\text{C}}_6{\text{H}}_6$).

Counting Methylene Groups

After this epic journey, we must answer the actual question asked by the examiners: What is the number of methylene ($-{\text{CH}}_2-$) groups in the final product?

Let's look closely at the structure of benzene. It is a perfect hexagon where every single carbon atom is $s{p}^2$ hybridized. Because of the alternating double bonds (or the delocalized pi electron cloud), each carbon is bonded to exactly two other carbons and exactly one hydrogen atom.

This means benzene consists entirely of $-\text{CH}-$ groups. There are absolutely zero $-{\text{CH}}_2-$ (methylene) groups present in the molecule.

Therefore, our final answer is 0.

This question is a beautiful test of your ability to chain multiple concepts together. It requires you to know ozonolysis, oxidation, Kolbe electrolysis, and aromatization. But more importantly, it requires you to keep a cool head and track your carbon atoms carefully from start to finish. Keep practicing, and soon these reaction sequences will feel like second nature!