Animated Solution for Chemistry - Organic Compounds: The major product of the following reaction is

A Journey from Phenol to Benzofuran

Welcome to a beautiful organic synthesis problem! This question is a fantastic test of your ability to predict chemoselectivity and visualize intramolecular reactions. Let's carefully analyze our starting material, $4$-bromo-$2$-(2-bromopropyl)phenol.

We have a benzene ring with three distinct reactive sites: a bromine atom directly attached to the ring, a phenolic $-OH$ group, and a $2$-bromopropyl side chain. The key to solving this problem is taking it one reagent at a time and asking yourself, "Who is the most vulnerable target here?"

Selective Substitution ($S_{N}2$)

Our first reagent is aqueous $KOH$, which provides hydroxide ions ($O{H}^{-}$). These are excellent nucleophiles. Now, where will they attack?

The bromine attached directly to the benzene ring is highly unreactive. Thanks to resonance, the lone pairs on the bromine delocalize into the aromatic ring, giving the $C-Br$ bond partial double bond character. It's locked in tight.

However, the bromine on the alkyl side chain is a perfect target for an $S_{N}2$ reaction. The hydroxide ion attacks the secondary carbon from the back, kicking out the bromide ion. This smoothly converts our alkyl bromide into a secondary alcohol, yielding $4$-bromo-$2$-(2-hydroxypropyl)phenol. Notice how the rest of the molecule remains completely untouched.

Targeted Oxidation

Next, we introduce chromium trioxide in an acidic medium ($Cr{O}_3/H^{+}$), also known as Jones reagent. This is a classic, powerful oxidizing agent. Its job is to look for oxidizable groups, and it finds our newly formed secondary alcohol right away.

The Jones reagent oxidizes the secondary alcohol, stripping away hydrogens to form a carbon-oxygen double bond. We now have a ketone on our side chain! You might wonder, "Could the $Cr{O}_3$ oxidize the phenolic $-OH$?" While phenols can be oxidized under harsh conditions, the secondary alcohol is far more susceptible. The aromatic ring provides significant stability to the phenol, protecting it from simple oxidation here. Our intermediate is now $1$-(5-bromo-2-hydroxyphenyl)propan-2-one.

The Grand Finale - Intramolecular Cyclization

Now for the grand finale! We add sulfuric acid ($H_{2}S{O}_4$) and apply heat. The acid protonates the ketone, making its carbonyl carbon highly electrophilic.

Right next door, we have our phenolic oxygen with its lone pairs, just waiting for an opportunity. It swings around and attacks the carbonyl carbon. Because the carbonyl carbon is exactly five atoms away from the phenolic oxygen, this intramolecular attack forms a highly favored five-membered ring, specifically a hemiketal.

But we are under strongly acidic and heated conditions. This means the system won't stop at the hemiketal stage. To gain ultimate stability, the protonated hydroxyl group leaves as a water molecule.

The loss of water creates a new carbon-carbon double bond, giving us a fully conjugated, highly stable benzofuran ring system. If you trace the carbon skeleton carefully, you'll see that the methyl group remains attached to the carbon that was once the carbonyl carbon. In the final benzofuran structure, this is designated as $C-2$.

And there we have it, our final major product: $5$-bromo-$2$-methylbenzofuran!