Animated Solution for Chemistry - Organic Chemistry: The major product formed in the following reaction is

The journey to solving this organic chemistry puzzle is a beautiful exercise in recognizing not just the primary reaction, but the subtle side-quests that happen along the way. Let's dive into the molecular world of Tryptophan and see how it transforms!

We are given a molecule of Tryptophan, an essential amino acid. If we look closely at its structure, we can identify three key features: 1. An indole ring (a bicyclic structure containing a nitrogen atom). 2. A carboxylic acid group ($-COOH$). 3. A primary amino group ($-N{H}_2$).

The reagents provided are thionyl chloride ($SOC{l}_2$) and methanol ($C{H}_{3}OH$). Our goal is to determine the major product when these reagents interact with our starting molecule.

The first reagent, $SOC{l}_2$, is a classic tool in organic chemistry. Its primary job is to convert carboxylic acids into highly reactive acid chlorides.

When the $-COOH$ group reacts with $SOC{l}_2$, the hydroxyl group is replaced by a chlorine atom:

Notice the byproducts here: sulfur dioxide ($S{O}_2$), which bubbles away as a gas, and hydrogen chloride ($HCl$). This $HCl$ is crucial for the next part of our story.

Acid-Base Chemistry

Here is where many students make a silly mistake! They immediately proceed to the next step without considering the environment. We just generated $HCl$, a strong acid, right in the same flask as our molecule.

Our molecule contains a basic amino group ($-N{H}_2$). Acid-base reactions are typically the fastest reactions in organic chemistry. Before anything else can happen, the $HCl$ will protonate the basic amino group:

This forms a hydrochloride salt, often written as $-N{H}_2\cdot HCl$.

But wait, what about the nitrogen in the indole ring? The lone pair of electrons on the indole nitrogen is delocalized into the ring to maintain its aromaticity. Because these electrons are busy keeping the ring stable, they are not available to accept a proton. Therefore, the indole nitrogen remains unprotonated.

Now we have an acid chloride intermediate, and we introduce our second reagent: methanol ($C{H}_{3}OH$).

Acid chlorides are incredibly reactive electrophiles. The oxygen atom of methanol acts as a nucleophile, attacking the carbonyl carbon of the acid chloride. The chloride ion is an excellent leaving group and is quickly kicked out.

This nucleophilic acyl substitution converts the acid chloride into a methyl ester:

Combining all these steps, our final major product has the carboxylic acid converted to a methyl ester ($-COOC{H}_3$) and the amino group converted to its hydrochloride salt ($-N{H}_2\cdot HCl$). The indole ring remains completely untouched.

This elegant sequence of activation, protonation, and substitution leads us directly to the correct answer!