Animated Solution for Chemistry - Organic Chemistry: Scheme 1 and 2 describe the conversion of P to Q and R to S, respectively. Scheme 3 describes the synthesis of T from Q and S. The total number of Br atoms in a molecule of T is ________.

Welcome to this beautiful multi-step organic synthesis problem from JEE Advanced. Our ultimate goal is to find the total number of bromine atoms in the final product, T. To do this, we need to decode three interconnected reaction schemes. Let's break them down logically, starting with Scheme 1.

The Transformation of Aniline

In Scheme 1, our starting material P is aniline ($Ph-N{H}_2$). When we treat aniline with excess bromine water ($B{r}_2/H_{2}O$), the highly activating amino group strongly directs the incoming electrophiles to all ortho and para positions. This rapid electrophilic aromatic substitution gives us 2,4,6-tribromoaniline.

Next, we use sodium nitrite and hydrochloric acid ($NaN{O}_2,HCl$) at $273\text{ K}$. This is the classic diazotization reaction, converting the amino group into a diazonium salt ($-N_2^{+}C{l}^{-}$). Immediately after, we apply the Sandmeyer reaction using copper cyanide ($CuCN/KCN$), which replaces the diazonium group with a nitrile group ($-CN$), yielding 2,4,6-tribromobenzonitrile.

Now, acidic hydrolysis of the nitrile group under heating ($H_3{O}^{+},\Delta$) converts it into a carboxylic acid. Finally, reacting this acid with thionyl chloride ($SOC{l}_2$) in the presence of pyridine gives us an acyl chloride. This is our major product Q: 2,4,6-tribromobenzoyl chloride. Notice carefully, Q carries exactly three bromine atoms.

The Journey of Benzene

Moving on to Scheme 2, we start with benzene, labeled as R. Treating it with oleum sulfonates the ring to form benzenesulfonic acid ($Ph-S{O}_{3}H$). Fusing this intermediate with sodium hydroxide ($NaOH$) at high temperatures, followed by acidification ($H^{+}$), is the standard industrial method to synthesize phenol ($Ph-OH$).

Phenol is then treated with bromine in carbon disulfide ($B{r}_2,C{S}_2$) at a low temperature ($273\text{ K}$). Because carbon disulfide is a non-polar solvent, it restricts the ionization of phenol, leading to monobromination rather than polybromination. Due to steric hindrance at the ortho position, the major product S is para-bromophenol, which contains exactly one bromine atom.

The Grand Finale

Finally, let's look at Scheme 3. We react para-bromophenol (S) with sodium hydroxide ($NaOH$) to generate a highly nucleophilic phenoxide ion.

This phenoxide ion then attacks the electrophilic carbonyl carbon of our acyl chloride Q. The chloride ion is kicked out as a leaving group, resulting in a nucleophilic acyl substitution (specifically, the Schotten-Baumann reaction) that forms an ester. This massive ester molecule is our final product T.

The question asks for the total number of bromine atoms in a molecule of T. Let's count them up. We have three bromine atoms originating from the benzoyl part, which came from Q. And we have one bromine atom on the phenoxy part, which came from S.

Three plus one gives us a total of 4 bromine atoms. And that is our final, elegant answer.