Animated Solution for Chemistry - s and p-Block Elements: The reaction of $H_3{N}_3{B}_{3}C{l}_3$ (A) with $LiB{H}_4$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to $H_3{N}_3{B}_3(Me{)}_3$. Compounds (B) and (C) respectively, are

A Borazine Reaction Sequence

Imagine a molecule that looks exactly like benzene, behaves somewhat like benzene, but isn't benzene at all. Welcome to the fascinating world of borazine ($B_3{N}_3{H}_6$), often dubbed "inorganic benzene." In this problem, we embark on a chemical journey starting from a substituted borazine derivative and explore two distinct reaction pathways that reveal the beautiful reactivity of this inorganic ring.

The Starting Material

Our starting material, Compound (A), is given as $H_3{N}_3{B}_{3}C{l}_3$. If we draw its structure, we see a six-membered ring consisting of alternating boron and nitrogen atoms. Because nitrogen is more electronegative than boron, the B-N bonds are polar.

Furthermore, nitrogen has a lone pair of electrons, while boron has an empty p-orbital. This allows for $p\pi -p\pi$ back-bonding, creating a delocalized electron cloud similar to the $\pi$-system in benzene. In Compound (A), the three hydrogen atoms are attached to the nitrogen atoms, and the three chlorine atoms are attached to the boron atoms. This specific arrangement makes the boron atoms highly susceptible to nucleophilic attack.

Reduction to Borazine

The first reaction treats Compound (A) with lithium borohydride ($LiB{H}_4$) in tetrahydrofuran (THF).

Lithium borohydride is a potent source of nucleophilic hydride ions ($H^{-}$). When it encounters the electron-deficient boron atoms in our ring, it readily substitutes the electronegative chlorine atoms. The result? All three chlorines are replaced by hydrogens, yielding $B_3{N}_3{H}_6$. This molecule is borazine, the famous inorganic analog of benzene. Thus, we have successfully identified Compound (B).

Alkylation via Grignard

Now, let's shift our focus to the second reaction. Compound (A) reacts with an unknown Compound (C) to produce $H_3{N}_3{B}_3(Me{)}_3$.

Looking at the product, we can see that the three chlorine atoms on the boron have been replaced by methyl ($-C{H}_3$) groups. To achieve this transformation, we need a reagent that can deliver a nucleophilic methyl group (a methyl carbanion, $C{H}_3^{-}$).

Among the standard organometallic reagents, Grignard reagents are the classic choice for this job. Methyl magnesium bromide ($C{H}_{3}MgBr$, or $MeMgBr$) perfectly fits the bill. The nucleophilic methyl group attacks the electrophilic boron, displacing the chloride ion. Therefore, Compound (C) must be $MeMgBr$.

By carefully analyzing the substitution patterns on the borazine ring, we deduced that Compound (B) is borazine and Compound (C) is $MeMgBr$. This logical deduction leads us straight to the correct option. It is a brilliant reminder of how inorganic rings mimic organic chemistry while maintaining their unique polar characteristics!