Animated Solution for Chemistry - Organic Chemistry: The species which can best serve as an initiator for the cationic polymerization is

To truly master polymer chemistry, we must understand the fundamental mechanisms that drive chain growth. In cationic polymerization, the growing polymer chain is not a radical or an anion, but a positively charged species—a carbocation.

Because the active center is positively charged, the monomer molecules that add to the chain must be electron-rich. Alkenes with electron-donating groups (like isobutylene) are perfect candidates because their $\pi$-bonds are dense clouds of polarizable electrons.

Every great chain reaction needs a spark. For cationic polymerization, this spark must be an initiator capable of pulling electrons away from the monomer's $\pi$-bond to create that very first carbocation.

If the monomer is electron-rich, the initiator must be electron-deficient. In the language of chemistry, an electron-pair acceptor is known as a Lewis acid. The ideal initiator is a species that is "hungry" for electrons and can stabilize the resulting complex.

Let's critically examine the options provided in the question:

1. ${\text{LiAlH}}_4$: This is a classic, powerful reducing agent. It acts as a source of hydride ions ($H^{-}$). Since hydride ions are negatively charged and electron-rich, they act as nucleophiles. They would initiate anionic polymerization, not cationic.

2. ${\text{HNO}}_3$: While nitric acid is a strong Brønsted-Lowry acid and can provide a proton ($H^{+}$) to start the chain, it leaves behind the nitrate ion ($N{O}_3^{-}$). This anion is a decent nucleophile and will quickly attack the newly formed carbocation, terminating the chain prematurely before a high-molecular-weight polymer can form.

3. $\text{BaLi}$: This is an unusual alloy/complex and does not possess the standard properties required to initiate cationic chain growth.

Aluminum Chloride

This brings us to ${\text{AlCl}}_3$. Let's look at its electronic structure. The central Aluminum atom is bonded to three Chlorine atoms, giving it a total of only $6$ valence electrons.

Because its octet is incomplete, Aluminum has an empty p-orbital. This makes ${\text{AlCl}}_3$ a highly potent Lewis acid. When an electron-rich alkene approaches, the empty orbital of Aluminum eagerly accepts the $\pi$-electrons from the double bond. This interaction successfully generates the initial carbocation without leaving behind a highly nucleophilic counter-ion that would kill the reaction.

Therefore, ${\text{AlCl}}_3$ serves as an excellent initiator for cationic polymerization, making option (c) the correct answer.