Animated Solution for Chemistry - Organic Chemistry: For the reaction sequence given below, the correct statement(s) is(are) (In the options, X is any atom other than carbon and hydrogen, and it is different in P, Q and R)

Welcome to an exciting journey through the world of organic chemistry! Today, we are going to unravel a fascinating reaction sequence that beautifully ties together core organic reactions with fundamental periodic properties. I know reaction schemes can sometimes look like a tangled web, but let's take a breath and break it down step by step. Imagine you are a detective, and each reagent is a clue leading you to the final molecular structure.

Our journey begins with a simple, elegant molecule: cyclohexene. We introduce it to hydrogen bromide ($HBr$). What happens next? This is a classic electrophilic addition reaction. The double bond, rich in electrons, reaches out and grabs the hydrogen, leaving the bromide ion to attack the resulting carbocation. Because cyclohexene is perfectly symmetrical, we don't even need to worry about Markovnikov's rule here! The result is our first product, P, which is bromocyclohexane. So, for product $P$, our mystery halogen $X$ is bromine ($Br$).

Now, the plot thickens as our product $P$ takes two different paths.

On the first path, $P$ undergoes the Finkelstein reaction. This is a named reaction you definitely want to keep in your arsenal. By treating bromocyclohexane with sodium iodide ($NaI$) in acetone, we perform a neat halogen exchange. The iodine swaps places with the bromine, giving us iodocyclohexane. This is our product Q, where the halogen $X$ is iodine ($I$).

On the second path, $P$ is subjected to the Swarts reaction. This is another halogen exchange, but it's specifically designed to introduce fluorine, which is notoriously difficult to add directly. By using heavy metal fluorides like silver fluoride ($AgF$), the bromine is replaced by fluorine, yielding fluorocyclohexane. This is our product R, where the halogen $X$ is fluorine ($F$).

Now that we have unmasked $P$, $Q$, and $R$, let's evaluate the given options by looking at the periodic trends of our halogens: fluorine, bromine, and iodine.

Bond Length: As we travel down the halogen group from fluorine to iodine, the atomic size increases significantly. A larger atom means a longer bond. Therefore, the carbon-halogen bond length follows the order $C-F<C-Br<C-I$. Translating this to our products, the order is $R<P<Q$. Option (A) suggests the opposite, so we can confidently cross it out.

Bond Enthalpy: Here is a crucial concept—bond strength is inversely proportional to bond length. The short $C-F$ bond is incredibly strong and requires a lot of energy to break. Conversely, the long $C-I$ bond is much weaker. Thus, the bond enthalpy order is $C-F>C-Br>C-I$, which means $R>P>Q$. This perfectly matches Option (B)!

$S_{N}2$ Reactivity: For a successful $S_{N}2$ reaction, we need a stellar leaving group. The rule of thumb is that weaker bases make better leaving groups. Since hydroiodic acid ($HI$) is a much stronger acid than hydrobromic acid ($HBr$) and hydrofluoric acid ($HF$), the iodide ion is the weakest base and the best leaving group. Therefore, the reactivity order is $C-I>C-Br>C-F$, or $Q>P>R$. Option (C) is incorrect.

$p{K}_a$ of Conjugate Acids: Finally, let's look at the acidity of the conjugate acids: $HF$, $HBr$, and $HI$. We already established that $HI$ is the strongest acid, which means it has the lowest $p{K}_a$ value. $HF$, being the weakest, has the highest $p{K}_a$. The $p{K}_a$ order is $HF>HBr>HI$, meaning $R>P>Q$. Option (D) gets this wrong.

After meticulously analyzing the chemical structures and their physical properties, we have found our answer. The only statement that holds true is the one describing the bond enthalpy. This problem is a fantastic reminder of how interconnected organic chemistry is with the fundamental principles of the periodic table. Keep practicing, stay curious, and you'll master these concepts in no time!