Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Welcome to a classic organic chemistry roadmap problem! These types of questions are fantastic because they test your knowledge of multiple reagents and reaction mechanisms in a single, interconnected sequence. Let's break down this journey step by step.

Our journey begins with a molecule having the molecular formula $C_8{H}_6$. The first thing we should always do is calculate the Degree of Unsaturation (DU) to get a hint about its structure.

The formula for DU is $C+1-\frac{H}{2}$. Plugging in our values, we get $8+1-\frac{6}{2}=6$.

A DU of 6 is highly indicative of an aromatic ring (which accounts for a DU of 4) and an additional alkyne group (which accounts for a DU of 2). Given the carbon count, the most logical structure is phenylacetylene ($Ph-C\equiv CH$).

The Journey to Compound X

The first pathway subjects phenylacetylene to $H_2$ in the presence of $Pd-BaS{O}_4$. This is the famous Lindlar's catalyst. Its job is to perform a partial reduction, taking the alkyne and stopping exactly at the alkene stage. This transforms our phenylacetylene into styrene ($Ph-CH=C{H}_2$).

Next, styrene undergoes hydroboration-oxidation using $B_2{H}_6$ followed by $H_2{O}_2$ in a basic medium. This reaction is known for adding water across a double bond following anti-Markovnikov's rule. The hydroxyl ($-OH$) group will attach to the less sterically hindered, less substituted carbon atom.

This leads us straight to 2-phenylethanol ($Ph-C{H}_2-C{H}_2-OH$). Looking at the options for Question 11, this perfectly matches Option (C).

The Journey to Compound Y

Now let's trace the second pathway. Phenylacetylene is treated with $H_{2}O$, $HgS{O}_4$, and $H_{2}S{O}_4$. This is the Kucherov reaction, a classic method for the hydration of alkynes. Unlike hydroboration, this reaction follows Markovnikov's rule. The intermediate enol rapidly tautomerizes to form a stable ketone. In our case, the product is acetophenone ($Ph-CO-C{H}_3$).

Acetophenone is then reacted with ethyl magnesium bromide ($EtMgBr$), a Grignard reagent. The nucleophilic ethyl group attacks the electrophilic carbonyl carbon. After hydrolysis, the ketone is converted into a tertiary alcohol: $Ph-C(OH)(C{H}_3)(C{H}_{2}C{H}_3)$.

The final step involves treating this tertiary alcohol with acid ($H^{+}$) and heat. This triggers an acid-catalyzed dehydration, eliminating a water molecule to form an alkene. According to Saytzeff's rule, the major product will be the most highly substituted and thermodynamically stable alkene.

By removing a proton from the adjacent $-C{H}_2-$ group of the ethyl chain, we form a double bond that is not only highly substituted but also in direct conjugation with the benzene ring, providing immense stability. The resulting major product is 2-phenyl-2-butene ($Ph-C(C{H}_3)=CH-C{H}_3$).

Checking the options for Question 12, this structure corresponds to Option (D).

By systematically applying the rules for each reagent, we've successfully navigated this organic roadmap!