Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Welcome to a beautiful matrix match problem! We are presented with a table containing four starting materials, four sets of reagents, and four reaction types. Our objective is to find the correct combinations for three distinct chemical scenarios. Let's break them down one by one and uncover the organic chemistry magic happening behind the scenes.

The Quest for Benzoic Acid

Our first goal is to synthesize Benzoic Acid ($Ph-COOH$). We must look closely at our starting materials to find a suitable candidate. Acetophenone ($Ph-COC{H}_3$) immediately stands out because it possesses a methyl ketone group attached directly to the benzene ring. This structural feature is a massive hint!

Do you remember what methyl ketones love to do? They are prime candidates for the Haloform reaction. When Acetophenone is treated with sodium hydroxide and bromine ($NaOH/B{r}_2$), the methyl group is oxidatively cleaved off.

The reaction proceeds as follows:

This reaction produces sodium benzoate and bromoform ($CHB{r}_3$). If we subsequently acidify the sodium benzoate, we obtain our desired benzoic acid! Therefore, the correct combination for question thirteen is Acetophenone (II), $NaOH/B{r}_2$ (i), and the Haloform reaction (S). This perfectly matches option (D).

Hunting for Radicals

Moving on to question fourteen, we are searching for a reaction that proceeds strictly through a radical mechanism. What is the classic signature of a radical mechanism in organic chemistry? It is the presence of an initiator like light (denoted by $h u$) or peroxides.

Looking at our table, Toluene ($Ph-C{H}_3$) has benzylic hydrogens. These hydrogens are highly susceptible to radical attack because the resulting benzylic radical is resonance-stabilized. Right there in column two, we see bromine paired with light ($Br_2/h u$)!

The light homolytically cleaves the bromine molecule into two highly reactive bromine radicals. These radicals then abstract a benzylic hydrogen from toluene, initiating a chain reaction. This free radical substitution yields benzyl bromide ($Ph-C{H}_{2}Br$).

Thus, the correct combination is Toluene (I), $Br_2/h u$ (ii), and Substitution (R). This matches option (A).

The Stereochemical Trap

Finally, we arrive at question fifteen. This one is a brilliant conceptual trap! We need a reaction that gives two different carboxylic acids. How is it possible to get two different acids from a single, clean reaction? The secret lies in stereochemistry. If a reaction creates a rigid double bond alongside a carboxylic group, we can obtain cis and trans geometrical isomers!

Let's focus on Benzaldehyde ($Ph-CHO$). It is an aromatic aldehyde with no alpha-hydrogens. When treated with an acid anhydride and its corresponding alkali metal salt—specifically acetic anhydride ($(C{H}_{3}CO{)}_{2}O$) and sodium acetate ($C{H}_{3}COOK$)—it undergoes the famous Perkin condensation.

The Perkin condensation yields an $\alpha ,\beta$-unsaturated aromatic acid. In this specific case, it forms Cinnamic acid ($Ph-CH=CH-COOH$).

Because of the restricted rotation around the newly formed carbon-carbon double bond, Cinnamic acid exists as two distinct geometrical isomers: Cis-Cinnamic acid and Trans-Cinnamic acid. These are physically and chemically two different carboxylic acids! Therefore, the correct combination is Benzaldehyde (III), Acetic anhydride/Sodium acetate (iii), and Condensation (P). This matches option (B).

Phenol's Destiny

Just to complete our matrix and satisfy our chemical curiosity, what happens to Phenol? When phenol is treated with sodium hydroxide and carbon dioxide ($NaOH/C{O}_2$), it undergoes the Kolbe-Schmitt reaction. This is an electrophilic aromatic substitution (specifically, a carboxylation) that yields Salicylic acid, the famous precursor to Aspirin. And with that, we have perfectly decoded the entire matrix!