Which of the following graph is correct between log P$_{CO_2}$ vs log X$_{CO_2}$?[given P$_{CO_2}$ = Partial Pressure of CO$_2$, X$_{CO_2}$ = Mole fraction of CO$_2$ in solution]
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The equation log P = log K + log X represents a linear relationship, where log P$_{CO_2}$ vs log X$_{CO_2}$ should yield a straight line.
Step 1: Identify Reaction Flow.
- Reagent KOH/Br$_2$ is the **Hoffmann Bromamide Degradation**. This implies the starting material X is an **Amide**.
- Amide X $\to$ Amine P (via Hoffmann).
- P $\to$ Diazonium salt Q (via Diazotisation).
- Q $\to$ Nitrile R (via Sandmeyer/CuCN).
- R $\to$ Carboxylic Acid S (via Hydrolysis).
- Also, X (Amide) $\to$ S (Acid) via direct hydrolysis. This is consistent.
Step 2: Identify Structure T.
S reacts with KMnO$_4$/H$^+$ to form T. T has "two types of hydrogen".
If S is a substituted benzoic acid, oxidation of alkyl side chains occurs.
Step 3: Analyze Compound X.
Formula C$_8$H$_9$NO. An aromatic amide. Common structure: Methylbenzamide ($CH_3-C_6H_4-CONH_2$) or N-methylbenzamide or acetanilide types.
Given X is an amide that undergoes Hoffmann, it must be a primary amide ($R-CONH_2$).
If X is a methyl-benzamide, hydrolysis yields methyl-benzoic acid (S).
Oxidation of S (methyl-benzoic acid) yields Terephthalic or Isophthalic or Phthalic acid (Dicarboxylic acid).
The final product T must have high symmetry to have only "two types of hydrogen" (e.g., benzene ring protons and acid protons).
Terephthalic acid (Para isomer) fits well: 4 aromatic protons (identical by symmetry) and 2 acidic protons (identical). Total 2 types.
Conclusion: X is likely p-methylbenzamide or similar structure leading to para-substituted product. Matching Option (3).
