Problem 22 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An organic compound containing \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\) exists in two isomeric forms \(\mathrm{A}\) and \(\mathrm{B}\). An amount of \(0.108 \mathrm{~g}\) of one of the isomers gives on combustion \(0.308 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.072 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} . \mathrm{A}\) is insoluble in \(\mathrm{NaOH}\) and \(\mathrm{NaHCO}_{3}\) while \(\mathrm{B}\) is soluble in \(\mathrm{NaOH}\). A reacts with conc. HI to give compounds \(C\) and D. C can be separated from D by ethanolic \(\mathrm{AgNO}_{3}\) solution and \(\mathrm{D}\) is soluble in \(\mathrm{NaOH}\). B reacts readily with bromine water to give compound \(\mathrm{E}\) of molecular formula, \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{OBr}_{3}\). Identify, \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\) with justification and give their structures.

Step-by-Step Solution

Answer

A is anisole, B is phenol, C is methyl iodide, D is phenol, E is 2,4,6-tribromophenol.

Step 1: Determine the Empirical Formula from Combustion

From the combustion data of 0.108 g of an isomer forming 0.308 g of CO\(_2\) and 0.072 g of H\(_2\)O, we find the moles of C and H in the compound. For CO\(_2\), moles of C = \( \frac{0.308}{44} \approx 0.007 \) moles. This gives 0.007 moles of C. For H\(_2\)O, moles of H = \( 2 \times \frac{0.072}{18} \approx 0.008 \) moles. We find the empirical formula is 'close' to C\(_x\)H\(_y\)O\(_z\). However, we calculate mass C = 0.007 \(\times\) 12 = 0.084 g, mass H = 0.008 \(\times\) 1 = 0.008 g, leaving mass O = 0.108 - 0.084 - 0.008 = 0.016 g. Moles of O = \(\frac{0.016}{16}\) = 0.001. Hence, empirical ratio is C: H: O = 7: 8: 1.

Step 2: Identify Solubility Properties

Know the properties of isomers: A is insoluble in both NaOH and NaHCO\(_3\), suggesting it does not have an acidic or phenolic OH group. B is soluble in NaOH, suggesting it has an acidic/phnolic OH group and reacts with alkali.

Step 3: Analyze Reactivity with Concentrated HI

A reacts with concentrated HI to form compounds C and D. C can be separated from D by ethanolic AgNO\(_3\), indicating C is an alkyl or benzyl iodide. D is soluble in NaOH, indicating it could be an acidic compound such as a phenol.

Step 4: Analyze Reactivity with Bromine Water

Isomer B reacts with bromine water to form compound E with the formula C\(_7\)H\(_5\)OBr\(_3\). This points towards compound B being a phenol due to classic phenol-based bromination resulting in tribromo-substitution at positions ortho and para to the OH group.

Step 5: Identify and Justify Compounds

Based on the empirical formula C\(_7\)H\(_8\)O and reactivity:- A is anisole (C\(_6\)H\(_5\)OCH\(_3\)), not soluble in NaOH.- C is methyl iodide (CH\(_3\)I), reacts with ethanolic AgNO\(_3\). - D is phenol (C\(_6\)H\(_5\)OH), reacts with NaOH.- B is phenol, soluble in NaOH.- E is 2,4,6-tribromophenol (C\(_6\)H\(_2\)OHBr\(_3\)).

Key Concepts

Isomerism

Isomerism refers to a fascinating phenomenon in organic chemistry where compounds share the same molecular formula but differ in atom arrangements. These variations lead to significantly different chemical properties. In the case of this problem, we have two isomeric forms labeled as A and B. Both have the same general molecular formula, which, from the combustion analysis, turns out to be C\(_7\)H\(_8\)O.
Isomerism is especially important because it shows how changes in the structure affect properties like solubility and reactivity. For instance, in our exercise, Isomer A is insoluble in NaOH, while Isomer B is soluble. Such differences often indicate different functional groups or structural arrangements, which can significantly alter how these molecules interact with other chemicals.

Combustion Analysis

Combustion analysis is an experimental technique used to determine the composition of a compound, typically organic, by burning the compound and analyzing the resulting products. In this exercise, 0.108 g of the isomer is combusted, yielding 0.308 g of CO\(_2\) and 0.072 g of H\(_2\)O.
This process involves calculating the moles of carbon and hydrogen from the masses of CO\(_2\) and H\(_2\)O produced:

• Moles of carbon: calculated by dividing the mass of CO\(_2\) produced by the molar mass of CO\(_2\).
• Moles of hydrogen: derived from the mass and stoichiometry of H\(_2\)O.

The remaining mass difference gives us the moles of oxygen. Thus, combustion analysis efficiently provides the empirical formula, which can be further verified through additional structural and reactivity insights.

Phenol Bromination

Phenol bromination is a specific chemical reaction where phenol derivatives react with bromine, usually in water, to produce brominated phenols. This reaction indicates the presence of a phenolic group, a common feature in compounds like Isomer B.
When Isomer B reacts with bromine water, it forms compound E, which has the formula C\(_7\)H\(_5\)OBr\(_3\). The tribrominated pattern suggests substitution occurs at both ortho and para positions relative to the hydroxyl group, which is a classic characteristic of phenol's reactions.
These details help confirm the identity of B as phenol, which allows for predictable reactions with reagents like bromine, leading to useful structural insights.

Solubility and Reactivity

Solubility and reactivity are key properties that help distinguish between isomers. In this exercise, Isomer A's insolubility in NaOH and NaHCO\(_3\) compared to Isomer B's solubility in NaOH provides clues about their functional groups.
Typically, solubility behavior in such bases suggests the presence of acidic functional groups in Isomer B, like a phenolic OH group, which can react with NaOH to form soluble salts. Conversely, the lack of reactivity in Isomer A implies the absence of such reactive groups, pointing more towards a structure like ether in anisole (C\(_6\)H\(_5\)OCH\(_3\)), which does not ionize similarly.
Such insights not only assist in characterization but also in predicting how these compounds might behave in different chemical environments.

Empirical Formula Determination

Empirical formula determination involves finding the simplest whole-number ratio of elements in a compound. In the case study, the empirical formula was derived from combustion analysis data. After calculating the moles of C, H, and O, these numbers were used to establish a ratio.
For example:

• The empirical formula C:H:O was calculated as 7:8:1 from moles derived from combustion products.

The empirical formula serves as a stepping stone for identifying compounds' molecular formulas and understanding their isometric nature, as it reflects the actual number of atoms involved in the molecular structure, thus contributing to differentiating the two isomers under investigation.