Problem 103 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An unknown compound of carbon, hydrogen and oxygen contains \(69.77 \%\) carbon and \(11.63 \%\) hydrogen and has a molecular weight of 86\. It does not reduce Fehling solution, but forms a bisulphite addition compound and gives a positive iodoform test. What are the possible structures for the unknown compound?

Step-by-Step Solution

Answer

Possible structures are 2-pentanone and 3-pentanone.

Step 1: Determine the empirical formula

The compound consists of carbon (69.77%), hydrogen (11.63%), and the remainder oxygen. First, let's calculate the oxygen percentage: \[ 100 ext{%} - 69.77 ext{%} - 11.63 ext{%} = 18.60 ext{%}. \]Next, assume 100 g of the compound to find moles of each element:- Moles of C: \( 69.77/12.01 = 5.81 \)- Moles of H: \( 11.63/1.008 = 11.54 \)- Moles of O: \( 18.60/16.00 = 1.16 \).Divide all by the smallest number of moles, 1.16, to find the simplest ratio:- C: \( 5.81/1.16 \approx 5 \)- H: \( 11.54/1.16 \approx 10 \)- O: \( 1.16/1.16 = 1 \).Empirical formula is \( \text{C}_5\text{H}_{10}\text{O} \).

Step 2: Determine the molecular formula

To find the molecular formula, compare the empirical formula mass with the molecular mass:- Empirical formula mass \( = (5*12.01) + (10*1.008) + (16.00) = 86.14 \).- Molecular weight is given as 86.Therefore, the molecular and empirical formulas are the same, as 86.14 is very close to 86.

Step 3: Analyze the chemical tests

The compound does not reduce Fehling solution, suggesting it is not an aldehyde. It forms a bisulfite addition compound, which indicates the presence of a carbonyl group (ketone or aldehyde). It gives a positive iodoform test, implying a methyl ketone structure.

Step 4: Determine possible structures

Given the information, we have a methyl ketone with five carbons: 1. 2-Pentanone (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \))2. 3-Pentanone (\( \text{CH}_3\text{CH}_2\text{CO}\text{C}_2\text{H}_5 \))These are possible structures that match all the given criteria and tests.

Key Concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It's a fundamental concept in chemistry, often used as a starting point to determine more specific chemical formulas. To find an empirical formula, you do the following:

• Convert the percentage of each element into grams, assuming you have 100 grams of the compound. This makes it easy to translate percentages directly into gram amounts.
• Calculate the number of moles of each element using their molar masses. For carbon, use 12.01 g/mol; for hydrogen, 1.008 g/mol; and for oxygen, 16.00 g/mol.
• Find the simplest ratio of moles by dividing each value by the smallest number among them.

In our example, this process revealed an empirical formula of \( \text{C}_5\text{H}_{10}\text{O} \). This indicates that in the simplest form, the molecule consists of five carbon atoms, ten hydrogen atoms, and one oxygen atom. Further analysis can compare this to the molecular formula, using the molecular weight.

Iodoform Test

The iodoform test is a chemical reaction used predominantly to identify methyl ketones or alcohols with a methyl group adjacent to the carbonyl carbon. The reaction produces a yellow precipitate of iodoform (\(\text{CHI}_3\)), indicating a positive result. Here's a quick rundown of the process:

• Methyl ketones and certain secondary alcohols react with iodine in a basic solution.
• The presence of a positive test is shown by the formation of a pale yellow precipitate.

In the problem context, the unknown compound gives a positive iodoform test. This confirms the presence of a methyl ketone, as it fits the criteria required for a positive response. The test's specificity for certain structural features makes it particularly useful in narrowing down possible molecular structures.

Methyl Ketone

Methyl ketones are a group of organic compounds containing the \(\text{-COCH}_3\) group. This distinct structural feature is what makes methyl ketones recognizable. They are notable for their reactivity and what tests can distinguish them from other ketones or aldehydes. Here's why they matter:

• The carbon-carbon bond adjacent to the carbonyl group allows for specific reactions, like the iodoform test.
• They are distinct from other ketones because their defining feature is the methyl group bound to the carbonyl carbon.

Given the problem data, the solution suggests the compound could be a methyl ketone, supported by both the positive iodoform test and the inability to reduce Fehling solution, which rules out aldehydes. Thus, examples like 2-pentanone and 3-pentanone align well with the features identified.

Fehling Solution

Fehling solution is a chemical reagent used to differentiate between water-soluble carbohydrates and ketone-functional groups, or to test for reducing sugars. The test involves the following steps:

• Fehling solution consists of two separate solutions, usually labeled as Fehling's A and Fehling's B. They are mixed in equal parts just before the test.
• In the presence of an aldehyde, a red precipitate of copper(I) oxide will form, indicating a positive test. Ketones do not yield this result, except for some specific cases.

In our textbook problem, the unknown compound not reducing Fehling solution suggests the absence of an aldehyde group. This aligns with the identification of the compound as a keto derivative, fitting the structural deductions made through other tests like the iodoform test.