Problem 119 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

A chemist treated a compound \(\mathrm{X}\) with \(\mathrm{NaOH}\) in presence of acetone as solvent. However, he recovered the starting material as such, and instead isolated a small amount of the product A. The product A was shown to have \(\mathrm{C}\), \(\mathrm{H}\) and \(\mathrm{O}\) and it had a molecular weight of \(116 \mathrm{~g} / \mathrm{mol}\). It gave a positive iodoform test and was found to be identical with a compound obtained by the aldol self-condensation of acetone. Although the product A did not discharge colour of bromine in \(\mathrm{CCl}_{d^{\prime}}\) its dehydration product \(B\) with hot sulphuric acid discharged bromine dissolved in \(\mathrm{CCl}_{4}\).

Step-by-Step Solution

Answer

Product A is diacetone alcohol, and product B is mesityl oxide.

Step 1: Identify Key Observations

A compound X was treated with NaOH and acetone, recovering mostly X but also obtaining a small amount of product A. Product A contains C, H, and O, with a molecular weight of 116 g/mol, gives a positive iodoform test, and matches the product from aldol self-condensation of acetone.

Step 2: Understand the Iodoform Test

The iodoform test is positive for compounds such as methyl ketones, secondary alcohols with a hydroxyl group two carbons away from a methyl group, or aldehydes with a similar configuration. Since A gives a positive result, it indicates it might be a methyl ketone.

Step 3: Analyze the Aldol Condensation

The aldol condensation of acetone involves two acetone molecules reacting to form a larger molecule. The structural result is 4-hydroxy-4-methylpentan-2-one (diacetone alcohol) with the molecular formula C6H12O2, matching A's description.

Step 4: Calculate the Molecular Weight

The molar mass of diacetone alcohol, C6H12O2, is: \[ (6 \, \text{C} \times 12.01) + (12 \, \text{H} \times 1.008) + (2 \, \text{O} \times 16.00) = 116 \, \text{g/mol} \] This matches the given molecular weight for product A.

Step 5: Dehydration Process

When product A is dehydrated with hot sulfuric acid, it forms product B, which discharges bromine in CCl4, indicating the presence of a double bond in B. The dehydration converts diacetone alcohol to mesityl oxide.

Key Concepts

Iodoform Test

The iodoform test is a chemical reaction used to identify certain functional groups in organic compounds. It is particularly useful in detecting the presence of methyl ketones and specific alcohols. In this test, when a compound reacts with iodine and a base, usually sodium hydroxide (\(\text{NaOH}\)\u000a), a yellow precipitate of iodoform (\(\text{CHI}_3\)\u000a) is formed if the compound contains a methyl ketone or certain secondary alcohols.
For the iodoform test to be positive, the compound must have either:

• A methyl ketone group where the carbonyl group is adjacent to a methyl group.
• A secondary alcohol, with the OH group two carbons away from a reachable methyl group.

This reaction occurs because these structural motifs can form the required organoiodine intermediates that lead to iodoform precipitation. In our exercise, compound A demonstrated a positive iodoform test result, suggesting it contains the accessible methyl ketone structure.
Since product A matches the compound diacetone alcohol, it's the specific methyl group configuration that results in this positive test. The iodoform test not only confirms the presence of specific functional groups but also aids in narrowing down the possible structure of unknown substances.

Molecular Weight Calculation

Calculating molecular weight involves adding the atomic weights of all atoms present in a molecule. It is an essential step in understanding the composition and structure of an organic compound.
For the given compound A, identified as diacetone alcohol, its molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_2\).\u000a To calculate its molar mass:

• Carbon (\(\text{C}\)\u00a0): 6 atoms x 12.01 g/mol = 72.06 g/mol
• Hydrogen (\(\text{H}\)\u00a0): 12 atoms x 1.008 g/mol = 12.096 g/mol
• Oxygen (\(\text{O}\)\u00a0): 2 atoms x 16.00 g/mol = 32.00 g/mol

Add these values together to find the total molecular weight:
\[72.06 + 12.096 + 32.00 = 116.156\, \text{g/mol}\]This closely matches the molecular weight given for product A, confirming its identification as diacetone alcohol. Such precise calculations are crucial in ensuring the empirical accuracy of molecular formulas.
Understanding how to accurately calculate molecular weights allows scientists and students to analyze the chemical identity of compounds critically.

Dehydration Reaction Analysis

Dehydration reactions involve the removal of water (\(\text{H}_2\text{O}\)\u00a0) from a molecule, typically catalyzed by strong acids like sulfuric acid. These reactions often lead to the formation of alkenes or unsaturated compounds. In the case of diacetone alcohol, its dehydration produces mesityl oxide, which is a compound with a double bond.
During dehydration, the hydroxyl group (\(-OH\))\u000a and a hydrogen atom from an adjacent carbon atom are removed, resulting in the formation of a double bond between the adjacent carbon atoms. This conversion occurs in the presence of hot sulfuric acid:\u000a

• The hydroxyl group from one molecule is protonated and leaves as water.
• The neighboring hydrogen atom is then removed as a proton.

This leads to the association of the remaining carbon atoms via a double bond. In the experiment, the resulting compound B from the dehydration of A discharged bromine in \(\text{CCl}_4\),\u000a an indication of the presence of a double bond. This reaction makes use of the chemical affinity of bromine for unsaturated hydrocarbons, evidenced by its disappearance through the reaction.
These transformations illustrate how dehydration can profoundly modify the chemical properties and reactivity of organic compounds.