Problem 27 - A Complete Resource Book in Chemistry for JEE Main

Question

A yellow product is formed when an organic compound \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) is warmed with aqueous sodium carbonate and iodine solution. The product formed is (a) iodoform (b) sodium iodate (c) sodium iodide (d) adduct of \(\mathrm{NaI}\) and \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\)

Step-by-Step Solution

Answer

The yellow product is iodoform (Option a).

Step 1: Identify the Reaction

The reaction in question involves the formation of a yellow product when an organic compound, which is a primary, secondary, or tertiary alcohol, or a ketone ( C_3H_8O), is treated with iodine ( I_2) and sodium carbonate ( Na_2CO_3). This is indicative of the iodoform test, which is used to identify methyl ketones or secondary alcohols with a methyl group attached to the carbon bearing the hydroxyl group.

Step 2: Determine Possible Organic Functional Groups

Since the compound is C_3H_8O, it could potentially be propanol either primary (propan-1-ol) or secondary (propan-2-ol), or it could be acetone. Only propan-2-ol (isopropyl alcohol) or acetone can form iodoform with this test, as both have the requisite CH_3CO- group (in the case of acetone) or equivalent through secondary alcohol (in isopropanol).

Step 3: Identify Yellow Product Produced

The iodoform test is characterized by the formation of a yellow precipitate of iodoform ( CHI_3) when a methyl ketone or secondary alcohol is present. The presence of methyl group attached to the carbonyl carbon or secondary alcohol oxidizing into such a group during the reaction is key.

Step 4: Final Step: Conclude the Product Formed

Given the organic compound fits the criteria for the iodoform test, the yellow compound formed is indeed iodoform. None of the other options (b, c, or d) relate to a yellow compound in context with the described reaction.

Key Concepts

Organic Chemistry

Organic chemistry is the branch of chemistry that studies the structure, properties, composition, reactions, and synthesis of carbon-containing compounds. These are not limited to hydrocarbons but also include compounds with elements such as nitrogen, oxygen, sulfur, and more.
In our discussed exercise, organic chemistry helps us understand the interactions of different organic molecules. It features prominently in understanding concepts like functional groups, which are groups of atoms responsible for the characteristic reactions of a particular compound.

For instance, alcohols and ketones are organic compounds with specific functional groups that make them candidates for various chemical tests. These tests, such as the iodoform test, highlight the specific reactions formed due to these unique arrangements of atoms. Understanding these basic principles is essential for delving into more complex organic chemistry scenarios.

Methyl Ketones

Methyl ketones are a special category in organic chemistry, recognized by their unique compound structure featuring a carbonyl group (\(\text{C=O}\)) flanked by a methyl (\(\text{CH}_3\)) group. These might sound complex, but think of them as a family of chemicals where a little branch of carbon atoms sticks out from the more substantial body of the molecule.

This simple structure makes methyl ketones especially reactive and useful in chemical tests. For example, the iodoform test specifically helps identify the presence of methyl ketones. In this test, methyl ketones react with iodine in the presence of a base to form iodoform, noted by its characteristic yellow precipitate.

The importance of methyl ketones goes beyond just identification. These compounds play significant roles in chemical synthesis and are involved in creating a wide range of essential substances. Understanding methyl ketones is foundational for students aiming to grasp the iodoform test and its role in organic chemistry.

Secondary Alcohols

Secondary alcohols are a type of alcohol that are structurally defined by having the hydroxyl group (\(\text{-OH}\)) attached to a carbon atom that is itself bonded to two other carbon atoms. Think of them as being right in the middle - more crowded than primary but not quite as saturated as tertiary alcohols.

One of the interesting aspects of secondary alcohols is their ability to react in a way that mimics the behavior of ketones through oxidation. When participating in the iodoform test, these secondary alcohols must include a methyl group for the yellow iodoform precipitate to form.

A practical example discussed above is propan-2-ol, a secondary alcohol that, upon oxidation, simulates the behavior of a methyl ketone. This property makes secondary alcohols incredibly useful in differentiating compounds and understanding reaction pathways in organic chemistry. By grasping the role of secondary alcohols, one appreciates their role in broader chemical reactions and tests, particularly in identifying unknown compounds.