Question

Compound which gives 1 mole of \(\mathrm{CHI}_{3}\) per mole on treatment with \(\mathrm{NaOI}\) is (A) CC(C)=O (B) CC(=O)C(C)=O (C) (D) CC(=O)C(C)(C)C(C)=O

Short answer

The compound that gives 1 mole of CHI3 per mole on treatment with NaOI is (A) Propanone, which has the molecular structure CC(C)=O and contains a suitable CH3 group that can react with NaOI to produce CHI3.

Step-by-step solution

Write down the given information

We are given the chemical reaction of an unknown compound with sodium iodide (NaOI), which produces CHI3 as one of the products.

Analyze each compound for potential CH groups to react with NaOI

We want to identify the compound containing the suitable CH group that could be replaced by the I atom, resulting in the formation of CHI3. Look at each compound and their molecular structure: (A) CC(C)=O Propanone: C_(3)H_(6)O with one CH_3 and one CH_2. (B) CC(=O)C(C)=O Pentane-2,4-dione: C_(5)H_(8)O_2 with three CH_3. (C) No molecular structure is given, so we can't analyze this option. (D) CC(=O)C(C)(C)C(C)=O 2,4,6-Heptanetrione: C_(7)H_(12)O_3 with four CH_3.

Identify the compound that produces CHI3 upon reaction with NaOI

Among the given options, compound (A) has a suitable CH group (CH3) that could react with NaOI to produce CHI3. When treated with NaOI, the iodine would replace the CH3 group, forming CHI3. So, the answer is (A) Propanone.

Key concepts

Understanding Propanone

Propanone, also commonly known as acetone, is the smallest and simplest ketone, with the chemical formula \( C_3H_6O \). It's well-known in both academic and commercial settings for its solvent properties, widely used in nail polish removers and paint thinners.

Propanone's structure consists of a carbonyl group (a carbon double-bonded to oxygen) flanked by two methyl groups \( CH_3 \). This molecular setup is pivotal to its reactivity, particularly in the iodoform test. The presence of at least one methyl group attached to the carbonyl carbon is what allows propanone to participate in the reaction that leads to the formation of iodoform \( CHI_3 \).

An understanding of propanone's structure is crucial for students as it illustrates the type of carbon-hydrogen bonds that respond to the iodoform test. This knowledge assists in predicting the outcomes of various chemical reactions involving ketones, especially when analyzing their conversion to other compounds.

The Sodium Iodide Reaction

The sodium iodide reaction, an essential part of the iodoform test, is the interaction between a compound, like propanone, and sodium iodide \( NaI \) in the presence of an oxidizing agent like iodine. In essence, when a compound capable of forming iodoform is treated with \( NaOI \), it undergoes a series of halogenation and hydrolysis reactions.

In the presence of a base, the \( NaI \) provides iodide ions, which are then oxidized to elemental iodine. This iodine can then substitute the hydrogen atoms attached to the carbon next to the carbonyl group. This substitution is facilitated by the base, which also deprotonates the intermediate, eventually leading to \( CHI_3 \) formation if the structure permits.

For the reaction to yield \( CHI_3 \), it's necessary for the original compound to have methyl groups adjacent to the carbonyl, as we've seen with propanone. No reaction will take place with other types of hydrogen atoms, making the test specific for certain structures.

Formation of CHI3 (Iodoform)

The formation of CHI3, or iodoform, is a hallmark of the iodoform test, which serves as a qualitative method to ascertain the presence of a methyl ketone structure. In the context of the given exercise, a mole of CHI3 is generated from a mole of a compound containing the necessary CH group when treated with sodium iodide and iodine.

The chemical dance that leads to the creation of yellow, crystalline iodoform includes the substitution of the hydrogen atoms on the methyl group adjacent to a carbonyl group by iodine atoms. The final act is the release of iodoform, while the remainder of the original compound becomes a carboxylic acid or salt equivalent.

In the classroom and laboratory, the iodoform test has pedagogical value well beyond simply identifying methyl ketones. It provides a concrete example of nucleophilic substitution reactions and the influence of molecular structure on chemical reactivity. Therefore, understanding the intricacies of CHI3 formation deepens a student's comprehension of organic chemistry and reaction mechanisms.