Question

An organic compound \(\mathrm{A}\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}\right)\) neither decolourise bromine water nor changes the colour of acidic dichromate solution. A on heating with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces an alkene which on oxidative ozonolysis gives \(\mathrm{B}\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{3}\right)\), which gives an yellow precipitate with \(\mathrm{NaOH} / \mathrm{I}_{2}\). The most probable structure of \(\mathrm{A}\) is

Short answer

The most probable structure of compound \( A \) is cyclohexanol.

Step-by-step solution

Analyze Decolourisation Property

The fact that compound \( A \) neither decolourises bromine water nor changes the colour of acidic dichromate solution suggests that it does not contain any carbon-carbon double bonds or oxidizable alcohol groups. Therefore, it is most likely a saturated alcohol or ether.

Examine the Reaction With \( \mathrm{H}_2\mathrm{SO}_4 \)

When \( A \) is heated with \( \mathrm{H}_2\mathrm{SO}_4 \), it produces an alkene. This elimination reaction suggests that \( A \) might be an alcohol because alcohols can dehydrate, losing water to form alkenes.

Ozonolysis and Analysis of Product \( B \)

The alkene formed from \( A \) undergoes oxidative ozonolysis to form compound \( B \) \((\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_3)\), which gives a yellow precipitate with \( \mathrm{NaOH} / \mathrm{I}_2 \). This is indicative of a methyl ketone or a terminal alkene, suggesting the presence of a carbonyl group next to a methyl group.

Identify Functional Group in \( B \)

The formation of a yellow precipitate upon treatment of \( B \) with \( \mathrm{NaOH} / \mathrm{I}_2 \) (iodoform test) confirms that \( B \) contains a methyl ketone group. This results in a methyl ketone structure for part of \( B \).

Propose Structure for Compound \( A \)

Using the information that \( A \) is a saturated alcohol which dehydrates to form a compound that resembles a methyl ketone upon ozonolysis, \( A \) could be cyclohexanol. Cyclohexanol \((\mathrm{C}_6\mathrm{H}_{12}\mathrm{O})\) upon dehydration, forms cyclohexene. Upon ozonolysis, cyclohexene forms adipic acid or a similar compound, which presents the characteristic that decomposes to give a yellow precipitate.

Key concepts

Dehydration Reactions

When discussing organic chemistry, especially in the context of how compounds transform, dehydration reactions play a pivotal role. Here, dehydration refers to the removal of a molecule of water from a compound.

In the context of alcohols, dehydration is an elimination reaction where water is removed, resulting in the formation of an alkene. This is often facilitated by heating the alcohol in the presence of an acid such as sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)).

Key points about dehydration reactions include:

• They primarily affect alcohols, transforming them into alkenes.
• The reaction typically requires an acid catalyst and heat.
• This is a common method for producing simple alkenes from saturated alcohols.

The process is an example of an elimination reaction, since a small molecule (water) is eliminated during the reaction transformation.

Ozonolysis

Ozonolysis is a fascinating oxidation reaction used for cleaving alkenes or alkynes using ozone (\( \mathrm{O}_3 \)). This reaction results in the formation of carbonyl compounds like aldehydes, ketones, or carboxylic acids.

In our exercise, ozonolysis helps us determine the structure of compound \( \mathrm{A} \) because the alkene produced via dehydration undergoes this oxidative cleavage.

Several significant points about ozonolysis are:

• The process involves the breaking of carbon-carbon double bonds.
• It converts the double bonds into carbonyl groups after reductive work-up.
• Ozonolysis aids in determining the position and presence of double bonds in a compound.

Thus, ozonolysis is a powerful tool in organic chemistry for structural determination and synthesizing carbonyl compounds.

Iodoform Test

The iodoform test is a qualitative test in organic chemistry used to detect methyl ketone groups. Compounds that give a positive iodoform test generate a yellow precipitate—iodoform (\( \mathrm{CHI}_3 \)).

In our exercise, compound \( \mathrm{B} \), which results after ozonolysis, shows a positive iodoform test. This strongly indicates the presence of a methyl ketone group.

Some crucial aspects of the iodoform test include:

• It is specific for detecting methyl groups adjacent to carbonyls.
• Positive results manifest as a yellow precipitate.
• This test helps in differentiating between various types of carbonyl-containing compounds.

Consequently, the iodoform test is a useful diagnostic tool in identifying methyl ketone moieties in an organic compound.

Saturation in Organic Compounds

Saturation refers to organic molecules that contain only single bonds between carbon atoms. In contrast, unsaturated compounds have one or more double or triple bonds.

In the exercise, compound \( \mathrm{A} \) does not decolourise bromine water, indicating it lacks double bonds, typical of saturated compounds.

Some essential points to know about saturation include:

• Saturated compounds are generally more stable due to the presence of single bonds only.
• They do not participate easily in addition reactions, like those with bromine.
• Saturation indicates a maximum number of hydrogen atoms bonded to the carbon skeleton.

Understanding saturation in organic compounds is crucial because it dictates reactivity and properties of the substance in chemical reactions.