Question

An unknown compound A rapidly decolorized a solution of bromine in carbon tetrachloride. When A was subjected to ozonolysis, the products were butanone and propanal. What might be the structure of \(\mathrm{A}\) ?

Short answer

The probable structure of compound A is \(CH_3CH_2CH=CHCH_3\), as it decolorizes a bromine solution and forms butanone and propanal after ozonolysis.

Step-by-step solution

Identify the products of ozonolysis

According to the given information, butanone and propanal are the products of ozonolysis. Their molecular structures are shown below: - Butanone: \(CH_3CH_2COCH_3\) - Propanal: \(CH_3CH_2CHO\)

Reconnect the products to form the original molecule

As ozonolysis cleaves the carbon-carbon double bond, we need to combine the two products in such a way that we can form the structure of compound A with a carbon-carbon double bond. Imagine that we took a carbon from the carbonyl group of butanone and a carbon from the aldehyde group of propanal, and reconnect them with a double bond to form the original molecule. The structure of compound A will be: - Compound A: \(CH_3CH_2CH=CHCH_3\)

Verify the structure using information about bromine decolorization

To confirm our proposed structure, it should decolorize bromine in carbon tetrachloride (i.e., participate in the addition reaction with bromine). Since compound A contains a carbon-carbon double bond, it will react with bromine to form 1,2-dibromoheptane. Thus, the probable structure of compound A is \(CH_3CH_2CH=CHCH_3\).

Key concepts

Ozonolysis

Ozonolysis is a powerful organic chemistry reaction used to break down alkenes, which are compounds containing carbon-carbon double bonds. This process employs ozone, a triatomic molecule comprised of three oxygen atoms (\(O_3\)), to cleave the double bond and form two carbonyl compounds.

The process begins with the reaction of ozone with the alkene, resulting in an unstable intermediate known as a molozonide. This quickly rearranges into a more stable structure called an ozonide. Upon treatment with a reducing agent such as zinc in acetic acid or dimethyl sulfide, the ozonide breaks down into carbonyl compounds—either aldehydes or ketones, depending on the structure of the original alkene.

In our given problem, ozonolysis of compound \( \mathrm{A} \) yields butanone and propanal. This indicates that the original compound \(\mathrm{A}\) contained a carbon-carbon double bond positioned in such a way that these two compounds are formed upon breaking of the double bond. Understanding ozonolysis helps in deducing the structure of unsaturated hydrocarbons.

Carbon-Carbon Double Bond

A carbon-carbon double bond is a core feature of unsaturated organic molecules known as alkenes. This bond consists of one sigma bond and one pi bond, making it shorter and stronger than a single carbon-carbon bond. The presence of a double bond imparts distinct chemical properties to the molecule, influencing its reactivity and stability.

In reactions such as ozonolysis and bromine decolorization, the double bond is the reactive site. It can undergo cleavage or addition reactions, like the formation of dibromo compounds when exposed to bromine. The location and geometry of carbon-carbon double bonds significantly affect the outcome of these reactions.

• Double bonds are planar, with atoms around the bond lying in the same plane.
• The pi bond formed by lateral overlap of p orbitals can easily be broken in reactions, providing alkenes with their characteristic reactivity.

With compound \( \mathrm{A} \), the double bond connects the carbon atoms in such a way that upon ozonolysis, butanone and propanal are produced. Understanding this linkage is crucial to deduce the original structure of the compound.

Bromine Decolorization

Bromine decolorization is a simple test for unsaturation in organic compounds, particularly alkenes. An alkene like compound \(\mathrm{A}\) reacts with bromine (\(\mathrm{Br}_2\)) dissolved in carbon tetrachloride (\(\mathrm{CCl}_4\)). Alkenes cause the red-brown color of bromine to disappear, indicating the presence of a carbon-carbon double bond.

When the bromine solution is added to an alkene, it rapidly adds across the double bond, resulting in the formation of a colorless dibromo compound. This reaction is a type of electrophilic addition where the double bond acts as a nucleophile attacking the bromine molecule, facilitating the reaction.

• The loss of color is a visual cue for the confirmation of an alkene, as saturated compounds do not react in the same way.
• This method provides an easy way to differentiate between alkanes (no reaction) and alkenes (bromine decolorization).

In the solved exercise, the rapid decolorization of bromine by compound \(\mathrm{A}\) confirms the presence of a carbon-carbon double bond, supporting its proposed structure as an alkene.