Question

Compound \(\mathrm{K}, \mathrm{C}_{5} \mathrm{H}_{10}\) decolored a solution of \(\mathrm{Br}_{2}\) in carbon tetrachloride. When \(\mathrm{K}\) was dissolved in cold, concentrated sulfuric acid and then heated with water, L resulted. L, \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), reacted with chromic acid \(\mathrm{H}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to give \(\mathrm{M}\), \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\). Both \(\mathrm{L}\) and \(\mathrm{M}\) gave positive iodoform tests. The reaction mixture of the iodoform test also produced isobutyric acid \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}\). Determine the structure of \(\mathrm{K}, \mathrm{L}\), and \(\mathrm{M}\).

Short answer

The structures of compounds K, L, and M are as follows: Compound K: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH}\mathrm{=} \ce{CH2} \) Compound L: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CHOH}\mathrm{-} \ce{CH3} \) Compound M: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CO}\mathrm{-} \ce{CH3} \)

Step-by-step solution

Identify the structure of compound K

Since compound K decolorizes a solution of \(Br_2\) in carbon tetrachloride, this indicates the presence of a carbon-carbon double bond in its structure (an alkene). With the formula \(C_5H_{10}\), the simplest possible alkene structure that has a methyl keto group (\(CH_3CO\)) is: $$ \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH}\mathrm{=} \ce{CH2} $$

Identify the structure of compound L

Compound K reacts with cold concentrated sulfuric acid (\(\ce{H2SO4}\)) followed by heating with water to give compound L (\(C_5H_{12}O\)). This is a typical reaction in which the double bond of the alkene gets converted to an alcohol group. The addition of the –OH group occurs anti-Markovnikov style because the reaction is in cold sulfuric acid: $$ \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CHOH}\mathrm{-} \ce{CH3} $$

Identify the structure of compound M

Compound L (\(C_5H_{12}O\)) reacts with chromic acid to form compound M (\(C_5H_{10}O\)). This reaction indicates the oxidation of the alcohol group in compound L to a carbonyl group in compound M: $$ \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CO}\mathrm{-} \ce{CH3} $$

Verify the Iodoform Test's result

Both compounds L and M give positive iodoform tests, indicating the presence of a methyl keto group (\(CH_3CO\)) in their structures. Based on their structures, L has an alcohol group attached to a terminal methyl group, while M has a terminal methyl keto group: Compound L: $$ \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CHOH}\mathrm{-} \ce{CH3} $$ Compound M: $$ \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CO}\mathrm{-} \ce{CH3} $$ Thus, the structures of compounds K, L, and M are consistent with the given information and reaction patterns: Compound K: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH}\mathrm{=} \ce{CH2} \) Compound L: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CHOH}\mathrm{-} \ce{CH3} \) Compound M: \( \ce{CH3}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CH2}\mathrm{-} \ce{CO}\mathrm{-} \ce{CH3} \)

Key concepts

Alkene Reactions

Alkenes are unsaturated hydrocarbons characterized by at least one carbon-carbon double bond. A hallmark reaction of alkenes is their ability to undergo addition reactions, where atoms or groups of atoms are added to the carbon atoms involved in the double bond.

This class of reactions is a staple in organic chemistry because it provides a route to convert alkenes into a wide variety of functional groups. The decolorization of bromine in carbon tetrachloride, as in the example of compound K, is an electrophilic addition reaction. The double bond in the alkene is nucleophilic and can attack the electrophilic bromine molecules, leading to the addition of bromine across the double bond. The reaction proceeds through the formation of a cyclic bromonium ion intermediate, followed by nucleophilic attack, which results in the dibromoalkane product.

Understanding the reactivity of alkenes is crucial because it opens a gateway to synthesize many products, with applications ranging from pharmaceuticals to polymers. Students often find visualizing the stepwise process of the reaction mechanism helpful in grasping these concepts.

Alcohol Oxidation

The oxidation of alcohols is a transformation widely studied in organic chemistry, involving the increase in the oxidation state of the carbon bearing the hydroxyl group. Primary alcohols can be oxidized to aldehydes and further to carboxylic acids, while secondary alcohols can be transformed into ketones.

In the given exercise, the conversion of compound L to M via chromic acid is an example of alcohol oxidation. The chromic acid acts as an oxidizing agent and converts the secondary alcohol present in compound L to a ketone in compound M. This process is typically characterized by the color change from orange to green due to the reduction of chromium from hexavalent to trivalent state.

Oxidation reactions are pivotal in organic synthesis, allowing chemists to alter functional groups and thus modify molecular properties. Students can benefit from focusing on the type of alcohol and the specific oxidizing agent used to predict the products formed in such reactions.

Iodoform Test

The iodoform test is a classical chemical test for identifying methyl ketones or secondary alcohols that can be oxidized to methyl ketones. The presence of the functional group CH3CO- in a molecule enables it to give a positive iodoform test.

A positive test is indicated by the formation of a yellow precipitate of iodoform (CHI3). When compounds L and M yield positive results, it signifies the presence of this functional group. In practice, the test involves adding iodine and a base (usually sodium hydroxide) to the compound in question. The iodine reacts with the methyl ketone or methyl secondary alcohol to form iodoform and carboxylic acid, as seen in the formation of isobutyric acid in this exercise.

The iodoform test is an excellent diagnostic tool for identifying specific structural features in organic molecules. Students often find performing this test in the lab an effective way to confirm the presence of a functional group predicted theoretically through reaction mechanisms.