Question

Predict the products of each precursor by the hydration of olefins method. (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}\) Propylene Ethene

Short answer

The products of the hydration of the given olefins are: (a) Propylene (\(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}\)) → Propan-2-ol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)) (b) Ethene (\(\mathrm{CH}_{2}=\mathrm{CH}_{2}\)) → Ethanol (\(\mathrm{CH}_{2}\mathrm{OH}\mathrm{CH}_{2}\mathrm{H}\))

Step-by-step solution

Identify the olefin structure

For each of the given olefins, identify the carbon-carbon double bond and the attached hydrogen atoms and other functional groups. (a) In Propylene, there is a double bond between the second and third carbon atoms, and the first carbon atom has three hydrogen atoms attached to it. (b) In Ethene, there is a double bond between the two carbon atoms, and each of the carbon atoms has two hydrogen atoms attached to it.

Apply Markovnikov's rule

According to Markovnikov's rule, when a nucleophile like water reacts with an alkene, the hydroxyl group (OH) will attach to the carbon with fewer hydrogen atoms, and the hydrogen will attach to the carbon with more hydrogen atoms. (a) For Propylene, the carbon with fewer hydrogen atoms is the third one, so the hydroxyl group will attach to it. (b) For Ethene, both carbon atoms have the same number of hydrogen atoms, so either carbon atoms can have the hydroxyl group attached to them.

Form the final product by adding the hydroxyl group and hydrogen to the respective carbon atoms

Add the hydroxyl group and hydrogen to the carbon atoms according to Markovnikov's rule. (a) For Propylene, the final product is \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\) (Propan-2-ol or Isopropanol). (b) For Ethene, the final product is \(\mathrm{CH}_{2}\mathrm{OH}\mathrm{CH}_{2}\mathrm{H}\) (Ethanol). In conclusion, the products of the hydration of the given olefins are: (a) Propylene (\(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}\)) → Propan-2-ol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)) (b) Ethene (\(\mathrm{CH}_{2}=\mathrm{CH}_{2}\)) → Ethanol (\(\mathrm{CH}_{2}\mathrm{OH}\mathrm{CH}_{2}\mathrm{H}\))

Key concepts

Markovnikov's Rule

Understanding Markovnikov's rule is fundamental when predicting the outcome of chemical reactions involving olefins, such as hydration. This empirical rule states that with the addition of a protic acid (HX) to an asymmetric alkene, the acidic hydrogen (H) attaches to the more substituted carbon (the carbon with more hydrogen atoms already attached), while the halide (X) or other nucleophile, attaches to the less substituted carbon (the carbon with fewer hydrogen atoms).

In the given exercise, this rule helps to predict that in propylene, since one carbon atom in the double bond has fewer hydrogens than the other, the hydroxyl group (OH) will add to this carbon atom following hydration of the alkene. For ethene, because the molecule is symmetrical, the addition of a hydroxyl group will yield the same result regardless of which carbon atom the group attaches to, leading to the production of ethanol.

Alkene Reaction Mechanisms

The mechanisms behind alkene reactions, particularly hydration, involve multiple steps where the reactants undergo different transitions to become the final products. The typical mechanism for an alkene reaction, like the hydration reaction described, starts with the alkene undergoes protonation, which leads to the formation of a more stable carbocation intermediate.

Subsequently, a nucleophile, often water in the case of hydration, attacks the carbocation, leading to a new bond formation. In the hydration of olefins, the water molecule provides the hydroxyl group that attaches to the carbocation, forming the alcohol. A crucial aspect of these reaction mechanisms is to consider both the stability of the carbocation that forms and the regioselectivity of the addition, which is predicted by Markovnikov's rule.

Organic Chemistry Synthesis

In organic chemistry, synthesizing complex molecules often requires a step-by-step procedure, where chemists transform simpler molecules into more complex ones using a series of chemical reactions. It is necessary for establishing the specific sequence of steps to achieve the desired transformations while preserving the functionality of the molecules involved.

Synthesis involving olefins, such as our example of hydration, is a common reaction in which a simple molecule (an alkene) is transformed into a more complex molecule (an alcohol). Understanding the properties of alkenes and the types of reactions they can undergo is pivotal for performing successful organic syntheses. Hydration is just one example of how functional groups can be strategically added to an organic molecule to increase its complexity.

Functional Groups in Organic Molecules

Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. They are the reactive parts of molecules that give organic compounds their properties and reactivity.

In the context of the hydration of olefins, the focus is on alkenes and alcohols. Alkenes contain a carbon-carbon double bond, which is a functional group that is reactive due to its electrons being easily accessible. In the hydration reaction, the addition of a hydroxyl group (–OH) transforms the alkene into an alcohol, another functional group. Alcohols are characterized by the presence of this hydroxyl group, which imparts different chemical properties and allows for further chemical modifications. Mapping the changes in functional groups is essential for understanding organic chemistry's vast matrix of reactions, including synthesis and reactivity.