Question

Draw structural formulas of all chloroalkanes that undergo dehydrohalogenation when treated with \(\mathrm{KOH}\) to give each alkene as the major product. For some parts, only one chloroalkane gives the desired alkene as the major product. For other parts, two chloroalkanes may work.

Short answer

Answer: The chloroalkanes that undergo dehydrohalogenation to give alkenes as the major product are chloroethane, 1-chloropropane, 2-chloropropane, 1-chlorobutane, 2-chlorobutane, and 2-chloro-2-methylpropane. Their structural formulas are as follows: - Chloroethane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}\) - 1-Chloropropane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\) - 2-Chloropropane: \(\mathrm{CH}_{3}\mathrm{CHClCH}_{3}\) - 1-Chlorobutane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\) - 2-Chlorobutane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHClCH}_{3}\) - 2-Chloro-2-methylpropane: \(\mathrm{CH}_{3}\mathrm{C}(\mathrm{CH}_{3})\mathrm{CH}_{2}\mathrm{Cl}\)

Step-by-step solution

Understanding Dehydrohalogenation

Dehydrohalogenation is an elimination reaction during which a halogen (in this case, chlorine) and a hydrogen atom are removed from adjacent carbons in a molecule, forming an alkene. When treated with KOH, chloroalkanes can undergo this reaction. The product distribution depends on the structure of the chloroalkane and Zaitsev's rule, which states that in an elimination reaction, the more substituted alkene (the one with more alkyl groups attached to the double bond) will be the major product.

Identifying Chloroalkanes That Give Alkene as Major Product

To identify which chloroalkanes undergo dehydrohalogenation to give each alkene as the major product, we can work backwards by first identifying the potential alkene products and then determining which chloroalkanes would be most likely to form these alkenes upon reaction with KOH.

Listing Possible Alkene Products

We will consider the smallest alkenes and their corresponding chloroalkanes: - Ethene (\(\mathrm{CH}_{2}=\mathrm{CH}_{2}\)): The only chloroalkane that can undergo dehydrohalogenation to give ethene as a product is chloroethane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}\)). - Propene (\(\mathrm{CH}_{2}=\mathrm{CHCH}_{3}\)): The chloroalkanes that can undergo dehydrohalogenation to give propene as a product are: 1. 1-Chloropropane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\)) 2. 2-Chloropropane (\(\mathrm{CH}_{3}\mathrm{CHClCH}_{3}\)) - Butene (\(\mathrm{CH}_{2}=\mathrm{CHCH}_{2}\mathrm{CH}_{3}\)) and (\(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CHCH}_{3}\)): The chloroalkanes that can undergo dehydrohalogenation to give butene as the major product are: 1. 1-Chlorobutane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\)) 2. 2-Chlorobutane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHClCH}_{3}\)) 3. 2-Chloro-2-methylpropane (\(\mathrm{CH}_{3}\mathrm{C}(\mathrm{CH}_{3})\mathrm{CH}_{2}\mathrm{Cl}\))

Drawing Structural Formulas of the Identified Chloroalkanes

We have identified the chloroalkanes that would yield ethene, propene, and butene as the major product upon dehydrohalogenation. The structural formulas of these chloroalkanes are as follows: - Chloroethane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}\) - 1-Chloropropane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\) - 2-Chloropropane: \(\mathrm{CH}_{3}\mathrm{CHClCH}_{3}\) - 1-Chlorobutane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Cl}\) - 2-Chlorobutane: \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHClCH}_{3}\) - 2-Chloro-2-methylpropane: \(\mathrm{CH}_{3}\mathrm{C}(\mathrm{CH}_{3})\mathrm{CH}_{2}\mathrm{Cl}\)

Key concepts

Organic Chemistry Reactions

In organic chemistry, reactions are the backbone of synthesizing and transforming molecules. In the textbook exercise, dehydrohalogenation is featured, which is an essential type of organic reaction. Notably, dehydrohalogenation is an elimination reaction in which a halogen atom and a hydrogen atom are removed from adjacent carbon atoms within a molecule. This removal typically results in the formation of an alkene, a type of hydrocarbon with a carbon-carbon double bond.

Different reagents can induce dehydrohalogenation. In this context, potassium hydroxide (KOH) is used, which is a strong base capable of abstracting a proton, facilitating the loss of a halogen and the formation of the double bond in the alkene. This kind of reaction is not only a cornerstone in alkene synthesis but also a fundamental transformation in the broader scope of organic synthesis.

Structural Formulas

Understanding structural formulas is crucial for visualizing the organization of atoms within a molecule. These formulas convey not only the connectivity of atoms but also provide insight into the molecule's geometry and isomerism. In the exercise, the primary task is to draw the structural formulas for chloroalkanes that can be converted into alkenes through dehydrohalogenation.

The importance of this skill cannot be overstated, as different structural arrangements can lead to vastly different chemical behaviors. Chloroalkanes, compounds consisting of carbon chains with chlorine atoms, are depicted with lines (bonds) and letters (elements). Each variation in structure offers a different avenue for reaction, influencing which product forms as the major alkene output. Mastery of structural formulas equips students with the ability to predict and understand the outcomes of organic reactions.

Alkene Synthesis

Alkene synthesis is an important aim in organic chemical reactions, as alkenes serve as fundamental intermediates for a variety of complex molecules. Through dehydrohalogenation, chloroalkanes shed their halogen atoms and hydrogen to form these versatile double bonds. For example, converting chloroethane to ethene or 2-chloro-2-methylpropane to butene represents alkene synthesis.

In these reactions, the base (KOH) is pivotal as it abstracts a proton (H+) adjacent to the carbon-halogen bond, which in turn facilitates the elimination of the halogen and the creation of the double bond. This versatility in synthetically manipulating molecular structures to produce alkenes is a pivotal aspect of organic chemistry, opening doors to more elaborate synthesis pathways involving polymers, pharmaceuticals, and other organically-based materials.

Zaitsev's Rule

Zaitsev's rule is a guiding principle in elimination reactions like dehydrohalogenation. It predicts that the most substituted alkene—meaning the alkene with the most alkyl groups attached to the carbon atoms of the double bond—will usually form as the major product. This occurs because the more substituted alkene typically has a lower energy and is therefore more stable.

When applying Zaitsev's rule, as in the exercise, one must consider all possible alkenes that could form from a given chloroalkane and then identify which is most substituted. In the case of 2-chloropropane, for example, propene is the major product over the less substituted methyl ethene. Zaitsev's rule is an invaluable heuristic in organic chemistry as it helps predict the outcome of elimination reactions, which is critical for efficient and targeted synthesis strategies.