Question

Which compound would undergo dehydrohalogenation with strong base to give the alkene shown below as the only alkene product? $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} $$ (A) 1-chloropentane (B) 2 -chloropentane (C) 3 -chloropentane (D) 1-chloro-2-methylbutane

Short answer

The compound that will undergo dehydrohalogenation with strong base to give the required alkene (CH3-CH2-CH=CH-CH3) as the only alkene product is (B) 2-chloropentane.

Step-by-step solution

Option (A): 1-chloropentane

For 1-chloropentane, the dehydrohalogenation will occur by removing the chlorine atom and a hydrogen atom from the carbon next to it. This will result in the formation of an alkene with a double bond between C1 and C2. However, the required alkene has a double bond between C2 and C3, therefore 1-chloropentane is not the correct choice.

Option (B): 2-chloropentane

For 2-chloropentane, dehydrohalogenation will happen by removing the chlorine atom and a hydrogen atom from the adjacent carbon (C3). This will result in the formation of an alkene with a double bond between C2 and C3, which exactly matches the given alkene. Thus, 2-chloropentane is the correct choice for this exercise.

Option (C): 3-chloropentane

For 3-chloropentane, the dehydrohalogenation will occur by removing the chlorine atom and a hydrogen atom from either the C4 or C2 carbon, resulting in an alkene with a double bond between C2 and C3 or between C3 and C4. However, the given alkene has a double bond between C2 and C3, and since there are two possible alkenes that can form, 3-chloropentane cannot be the sole product. Thus, 3-chloropentane is not the correct choice.

Option (D): 1-chloro-2-methylbutane

For 1-chloro-2-methylbutane, dehydrohalogenation will happen by removing the chlorine atom and a hydrogen atom from the adjacent carbon (C2). This will result in an alkene with a double bond between C1 and C2. However, the required alkene has a double bond between C2 and C3, therefore 1-chloro-2-methylbutane is not the correct choice. Based on the analysis, the correct choice for this exercise is:

Answer

(B) 2-chloropentane

Key concepts

Organic Chemistry

Organic chemistry is the scientific study of the structure, properties, composition, reactions, and synthesis of organic compounds that contain carbon atoms. Understanding the complex nature of these compounds and how they react is foundational to organic chemistry.

Dehydrohalogenation reactions are a classic example of organic transformations. They involve the elimination of a hydrogen halide (HX, where X is a halogen) from a haloalkane (alkyl halide), resulting in the formation of an alkene. This type of reaction is important for the synthesis of alkenes from more readily available alkyl halides.

The process typically requires a strong base to initiate the elimination, and the outcome of the reaction can depend heavily on the structure of the starting material, as is well demonstrated by the textbook problem regarding the formation of a specific alkene from various haloalkanes.

Alkene Formation

Alkenes are hydrocarbons containing at least one carbon-carbon double bond, and they are a central feature in various chemical reactions. The formation of alkenes is commonly achieved through elimination reactions, where dehydrohalogenation is a prime method for synthesizing these compounds.

In the context of the provided exercise, dehydrohalogenation selectively removes a halogen atom and a hydrogen from adjacent carbon atoms in a haloalkane, resulting in the formation of a pi bond between those carbons. This step of constructing the double bond is critical—its location determines the properties and reactivity of the resultant alkene. When analyzing potential reactants for alkene synthesis, chemists apply Zaitsev's rule, which predicts the more substituted alkene as the major product. However, in exercises like the one provided, regioselectivity can be determined to yield a single, specific alkene product.

Haloalkane Reactions

Haloalkanes, or alkyl halides, are compounds containing a halogen atom attached to an aliphatic carbon chain. These molecules are versatile in organic synthesis due to their reactivity, especially in nucleophilic substitution and elimination reactions.

In dehydrohalogenation reactions specifically, the base abstracts a hydrogen atom (usually the most accessible one) from a carbon adjacent to the one holding the halogen, leading to the formation of a double bond after the elimination of the hydrogen halide. The detailed mechanism involves the formation of a carbanion intermediate or a concerted E2 mechanism depending on the reaction conditions and the structure of the haloalkane.

The position and nature of the halogen in the haloalkane greatly influence which alkene product is formed, as demonstrated in the textbook solution. In the case of 2-chloropentane, the base-induced elimination uniquely leads to the desired product, showcasing the importance of understanding how haloalkane structure can direct the course of a dehydrohalogenation reaction.