Question

Which of the following is not a \(\beta\) elimination? (a) dehydration of alcohols (b) dehydrohalogenation of alkyl halides (c) dehalogenation of vicinal dibromides (d) dehydrohalogenation of chloroform

Short answer

(a) Dehydration of alcohols (b) Dehydrohalogenation of alkyl halides (c) Dehalogenation of vicinal dibromides (d) Dehydrohalogenation of chloroform Answer: (d) Dehydrohalogenation of chloroform does not involve a β elimination reaction.

Step-by-step solution

(a) Dehydration of alcohols

Dehydration of alcohols involves the removal of a water molecule (H\(_2\)O) from the alcohol molecule. This reaction usually occurs under acidic conditions and is a \(\beta\) elimination reaction because a hydrogen atom on the \(\beta\) carbon is removed along with the hydroxyl group on the \(\alpha\) carbon.

(b) Dehydrohalogenation of alkyl halides

Dehydrohalogenation of alkyl halides involves the removal of a hydrogen atom and a halogen atom (such as Cl, Br, or I) from adjacent carbon atoms (the \(\alpha\) and \(\beta\) carbons) in the alkyl halide molecule. This reaction is also a \(\beta\) elimination process because it eliminates a hydrogen atom from the \(\beta\) carbon and a halogen atom from the \(\alpha\) carbon.

(c) Dehalogenation of vicinal dibromides

Dehalogenation of vicinal dibromides involves the removal of two halogen atoms (in this case, bromine) from two adjacent carbon atoms (the \(\alpha\) and \(\beta\) carbons) in the dibromide molecule. As the two halogen atoms are removed from the adjacent carbons, a \(\pi\) bond is formed between them. This reaction falls under the \(\beta\) elimination because it involves the removal of groups from adjacent carbons, although in this case, both groups are halogens.

(d) Dehydrohalogenation of chloroform

Chloroform (CHCl\(_3\)) is a molecule with three chlorine atoms and one hydrogen atom bonded to a central carbon atom. Dehydrohalogenation of chloroform involves removing a hydrogen atom and a chlorine atom from the molecule, which results in the formation of a dichlorocarbene (CCl\(_2\)). This reaction does not involve the elimination of groups from adjacent carbon atoms, as chloroform only has one carbon atom. Hence, this reaction does not fit into the category of \(\beta\) elimination. Based on the analysis of each option, we can conclude that:

Answer

Option (d) Dehydrohalogenation of chloroform is not a \(\beta\) elimination reaction.

Key concepts

Dehydration of Alcohols

Dehydration of alcohols is an important reaction that involves removing a water molecule from an alcohol. This process typically happens under acidic conditions, using acids such as sulfuric acid or phosphoric acid. The removal of water, or dehydration, occurs through

• the loss of an -OH group, which is replaced by a double bond, forming an alkene,
• the elimination of a hydrogen atom from a neighboring carbon atom, often referred to as the \(\beta\) carbon.

This makes the reaction a \(\beta\) elimination process. In essence, it's transforming an alcohol into an alkene by removing water. Here's how it works: First, the hydroxyl group (-OH) on the \(\alpha\) carbon snags a hydrogen ion (H\(^+\)) from the acid, becoming a better leaving group. Then, the molecule loses the newly protonated water molecule, creating a positively charged carbocation. Finally, a base (often the conjugate base of the original acid) takes the hydrogen from the \(\beta\) carbon, formulating a double bond between the \(\alpha\) and \(\beta\) carbons — resulting in an alkene and completing the \(\beta\) elimination.

Dehydrohalogenation

Dehydrohalogenation is quite similar to dehydration, but instead of losing water, this process involves the removal of a hydrogen atom and a halogen (like Cl, Br, or I) from adjacent carbons on an alkyl halide molecule. By focusing on the position of these elements:

• The halogen is attached to the \(\alpha\) carbon,
• The hydrogen is attached to the \(\beta\) carbon.

This process is an exemplary \(\beta\) elimination as it results in the formation of a double bond between the \(\alpha\) and \(\beta\) carbons upon the elimination of the halogen and hydrogen atoms. Usually, strong bases such as sodium or potassium hydroxide help facilitate this reaction. They not only abstract the \(\beta\) hydrogen but also push the reactions toward forming the double bond. As a result, new alkenes are created, transforming the original alkyl halide.

Dehalogenation of Vicinal Dibromides

Dehalogenation of vicinal dibromides takes the classic \(\beta\) elimination concept into another direction, focusing on molecules with two bromine atoms on adjacent (vicinal) carbons. Here's what occurs during this intriguing reaction:

• Both bromine atoms are eliminated simultaneously from the \(\alpha\) and \(\beta\) carbons.
• This elimination results in the formation of a \(\pi\) bond, thus forming an alkene.

In a nutshell, it's all about converting vicinal dibromides into alkenes by getting rid of the two neighboring bromine atoms. The reaction utilizes zinc dust or similar reagents to dislodge the bromines. Unlike dehydrohalogenation, which involves a hydrogen and a halogen, dehalogenation clears out two halogens without involving hydrogens.This process underscores the flexible nature of \(\beta\) eliminations, showcasing how the removal of elements from adjacent carbons can vary based on the starting material.