Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the reaction in which Hofmann product could be the major product? (a) 2 - chlorohexane \(\frac{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}}{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}\) (b) 2 - fluorohexane \(\frac{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}}{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}\) (c) 2- iodohexane \(\frac{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}}{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}\) (d) 2 - bromohexane \(\frac{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}}{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}\)

Step-by-Step Solution

Answer

Hofmann product is most likely with 2-fluorohexane (b).

Step 1: Understanding the Hofmann Elimination

The Hofmann elimination leads to the formation of the least substituted alkene, also known as the Hofmann product. It typically occurs with bulky bases or poor leaving groups.

Step 2: Identifying Good and Poor Leaving Groups

In the given options, chloride, bromide, and iodide are common leaving groups. Fluoride is a poor leaving group compared to the others due to its strong bond with carbon.

Step 3: Analyze the Reactions

Examine each reaction: (a) 2-chlorohexane can lead to a Hoffman or Saytzeff product with EtO-; (b) 2-fluorohexane has a favorable condition for Hofmann elimination because of the F- being a poor leaving group; (c) 2-iodohexane will likely favor Saytzeff product due to I- being a good leaving group; (d) 2-bromohexane will also favor Saytzeff elimination due to Br- being a good leaving group.

Step 4: Conclusion on Major Product Formation

Because a poor leaving group is more likely to produce the Hofmann product, 2-fluorohexane (b) is the scenario where Hofmann elimination is the most likely major reaction.

Key Concepts

Leaving Groups in Organic Chemistry

In organic chemistry, leaving groups play a crucial role in determining the outcome of elimination reactions. A leaving group is an atom or group of atoms that is detached from the parent molecule during a chemical reaction. The nature of the leaving group can strongly influence the reaction pathway.

Here are some key points about leaving groups:

• Good leaving groups are typically stable anions or neutral molecules once they detach from the substrate molecule.
• Common good leaving groups include halides like iodide (I-), bromide (Br-), and chloride (Cl-), listed in decreasing order of leaving ability.
• Poor leaving groups, like fluoride (F-), hold a much stronger bond to the carbon atom due to their high electronegativity and small size, making them less prone to leave.

In reactions involving elimination, the type of leaving group directly influences whether the reaction will proceed through a Hofmann or a Saytzeff pathway. A poor leaving group often leads to Hofmann elimination, where the least substituted alkene forms as the major product.

Product Formation in Elimination Reactions

Elimination reactions produce alkenes by removing atoms or groups from a substrate molecule. The product of such reactions depends on various factors, including the leaving group, the base used, and the presence of bulky substituents.

In an elimination reaction:

• A base abstracts a proton from the molecule, often in the vicinity of the leaving group, which eventually leaves the substrate, forming a double bond.
• The position on the molecule where the double bond forms influences whether the product will be a major or minor product, depending on its substitution pattern.
• Typically, the more substituted alkene (Saytzeff product) is the major product because of greater thermodynamic stability, unless specific conditions favor the formation of the least substituted alkene (Hofmann product).

Therefore, understanding the elements at play, such as leaving group ability and base strength, helps predict the outcome and major product in elimination reactions.

Saytzeff Rule in Elimination Reactions

The Saytzeff rule is an important concept in elimination reactions, guiding which alkene will be the major product. According to the Saytzeff rule, the more substituted alkene is generally favored and more stable due to the distribution of positive electron density.

Here are some details about the Saytzeff rule:

• It predicts that the most stable alkene, with the highest number of alkyl substituents, will be the major product in most elimination reactions.
• This rule applies prominently when good leaving groups and small bases are present, as they help facilitate the elimination of a proton leading to the formation of the more substituted double bond.
• Exceptions to the Saytzeff rule occur primarily when poor leaving groups or bulky bases are involved, which can direct the reaction towards producing the less substituted Hofmann product.

Understanding the Saytzeff rule allows chemists to anticipate and control the molecular architecture of the resulting alkene during an elimination reaction.