Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

Dehydrobromination (-HBr) of the following in increasing order is (1) CCCCBr \((2)\) CCCC(C)Br (3) (a) \(1<2<3\) (b) \(3<2<1\) (c) \(2=3<1\) (d) \(3<1<2\)

Step-by-Step Solution

Answer

The increasing order of dehydrobromination is (1) < (2), hence the order is 1 < 2.

Step 1: Understand the Reaction Mechanism

Dehydrobromination involves the removal of a hydrogen bromide (HBr) molecule from an alkyl bromide, leading to the formation of an alkene. The process typically proceeds through an E2 elimination mechanism, which involves a strong base abstracting a proton from the β-carbon, resulting in the removal of a bromine atom from the α-carbon.

Step 2: Analyze the Substrate Structure

Review the structures of each compound to determine the substituents around the bromine atom. - (1) CCCCBr: 1-bromobutane - (2) CCCC(C)Br: 2-bromo-2-methylbutane

Step 3: Order of Reactivity in Dehydrobromination

The rate of dehydrobromination is influenced by steric hindrance and stability of the resulting alkene. More substituted alkenes are more stable due to hyperconjugation and the inductive effect. The sequence typically follows Zaitsev's rule: the most substituted alkene is favored. - Compound 1 forms a less substituted alkene when dehydrobrominated. - Compound 2 forms a more substituted and therefore more stable alkene.

Step 4: Determine the Increasing Order

Based on the stability of the resulting alkenes, arrange the compounds according to their propensity to undergo dehydrobromination. - Compound 2 (2-bromo-2-methylbutane) reacts faster than Compound 1 (1-bromobutane) because it leads to a more substituted and stable alkene.

Key Concepts

E2 elimination mechanism

The E2 elimination mechanism is a critical concept in organic chemistry, particularly when discussing reactions like dehydrobromination. This one-step bimolecular reaction is known for its simultaneous removal of a leaving group (like a halogen, such as bromine) and the removal of an adjacent hydrogen atom. The process kicks off when a strong base abstracts a proton from the β-carbon of the substrate. At the same time, the leaving group detaches from the α-carbon. This simultaneous action results in the formation of a double bond, hence creating an alkene.

• Concerted Mechanism: Both removal of the proton and the leaving group happen in a single, concerted step. This is why it's called bimolecular—there are two molecular entities involved in the rate-determining step.
• Role of the Base: The strength and concentration of the base can affect the rate of the E2 reaction. Typically, a stronger base speeds up the reaction.
• Stereochemistry: The preferred geometry is anti-periplanar. This means the hydrogen and the leaving group should be on opposite sides of the molecule for optimal overlap of orbitals, facilitating the double bond formation.

The E2 mechanism is favored for its efficiency and speed, especially in less hindered primary and secondary substrates.

Zaitsev's rule

Zaitsev's rule plays a fundamental role in determining the outcome of elimination reactions like dehydrobromination. The rule states that when there is a possibility of forming more than one alkene during an elimination reaction, the most substituted alkene will predominate. This is because more substituted alkenes are generally more stable owing to the hyperconjugation and the inductive effects that stabilize the alkene.

• Substitution and Stability: Alkenes with more alkyl groups attached to the double-bonded carbons are more stable. More alkyl groups mean more electron-donating groups, which enhance stability through hyperconjugation.
• Predicts Major Product: According to Zaitsev’s rule, the product distribution is skewed towards the formation of the most stable, more substituted alkene.
• Exceptions: While Zaitsev's rule is generally applicable, in some cases such as when steric hindrance or particular reaction conditions apply, a less substituted alkene might be the major product, known as the Hofmann product.

Understanding Zaitsev's rule is crucial when predicting which alkene product will be favored in an elimination reaction.

alkyl bromide substrates

Alkyl bromide substrates are key players in dehydrobromination reactions, acting as the starting materials from which HBr is eliminated to form alkenes. Their structure significantly influences the course and outcome of the elimination reaction.

• Primary, Secondary, and Tertiary: Classified based on the carbon atom to which the bromine atom is attached: primary (connected to one other carbon), secondary (connected to two carbons), and tertiary (connected to three carbons).
• Reactivity: Tertiary bromides typically undergo elimination faster than secondary, followed by primary, due to the stability of the resulting carbocations and alkenes.
• Steric Hindrance: Bulky alkyl groups can hinder the approach of the base, affecting the reaction rate and sometimes leading the reaction pathway towards an E1 mechanism in highly substituted systems.

When evaluating alkyl bromide substrates, considering their substitution level and steric effects is essential for predicting the outcome of dehydrobromination.