Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the following reactions: (1) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}(\mathrm{OH}) \mathrm{CH}_{3} \quad\) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \quad \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow}\) (3) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \quad \frac{\left(\mathrm{CH}_{3}\right)_{3} \mathrm{O}^{-\mathrm{K}} \mathrm{K}^{\oplus}}{\longrightarrow}\) (4) \(-\mathrm{CH}_{2}-\mathrm{CHO} \stackrel{\Delta}{\longrightarrow}\) Which of these reaction(s) will not produce Saytzeff product? (a) (1), (3) and (4) (b) (4) only (c) (3) only (d) (2) and (4)

Step-by-Step Solution

Answer

(d) (2) and (4)

Step 1: Understanding Saytzeff's Rule

Saytzeff's Rule, often encountered in elimination reactions, states that the more substituted alkene will generally predominate. This is due to the stability conferred by alkyl groups acting via hyperconjugation and inductive effects.

Step 2: Analyzing Reaction 1

In Reaction 1, an alcohol undergoes dehydration using conc. \\( \mathrm{H}_2\mathrm{SO}_4 \). This typically follows E1 mechanism, where the more substituted, and therefore more stable, alkene according to Saytzeff's rule, is usually the major product.

Step 3: Analyzing Reaction 2

In Reaction 2, an alkyl halide undergoes elimination using alcoholic KOH, typically following the E2 mechanism. The result is the Saytzeff product, which is the more substituted alkene.

Step 4: Analyzing Reaction 3

Reaction 3 involves dehydrohalogenation using a sterically hindered base, \\( \mathrm{KOtBu} \). This often leads to the formation of a less substituted alkene, known as the Hofmann product, rather than following Saytzeff's rule.

Step 5: Analyzing Reaction 4

Reaction 4 involves heat, which implies a possible decarboxylation or similar reaction where Saytzeff’s rule can be irrelevant if no elimination of \\( \mathrm{H}_2 \) occurs. Thus, it doesn't generate a Saytzeff product, as it may not yield an alkene at all and isn't concerned with alkene substitution.

Step 6: Compare Results with Options

Reactions that do not produce Saytzeff products: - Reaction 3 produces Hofmann products due to the bulky base. - Reaction 4 likely involves decarboxylation which may not produce an alkene. Checking options: (a) (1), (3), (4) - Incorrect (b) (4) only - Incorrect (c) (3) only - Incorrect (d) (2) and (4) - Correct, since Reaction 3 also doesn't produce Saytzeff but is not listed here.

Key Concepts

Elimination reactions

Elimination reactions are a fundamental category of organic reactions where a molecule loses atoms or groups of atoms to form double or triple bonds. Typically, they result in the formation of alkenes or alkynes from more saturated molecules like alkanes, halides, or alcohols.
These reactions are crucial in organic synthesis, as they help form new double bonds, increasing the molecular complexity and reactivity. There are two main types of elimination reactions:

• Dehydrohalogenation: This occurs when an alkyl halide loses a hydrogen halide (H-X) to form an alkene.
• Dehydration: Involves the removal of water from an alcohol to yield an alkene.

Factors like the choice of base, the nature of the leaving group, and reaction conditions, such as heat and solvent, can influence the outcome of elimination reactions significantly. These factors determine whether the reaction follows the Saytzeff rule or produces a Hofmann product, impacting the double bond's location and stability.

Hofmann product

The Hofmann product is an outcome of elimination reactions where a bulky base removes a proton, leading to the formation of a less substituted alkene. Unlike the more common Saytzeff product, which favors the creation of the more substituted and thus more stable alkene, the Hofmann product results when sterically hindered bases are used.
Hofmann elimination typically involves conditions where:

• A large, bulky base is present, such as potassium tert-butoxide (KOtBu).
• Reactivity is controlled more by steric effects than by the ideal formation of the most substituted double bond.

In reaction 3 from the original exercise, KOtBu promotes the formation of the Hofmann product, pointing to its significance in synthetic strategies where less-substituted alkenes are desirable. Understanding when and why a Hofmann product will form over a Saytzeff product is crucial for mastering organic synthesis.

Alkene stability

Alkene stability is a major consideration in elimination reactions, influencing which products are favored under certain conditions. The stability of alkenes is generally assessed based on substitution and the presence of stabilizing interactions such as hyperconjugation and resonance.
Key factors contributing to alkene stability include:

• Substitution: More highly substituted alkenes (like tetrasubstituted ones) tend to be more stable due to hyperconjugation. Alkyl groups can donate electron density through sigma bonds, stabilizing the positive charge of the pi bond.
• Conjugation: Alkenes conjugated with other double bonds or aromatic systems gain further stability through delocalization of electrons.
• Sterics: Lower steric hindrance around the double bond increases stability.

In most elimination reactions, because of these factors, the more substituted alkene tends to be the preferred product, following Saytzeff’s rule. Understanding alkene stability helps predict the outcomes of elimination reactions accurately.

E1 and E2 mechanisms

The E1 and E2 elimination mechanisms describe different pathways for the formation of alkenes, categorized based on the kinetics of the reaction and the conditions required.
E1 Reactions:

• Occur in two steps and involve a carbocation intermediate.
• Typically seen in reactions like the dehydration of alcohols under acidic conditions.
• The rate of the reaction depends only on the concentration of the substrate, making it a unimolecular reaction.

E2 Reactions:

• Single-step concerted reactions where the base abstracts a proton while the leaving group departs simultaneously.
• Often occur with strong bases, such as in the removal of halogens from alkyl halides (as seen with alcoholic KOH).
• Rate depends on both the substrate and the base concentration, thus bimolecular.

The choice between these mechanisms influences both the regiochemistry and stereochemistry of the product. E1 often leads to more substituted, Saytzeff products, while E2 can favor less substituted, Hofmann products in cases of sterically hindered bases.