Question

2-chloro-2-methyl pentane on reaction with sodium methoxide in methanol yields. (a) CCCC(C)(C)OC (b) C=C(C)CCC (c) CCC=C(C)C (a) (a) and (c) (b) c only (c) (a) and (b) (d) all of these

Short answer

The reaction primarily yields (b) and (c); thus, the answer is (d) all of these.

Step-by-step solution

Understanding the Reaction Type

The given compound, 2-chloro-2-methylpentane, indicates a reaction involving an alkyl halide. Sodium methoxide in methanol is a strong base that can promote both substitution and elimination reactions. We should consider both pathways.

Analyzing Substitution Potential

The tertiary carbon of 2-chloro-2-methylpentane is typically unfavorable for S_N2 reactions due to steric hindrance. S_N1 is also unlikely given the strong base present, which favors elimination.

Evaluating the Elimination Reaction

A strong base like sodium methoxide is likely to promote an E2 elimination reaction. In an E2 mechanism, the base abstracts a proton (β-hydrogen), leading to the formation of a double bond.

Identifying Possible Elimination Products

Identify β-hydrogens on the neighboring carbons to the tertiary carbon. For 2-chloro-2-methylpentane, eliminating a β-hydrogen from different positions will form different alkenes.

Predicting Major Product

Between E2 possible products: 1. Elimination of a β-hydrogen from carbon-3 leads to 2-methylpent-2-ene: - - (CH₃)C=CH-CH₂-CH₃ (Option c) 2. Eliminating a β-hydrogen from carbon-1 will yield 2-methyl-1-pentene: - - - - - CH₂=C(CH₃)-CH₂-CH₂-CH₃ (Option b) The more substituted alkene (2-methylpent-2-ene, Option c) is usually the major product due to Zaitsev's rule.

Evaluating Given Options

Examine the SMILES strings to identify the options corresponding to the predicted products: - Option (b): - - - C=C(C)CCC corresponds to 2-methyl-1-pentene. - Option (c): - - CCC=C(C)C corresponds to 2-methylpent-2-ene (major product). Hence, both (b) and (c) match the product predictions from E2 elimination. Only option (d) includes both isomeric products.

Key concepts

Alkyl Halides

Alkyl halides are organic compounds where a halogen atom (such as chlorine, bromine, or iodine) is bonded to an alkyl group. In this exercise, the compound in question is 2-chloro-2-methylpentane, which features a chlorine atom attached to a carbon atom in a hydrocarbon chain.
This particular structure is a tertiary alkyl halide because the chlorine is attached to a tertiary carbon, which in turn is bonded to three other carbon atoms.

Due to the presence of the halogen, alkyl halides are highly reactive, making them suitable candidates for various organic reactions including substitution and elimination.

• Substitution vs. Elimination: Substitution involves the replacement of the halogen with another group, while elimination leads to the removal of the halogen and formation of a double bond.
• Tertiary Alkyl Halides: These generally favor elimination over substitution, particularly in the presence of strong bases like sodium methoxide.

The 2-chloro-2-methylpentane in this exercise is prone to elimination, as we explore further with the E2 mechanism.

E2 Mechanism

The E2 mechanism is a one-step elimination reaction commonly observed with alkyl halides, especially in the presence of a strong base. In our exercise, sodium methoxide acts as the strong base which initiates this mechanism.

During the E2 process, the base removes a β-hydrogen from a carbon atom adjacent to the carbon bearing the halogen. Simultaneously, the electrons from this hydrogen-carbon bond form a double bond, and the halogen is expelled as a leaving group.

• Concerted Mechanism: E2 reactions are concerted, meaning the breaking of bonds and formation of the double bond occur in a single step without intermediates.
• Stereospecificity: The hydrogen abstracted must be anti-periplanar (opposite side) to the leaving group, which dictates the stereochemistry of the product.
• Substrate Effect: Tertiary substrates, like 2-chloro-2-methylpentane, are highly suited for E2 due to less steric hindrance affecting the base's approach.

In this process, different β-hydrogens can be removed, leading to multiple elimination products as seen in this problem.

Zaitsev's Rule

Zaitsev's Rule is a guiding principle used to predict the major product in elimination reactions, particularly those following the E2 mechanism. The rule states that the more substituted alkene (the one with more carbon groups attached to the double-bonded carbons) will usually be the major product.

This rule applies to our reaction as follows:

• More Substituted Alkene: Among the possible elimination products of 2-chloro-2-methylpentane, 2-methylpent-2-ene (Option c) is more substituted than 2-methyl-1-pentene (Option b).
• Stability of the Alkene: More substituted alkenes are more stable due to hyperconjugation and the inductive effect, making them the preferred product.
• Predicting Product: The exercise demonstrates that applying Zaitsev's Rule accurately predicts that 2-methylpent-2-ene, the more substituted product, is indeed the major product observed.

Zaitsev's Rule is an invaluable tool in organic chemistry for understanding how structural features influence the outcome of elimination reactions.