Question

Which of the following is correctly matched? (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{N}\left(\mathrm{CH}_{2}\right)_{3}^{\circ} \mathrm{OH} . \mathrm{E}_{2}\) reaction (b) CC(C)(C)Br E_reaction (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}, \mathrm{E}_{1}\) reaction (d) CC(C)(C)O \(\mathrm{E}_{1}\) cb reaction

Short answer

Choice (d) is correctly matched with E1 reaction for tert-butyl alcohol.

Step-by-step solution

Identify Reactions and Compounds

To solve this, first recognize what the reactions and compounds in each option are. The options contain different reactions to analyze, such as E1, E2, and some structural descriptions in SMILES notation.

Option Analysis: Choice (a)

Option (a) shows a structure with an E2 reaction, typically involving a strong base and leading to the formation of a double bond by removing a leaving group and a proton from adjacent carbon atoms. However, it implies an incorrect reaction setup, as the structure suggests more substitution reactions would occur than elimination.

Option Analysis: SMILES Notation in (b)

Option (b) uses SMILES notation to describe tert-butyl bromide, a compound noted for undergoing E1 reactions typically, rather than a general "E_reaction." This makes it unclear and not specifically matched.

Option Analysis: Choice (c)

Option (c) describes 1-propanol undergoing an E1 reaction. E1 reactions generally involve the formation of a carbocation intermediate, and 1-propanol would typically not undergo E1 without strong acidic conditions to facilitate such a mechanism. Hence, it's also incorrectly matched.

Option Analysis: SMILES Notation in Option (d)

Option (d) includes a SMILES representation for tert-butyl alcohol and mentions an E1 reaction. Tert-butyl alcohol can undergo dehydration via an E1 mechanism to form isobutylene in the presence of an acid, making this option correctly matched.

Key concepts

Elimination Reactions

Elimination reactions are a fundamental concept in organic chemistry. They involve the removal of two atoms or groups from a molecule, typically resulting in the formation of a double bond. This reaction type is categorized mainly into E1 and E2 reaction mechanisms.
In elimination reactions, the molecule often loses a leaving group along with a hydrogen atom from an adjacent carbon. The goal is to create a new bond, often a pi bond like a double or triple bond, which increases the molecule's unsaturation.
Elimination reactions are essential in organic synthesis, enabling chemists to convert alkanes into alkenes or alkynes. There are various factors that can drive which type of elimination reaction occurs, such as the base strength and the structure of the substrate. Knowing how to distinguish between E1 and E2 is crucial when predicting reaction products.

E1 Reaction Mechanism

The E1 reaction mechanism is a two-step elimination process. It is vital to understand the nature of intermediates formed in E1 reactions, which significantly impacts the reaction's progression and outcome. Here's how it works:

• Step 1: Formation of a Carbocation - The first step in an E1 reaction is the departure of a leaving group, leading to the formation of a carbocation. This step is slow and rate-determining.
• Step 2: Deprotonation - A base removes a hydrogen atom from an adjacent carbon to form a double bond, completing the reaction.

The stability of the carbocation is pivotal in E1 reactions. Tertiary carbocations are more favored due to their enhanced stability from hyperconjugation and inductive effects. E1 reactions are often carried out with weak bases and require acidic conditions to proceed efficiently.
This mechanism is often seen in scenarios where alcohols undergo dehydration, particularly for tertiary alcohols like tert-butyl alcohol, which convert to alkenes.

SMILES Notation

SMILES (Simplified Molecular Input Line Entry System) notation is a way to represent chemical structures using short ASCII strings. Understanding SMILES is crucial for studying organic reactions, as it allows chemists to easily convey complex structures. Here's a basic guide to interpreting SMILES:

• Atoms: Atoms are represented by their chemical symbols (e.g., C for carbon)."
• Branches: Branches are enclosed in parentheses and are part of larger structures.
• Double & Triple Bonds: These types of bonds are represented by "/" and "=" symbols.

Take the example CC(C)(C)Br, which describes tert-butyl bromide. This SMILES string shows a carbon chain (C-C-C) with a bromine attached. Recognizing these structures helps in anticipating reaction pathways and product formation.
SMILES is widely used in databases and computational chemistry, making it indispensable for modern chemical analysis.

Tertiary Alcohol Dehydration

Tertiary alcohols, unlike primary and secondary alcohols, undergo elimination reactions under acidic conditions, which often lead to dehydration. This process involves the conversion of an alcohol group into a double bond while eliminating water.
The dehydration of tertiary alcohols like tert-butyl alcohol follows an E1 mechanism. Here’s a breakdown:

• Carbocation Formation: In an acidic environment, the hydroxyl group (2 OH) converts into water, a good leaving group, which departs to generate a more stable carbocation.
• Double Bond Formation: A base removes a hydrogen atom from an adjacent carbon, forming a carbon-carbon double bond. This results in the creation of either an alkene like isobutylene.

This reaction is specific due to the stability of the tertiary carbocation, which is needed for the E1 mechanism to proceed efficiently.
Understanding tertiary alcohol dehydration is crucial for predicting reaction conditions and outcomes in synthesis, making it a key topic in organic chemistry studies.