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Write the structure of the products that you expect from the reaction of 1 -methylcyclohexanol with (a) concentrated \(\mathrm{HCl}\) and (b) concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Compare and contrast the mechanisms of the two processes. (Hint: Compare the relative nucleophilicity of \(\mathrm{Cl}^{-}\) and \(\mathrm{HSO}_{4}^{-}\). Caution: When writing mechanisms, use "arrow pushing" to depict electron flow; write out every step separately; formulate complete structures, including charges and relevant electron pairs; and draw explicit reaction arrows to connect starting materials or intermediates with their respective products. Don't use shortcuts, and don't be sloppy!)

Short Answer

Expert verified
With \( \text{HCl} \), the product is 1-chloro-1-methylcyclohexane (substitution). With \( \text{H}_2\text{SO}_4 \), the product is 1-methylcyclohexene (elimination).

Step by step solution

01

Title - Identify the reactant

The reactant is 1-methylcyclohexanol, which is a secondary alcohol with the structure of a cyclohexane ring substituted with a hydroxyl group (OH) and a methyl group (CH3) at the same carbon.
02

Title - Reaction with concentrated HCl

Concentrated \(\text{HCl}\) acts as a source of \(\text{Cl}^{-}\) nucleophile. The reaction proceeds via an SN1 mechanism because it involves the formation of a stable carbocation intermediate.
03

Title - Mechanism of reaction with HCl

1. Protonation of the hydroxyl group by \(\text{HCl}\) to form a better leaving group (water).2. Loss of water molecule, generating a methylcyclohexyl carbocation.3. Nucleophilic attack by \(\text{Cl}^{-}\) on the carbocation to form 1-chloro-1-methylcyclohexane.
04

Title - Reaction with concentrated H2SO4

Concentrated \(\text{H}_2\text{SO}_4\) acts as a dehydrating agent, promoting the formation of an alkene via an E1 elimination mechanism.
05

Title - Mechanism of reaction with H2SO4

1. Protonation of the hydroxyl group by \(\text{H}_2\text{SO}_4\) to form a better leaving group (water).2. Loss of water molecule, generating a methylcyclohexyl carbocation.3. Elimination of a proton from an adjacent carbon, forming a double bond (alkene). The major product is 1-methylcyclohexene.
06

Title - Compare nucleophilicity

\(\text{Cl}^{-}\) is a better nucleophile than \(\text{HSO}_4^{-}\). Therefore, in the presence of HCl, a substitution product is formed, while H2SO4, being a poor nucleophile and a strong acid, leads to elimination.
07

Title - Draw structures of products

The product of reaction with HCl is 1-chloro-1-methylcyclohexane. The product of reaction with H2SO4 is 1-methylcyclohexene.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

1-methylcyclohexanol
1-methylcyclohexanol is an organic compound characterized by a cyclohexane ring, which is a six-membered carbon ring structure. At one of the carbon atoms in this ring, there is a hydroxyl group (-OH) attached, making it an alcohol. Additionally, this same carbon atom has a methyl group (-CH3) attached. This arrangement means that 1-methylcyclohexanol is a secondary alcohol, as the hydroxyl group is connected to a carbon atom that is attached to two other carbon atoms.
SN1 Mechanism
The SN1 mechanism stands for 'Substitution Nucleophilic Unimolecular.' It occurs in two main steps:
1. **Formation of Carbocation:** The first step involves the loss of a leaving group (e.g., water in the case of 1-methylcyclohexanol reacting with HCl) to form a positively charged carbocation intermediate.
2. **Nucleophilic Attack:** A nucleophile (e.g., Cl鈦) then attacks the carbocation to form the final product.
The SN1 mechanism is favored in situations where the carbocation intermediate is particularly stable. The structure of 1-methylcyclohexanol, when reacting with HCl, supports this mechanism as it results in a relatively stable carbocation. The formation of this stable intermediate is crucial because it determines the reaction's feasibility.
E1 Elimination
The E1 elimination mechanism is a two-step process often favored by strong acids like H鈧係O鈧 (sulfuric acid). It leads to the formation of alkenes.
1. **Formation of Carbocation:** Similar to the SN1 mechanism, the first step involves the formation of a carbocation intermediate through the loss of a leaving group (usually water, when an alcohol reacts with H鈧係O鈧).
2. **Proton Elimination:** A proton is then removed from a carbon atom adjacent to the carbocation. This step involves the elimination of a proton to form a double bond, resulting in the formation of 1-methylcyclohexene from 1-methylcyclohexanol.
The E1 mechanism is influenced by the stability of the carbocation and the nature of the leaving group.
Nucleophilicity
Nucleophilicity refers to a molecule's ability to donate a pair of electrons to an electrophile during a chemical reaction.
- **Cl鈦 (Chloride Ion):** A very good nucleophile due to its high electron density and ability to easily donate electrons to form bonds.
- **HSO鈧勨伝 (Hydrogen Sulfate Ion):** A much poorer nucleophile since it is relatively larger and less capable of effectively donating electrons compared to Cl鈦.
In the context of the given reactions, concentrated HCl provides Cl鈦 ions that can readily attack the carbocation intermediate formed from 1-methylcyclohexanol, leading to a substitution product through the SN1 mechanism. Conversely, H鈧係O鈧 promotes elimination rather than nucleophilic substitution due to its weak nucleophilicity, thus forming an alkene product.
Carbocation Stability
Carbocation stability is a crucial aspect determining the feasibility of both SN1 and E1 reactions.
- **Tertiary Carbocations:** These are the most stable due to the inductive and hyperconjugative effects from the surrounding alkyl groups.
- **Secondary Carbocations:** Relatively less stable than tertiary but more stable than primary carbocations.
- **Primary Carbocations:** These are the least stable and rarely form unless under very specific conditions.
In the reaction involving 1-methylcyclohexanol, the carbocation formed is secondary. The presence of the methyl group provides some stabilizing effect through hyperconjugation and inductive effects, making the carbocation sufficiently stable for both SN1 and E1 mechanisms. This stability is essential for the reaction to proceed efficiently, whether forming a substitution product with HCl or an elimination product with H鈧係O鈧.

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Most popular questions from this chapter

Sugars, being polyhydroxylic compounds (Chapter 24), undergo reactions characteristic of alcohols. In one of the later steps in glycolysis (the metabolism of glucose), one of the glucose metabolites with a remaining hydroxy group, 2 -phosphoglyceric acid, is converted into 2-phosphoenolpyruvic acid. This reaction is catalyzed by the enzyme enolase in the presence of a Lewis acid such as \(\mathrm{Mg}^{2+}\). (a) How would you classify this reaction? (b) What is the possible role of the Lewis acidic metal ion?

Identify the nucleophile in the following reaction: $$ \mathrm{RX}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ROH}+\mathrm{H}^{+} \mathrm{X}^{-} $$ (a) \(\mathrm{X}^{-}\) (b) \(\mathrm{H}^{+}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{ROH}\) (e) \(\mathrm{RX}\)

We shall encounter later many reactions that require catalytic base, for example, catalytic sodium methoxide in methanol. Assume that you would like to make such a solution containing \(10 \mathrm{mmol}\) of \(\mathrm{NaOCH}_{3}\) in 1 liter of \(\mathrm{CH}_{3} \mathrm{OH}\). Would it be all right simply to add \(10 \mathrm{mmol}\) of \(\mathrm{NaOH}\) to the solvent?

Give detailed mechanisms and final products for the reaction of 3-methyl-2-pentanol with each of the reagents that follow. (a) \(\mathrm{NaH}\) (b) Concentrated \(\mathrm{HBr}\) (c) \(\mathrm{PBr}_{3}\) (d) \(\mathrm{SOCl}_{2}\) (e) Concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at \(130^{\circ} \mathrm{C}\) (f) Dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\)

The following ethers can, in principle, be made by more than one Williamson ether synthesis. Consider the relative merits of your approaches. (a) 1-Ethoxybutane; (b) 2-methoxypentane; (c) propoxycyclohexane; (d) 1,4-diethoxybutane. [Cautions: Alkoxides are strong bases. What is wrong with 4 -bromo-1-butanol as a starting material for (d)?]

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