You were told in [SECTION 7.11] that is best to use a methyl halide or a primary alkyl halide for the reaction of an acetylide ion with an alkyl halide. Explain why this is so.
Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
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Bruice 8th Edition
Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
Problem 39c
Bruice 8th Edition
Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
Problem 39cChapter 10, Problem 39c
Which reacts faster in an SN1 reaction?

Verified step by step guidance1
Step 1: Understand the SN1 reaction mechanism. SN1 reactions proceed via a two-step mechanism: (1) formation of a carbocation intermediate by the departure of the leaving group, and (2) nucleophilic attack on the carbocation. The rate-determining step is the formation of the carbocation, so the stability of the carbocation is a key factor in determining the reaction rate.
Step 2: Evaluate the structure of the reactants. Identify the type of carbon (primary, secondary, or tertiary) attached to the leaving group. Tertiary carbocations are more stable than secondary, which are more stable than primary, due to hyperconjugation and inductive effects.
Step 3: Consider the leaving group. A good leaving group facilitates the formation of the carbocation. Leaving groups such as halides (e.g., Cl⁻, Br⁻, I⁻) or sulfonates (e.g., tosylate) are commonly involved in SN1 reactions. Compare the leaving groups of the reactants to determine which one is better.
Step 4: Assess the solvent. SN1 reactions are favored in polar protic solvents (e.g., water, alcohols) because they stabilize the carbocation intermediate and the leaving group. Check if the reaction conditions include such solvents.
Step 5: Combine all factors. Based on the stability of the carbocation, the quality of the leaving group, and the solvent conditions, determine which reactant will react faster in the SN1 reaction. The reactant that forms the most stable carbocation and has the best leaving group will typically react faster.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
SN1 Reaction Mechanism
The SN1 (Substitution Nucleophilic Unimolecular) reaction mechanism involves two main steps: the formation of a carbocation intermediate and the subsequent nucleophilic attack. The rate of an SN1 reaction depends primarily on the stability of the carbocation formed, as the first step is the rate-determining step. More stable carbocations, such as tertiary carbocations, react faster than less stable ones.
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Carbocation Stability
Carbocation stability is influenced by the degree of alkyl substitution and resonance. Tertiary carbocations are the most stable due to hyperconjugation and inductive effects from surrounding alkyl groups. In contrast, primary carbocations are the least stable, making them less likely to form in SN1 reactions. Understanding the stability of carbocations is crucial for predicting the reactivity in SN1 mechanisms.
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Nucleophilicity
Nucleophilicity refers to the ability of a nucleophile to donate an electron pair to an electrophile. In SN1 reactions, the nucleophile attacks the carbocation after its formation. While the strength of the nucleophile can influence the overall reaction rate, in SN1 reactions, the rate is primarily determined by the stability of the carbocation rather than the nucleophile's strength.
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Related Practice
Textbook Question
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Textbook Question
Draw the substitution and elimination products for the following reactions, showing the configuration of each product:
c. 1-chloro-1-methylcyclohexane + CH3O−
d. 1-chloro-1-methylcyclohexane + CH3OH
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Textbook Question
Which reacts faster in an SN2 reaction?
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Textbook Question
Explain why only a substitution product and no elimination product is obtained when the following compound reacts with sodium methoxide:
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Textbook Question
Which reacts faster in an E1 reaction?
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Textbook Question
How does the ratio of substitution product to elimination product formed from the reaction of propyl bromide with CH3O− in methanol change if the nucleophile is changed to CH3S−?
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