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Chapter 6: Problem 7

Identify the compound in each of the following pairs that reacts with sodium iodide in acetone at the faster rate: (a) 1-Chlorohexane or cyclohexyl chloride (b) 1-Bromopentane or 3-bromopentane (c) 2-Chloropentane or 2 -fluoropentane (d) 2-Bromo-2-methylhexane or 2 -bromo-5-methylhexane (e) 2-Bromopropane or 1-bromodecane Sample Solution (a) Compare the structures of the two chlorides. 1-Chlorohexane is a primary alkyl chloride; cyclohexyl chloride is secondary. Primary alkyl halides are less crowded at the site of substitution than secondary ones and react faster in substitution by the \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. 1-Chlorohexane is more reactive.

Short Answer

Expert verified
1-Chlorohexane, 1-Bromopentane, 2-Chloropentane, 2-Bromo-5-Methylhexane, 1-Bromodecane.

Step by step solution

01

Understanding the Reaction Mechanism

Sodium iodide in acetone is a classic test for the SN2 reaction, in which the nucleophile (iodide) attacks the carbon atom bonded to the halogen, displacing it. This reaction proceeds faster with less hindered (less substituted) alkyl halides.
02

Step (a): Analyze Reactivity of 1-Chlorohexane vs Cyclohexyl Chloride

1-Chlorohexane is a primary alkyl chloride, while cyclohexyl chloride is a secondary alkyl chloride. Primary alkyl halides react faster in SN2 reactions due to less steric hindrance at the reaction site. Thus, 1-chlorohexane reacts faster.
03

Step (b): Compare 1-Bromopentane and 3-Bromopentane

1-Bromopentane is a primary alkyl bromide, and 3-bromopentane is a secondary alkyl bromide. Primary halides are less sterically hindered and generally react faster in SN2 reactions. Therefore, 1-bromopentane reacts faster.
04

Step (c): Evaluate 2-Chloropentane vs 2-Fluoropentane

The leaving group ability determines the compound reactivity in SN2. Chloride is a much better leaving group than fluoride due to its larger size and ability to stabilize negative charge. Therefore, 2-chloropentane reacts faster.
05

Step (d): Consider 2-Bromo-2-Methylhexane vs 2-Bromo-5-Methylhexane

2-Bromo-2-methylhexane is tertiary, whereas 2-bromo-5-methylhexane is secondary. Tertiary halides react slower in SN2 due to more steric hindrance, so the secondary halide, 2-bromo-5-methylhexane, reacts faster.
06

Step (e): Determine Reactivity of 2-Bromopropane vs 1-Bromodecane

2-Bromopropane is a secondary halide, while 1-bromodecane is a primary halide. As primary halides experience less steric hindrance, 1-bromodecane reacts faster in the SN2 mechanism.

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Key Concepts

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

Primary Alkyl Halides
Primary alkyl halides are compounds where the carbon atom bonded to the halogen is attached to only one other carbon atom. These compounds are crucial in SN2 reactions. Why? Because their structure leads to less steric hindrance, allowing the nucleophile easier access to the carbon-halogen bond. This means primary alkyl halides generally react faster in SN2 reactions. For example, 1-chlorohexane is primary and reacts faster than a secondary alkyl halide like cyclohexyl chloride. This lower level of congestion at the reaction site facilitates a smoother path for the nucleophilic attack, contributing to their rapid reaction speed.
Secondary Alkyl Halides
Secondary alkyl halides have the carbon atom holding the halogen bonded to two other carbon atoms. They are more sterically hindered compared to primary alkyl halides. As a result, the SN2 reactions involving secondary alkyl halides usually occur more slowly than those involving primary ones. A real-world example is 3-bromopentane, a secondary alkyl bromide, which reacts slower than 1-bromopentane. Once more carbon atoms are attached to the reaction center, the space becomes crowded, making it harder for nucleophiles like iodide to reach and displace the leaving group.
Leaving Group Ability
The ability of a leaving group to depart from the carbon it is bound to greatly impacts the rate and success of SN2 reactions. A good leaving group is typically a halogen atom that can stabilize the negative charge it carries after leaving. One great example is the chloride ion: it is bigger and can better stabilize the extra electrons it gets during the reaction than smaller halogens like fluoride. That's why 2-chloropentane reacts faster than 2-fluoropentane in an SN2 mechanism. Remember, stronger leaving groups lead to faster and more efficient reactions.
Steric Hindrance
Steric hindrance refers to the prevention of chemical reactions due to the spatial arrangement of atoms within a molecule. In the context of SN2 reactions, greater steric hindrance means that the nucleophile has a harder time reaching the carbon center to displace the leaving group. Primary alkyl halides have less steric hindrance, so they participate in SN2 reactions more readily than secondary or tertiary ones. Consider how 1-bromodecane is significantly less hindered and thus reacts faster than 2-bromopropane. The arrangement and number of carbon atoms around the reaction center drastically affect how swiftly and successfully the reaction proceeds.

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Most popular questions from this chapter

Write an equation, clearly showing the stereochemistry of the starting material and the product, for the reaction of \((S)-1\) -bromo-2-methylbutane with sodium iodide in acetone. What is the configuration \((R\) or \(S)\) of the product?

Unlike protic solvents, which solvate anions, polar aprotic solvents form complexes with cations better than with anions. Use a dashed line to show the interaction between dimethyl sulfoxide with a cation, using sodium azide \(\left(\mathrm{Na} \mathrm{N}_{3}\right)\) as the source of the cation.

If the temperature is not kept below \(25^{\circ} \mathrm{C}\) during the reaction of primary alcohols with \(p\) -toluenesulfonyl chloride in pyridine, it is sometimes observed that the isolated product is not the desired alkyl \(p\) -toluenesulfonate but is instead the corresponding alkyl chloride. Suggest a mechanistic explanation for this observation.

(a) Suggest a reasonable series of synthetic transformations for converting trans2-methylcyclopentanol to cis-2-methylcyclopentyl acetate. (b) How could you prepare cis-2-methylcyclopentyl acetate from 1-methylcyclopentanol?

Give the mechanistic symbols \(\left(\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2\right)\) that are most consistent with each of the following statements: (a) Methyl halides react with sodium ethoxide in ethanol only by this mechanism. (b) Unhindered primary halides react with sodium ethoxide in ethanol mainly by this mechanism. (c) The substitution product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. (d) Reactions proceeding by this mechanism are stereospecific. (e) Reactions proceeding by this mechanism involve carbocation intermediates. (f) This mechanism is most likely to have been involved when the products are found to have a different carbon skeleton from the substrate. (g) Alkyl iodides react faster than alkyl bromides in reactions that proceed by these mechanisms.
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