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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

Short Answer

Expert verified
(a) 1-chlorohexane, (b) 1-bromopentane, (c) 2-chloropentane, (d) 2-bromo-5-methylhexane, (e) 1-bromodecane.

Step by step solution

01

Understand the Reaction Mechanism

The reaction of sodium iodide (NaI) in acetone is an SN2 reaction, wherein the iodide ion acts as a nucleophile and attacks the carbon bonded to the halogen, leading to the formation of carbon-iodine and the expulsion of the original halogen as a leaving group. SN2 reactions are favored by primary alkyl halides and less hindered substrates.
02

Analyze Pair (a) 1-Chlorohexane or Cyclohexyl Chloride

1-chlorohexane is a primary alkyl halide, while cyclohexyl chloride is a secondary alkyl halide. Primary alkyl halides react faster in SN2 reactions due to less steric hindrance. Therefore, 1-chlorohexane reacts faster with sodium iodide in acetone.
03

Analyze Pair (b) 1-Bromopentane or 3-Bromopentane

1-bromopentane is a primary alkyl halide, while 3-bromopentane is a secondary alkyl halide. Primary haloalkanes have less steric hindrance and thus react faster in SN2 reactions. Therefore, 1-bromopentane reacts faster with sodium iodide in acetone.
04

Analyze Pair (c) 2-Chloropentane or 2-Fluoropentane

Fluorine is a poorer leaving group compared to chlorine, due to its higher electronegativity and stronger bond to carbon. Thus, 2-chloropentane, despite being secondary, reacts faster than 2-fluoropentane with sodium iodide in acetone.
05

Analyze Pair (d) 2-Bromo-2-methylhexane or 2-Bromo-5-methylhexane

2-bromo-2-methylhexane is a tertiary alkyl halide, whereas 2-bromo-5-methylhexane is a secondary alkyl halide. Tertiary alkyl halides are less favorable for SN2 reactions due to high steric hindrance. Therefore, 2-bromo-5-methylhexane reacts faster with sodium iodide in acetone.
06

Analyze Pair (e) 2-Bromopropane or 1-Bromodecane

2-bromopropane is a secondary alkyl bromide, while 1-bromodecane is a primary alkyl bromide. Primary alkyl bromides react faster in SN2 reactions due to lesser steric hindrance, so 1-bromodecane reacts faster with sodium iodide in acetone.

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Key Concepts

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

Nucleophilic Substitution
Nucleophilic substitution is a fundamental type of chemical reaction. In this process, a nucleophile, which is an electron-rich species, replaces a leaving group attached to a carbon atom. This is common in organic chemistry, especially in saturated hydrocarbons. The most studied nucleophilic substitution reactions are the SN1 and SN2 reactions.
  • SN2 reactions are bimolecular, meaning two reactant species are involved in the rate-determining step.
  • These reactions proceed in a single step without any intermediates.
  • Typically, SN2 reactions feature a direct attack of the nucleophile, resulting in an inversion of spacial configuration, known as Walden inversion.
Understanding nucleophilic substitution is essential for predicting how molecules will interact in different chemical environments.
Primary Alkyl Halide
A primary alkyl halide is an organic compound where a halogen atom is bonded to a primary carbon atom. This means the carbon with the halogen is only attached to one other carbon atom. Because of their structure, primary alkyl halides are excellent substrates for SN2 reactions.
  • They provide less steric hindrance compared to secondary and tertiary alkyl halides, allowing nucleophiles to easily attack the carbon bearing the halogen.
  • Examples include compounds like 1-bromopentane and 1-chlorohexane.
  • These types of alkyl halides favor clean and rapid SN2 reactions.
This property makes primary alkyl halides a popular choice for syntheses involving nucleophilic substitutions.
Leaving Group
In nucleophilic substitution reactions, the "leaving group" is an atom or group that is replaced by the nucleophile. A good leaving group can stabilize the negative charge it acquires after breaking away from the main molecule.
  • Iodide is considered an excellent leaving group due to its large size and ability to disperse the negative charge effectively.
  • Other good leaving groups include bromide and chloride, though they are not as effective as iodide.
  • Fluorine, on the other hand, is a poor leaving group because it forms a very strong bond with carbon, making it difficult to displace.
The effectiveness of the leaving group is crucial in determining the rate and feasibility of the substitution reaction.
Steric Hindrance
Steric hindrance refers to the slowing down of chemical reactions due to the physical size of groups within a molecule. In SN2 reactions, which require direct approach to the reacting center, steric hindrance can significantly influence reaction speed.
  • Primary alkyl halides are less sterically hindered than secondary or tertiary alkyl halides, making them more reactive in SN2 scenarios.
  • The more bulky groups around the reaction center, the slower the SN2 reaction because the nucleophile finds it harder to reach the carbon atom bonded to the leaving group.
  • Understanding steric hindrance is key to predicting the outcomes and speeds of nucleophilic substitution reactions.
It's a critical concept when planning chemical syntheses and ensuring reactions proceed efficiently.

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Most popular questions from this chapter

The hydrolysis of sulfonates of 2 -octanol is stereospecific and proceeds with complete inversion of configuration. Write a structural formula that shows the stereochemistry of the 2-octanol formed by hydrolysis of an optically pure sample of \((\mathrm{S})-(+)-1\) -methylheptyl \(\mathrm{p}\) toluenesulfonate, identify the product as \(\mathrm{R}\) or \(\mathrm{S}\), and deduce its specific rotation.

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. (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.

Identify the compound in each of the following pairs that reacts at the faster rate in an \(\mathrm{S}_{\mathrm{N}} 1\) reaction: (a) Isopropyl bromide or isobutyl bromide (b) Cyclopentyl iodide or 1-methylcyclopentyl iodide (c) Cyclopentyl bromide or 1 -bromo-2,2-dimethylpropane (d) tert-Butyl chloride or tert-butyl iodide

Would you expect the 2-octanol formed by \(\mathrm{S}_{\mathrm{N}} 2\) hydrolysis of \((-)-2\) -bromooctane to be optically active? If so, what will be its absolute configuration and sign of rotation? What about the 2 -octanol formed by hydrolysis of racemic 2 -bromooctane?

Reaction of ethyl iodide with triethylamine \(\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}:\right]\) yields a crystalline compound \(\mathrm{C}_{8} \mathrm{H}_{20} \mathrm{NI}\) in high yield. This compound is soluble in polar solvents such as water but insoluble in nonpolar ones such as diethyl ether. It does not melt below about \(200^{\circ} \mathrm{C}\). Suggest a reasonable structure for this product.
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