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Chapter 13: Problem 268

Predict the product(s) of the \(\mathrm{S}_{\mathrm{N}} 2\) reaction of each of the following nuclephiles with \((\pm)-2\) -iodooctane.

Short Answer

Expert verified
In order to predict the product(s) of the $\mathrm{S}_{\mathrm{N}} 2$ reaction of each nucleophile (not provided) with (±)-2-iodooctane, follow these steps: 1. Identify the nucleophile and electrophile ((±)-2-iodooctane). 2. Identify the leaving group (I-). 3. Determine the site of nucleophilic attack (carbon atom bonded to iodine). 4. Predict the reaction product(s) based on the SN2 mechanism, considering an inversion of stereochemistry at the carbon center. 5. Consider possible side products and reaction stereochemistry, taking into account the presence of both R and S enantiomers in the starting material (±)-2-iodooctane.

Step by step solution

01

Identify the nucleophiles and electrophile

First, we need to identify the nucleophiles given in this exercise. Unfortunately, they are not provided. However, we can assume that we have a set of typical nucleophiles, such as hydroxide (OH-) or cyanide (CN-), for example. The electrophile is provided in the form of (±)-2-iodooctane.
02

Identify the leaving group

In this reaction, the leaving group is the iodide ion (I-), which will be released once the nucleophile attacks the electrophile.
03

Determine the attack site on the electrophile

The site of nucleophilic attack in this reaction will be the carbon atom bonded to the iodine atom. This is because the nucleophile will have to displace the leaving group (I-) in order to form the product.
04

Predict the reaction products based on SN2 mechanism

When a nucleophile attacks (±)-2-iodooctane, the SN2 mechanism will result in an inversion of the stereochemistry at the carbon center, which was previously bonded to iodine. Therefore, the product will have the opposite configuration (R or S) to the starting material. Predict the product(s) for each of the nucleophiles given, considering their specific reactions with (±)-2-iodooctane. (Note: Since the nucleophiles themselves are not provided, we cannot give specific examples here.)
05

Consider possible side products and reaction stereochemistry

Since some side products may be possible, always consider them in your analysis. However, for the purpose of this exercise focusing on SN2 reactions, we will not delve into side product prediction. Furthermore, be aware that the stereochemistry of the starting material, (±)-2-iodooctane, indicates presence of both R and S enantiomers. Thus, reaction with nucleophiles will yield diastereomeric products with unique stereochemistry depending on the starting enantiomer.

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Most popular questions from this chapter

(a) What product would be formed if the reaction of cis- 4-bromocyclohexanol with \(\mathrm{OH}^{-}\) proceeded with inversion? (b) Without inversion? (c) Is it always necessary to use optically active compounds to study the stereochemistry of substitution reactions?

Each of the following might have been synthesized by an \(\mathrm{S}_{\mathrm{N}} 2\) reaction. Suggest a combination of substrate and nucleophile which could have led to their production.

(a) In the liquid form, tert-buty1 fluoride and isopropyl fluoride gave the following nmr spectra; tert-butyl fluoride: doublet, \(\delta 1.30, \mathrm{~J}=20 \mathrm{~Hz}\) isopropyl fluoride: two doublets, \(\delta 1,23,6 \mathrm{H}\), \(\mathrm{J}=23 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) two multiplets, \(\delta 4.64,1 \mathrm{H}\) \(\mathrm{J}=48 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) How do you account for each of these spectra? (b) When the alkyl fluorides were dissolved in liquid \(\mathrm{SBF}_{5}\), the following nmr spectra were obtained* tert-butyl fluoride: singlet, \(\delta 4.35\) isopropyl fluoride: doublet, \(\delta 5.06,6 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) multiplet, \(\delta 13.5,1 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) To what molecule is each of these spectra due? (Hint: What does the disappearance of just half the peaks observed in part (a) suggest?) Is the very large downfield shift what you might have expected for molecules like these?

Explain why \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+}\) is neither a nucleophile nor an electrophile.

(a) Draw the structures of ethyl, n-propy1, isobutyl, and neopentyl bromides. These structures can be considered methyl bromide with one of its hydrogens replaced by various alky1 groups \(\left(\mathrm{GGH}_{2} \mathrm{Br}\right)\). What is the group \(\mathrm{G}\) in each case? (b) The relative rates of reaction (with ethoxide ion) are roughly: methyl bromide, \(100 ;\) ethyl bromide, \(6 ; \mathrm{n}\) -propyl bromide, \(2 ;\) isobutyl bromide, \(0.2 ;\) neopentyl bromide, \(0.00002\). What is the effect of the size of the group \(G\) attached to carbon bearing the halogen?
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