CHEM 2211 Lecture Notes - Lecture 2: Leaving Group, Nucleophilic Aromatic Substitution, Silver Nitrate

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burgundymoose342

27 Feb 2018

School

University of Georgia

Department

Chemistry

Course

CHEM 2211

Professor

Hubbard

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Document Summary

The purpose of this lab is to investigate how the identity of the leaving group, and the structure of the electrophile, and the polarity of the solvent system each affect both types of substitution reactions. The first portion of the experiment focuses on the effect of substrate substitution and leaving group identity under sn1 conditions. The second part of the experiment deals with the effect of substrate substitution and leaving group identity under sn2 conditions. The third part of the experiment deals with solvent polarity as it effects sn1 reactions. Nucleophilic substitution is characterized by the attack of a nucleophile on an electrophilic species and results in the displacement of the leaving group. The two mechanisms through which this can occur is unimolecular nucleophilic substitution and bimolecular nucleophilic substitution. A bimolecular reaction occurs in a single step while a unimolecular reaction occurs in two steps.

Related Questions

Nucleophilic Substitution â Synthesis of an Alkyl Halide

Introduction

Alkyl halides can be prepared from alcohols by reactions with hydrogen halides (HCl, HBr, or HI) via nucleophilic substitution. In this type of reaction, the nucleophile displaces a leaving group from a carbon atom of an organic substrate (here the alcohol once protonated). Both electrons of the new bond to the carbon are provided by the nucleophile while the leaving group departs with both electrons of its bond to the substrate. There are two limiting mechanisms that are most common in describing nucleophilic substitution reactions at sp3 carbon centres : (1) a direct displacement (SN2) process and (2) a process that generates an intermediate carbocation (SN1). Which mechanism is a better description depends on the structure of the molecule containing the leaving group, the nature of the leaving group, the size of the nucleophile, and the polarity of the solvent. This experiment is a near-perfect example of an SN1 substitution. H3C OH CH3 H3C + HCl H3C Cl CH3 H3C + H2O (Equation 1) In this reaction, you will prepare 2-chloro-2-methylpropane (t-butyl chloride) from 2- methyl-2-propanol (t-butyl alcohol) using HCl as the strong acid and chloride source. Note that HCl first activates the alcohol in an acid/base reaction before the SN1 reaction proceeds. The final product is characterized by its yield, boiling point, and density. The presence of a halide atom is confirmed with a silver nitrate test.

1) State which of the electrophiles given below will react preferentially by i) SN1, ii) by SN2, or iii) capable of reacting by either of the two mechanisms depending on the given conditions. How can you affect those conditions to favour SN1 or SN2? Reason your predictions based on the structures of the compounds. Br-CH3, Br-CH2CH3, Br-CH(CH3)2, Br-C(CH3)3, Br-CH2-C5H6; C5H6 = phenyl

2) Rank the nucleophiles given below based on high, mediocre, and low nucleophilicity and reason why based on their structures. HO-CH3, H2N-CH3, KO-CH3, KO-C(CH3)3

3) Name two solvents that are commonly used for SN2 reactions. because they considerably slow down SN1 reactions.

4) Can you ever have only SN2 or only SN1?

5) Iodine is a better leaving group than bromine. But iodine is a better nucleophile than bromine. Why is that?

6) Using orbitals, draw the transition state of the SN2 reaction between sodium cyanide and bromomethane. Which orbitals are interacting with which? Why is the bond breaking? Why is the orientation important? Is there another angle besides backside attack that could theoretically work based on orbital geometry? Why doesnât this happen?

7) Name four reagents you can use to dry solvent. Why do we use magnesium sulfate in this lab instead of them?

8) Draw a mechanism with all the proper arrows for the reaction from this experiment.

9) What are we removing with the water wash? Why donât we just add the sodium bicarbonate directly to the reaction mixture instead of doing a water wash?

Questions:

1) In the tests with sodium iodide in acetone and silver nitrate in ethanol, why should 2-bromobutane react faster than 2-chlorobutane?

2) Why is benzyl chloride reactive in both tests, whereas bromobenzene is unreactive?

3) When benzyl chloride is treated with sodium iodide in acetone, it reacts much faster than 1-chlorobutane, even though both compounds are primary alkyl chlorides. Explain this rate difference.

4) 2-Chlorobutane reacts much more slowly than 2-chloro-2-methylpropane in the silver nitrate test. Explain this difference in reactivity.

5) Bromocyclopentane is more reactive than bromocyclohexane when heated with sodium iodide in acetone. Explain this difference in reactivity.

CHEM 211 Biological Chemistry Lab #8-Factors effecting the reactions of alkylhalides A. Experimental objectives Examine the effects of leaving group, alkyl group structure and solvent polarity on the reaction of alkyl halides Examine the effects of solvent polarity on the reaction of a tertiary alkyl halide with sodium hydroxide To examine SN1 and SN2 reaction conditions To learn the most two important factors that effects the reactivities of alkyl halides in nucleophilic substitution reaction B. Introduction The reactivities of alkyl halides in nucleophilic substitution reactions depend on two important factors: reaction conditions and substrate structure. The reactivities of the several substrate types will be examined under both SN1 and SN2 reaction conditions in this experiment A reagent composed of sodium iodide or potassium iodide dissolved in acetone is useful in classifying alkyl halides according to their reactivity in an SN2 reaction. Iodide is an excellent nucleophile, and acetone is a nonpolar solvent. Thetendency to form a precipitate increases the completeness of the reaction. Sodiumiodide and potassium iodide are soluble in acetone, but the corresponding bromidesand chlorides are not soluble. Consequently, as bromide ion or chloride ion is produced, the ion is precipitated from the solution. According to LeChatelier' s Principle, the precipitation of a product from the reaction solution drives the equilibrium toward the right such is the case in the reaction described here R-Cl Na RI NaCl (s) R-Br Na RI-NaBr (s) A reagent composed of silver nitrate dissolved in ethanol is useful in classifying alkyl halides according to their reactivity in an SN1 reaction. Nitrate ion is a poor nucleophile and ethanol is a moderately powerful ionizing solvent. The silver ion, because of its ability to coordinate the leaving halide ion to form a silver halide precipitate ,greatly assists the ionization of the alkyl halide. Again a precipitate as one of the reaction products also enhances the reaction C2H5OH R OC2H5 CI Ag AgCl (s) Page 1 of 5

CHEM 2211 Lecture Notes - Lecture 2: Leaving Group, Nucleophilic Aromatic Substitution, Silver Nitrate

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