CHM 25500 Lecture Notes - Lecture 31: Electrophile, Nucleophilic Substitution, Nucleophile

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Chm 255 - lecture 31 - substitution vs. Substitution and elimination reactions often occur at the same time and compete with one another. The ratio of substitution versus elimination products therefore depends on the relative rates of the reactions. There are several factors that determine which reactions (substitution or elimination or both) will occur as well as the relative rates of the reactions. Methyl halides only go through substitution because they do not have beta carbons or beta hydrogens. Primary alkyl halides only go through sn2 reactions and e2 reactions, e2 reactions with primary alkyl halides only taking place in the presence of a strong base and heat. For secondary and tertiary alkyl halides, the type of reaction is determined by other factors. If the nucleophile is also a strong base, elimination reactions are favored. If the nucleophile is not a strong base, then mostly substitution reactions will occur, however there may be some elimination reactions occurring.

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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?

Â¥Write a brief (less than 150 words, 3 or 4 sentences) of the aim of the experiment

Â¥Address question of what was done, how it was done, and what compounds were used.

Â¥Include Chemical Reaction where relevant.

See text below. Please provide Purpose following the steps above. Thank you.

Alcohols R-O-H are characterized by the presence of the hydroxyl group

Because of intermolecular co-association, alcohols show higher boiling points

than aldehydes, ethers, or ketones of similar mass. Among alcohols, boiling

points increase with increased surface area. Increased surface area occurs as

more carbons are added to the alkyl portion or with decreased branching of th

alkyl group. Thus, 1-propanol has a higher boiling point than ethanol

(increased C#). 1-propanol also has a higher boiling point than 2-propanol

(decreased branching)

The solubility of alcohols in water decreases as the surface area of the

alkyl (nonpolar) portion increases. Thus, alcohols become less soluble in water

as the number of carbons in the alkyl group increases. When comparing

alcohols that contain the same number of carbons, branching of the alkyl group

decreases its surface area and thus increases the water solubility of the alcohol.

1-Propanol is slightly soluble in water while 2-propanol is quite soluble

The limit of water solubility is reached at about 4 or 5 carbons in the

alkyl group. However, it is possible to "salt out" or separate a water-soluble

alcohol from the water it is mixed with by the addition of certain salts such as

K2CO3. If an alcohol-water mixture is saturated with such a salt, the anion

selectively forms strong ion-dipole bonds to water, leaving the alcohol as a

separate layer. Thus, water is separated from the alcohol because the ionic salt

bonds more strongly with water than with the alcohol. There are twice as many

ion-dipole bonds possible with water than with the alcohol.

Alcohols are relatively acidic organic molecules, being weaker acids than

water, yet stronger than acetylene.

Acidity: water > alcohol > acetylene > ammonia > alkanes

Alcohols may be converted to sodium alkoxides by the action of metallic sodium

under anhydrous conditions. Primary alcohols react more readily tharn

secondary or tertiary alcohols due to the relative stability of the alkoxide ions in

the solvent alcohol.

R-o:-

Alcohols undergo nucleophilic substitution reactions with ease under

acidic conditions. The relative ease of such substitution for primary, secondary

and tertiary alcohols serves as a means of classification. Upon treatment with

concentrated HC1, tertiary alcohols are rapidly converted to alkyl chlorides

Secondary alcohols require warming in a boiling water bath for 15 minutes

Primary alcohols do not react at all under these conditions. The alkyl chlorides

formed by this reaction are insoluble in the concentrated HCl and form an

easily detectable second layer

R-O-H HC1 >

Ð-Ð-Ð

Complications arise from the fact that benzylic and allylic alcohols also

react immediately, but yield products that are soluble in the concentrated HCI

Alcohols are easily oxidized by any of these strong oxidizing agents:

dilute KMnO4, NajCr.07 in acid, or CrO3 (chromic anhydride)

acidified sodium dichromate is somewhat helpful in the classification of

alcohols. Prim

ary and secondary alcohols are oxidized rapidly within several

minutes, while tertiary alcohols do not react under these conditions

Oxidation occurs more easily when the carbon bearing the -OH group

bonded to at least one atom of hydrogen. The complex mechanism of

involves the transfer of a hydride ion H:- as a critical step.

hydrogen atom is absent (as in tertiary alcohols), no oxidation can occur

directly. Tertiary alcohols may be oxidized indirectly through the formation of

Alcohols which contain the partial structure C- CH3 give positive

Ð¾Ð½

results in the iodofora reaction. When the alcohol is dissolved in water and

KOH, the addition of a solution of 12 in KI yields a yellow precipitate of

iodoform, CHI3, along with the oxidized portion of the alcohol

Hypoiodite ion is produced by the reaction between molecular iodine and

hydroxide ion:

I2 + 4(OH) -

The hypoiodite ion in turn oxidizes an alcohol with the proper structure

Once this oxidation has occurred, additional hypoiodite ion continues to

react with the aldehyde or ketone produced:

CH,-C

3 (0I)-IC-c3 (OH)-

resulting in the final substitution reaction to give iodoform and the anion of a

carboxylic acid containing one carbon atom less than the original alcohol:

128

lodoform precipitates as fine yellow crystals. This precipitate, a result of a

specific oxidation followed by a substitution, is an indication of the partial

structure of the original alcohol.

Simple alcohols of up to ten carbon atoms may be detected by the

ammonium hexanitrocerrata(lV) reagent, (NH4Ce(NO,). The hexanitro-

cerrate(IV) ion has a distinctive orange-red color. When an alcohol, ROH is

added, one or more OR- groups replace the nitro groups as ligands in the

complex ion. The new complex ion formed has a distinctive color. The general

equation for this reaction is:

This test is positive for aliphatic alcohols up to approximately ten carbons in

length. Phenols or larger alcohols yield insoluble mixtures

Alcohols react with acids or with acid derivatives to form esters. This

characteristic reaction is useful in the formation of specific esters such as ethyl

acetate, acetylsalicylic acid (aspirin) and methyl salicylate (oil of wintergreen). In

this experiment, methyl salicylate will be prepared by reacting methyl alcohol

with salicylic acid in the presence of a trace of sulfuric acid as the catalyst

Lower molecular weight alcohols such as ethanol, methanol, and

isopropyi alcohol absorb water on exposure to moist air or are prepared as

azeotropic mixtures containing water. The presence of water in alcohols may be

detected in several ways. If a small lump of CaC2 is added to any alcohol that

contains a trace of water, a reaction occurs that produces immediate bubbles of

acetylene gas:

Thus, evolution of acetylene gas upon the addition of calcium carbide indicates

the presence of water in the alcohol. Another way to test for the presence of

water in alcohols involves the use of anhydrous cupric sulfate powder (CuSO4)

Anhydrous cupric sulfate is a very pale, almost white powder. In the presence of

water, deep blue crystals of cupric sulfate pentahydrate,

CuSO4-5H20, are formed. Thus, anhydrous cupric sulfate indicates the

presence of water in an alcohol by a simple color change.

Commercial ethanol is produced as an azeotropic mixture that contains

95% ethanol and 5% water. Absolute alcohol contains no water. Many reagent

grade alcohols are composed of a nonpotable mixture of 95% ethanol and 5%

isopropyl alcohol. Thus, absolute ethanol and reagent alcohols are both water

free and give negative results when tested for the presence of water

EXPERIMENT

Williamson Ether Synthesis

Reading and Preparation

TaskSource

115

Safety

Chemwatch

(website)

Background

In todayâs laboratory class you will preparetert

-butyl methyl ether from potassiumtert

butoxide

and iodomethane using the Williamson ether synthesis according to Scheme 1. This is classified

as a substitution reaction of an alkyl halide, where iodine

iodomethane is replaced by a

tert

butoxide group.

Scheme 1

Substitution reactions of alkyl halides can occur by either an SN1 or S

N2 mechanism. In the SN1process, the rate-determining step in the reaction is breaking of the carbon-halogen bond and the

formation of a carbocation intermediate (Scheme 2).Scheme 2Such reactions follow first order kinetics because the rate of the reaction only depends on the

decomposition of one molecule and is therefore only affected by the concentration of the alkylhalide. In the SN2 mechanism,

the rate-determining step is the displacement of the halide by an attacking nucleophile (Scheme 3).

Scheme 3

This reaction follows second order kinetics because the rate depends on the interaction between

two molecules and thus the concentration of both the alkyl halide and thetert

butoxide

nucleophile. Nucleophilic substitution reactions are discussed in McMurry,

Chapter 11, pages

HOC(CH

)

372

395. You will follow the progress of the reaction between potassium

tert

butoxide and

iodomethane in order to determine the mechanism of this reaction. If the reaction follows first

order kinetics, then the reaction occurs by the

1 mechanism. If the reaction follows second

order kinetics, then the reaction occurs by the S

2 mechanism.

The progress of the reaction is monitored by following the change in concentration of the

reactants over time. For this experiment, aliquots of the reaction mixture are withdrawn every

10 minutes for the first hour and every 20 minutes for the second hour

. The aliquots of reaction

mixture are placed in cold water to stop the reaction. The unreacted

tert

butoxide ions present

reacts with water to liberate hydroxide ions (Scheme 4).

Scheme 4

+ H

COH + HO

Hydroxide ion is then t

itrated using standardised sulfuric acid and the concentration of the

tert

butoxide ion and iodomethane for each time can be calculated.

Notes

The concentration of the

potassium

tert

butoxide

stated on the stock bottle

cannot be relied

upon as the

solution begins decomposing once prepared. For this reason you will need to

take a 1.00 mL aliquot of the stock solution and quench this in cold water. This solution

will then need to be titrated against the standard H

solution to get the correct sto

concentration of

tert

butoxide

. Note also this is

not

the concentration of the (H

COK at

time zero in the reaction mixture

the (H

COK solution has been mixed with the CH

solution in the reaction flask. Thus an appropriate calculation needs t

o be done to get the

COK concentration in the reaction mixture at time zero.

The concentration stated on the bottle for the stock CH

I solution can be used but again note

that the time zero concentration of the CH

I in the reaction mixture is not th

is value.

Remember the CH

I solution is mixed with (H

COK solution in the reaction flask so an

appropriate calculation needs to be carried out to get the time zero concentration for the

Record the initial concentrations of your two solutions and

the standard sulfuric acid as

shown below and d

raw

table in your laboratory notebook

or prepare an excel

spreadsheet

Initial Concentration of (H

COK in (H

COH

Initial Concentration of CH

I in (H

COH

Concentration of standardised H

You will use the data calculated in the table to draw 2 graphs. A plot or time versus log [CH

will give a straight line if the reaction is first order.

A plot of time versus log ([(H

] / [CH

I]) will give a straight line if the reaction is second

order.

Experimental

Note that the melting point of (H

COH is 23

C. It may solidify if the room

temperature is cool, so keep all solutions warm.

Work in pairs for this experiment.

In separate conical flasks, place iodomethane in

tert

butanol (200.0 mL) and potassiumtert

butoxide intert

butanol (20.0 mL). If not already warmed, place the separate solutions in a water

bath (~40

C) for a few minutes. Start

timing on

the stop

watch as you quickly combine the

solutions, swirling to mix them thoroughly. Place the reaction vessel in the water bath to

maintain constant temperature. Prepare a conical flask containing cold water (20.0 mL). After

10 minutes pipette an aliquot of th

e reaction mixture (10.0 mL) and run it into the cold water to

quench the reaction. Later you can add phenolphthalein indicator and titrate the liberated

hydroxide ion using the standardised sulfuric acid. After removal of each aliquot the pipette

should

be thoroughly rinsed with water and acetone. The pipette can be dried if necessary by

blowing air through it using the house compressed air and appropriate tubing. Remove aliquots

every 10 minutes for the first hour and every 20 minutes for the second h

our. Write the sulfuric

acid titres in your table, carry out all necessary calculations and plot the graphs.

Questions

By what mechanism would you expect the reaction between potassium

tert

butoxide

and iodomethane to proceed? Explain your answer and

write the curved arrow

mechanism.

Do your results support your prediction? If not, offer some possible explanations.

Provide a sample calculation of each of the cells in the spreadsheet for one time

interval. Show the formulae used for each calculation

CHM 25500 Lecture Notes - Lecture 31: Electrophile, Nucleophilic Substitution, Nucleophile

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