Log in
Chemistry
1
answer
0
watching
584
views
6 Nov 2019

How many of the following responses are true?

1. SN2 leads to a racemization of sterechemistry due to the planar carbocation intermediate
2. A primary alkyl halide will tend to react through an SN2 mechanism because there is little hindrance to backside attack and a primary carbocation is not very stable
3. 1-bromopentane will undergo substitution through an SN1 mechanism
4. If R-3-bromo-3-methylhexane is allowed to react with OH- the product will be a 50%/50% mixture of R-3-methyl-3-hexanol and S-3-methyl-3-hexanol
5. SN1 leads to a racemization of sterechemistry due to the nucleophile attacking from the backside and the leaving group leaving

Watch
1
answer
0
watching
584
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Collen Von
Collen VonLv2
9 Oct 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Basic Chemistry
5 Edition,
Timberlake
ISBN: 9780134138046
Principles of Chemistry Molecular Approach
4th Edition,
Tro
ISBN: 9780134112831
Chemistry: Structure and Properties
2nd Edition,
Tro
ISBN: 9780134293936
Principles of Chemistry Molecular Approach
3rd Edition, 2014
Tro
ISBN: 9780321971944
Chemistry: A Molecular Approach
3rd Edition,
Tro
ISBN: 9780321809247
Chemistry: A Molecular Approach
5th Edition,
Tro
ISBN: 9780134874371
Principles of Chemistry: A Molecular Approach
4th Edition,
Tro
ISBN: 9780134895741
Chemistry: The Central Science
14th Edition, 2017
Brown
ISBN: 9780134414232

Related questions

Nucleophilic Substitution – 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 doesn’t 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 don’t 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.

47

EXPERIMENT

9

Williamson Ether Synthesis

Reading and Preparation

TaskSource

S

-

115

Safety

Chemwatch

(website)

Background

In today’s 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

in

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

C

H

I

H

H

+

C

H

3

C

O

H

3

C

H

3

C

C

H

H

H

C

H

3

C

O

H

3

C

H

3

C

K

+

KI

HOC(CH

3

)

3

C

H

H

H

+

I

C

H

I

H

H

C

H

I

H

H

C

H

3

C

O

H

3

C

H

3

C

C

H

3

C

O

H

3

C

H

3

C

C

H

H

H

+

I

48

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

S

N

1 mechanism. If the reaction follows second

order kinetics, then the reaction occurs by the S

N

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

3

C)

3

CO

̄

+ H

2

O

→

(H

3

C)

3

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

:

1.

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

2

SO

4

solution to get the correct sto

ck

concentration of

tert

-

butoxide

. Note also this is

not

the concentration of the (H

3

C)

3

COK at

time zero in the reaction mixture

–

the (H

3

C)

3

COK solution has been mixed with the CH

3

I

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

o be done to get the

(H

3

C)

3

COK concentration in the reaction mixture at time zero.

2.

The concentration stated on the bottle for the stock CH

3

I solution can be used but again note

that the time zero concentration of the CH

3

I in the reaction mixture is not th

is value.

Remember the CH

3

I solution is mixed with (H

3

C)

3

COK solution in the reaction flask so an

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

CH

3

I.

Record the initial concentrations of your two solutions and

the standard sulfuric acid as

shown below and d

raw

a

table in your laboratory notebook

or prepare an excel

spreadsheet

.

Initial Concentration of (H

3

C)

3

COK in (H

3

C)

3

COH

Initial Concentration of CH

3

I in (H

3

C)

3

COH

Concentration of standardised H

2

SO

4

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

3

I]

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

A plot of time versus log ([(H

3

C)

3

CO

̄

] / [CH

3

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

order.

49

Experimental

*

Note that the melting point of (H

3

C)

3

COH is 23

o

C

–

26

o

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

o

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

1.

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.

2.

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

3.

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

interval. Show the formulae used for each calculation

Weekly leaderboard

Home
Homework Help 3,900,000
Chemistry 830,000
Start filling in the gaps now
Log in