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11 Dec 2019

Using the isothermal transformation diagram for a 0.45 wt% C steel alloy (Figure below), determine the final microstructure (in terms of just the microconstituents present) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 845C (1550F), and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure.

(a) Rapidly cool to 250C (480F), hold for 103 s, then quench to room temperature.

(b) Rapidly cool to 700C (1290F), hold for 30 s, then quench to room temperature.

(c) Rapidly cool to 400C (750F), hold for 500 s, then quench to room temperature.

(d) Rapidly cool to 700C (1290F), hold at this temperature for 105 s, then quench to room temperature.

(e) Rapidly cool to 650C (1200F), hold at this temperature for 3 s, rapidly cool to 400C (750F), hold for 10 s, then quench to room temperature.

(f) Rapidly cool to 450C (840F), hold for 10 s, then quench to room temperature.

(g) Rapidly cool to 625C (1155F), hold for 1 s, then quench to room temperature.

(h) Rapidly cool to 625C (1155F), hold at this temperature for 10 s, rapidly cool to 400C (750F), hold at this temperature for 5 s, then quench to room temperature.

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Related questions

Using the supplied isothermal transformation diagram for an iron–carbon alloy of eutectoid composition, specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time–temperature treatments. In each case assume that the specimen begins at 760°C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure.

Cool rapidly to 670°C, hold for 100 s, then observe at that instant and that temperature using a high temperature microscope or similar device

(a) 100% austenite

(b) 100% martensite

(c) 50% pearlite/50% austenite

(d) 50% pearlite/50% martensite

(e) bainite

Cool rapidly to 670°C, hold for 100 s, then quench to room temperature, and observe at room temperature

(a) 100% austenite

(b) 100% martensite

(c) 50% pearlite/50% austenite

(d) 50% pearlite/50% martensite

(e) bainite

Cool specimen to 700 °C and hold there for 100 hours

(a) Tempered martensite

(b) 100% spheroidite

(c) 100% coarse pearlite

(d) ) 50% pearlite/50% martensite

(e) austenite

Rapidly cool to 640 °C, hold for 10 s, rapidly cool to 350°C, hold for 1000 s, then quench to room temperature.

(a) 25% martensite/25% bainite/50% pearlite

(b) 50% martensite/50% bainite

(c) 100% bainite

(d) 50% pearlite/50% bainite

(e) mostly martensite

Rapidly cool (to room temperature) in less than 1 second.

(a) 25% martensite/25% bainite/50% pearlite

(b) 50% martensite/50% bainite

(c) 100% bainite

(d) 50% pearlite/50% bainite

(e) mostly martensite

Rank the microconstituents in order of increasing hardness

(a) Bainite, coarse pearlite, fine pearlite, martensite, spheroidite

(b) Coarse pearlite, bainite, spheroidite, fine pearlite, martensite

(c) Spheroidite, coarse pearlite, fine pearlite, bainite, martensite

(d) Martensite, coarse pearlite, bainite, fine pearlite,

(e) Spheroidite, martensite, coarse pearlite, bainite, ice cream

Which of the following statements about untempered (ie. freshly produced) martensite is false:

(a) it is very hard and well suited for use in tooling and other high strength applications

(b) it is normally too hard for safe use in any application

(c) it needs to be tempered to a lower hardness as soon as possible

(d) it is full of elastic strain and is very brittle

(e) its hardness increases as carbon content increases

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

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