Chemistry Lab Report 06 (Long)

Chemistry Lab Report 06

Name : Lam Wai Oi

Class : 7S

Class No : 04

Date : 17th Dec, 07

Title : A practical study of some group Ⅱ elements

Objective : To study some of the properties of the elements of Group Ⅱ

and their compounds.

Theory :

The s-block of the periodic table contains the most reactive and, in

chemical terms, the most typically metallic elements. All the elements in

Group Ⅰ are highly reactive, but those in Group Ⅱ are slightly less so and

show a rather more obvious trend in reactivity.

Procedure :

A. Reaction of the elements with water

1. Put a small piece of calcium metal into a large beaker of cold

water.

2. Observed the reaction and identify the products.

3. Repeated step 1 and 2 by using a small piece of clean magnesium

ribbon and then barium metal.

4. The reaction of magnesium with water was very slow.

Investigated the reaction further by setting up the experiment

shown in figure.

B. Acid-base character

1. Placed about 0.01g of magnesium oxide, calcium hydroxide and

barium hydroxide in three separate test tubes.

2. Added 10cm3 distilled water to each tube and shake.

3. Added 2 drops of universal indicator solution to each tube and

mix.

4. Recorded the pH values indicated for the three tubes.

C. Hydrolysis of the chlorides

1. Strongly heated about 1cm depth of each of the hydrated

chlorides of magnesium, calcium and barium in separate, dry

hard-glass test tubes in the fume cupboard.

2. Tested for the evolution of hydrogen chloride.

D. Thermal stability of the carbonates

1. Strongly heated about 1cm depth of each of the dry carbonates of

magnesium, calcium and barium separately in the apparatus

shown in figure.

2. Noted how rapidly gas is evolved, and the extent to which the

lime-water becomes milky.

3. Removed the tube from the lime-water before heating has

stopped.

4. Recorded the time that lime-water became milky.

E. Solubility of some compounds of Group Ⅱ elements

1. Put 2cm3 of a 0.1M solution of magnesium chloride, calcium

chloride and barium chloride under investigation in separate test

tubes.

2. Added an equal volume of a 1.0M solution of sodium hydroxides,

and mix.

3. Noted whether a precipitate was formed, if so, how dense it was.

4. Repeated the experiment twice, using first a 1.0M solution of

sodium sulphate and then a 1.0M solution of sodium carbonate,

instead of the sodium hydroxide.

Results :


Element

s

Fume produced Gases evolved Solution changed

Ca(s) white colourless became milky

Mg(s) - - -

Mg(powder) - - -


Elements pH value

MgO 7

Ca(OH)2 10

Ba(OH)2 10


Elements Had gas evolved or not

MgCl2 No

CaCl2 No

BaCl2 No


Elements Time of lime-water turned to milky

MgCO3 55s

CaCO3 97s

BaCO3 -


Elements Mg2+ Ca2+ Ba2+

OH- Translucent solid

occurred

White precipitate

occurred, small

colourless layer on

the above

Little amount of

translucent

precipitate occurred

SO42- No observation White precipitate

occurred, colourless

layer on the above

Had white

precipitate

CO32- Translucent

precipitate on the

above

Had white solid Large amount of

white precipitate

occurred

Discussion :


Those two metals, calcium and magnesium react with water

theoretically.

Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)

Mg(s) + H2O(l) → MgO(s) + H2(g)

However, even reacted with hot water, magnesium had no observable

changes in the reaction. The reaction may undergo too slow. But

calcium reacted with water readily and gave large amount of white fume.

This indicates that calcium is more reactive than magnesium.

In this experience, calcium and magnesium acted as reducing

agents. The hydrogen ions in water were reduced by the Group Ⅱ

elements became hydrogen gas.





Reference : http://tonny.earthppl.com/att/Chemistry%20Report%205.doc

http://www.rod.beavon.clara.net/bromobutane_prep.htm

http://chemistry.csudh.edu/faculty/jim/Preparation%20of

%201-bromobutane.doc

http://www.miracosta.cc.ca.us/home/dlr/210exp8.htm

http://tonny.earthppl.com/att/Chemistry%20Report%205.doc

In the experiment, 1-bromobutane was synthesized through the second order

mechanism in the present of concentrated sulphuric acid and sodium bromide. The

nucleophile in the experiment is bromide ion (Br-) while the leaving group is water.

Purification of the product was then obtained by extraction. The principle of

extraction as a purification method was based on the difference in solubility between

impurities and product.

The functional group inter-conversion of an alcohol into alkyl halide takes place by a

nucleophilic substitution reaction. In the experiment, the alkyl group was substituted

on the primary carbon, the reaction proceeds by an SN2 mechanism.

The primary alcohol, 1-butanol, was reacted with a mixture of sodium bromide and

the concentrated sulfuric acid and hooked up to a water-cooled reflux condenser. The

hydroxyl group of the 1-butanol was protonated by the sulphuric acid. Addition of the

strong acid formed an oxonium ion from the 1° alcohol OH group, changing a poor

leaving group (-OH) into a good one (H-O-H). The nucleophile, bromide, attacked the

primary carbon and water was released, forming 1-bromobutane. The water that

produced was then protonated, forming hydronium ion.

1-butanol sulfuric acid protonated alcohol

When sulphuric acid was added into the reaction mixture, cooling by means of an ice

bath was needed. The apparatus must also be swirled at the same time to ensure the

acid had reacted. This was to prevent the sulphuric acid from reacting too fast, as it

was an exothermic reaction, and might run away splashing the acid everywhere if the

reaction went too fast. Also the increased temperature might vaporise more quickly

the 1-butanol, which would decrease the yield. Also, hot sulphuric acid would cause

significant oxidation of the sodium bromide to bromine, which is useless in this

experiment. The yield of 1-bromobutane could therefore be affected adversely.

The need for reflux was because the reaction needed to go to completion, and as with

organic liquids it was often necessary to reflux because heat needed to be applied to

the reaction without loosing reactants through evaporation.

The reason for the multiple layers of products was that there were, several side

reactions taking place at the same time in the flask, they are the drawbacks to using

sulphuric acid. The acid could cause reactions such as the dehydration of alcohols

and/or ether formation to occur which lower the yields of the alkyl bromides.

(I) Sodium bromide reacted with sulphuric acid to form hydrogen bromide and

sodium hydrogen sulphate.

NaBr + H2SO4 → HBr + NaHSO4

(II) Hydrogen bromide was oxidised to bromine molecules as concentrated sulphuric

acid was a very good oxidising agent. The sulphuric acid reacted to form sulphur

dioxide gas.

HBr + H2 → SO4Br2 + 2SO2 (g)

(III) Hydrogen bromide dissociated and the bromide ion from it attacked the carbon

atom with the -OH function group in 1-butanol and displaced the -OH function group

forming a bromo function group and a hydroxide ion, which then associated itself

with another H3O+ ion to form water.

CH3CH2CH2CH2OH + Br- → CH3CH3CH2CH2Br + OH-

(IV) A molecule of sulphuric acid attacked the lone pair on an -OH function group,

releasing a molecule of water, and a mixture of butoxybutane and but-1-ene was

formed, along with the regenerated sulphuric acid.

CH3CH2CH2CH2OH + H2SO4 → CH3CH2CH=CH2 + H2O + H2SO4

2 CH3CH3CH2CH2OH + H2SO4 → CH3(CH2)3O(CH2)3CH3 + H2O + H2SO4

To identify the organic layer in the separatory funnel, carefully draw off about 0.25

mL of the lower layer into a test tube containing about 1 mL of water and mix

vigorously. If the liquid from the funnel dissolves in the water as you mix it, the lower

layer is aqueous.

1-butanol went through an SN2 reaction by using sodium bromide in water to create

hydrobromic acid in the presence of excess sulfuric acid. Thus, water was the leaving

group and bromide ion was the nucleophile in this in situ reaction. This displaced the

alcohols -OH group with a bromine creating 1-bromobutane. The excess sulfuric acid

pushed the reaction far to the right corresponding with LeChatlier’s principle which

states that more reactants yield more products.

The extractions further isolated the product, the substance which has higher density

was always the bottom layer and the lighter was on top. It turned out that the organic

layer was always on the bottom of each extraction. Anhydrous magnesium sulphate

was a very good drying solution. This allowed it to soak up the remaining water to a

great extent.

This reaction must be a SN2 reaction as the product was almost entirely 1-

bromobutane, if by an SN1 mechanism, the product should consist of mostly 2-

bromobutane derived from the sec-butyl cation formed via a 1,2-hydride shift.

Such an interpretation of the product distribution must be made with caution. The

presence of small amounts of 2-bromobutane in the reaction product did not

necessarily mean that it was produced from 1-butanol via an SN1 mechanism because

an alternative sequence of an E2 elimination of the protonated 1-butanol to give 1-

butene followed by HBr addition to the 1-butene would also produce 2-bromobutane.

Precautions were taken to prevent the highly inflammable and volatile alcohol from

catching fire or lost through evaporation.

The pressure of the separatory funnel must be periodically released to avoid the

stopper being pushed out and product being lost and sprayed. The pressure is due to

liberated carbon dioxide.

http://www.rod.beavon.clara.net/bromobutane_prep.htm

Why is the sulphuric acid added slowly, and why is

cooling and shaking needed?

Sulphuric acid when diluted with water gives out a great deal

of heat, enough sometimes to raise steam which would cause

dangerous splashing. Also, hot 50% sulphuric acid (which is

what is produced in the flask) will cause significant oxidation

of the sodium bromide to bromine, which is useless in this

experiment. The yield of 1-bromobutane could therefore be

affected adversely.

Why is a sand-bath used for heating?

The sand spreads the heating uniformly over the base of the

flask. This reduces the likelihood of cracking, and of unwanted

side reactions occurring (e.g. excessive oxidation either of

bromide ions to bromine or of the alcohol to carbon) owing to

hot-spots.

Why is the mixture heated for 45 minutes?

Most organic reactions are slow because of the need to break

strong covalent bonds.

Why is the mixture distilled at this stage?

The liquid 1-bromobutane is removed from the involatile

sodium salts (mostly sodium hydrogen sulphate at the end of

the reaction) and the much less volatile sulphuric acid. The 1-

bromobutane will be contaminated with water, unchanged

butan-1-ol, and some sulphuric acid carried over as tiny

droplets during the distillation.

What does shaking with water achieve?

Water will remove sulphuric acid and some of the butan-1-ol.

How do you decide which layer is to be kept?

On the basis of density; 1-bromobutane has a density of 1.276

g cm-3.

Why is concentrated hydrochloric acid added?

The acid protonates the butan-1-ol, giving an ionic species

that is much more soluble in water than the alcohol itself:

CH3CH2CH2CH2OH + H+ → CH3CH2CH2CH2OH2+

Why is the mixture shaken with sodium carbonate

solution?

This removes hydrochloric acid dissolved in the 1-

bromobutane:

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

Why must the pressure be periodically released?

To avoid the stopper being pushed out and product being lost

and sprayed all over you. The pressure is due to liberated

carbon dioxide.

What is the function of the calcium chloride?

Calcium chloride is a drying agent.

What if the mixture isn't clear?

Then it is not dry; the cloudiness is caused by tiny droplets of

water in the 1-bromobutane.

http://chemistry.csudh.edu/faculty/jim/Preparation%20of%201-bromobutane.doc

The overall reaction is:

H2SO4 + NaBr + CH3CH2CH2CH2OH CH3CH2CH2CH2Br + H2O + NaHSO4

http://www.miracosta.cc.ca.us/home/dlr/210exp8.htm

As you learned in lecture, this general class of reactions requires three things: a

nucleophile, an electrophile, and a leaving group. In order for the reaction to proceed

via the SN2 mechanism, the nucleophile should be in the "good" to "excellent" range,

the electrophile must be unhindered (methyl or 1°), and the leaving group should also

be in the "good" to "excellent" range.

CH3CH2CH2CH2-OH2+ + Br- → CH3CH2CH2CH2-Br + H2O

we see that the first two are satisfied (bromide is an excellent

nucleophile and the electrophile is a 1° alkyl group), but hydroxide

is a poor leaving group, due to its negative charge and its high

basicity.

The central question becomes, "how can we make OH- into a better leaving group?

We have a few choices: 1) react the alcohol with p-toluenesulfonyl chloride

(abbreviated TsCl), which will convert the —OH group into a sulfonic acid ester,

making it a much better leaving group--then react the ester with sodium bromide to

produce 1-bromobutane; 2) react the alcohol with phosphorus tribromide (PBr3),

which converts the —OH into a "—P(OH)X2" leaving group (where X is either —Br

or —OH), and which also produces free bromide ions which react with the

electrophile, replacing the new leaving group, all in one reaction mixture; or 3) using

a strong acid to protonate the —OH group in the presence of the bromide ion, which

changes the leaving group from hydroxide to water, and allows the bromide to react

with it in the same mixture.

Chemistry Lab Report 06 (Long)

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