HERMIN HARDYANTI UTAMI (111 304 0202)

CHEMISTRY DEPARTMENT

MATHEMATIC AND SCIENCE FACULTY

STATE UNIVERSITY OF MAKASSAR

2012

ORGANIC CHEMISTRY 2

COMPLETE REPORT

RATIFICATION PAGE

The complete report of Organic Chemistry II with the title of ―Synthetic

of Iodoform‖ which made by :

Name : Hermin Hardyanti Utami

ID : 111 304 0202

Group : VII

Class : ICP B

Have been checked by assistant and assistant coordinator. So, this report is

accepted.

Makkassar, December 2012

Assistant Coordinator

Fandi Ahmad, S.Pd.

Assistant

Sabaruddin

Known By,

Responsibility Lecturer

Dra. Hj. Ramdani, M.Si.

A. Tittle of Experiment

Synthetic of Iodoform

B. Purposes

At the last experiment, student can be understand about :

1. Work principles and crystallization technique of solid organic substance.

2. Haloform reaction.

3. Advantage of haloform reaction fo haloform synthetic and carboxylic acid

and to showing there are a methyl group and secondary alcohol.

C. Background

Functional group transformations allow the conversion of a functional group

to transformations an aldehyde or a ketone without affecting the carbon skeleton

of the molecule. Aldehydes can be synthesized by the oxidation of primary

alcohols, or by the reduction of esters, acid chlorides, or nitriles. Since nitriles can

be obtained from alkyl halides, this is a way of adding an aldehyde unit (CHO) to

an alkyl halide. Ketones can be synthesized by the oxidation of secondary

alcohols. Methyl ketones can be synthesized from terminal alkynes (Patrick,

2004:167).

According to Rasyid (2009:136-137) alcohol with one hydrogen that chain at

carbon as bring of carboxyl group can be oxidated become carbonyl compounds.

Primary alcohol that can be axidated with continue become to acid. Secondary

alcohol producing the ketone and tertiary alcohol is can’be oxidated:

H H OH

R C OH R C O R C O

H

Primary alcohol aldehyde acid

R’ R’

R C OH [O]

R C O

H

Secondary alcohol ketone

Most acidic protons are attached to heteroatoms like halogen, oxygen, and

nitrogen. Protons attached to carbon are not normally acidic but there are

exceptions. One such exception occurs with aldehydes or ketones when there is a

CHR2, CH2R or CH3 group next to the carbonyl group. The protons indicated are

acidic and are attached to the a (alpha) carbon. They are therefore called as

protons. A lone pair on the hydroxide oxygen forms a new bond to an proton.

Simultaneously, the C–H bond breaks. Both electrons of that bond end up on the

carbon atom and give it a lone pair of electrons and a negative charge (a

carbanion). However, carbanions are generally very reactive, unstable species that

are not easily formed. Therefore, some form of stabilisation is involved here

(Coulter, 2009:102 ).

According to Fescenden (1986:35-36) chaining of carbon hydrogen bonding

ussualy stable, nonpolar, and non acodic properties. But with there is one group

carbonyl happening alpha hidroden that has acid properties. If one hydrogen has

alpha position to two carbonyl group so its hydrogen is so acid so can be formed a

salt with sythetic of it compound with one alcogxyde. As for the reaction :

O O O

CH3 CCH3 CH3CCH2COCH2CH3

Acetone ethyl aceto acetic

pKa = 20 pKa = 11

According to Tim Doesen (2012:1) compound that contain the CH3CO group

or that produce this group, if occur the oxidation in one condition o one

experiment. Example acetaldehyde CH3CHO from ethanol CH3CH2OH react with

sodium hypochlorous forming iodofom.

O O

R C CH3 + 3NaOH R C CI3 + 3NaOH

O O

R C CH3 + NaOH R C O-Na

+ + CHI3

OH O

R C CH3 + NaOI R C CH3 + NaI + H2O

OH O

R C CH3 + 3NaOI R C O-Na

+ + CHI3 + H2O

According to Miller (1999:300) the radical initially produced by homolytic

decomposition of a dialkyl peroxide can undergo further scission. The rate of

scission depends on the temperature and the stability of the resulting radical. For

example, butoxy radicals decompose on heating to methyl radicals and acetone:

BuO• —> •CH3 +CH3COCH3

D. Chemical and Equipment

1. Chemical

a. Potassium Iodide (KI) solution

b. Acetone (CH3COCH3) solution

c. Sodium hypochlorous (NaOCl) solution

d. 1-propanol (CH3CH2CH2OH) solution

e. Dioxane (C4H8O2) solution

f. Sodium hydroxide (NaOH) 1% solution

g. Potassium Iodine Iodide (KI.I2) solution

h. Ethyl acetoacetic (CH3COH2COOC2H5) solution

i. Acetophenon (C6H5COCH3) solution

j. Aquadest (H2O)

k. Whatmann filter paper

l. Aluminium foil

m. Capillary tube

n. Matches

2. Equipment

a. Dropping pipette (10 pieces)

b. Electronical scale ( 1 piece )

c. Graduated cylinder 100 ml ( 1 piece )

d. Graduated cylinder 10 ml ( 2 pieces)

e. Stir bar ( 2 pieces)

f. Buchner funnel ( 1 piece )

g. Beaker glass 250 ml ( 2 pieces)

h. Oven ( 1 piece )

i. Erlenmeyer 500 ml ( 1 piece )

j. Thiele equipment ( 1 piece )

k. Bunsen burner ( 1 piece )

l. Test tube ( 6 pieces)

m. Test tube rack ( 1 piece )

n. Triangle ( 1 piece )

o. Asbestos gauze ( 1 piece )

p. Spray flask ( 1 piece )

q. Soft cloth and rough cloth ( 2 pieces)

E. Work Procedure

1. Synthetic of Iodoform

a. 6 g of KI was balance by electronical balance.

b. It was added by water 100 ml in beaker glass.

c. It was added by acetone 2 ml and closed by aluminium foil on acidic

cupboard.

d. It was added ba NaOCl 65 ml with slowly and stirred until the mixture

change color to greenish yellow solution.

e. In was let stand until 10th

minutes in acidic cupboard.

f. It was filtered by Buchner funnel.

g. It was rinsed by water until the precipitate in beaker glass is loss.

h. The residue (crystal) was took and it was dried in the oven.

i. It was balanced by electronical scale.

j. It was filled into capillary tube and the melting point was determined.

2. Iodoform Testing

a. 1-propanol test

5 drops of 1-propanol was dropped into test tube.

It was added by dioxane 5 ml.

It was added by 1 ml of NaOH 1% until form 2 layers.

It was added by some drops of KI.I2 until the color change to

brown.

It was added by 20th

drops of NaOH.

It was divided into two observation, namely heated and let stand.

b. Ethyl acetoacetic test

5 drops of ethyl acetoacetic was dropped into test tube.

It was added by dioxane 5 ml.

It was added by 1 ml of NaOH 1% until form 2 layers.

It was added by some drops of KI.I2 until the color change to

brown.

It was added by 20th

drops of NaOH.

It was divided into two observation, namely heated and let stand.

c. Acetophenon test

5 drops of acetophenon was dropped into test tube.

It was added by dioxane 5 ml.

It was added by 1 ml of NaOH 1% until form 2 layers.

It was added by some drops of KI.I2 until the color change to

brown.

It was added by 20th

drops of NaOH.

It was divided into two observation, namely heated and let stand.

F. Observation Result

No. Observation Result

1.

2.

3.

4.

5.

Synthetic of Iodoform :

KI(s) was balanced on electronical

balance.

It was added by water in beaker

glass.

It was added by acetone

It was added by NaOCl 5% drop

by drop and stirred.

It was let stand until 10th

minutes

- 6 g of KI

- 100 ml of water

- 2 ml of cetone

- 6 ml of NaOCl greenish

yellow mixture.

- Separated :

6.

7.

8.

9.

in cupboard.

It was filtrered by Buchner funnel.

The crystal dried in oven.

It was balanced on electronical

balance.

It was filled in capillary tube and

determined it melting point.

Transparent solution

Greenish yellow precipitate.

- Get yellow crystal.

- Pure crystal dry.

- 1.1 g

- 1200-124

0C

1.

2.

Iodoform Test :

1-propanol test

a. 1-propanol

b. It was added by dioxane

c. It was added by NaOH 1%

d. It was added by KI.I2

e. It was added by NaOH

f. It was divided into 2 part:

Heated

Let stand

Ethyl acetoacetic test

a. Ethyl acetoacetic

b. It was added by dioxane

- 5 drops of 1-propanol

- 5 ml of dioxane

- 1 ml of NaOH, form 2 layers:

Transparent solution

Turbidity solution

- Brown solution

- 20 drops of NaOH

- Transparent solution.

- Yellow dark solution.

- 5 drops of ethyl acetoacetic

- 5 ml of dioxane

3.

1.

c. It was added by NaOH 1%

d. It was added by KI.I2

e. It was added by NaOH

f. It was divided into 2 part:

Heated

Let stand

Acetophenon test

a. acetophenon

b. It was added by dioxane

c. It was added by NaOH 1%

d. It was added by KI.I2

e. It was added by NaOH

f. It was divided into 2 part:

Heated

Let stand

- 1 ml of NaOH, form 2 part:

Bubbles

Turbidity solution

- Brown solution

- 20 drops of NaOH

- Yellow solution.

- Yellow solution.

- 5 drops of acetophenon

- 5 ml of dioxane

- 1 ml of NaOH, form 2 layers:

Turbidity solution

Colorless solution

- Brown solution

- 20 drops of NaOH

- Yellow solution and white

precipitate.

- Yellow solution.

G. Data analysis

Known :

ρ acetone = 0.782 g/ml

Mw. Acetone = 58 g/mol

V. acetone = 2 ml

Mw. Iodoform = 373.79 g/mol

M. iodoform(obs) = 1.10 g

Problem :

Rendement……..?

Solution :

Mass acetone = (ρ x v) acetone

= 0,782 g/ml x 2 ml = 1.584 g

= 0.027 mol

Mol acetone = mol iodoform

Mass iodoform(the)= ( n x mw ) iodoform

= 0.027 mol x 393.79 g/mol = 10.63 g

= 10.34%

H. Discussion

a. Synthetic of Iodoform

In this experiment, that had been done of balancing the KI as the main

compound in iodoform synthetic that dissolved with H2O and added by acetone

and NaOCl with slowly. As the unction of KI to react with NaOClo to form KCl

and NaOI. As the function of NaOCl addition with slowly is to react with

perfectly.

KI + NaOCl KCl + NaOI

NaOI will be decompose in solution become to NaO+ and I

-. Addition of acetone

as function to reacting the ion I- with NaO

+ and H

+ at the solution is let stand. So

that formin CH3COONa solution (carboxylic salt) and CHI3 (iodoform) from

acetone compound. As the function of let stand on acidic cupboard is the

precipitate that precipitated in bottom of beaker glass is not evaporating to around

surrounding, as the reaction is :

1st

step

O O

CH3 C CH3 + KI + NaOCl CH3 C CH3 + KCl + NaOI

2nd

step

O O

CH3 C CH3 + NaOI CH3 C CH2I + NaOH

3rd

step

O O

CH3 C CH2I + NaOI CH3 C CHI2 + NaOH

4th

step

O O

CH3 C CHI2 + NaOI CH3 C CI3 + NaOH

5th

step

O O

CH3 C CI3 + NaOH CH3 C O-Na

+ + CHI3

Picture 1: iodoform solution that acidified in acidic cupboard

Solution with it precipitate filtered by Buchner funnel so that the crystal that

obtained more fast dry. Crystal then put into the oven so that the drying process

more fast and evaporating the left of water that contain in crystal. As the color of

iodoform crystal is yellow crystal (iodoform crystal). Crystal that obtained is nor

recrystallization because it has form an iodoform crystal reviewed from it color,

smell, and it melting point and the crystal weight is 1.10 grams and the rendement

is 10.34%, so if recrystallized again can be produce the less weight than 1.10

grams. As the melting point of iodoform crystal is 1200-124

0C. This is same with

theory that the melting point of iodoform is 1230C. least of rendement that

obtained because there is iodoform in filtrate and much of iodoorm that

evaporating at let stand in acidic cupboard and also the stirring way of solution at

the NaOCl addition.

Picture 2 : iodoform crystal

b. Iodorm testing

I. Closing

1. Conclusion

Based on the experiment that hade been done, so it may be conclude that :

a. The principle of isolation organic compound (EPMS) of phenyl

propanoid from galingale are maceration, separation, extraction, and

purification.

b. The isolation technique in galingale can be obtained with maceration

way, separation, and purification.

2. Suggestion

a. Student can be more careful in doing the experiment.

b. Student can me more compact with friends group.

BIBLIOGRAPHY

Barus, Rosbina. 2009. Amidasi Etil Para Metoksi Sinamat yang Diisolasi dari

Kencur. Tesis, Universitas Sumatra Utara.

Ekowati, Juni, dkk. 2010. Pengaruh Katalis pada Sintesis Asam Orto

Metoksisinamat dengan Materi Awal Arto Metoksi Benzaldehida dan Uji

Aktivitas Analgesiknya. Jurnal, Majalah Farmasi Airlangga, Vol.8, No.2,

Oktober 2010.

Fessenden, Ralph J. dan Joan S. Fessenden. 1982. Kimia Organik Edisi Ketiga.

Jakarta : Erlangga.

Kusumawati Idha dan Helmi Yusuf. 2011. Phospolipid Complex As A Carrier Of

Camempferia Galangal Rhizome Extract To Improve Its Analgecic

Activity. International Journal Of Pharmacy And Pharmaceutical Sciences

ISSN- 0975-1491 Vol 3, Suppl 1, 2011.

Nurdiansyah dan Abdi Radha. Efek Lama Maserasi Bubuk Kopra Terhadap

Rendemen, Densitas, dan Bilangan Asam Biodiesel yang Dihasilkan

dengan Metode Transesterifikasi In Situ. Penelitian Poltek Negeri

Pontianak.

Rasyid, Muhaidah. 2009. Kimia Organik I. Makasar : Badan Penerbit UNM.

Suyatno dan Emil Kurniati. Sintesis Senyawa Tabir Surya Eugenil Para Metoksi

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Question Answer

1. EPMS compound classified as based on its structure of benzene group and

alkyl which is the ester group derivate alkaloid to synthesis of amino acid and

binds cinnamic acid pathway.

2. Isolation of EPMS be masked and dried to ethyl para metoxhy cinnamic

compound contained in galingale can extracted well. In addition to the

teaching process can absorb the desired copound and galingale must be dry to

eliminate or reduce the water molecules contain in galingale.

3. Distillation temperature should be maintained arrounf to 800C because who

want to spent on distillation process is solvent ethanol. Where ethanol boils at

700-80

0C. if temperature less than or more than it temperature will be forming

or that out instead of ethanol but others compound.