Preparation of Iodoform

COMPLETE REPORT ORGANIC CHEMISTRY 2

HERMIN HARDYANTI UTAMI (111 304 0202)

CHEMISTRY DEPARTMENT MATHEMATIC AND SCIENCE FACULTY STATE UNIVERSITY OF MAKASSAR 2012

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 Assistant Coordinator

Assistant

Fandi Ahmad, S.Pd.

Sabaruddin

Known By, Responsibility Lecturer

Dra. Hj. Ramdani, M.Si.

2012

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 R

C

H OH

R

C

OH O

R

C

O

H Primary alcohol

aldehyde

R’ R

C

acid

R’ 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

CH3CCH3

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 R

C

O CH3

+ 3NaOH

R

O R

C

C

CI3

+ 3NaOH

O CH3

+ NaOH

R

OH R

C

C

O-Na+ +

CHI3

CH3 +

NaI +

O CH3

+ NaOI

R

C

H2O

OH R

C

O 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

Synthetic of Iodoform : 1.

KI(s) was balanced on electronical

-

6 g of KI

balance. 2.

It was added by water in beaker -

100 ml of water

glass. 3.

It was added by acetone

-

2 ml of cetone

4.

It was added by NaOCl 5% drop

-

6 ml of NaOCl

by drop and stirred. 5.

It was let stand until 10th minutes

yellow mixture. -

Separated :

greenish

in cupboard.



Transparent solution



Greenish yellow precipitate.

6.

It was filtrered by Buchner funnel.

-

Get yellow crystal.

7.

The crystal dried in oven.

-

Pure crystal dry.

8.

It was balanced on electronical balance.

-

1.1 g

9.

It was filled in capillary tube and

-

1200-1240C

a. 1-propanol

-

5 drops of 1-propanol

b. It was added by dioxane

-

5 ml of dioxane

c. It was added by NaOH 1%

-

1 ml of NaOH, form 2 layers:

determined it melting point. Iodoform Test : 1.

1-propanol test



Transparent solution



Turbidity solution

d. It was added by KI.I2

-

Brown solution

e. It was added by NaOH

-

20 drops of NaOH

f. It was divided into 2 part:

2.



Heated

-

Transparent solution.



Let stand

-

Yellow dark solution.

a. Ethyl acetoacetic

-

5 drops of ethyl acetoacetic

b. It was added by dioxane

-

5 ml of dioxane

Ethyl acetoacetic test

c. It was added by NaOH 1%

-

1 ml of NaOH, form 2 part: 

Bubbles



Turbidity solution

d. It was added by KI.I2

-

Brown solution

e. It was added by NaOH

-

20 drops of NaOH

f. It was divided into 2 part:

3.



Heated

-

Yellow solution.



Let stand

-

Yellow solution.

-

5 drops of acetophenon

-

5 ml of dioxane

-

1 ml of NaOH, form 2 layers:

Acetophenon test a. acetophenon 1.b. It was added by dioxane c. It was added by NaOH 1%



Turbidity solution



Colorless solution

d. It was added by KI.I2

-

Brown solution

e. It was added by NaOH

-

20 drops of NaOH

-

Yellow solution

f. It was divided into 2 part: 

Heated



Let stand

precipitate. -

G. Data analysis Known

:

Yellow solution.

and white

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 CH3

C

O CH3 + KI + NaOCl

CH3

C

CH3 + KCl

+ NaOI

2nd step O CH3

C

O CH3

+ NaOI

CH3

C

CH2I

+ NaOH

3rd step O CH3

C

O CH2I

+ NaOI

CH3

C

CHI2

+ NaOH

4th step O CH3

C

O CHI2

+ NaOI

CH3

C

CI3

+ NaOH

5th step O CH3

C

O 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 1200-1240C. 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 Sinamat Melalui Reaksi Esterifikasi. Penelititan Unversitas Negeri Surabaya. Tim Dosen. 2012. Penuntun Praktikum Kimia Organik II. Makassar :Jurusan Kimia FMIPA UNM.

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 700-800C. if temperature less than or more than it temperature will be forming or that out instead of ethanol but others compound.