Sabtu, 07 Juli 2012

FINAL EXAMINATION (NATURAL PRODUCT CHEMISTRY)



Name : Saely Natalia Sinaga
NIM : A1C109040


1.      
1   1. . Generally are solid while the oil is liquid phase, however both are equally trglyceride. Explain why the form of two different trglyceride, and point out important factors that determine the form of fat
      
     Answer : 


    Triglycerides in the fat oil
Can be either solid or liquid depending on the composition of the constituent fatty acids. Liquid vegetable oil because it contains unsaturated fatty acids, while animal fats are generally solid at room temperature because it contains unsaturated fatty acids.

    Fat oil
Triglycerides are composed of a mixture of esters of glycerol and long chain fatty acids. Fat if the oil will produce three molecular hydrolyzed long-chain fatty acids and a glycerol molecule.

    Triglycerides oil fatty
Derived from various sources have physico-chemical properties are different from each other that occurs because of differences in the number and types of esters contained therein.

    Triglycerides in the fat oil
Did not differ in their chemical composition differing only in shape or form. Triglycerides are called oils when melted at room temperature and is called solid if it freezes at room temperature.
There are many different kinds of fats, but each is a variation on the same chemical structure. All fats are derivatives of fatty acids and glycerol. The molecules are called triglycerides, which are triesters of glycerol (an ester being the molecule formed from the reaction of the carboxylic acid and an organic alcohol). As a simple visual illustration, if the kinks and angles of these chains were straightened out, the molecule would have the shape of a capital letter E. The fatty acids would each be a horizontal line; the glycerol "backbone" would be the vertical line that joins the horizontal lines. Fats therefore have "ester" bonds.
The properties of any specific fat molecule depend on the particular fatty acids that constitute it. Different fatty acids are composed of different numbers of carbon and hydrogen atoms. The carbon atoms, each bonded to two neighboring carbon atoms, form a zigzagging chain; the more carbon atoms there are in any fatty acid, the longer its chain will be. Fatty acids with long chains are more susceptible to intermolecular forces of attraction (in this case, van der Waals forces), raising its melting point. Long chains also yield more energy per molecule when metabolized.

   Saturated and unsaturated fats

A fat's constituent fatty acids may also differ in the C/H ratio. When all three fatty acids have the formula CnH(2n+1)CO2H, the resulting fat is called "saturated". Values of n usually range from 13 to 17. Each carbon atom in the chain is saturated with hydrogen, meaning they are bonded to as many hydrogens as possible. Unsaturated fats are derived from fatty acids with the formula CnH(2n-1)CO2H. These fatty acids contain double bonds within carbon chain. This results in an "unsaturated" fatty acid. More specifically, it would be amonounsaturated fatty acid. Polyunsaturated fatty acids would be fatty acids with more than one double bond; they have the formula, CnH(2n-3)CO2H and CnH(2n-5)CO2H. Unsaturated fats can be converted to saturated ones by the process of hydrogenation. This technology underpinned the development of margerine.
Saturated and unsaturated fats differ in their energy content and melting point. Since unsaturated fats contain fewer carbon-hydrogen bonds than saturated fats with the same number of carbon atoms, unsaturated fats will yield slightly less energy during metabolism than saturated fats with the same number of carbon atoms. Saturated fats can stack themselves in a closely packed arrangement, so they can freeze easily and are typically solid at room temperature. For example, animal fats tallow and lard are high in saturated fatty acid content and are solids. Olive and linseed oils on the other hand are highly unsaturated and are oily.


   Trans fats

There are two ways the double bond may be arranged: the isomer with both parts of the chain on the same side of the double bond (thecis-isomer), or the isomer with the parts of the chain on opposite sides of the double bond (the trans-isomer). Most trans-isomer fats (commonly called trans fats) are commercially produced. Trans fatty acids are rare in nature. The cis-isomer introduces a kink into the molecule that prevents the fats from stacking efficiently as in the case of fats with saturated chains. This decreases intermolecular forces between the fat molecules, making it more difficult for unsaturated cis-fats to freeze; they are typically liquid at room temperature. Trans fats may still stack like saturated fats, and are not as susceptible to metabolization as other fats. Trans fats may significantly increase the risk of coronary heart disease.
http://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Trimyristin-3D-vdW.png/250px-Trimyristin-3D-vdW.png





2. How a primary metabolite can ber converted into secondary metabolites. What is the basic idea and how the mechanism could describe


Answer: 


The basic idea a primary can be converted into secondary metabolites is The Building Block. The building blocks for secondary metabolites are derived from primary metabolism.The number of building blocks needed is surprisingly few. The most important building blocks employed in the biosynthesis of secondary metabolites are derived from:
1.Acetyl coenzyme A (acetyl-CoA)
2.Shikimic acid
3.Mevalonic acid
4.1-deoxyxylulose-5-phosphate
5.Amino acids

The Mechanism secondary metabolite, we can see in the bottom





3.  Hormone progesterone is essential for the survival of the pregnancy. These hormones are derived from a steroid biogenetically. Explain the logic of chemical reactions which may occur in the formation of progesterone


Answer : The Biosynthesis Progesterone


In mammals, progesterone , like all other steroid hormones, is synthesized frompregnenolone , which in turn is derived from cholesterol
Cholesterol  undergoes double oxidation to produce 20,22-dihydroxycholesterol . This vicinal diol is then further oxidized with loss of the side chain starting at position C-22 to produce pregnenolone . This reaction is catalyzed by cytochrome P450scc. The conversion of pregnenolone to progesterone takes place in two steps. First, the 3-hydroxyl group is oxidized to a keto group and second, the double bond is moved to C-4, from C-5 through a keto/enol tautomerization reaction.This reaction is catalyzed by 3beta-hydroxysteroid dehydrogenase/delta(5)-delta(4)isomerase.
Progesterone in turn (see lower half of figure to the right) is the precursor of the mineralocorticoid aldosterone, and after conversion to 17-hydroxyprogesterone (another natural progestogen) of cortisol and androstenedione. Androstenedione can be converted totestosterone, estrone and estradiol.
Pregenolone and progesterone can also be synthesized by yeast.









     4.       Many alkaloids are toxic to other organism. They often have pharmacological effects and are used as medications, are recreational drugs, or in entheogenic rituals. Describe in outline the process if biosynthesis of an alkaloid coumpound and describe the function groups which play an important role in the biological activities
Answer: 


One of alkaloid is toxic to organism is Nicotine



Nicotine, a plant defense alkaloid, is a pale yellow to dark brown liquid with a slight, fishy odor when warm.  Biologically, production of nicotine requires nicotinic acid (aka vitamin B3 or niacin) and an N-methyl-pyrrolinium cation, which is derived from ornithine (a non-protein amino acid derived from Krebs cycle intermediates) (1). (Figure 1)





Figure 1:  Synthesis of nicotine from nicotinic acid and ornithine precursors.


Laboratory Synthesis Of Nicotine














 

Jumat, 29 Juni 2012

LESSON PLAN

Alloca

    A.   Standard Competences
Understanding Properties Acid- Base Reaction
    B.   Basic Competences
To describe properties Acid–Base Titrations
    C.    Achievement Indicators Competences
a.     Cognitive Products
·         Explaining the  properties Acid- Base Titration
·         Explaining Neutralization Reactions and Neutralization Titration
b.    Cognitive Processes
·         Determining  the properties Acid Base Titration
·         Analyzing  the Neutralization Reactions  
·         Identifing the Neutralization Titration
c.      Psychomotor
·         Giving attention to the discussion group
d.    Affective
·         Deducing result that have learned about properties Acid- Base Titration and Neutralization Reactions based on own thinking
·         Answering the question about Acid Base Titration and Neutralization Reaction

   D.   Learning Objectives
a.         Knowledge: products
·         Students expected be able to explain the titration of Acid Base
·         Students expected be able to explain reaction of Neutralization
·         Student expected be able to explain titration of Neutralization


b.        Knowledge: the process
·         Students expected be able to determine properties Acid Base Reaction
·         Students expected be able to determine Neutralization Reactions
·         Students expected be able to analyzing Neutralization Titration
c.         Psychomotor
·         Students expected be able to pay attention to the discussion group
d.        Affective
·         Student expected be able to Deduce result that have learned about properties Acid- Base Titration and Neutralization Reactions based on own thinking
·         Student expected be able to Answer the question about Acid Base Titration and Neutralization Reaction
·           Student expected be able to give  opinion about properties Acid- Base Titration and Neutralization Reactions based on own thinking

    E.   Description of Learning Materials
Acid–Base Reactions: Neutralization Reactions
When a strong acid and a strong base solution are mixed, a neutralization reaction occurs, and the products do not have characteristics of either acids or bases. Instead, a neutral salt and water are formed. Look at the reaction below:
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

The anion from the acid (Cl) reacts with the cation from the base (Na+) to give a salt, and a salt is defined as any compound formed whose anion came from an acid and whose cation came from a base.

    F.    Learning Methods
a.       Learning approach            : the concept and process approach
b.      Learning model                 : STAD
c.       Methods                            : The method of discussion and information

   G.  Learning Activities

       Activity
Phase
Teacher Activities
Student activities
Allocation of Time  
     Introduction

Orientation
Teacher is conditioning student to focus and introducing about Titration Acid- Base
Student Concentrate and ready to study
10 minute
Apersepstion
Teacher ask to student about matery acid-base in last meeting
Student answer the question that teacher given
Motivation
Teachers arouse 
students' curiosity by telling me  about 
how Sodium chloride can beformed from two 
different compounds are 
concentrated HCl 
and NaOH compound
Student curious and pay attention
Giving Reference
Teacher explain about learning objective and source of learning relevant
Student listen the teacher explanation
Division of group learn
Teacher divide student in 5 group
Student start to sit on its group

      Core
Explanation
Teacher give the subject matter about Acid Base Titration and Neutralization Reaction
Student attention it


65 minute
Eksploration
The teacher asks the students to start talking about the Acid Base Titration and Neutralization Reaction that they have earned

Teacher give students the freedom to seek the source materials from anywhere
Teacher enable student to using learning media like virtual lab
Students start to disscuss in each groups and  seek the material about the Acid Base Titration and Neutralization Reaction
Elaboration

Teacher ask one group for presentating result of its discussion about the Acid Base Titration and Neutralization Reaction
Teacher ask to the other group to answer to result of from group which is presentation
Student presenting result of its discussion


Student from other group answered
Confirmation
Teacher explain about matery that have learned and clarify mistake in discussion
Student give attention clarification of teacher
Evaluation
Teacher ask student to doing  simple question from book about the Acid Base Titration and Neutralization Reaction that have been learned individually
Teacher ask one of member from the group to present one of the answer of question that determined by teacher in front of class
Student to do it
   
        Cover






Achievement
Teacher give assessment to each group which is presentating result of its discussion
Teachers also give a personal assessment to any student who asks and respond to discussions of other groups
Student pay attention



Student conclude matery which have been studied
15minute







Reflection
Teachers guide students and concluded the subject matter
The teacher gives the students the task to be done at home
 The teacher asks the students to re-learn the material that has been taught
Students pay attention

H. Learning Outcomes Assessment
1. Assessment Indicators
Aspect
Competency Achievement Indicators
Assessment Indicators
Form of Indicator
Type / Technical Assessment
1. Cognitive of Products

Explaining the  properties Acid- Base Titration

C2
Written
Test
Explaining Neutralization Reactions and Neutralization Titration
C2
Written
Test
2. Cognitive of Process
Determining  the properties Acid Base Titration
C3
Written
Test
Analyzing  the Neutralization Reactions 
C4
Written

Test
Identifing the Neutralization Titration
C3
Written
Test
2.Psychomotor
Giving attention to the discussion group
P2
Discuss

Observation
3. Affective
Deducing result that have learned about properties Acid- Base Titration and Neutralization Reactions based on own thinking
A1
Oral
Observation
Answering the question about Acid Base Titration and Neutralization Reaction
A2
Oral
Observation

2.      Assessment Tools
  1.    0.3 M HCl solution is titrated with NaOH solution, the titration end point is reached when10 ml of HCl solution requires 75 ml of NaOH solution. Determine the molarity of NaOH?
  2.    Determine the concentration of 20 ml of Ca (OH) 2 is titrated with 100 ml of 0.1 M HCl solution!

Key Answer
Determine the concentration of 20 ml of Ca (OH) 2 is titrated with 100 ml of 0.1 M HCl solution!
1. Given: MA = 0.3 M 
VB = 75 ml         VA = 10 ml 
nB = 1    nA = 1
Asked: MB?
Answer:
                             
                              MB = 0.04 M

         2.      Given: VA = 100 ml    VB =  20 ml
            MA = 0.1 M 
nB = 2 nA = 1
Asked: MB?
Answer: VA x MA x nA = VB x MB x nB
               100 x 0.1 x 1 = 20 x MB x 2
                MB = 0.25 M      
                                                                       
I.     Sources
o    Zumdahl,Zumdahl. 2007. Chemistry: Seventh Edition. USA. Houghton Mifflin Company.
o    Virtual laboratory macromedia flash