L.8 SAPONIFICATION

INTRODUCTION:

It is a reaction of a fatty acid with a strong base, which gives rise to a salt of fatty acid, called soap. This reaction is very important because the fatty acids are insoluble in water, the soap molecules form groups (mecel·les) that disperse in water. That is, even though they are insoluble form colloidal dispercions. The difference is due to the fact that the carboxyl group (COOH) of the fatty acid is ionized and the little group (COO + strong base) are ionized and cause very many radicals.

Our objective is prepare soap. 

MATERIALS:

- 600mL Beaker
- Spatula
-  Watch glass 
- Stirring rod
- 2 beakers 250mL
-electronic balance
- Gloves
- Safety goggles
- Hot plate

- 32g NaOH
- 90mL H2O
- 270mL Oil

PROCEDURE:

First we measured the grams of each substance with the balance, we took the beaker 250ml and we put 90 mL of water, and another beaker 250ml and we put 270 mL of oil. All this have led to a 600ml beaker, slowly pulling the oil and mixed, but be careful because the reaction is exothermic and it heats! Mix a good time, we put the 600mL on a hot plate for heat and this accelerate the reaction. Then we put a few drops of aroma and finally we put the mixture in a few evaporating dish. Now we have to wait for obtain the hard soap. 

Now when we have the flavor we wanted, we poured a box of porcelain. Now all that remains hard. And so how does the soap.

The reaction of saponification:

L.7 LIPIDS PROPERTIES

INTRODUCTION:

Lipids are heterogeneous group of compounds synthesized by organisms that are present in all biological tissues. Lipids are made in general of long chains of hydrocarbons with relatively little oxygen. As a result of this, they tend to be non-polar, meaning they do not dissolve in polar solvents such as water. 

In this experiment we test the solubility of different oils and how to identify lipids from a sample, Our objectives where: 

- Test the solubility of lipids.

- Identify lipids in liquids compounds

- Understand what are an emulsion and the effect of detergents.

MATERIAL:

- Test tube rack
- 250mL beaker
- Water
- 6 test tubes- Cellulose paper
- Dropper
- Scissors
- Glass rod

- Olive oil
- Soap
- Milk: full-cream, semi-skimmed and skimmed milk.
- Petroleum ether
- Ethanol
- Sudan III

PROCEDURE:

Solubility of some lipids:

- Clean and dry 3 test tubes. Label as W (water), E (ethanol) and PE (ether).

- Add 3 drops of oleic acid to 3 small test tubes.

- Add 1mL of water in the first test tubes (W).

- Add 1mL of ethanol in the second test tube (E).

- Add 1mL of petroleum ether in the third test tube (PE).

- Shake carefully each test tube and record solubility and observations.

Lipids identification: 

Translucent mark:

- Take 2 pieces of cellulose paper.

- Put 1 drop of water in the first piece. You will see a translucent spot. Wait for a while and observe what is happening.

- Put 1 drop of olive oil in the second piece of cellulose paper. You will see a translucent spot. Wait for a while and observe what is happening.

Sudan III dye:

Sudan III is a red fat soluble dye that is utilized in the identification of the presence of lipids, triglycerides and lipoproteins in liquids. 

- Take the W test tube of the first experiment and add 2 drops of Sudan III. 

- Prepare four test tubes: 3 with milk with different fat content, and soda. Add to drops of Sudan III and observe the results. 

Permanent emulsion:

- Take 250 mL beaker and put 100mL of water.

- Add 1 mL of olive oil. With a glass rod stir the mixture vigorously and let it stand for a few minutes.

- Make note of what is happening. 

- Add 2 drops of soap and stir the mixture again. Let it stand for a few minutes and notice the differences between both mixtures. 

RESULTS AND CONCLUSIONS:

- Lipids are insoluble in water and soluble in ether (organic compound). In the second test tube, the oil and ethanol didn't mix because the oil formed micelles.
- Sudan III stains lipids, but in our experiment didn't work.
- Lipids leave a translucent spot in cellulose paper. 

QUESTIONS:

1.- From your observation, wich compounds can dissolve lipids? 

Ether can dissolve lipids. 

2.- Do the oil and water mix? What can you conclude about the polarity of the oil if you know that water is polar? 

No, the oil is insoluble in water, the oil is non-polar. 

3.- Why is liquid the olive oil at room temperature? And why not the lard?

Oil is the fat acid insaturated. Because the lard is a fat acid saturated.

4.- Why does the lipid leave a translucent spot on paper? 

Because the lipid don't evaporate.

5.- Wich type of milk contains more lipids? Why?

 Full-cream milk, because they take away the semi-skimmed milk fats. Contain all the lipids.

6.- Did the oil and water mix when you added the soap?

No, it forms micelles that can separated oil from water. 

7.- What did the soap do to the fat? 

Separated in micelles.

8.- Can you think about process and locations were compounds like the soap would be important to an animal? 

Bile acids.

L.6 FEHLING'S TEST: REDUCING SUGARS

INTRODUCTION:

We start this experiment un 27th of october, but is not finished yet because we did it approximately the last 15 minutes of class. But this experiment is related with saccharides.
First of all, the fehling's solution is a chemical test to differentiate between reducing and non-reducing sugars. Is based on the reaction of a functional group of sugar molecules with Fehling's reagent. 

Fehling's A: is a blue aqueous solution of cooper(II) sulphate.

Fehling's B: clear and colourless solution of potassium sodium tartrate and sodium hydroxide. 

When both are mixed a deep colour solution can be seen. 

When a sugar has reducing ability. the mixture turns from deep blue to green colour suspension with a red precipitate. Some sugars are capable of reducing cooper II ions to cooper I ions. This reducing ability is useful in classifying sugars. When the sugar to be tasted is added to the Fehling's solution and the mixture is heated, some sugars can be oxidized and the Fehling's mixture can reduced. 

The goal is identify reducing sugars, comprehend redox reactions and understand the relation between structure ability of some sugars. 

MATERIALS:

- Test tube rack

- 10mL Pipet

- Distilled water

- 5 test tubes

- 5 Spatula

- Lactose

- Maltose

- Glucose

- Sucrose

- Starch

- Fehling's A and B solutions

- HCl

PROCEDURE:

First you will determine with sugars give a positive test with Fehling's reagent and then, by testing the reaction of some organic molecules containing only a single functional group, you should be able to deduce which functional group og sugars is reacting with Fehling's reagent.

1.  Take 5 test tubes and label: G ,M, S, L, ST.
2.  Put 2mL of water in each test tube.
3.  With different spatulas put a small amount of each sugar. Dissolve the suggar.
4.  Add 2mL of Fehling's A solution and then Fehling's B.
5.  Place each test-tube in a boiling water bath (250mL beaker on a hotplate stirrer).
6.  Observe what is happening.

Starch Hydrolysis: 

Hydrolysis is the reaction of a compound with water. As you know, starch is a polymer, consisting of many units of α-D-glucose covalently linked together.

1. Place 2mL of 1% starch in test tube and add o.5mL of 3M HCl. Mix and place this mixture in a boiling water bath for 10 minutes.
2. After 10 minutes, remove the tube from the waterbath and let it cool. Neutralize this solution with 1M NaOH and mix well. 
3. transfer 8-10 drops of this solutionto a small test tube.
4. Add 1mL of Fehling's A solution and 1mL of Fehling's B.
5. Heat for a few minutes in a boiling water bath.
6. Record your observations. compare the results of this test with your results for unhydrolyzed tarch in the step 1 of this experiment. 
7. You can test the absence of starch with iodine solution too. 

QUESTIONS:

1.-From your observations and the structures of the sugars given above, indicate which functional group in the sugar molecules reacts with Fehling's reagent.

The OH group is the one that reacts with Fehling's reagent because it is free and the sacharide have the reducing power.

2.-Compare the results you obtained for the Fehling's test of starch and Fehling's test of hydrolyzed starch. Explain your results.

When the starch is hydrolized it turns into glucose. But in the Fehling's experiment the starch doesn't have reducing power, the OH is not free. 

however, the glucose is a monosacharide and has the OH free, like the other solutions. 

3.-Would have you obtained a Fehling's positive test if you had hydrolyzed the sucrose? Why? 

 Glucose, have a free OH that's because they are monosacharide, this happens because we hydrolyzed fructose and we obtain glucose. 

4.-What does "reducing sugars" mean? 

A reducing sugar is any sugar that either has an aldehyde group or is capable of forming one in solution through  isomerism. The aldehyde functional group allows the sugar to act as a reducing agent, for example in the Tollen's test or Benedict's reagent, or the Maillard reaction important in the browning of many foods. The cyclic hemiacetal forms of aldoses can open to reveal an aldehyde and certain ketoses can undergo tautomerization to become aldoses. However, acetals, including those found in polysaccharide linkages, cannot easily become free aldehydes.

L5. SACCHARIDES PROPERTIES

INTRODUCTION:

On monday, we lerned more about saccharides. This are organic molecules consisting of C, H and O atoms. And are divided in three groups:

- Monosaccharides: Formed by a linear carbon chain, are the building blocks of oligo and polisaccharides. They have a one functional group: cetone or aldehyde. Pentoses and hexoses can swith from acyciclic to cyclic forms called furanoses and pyranoses. Like fructose, glucose or galactose.

- Oligosaccharides: Small polymer containing between 2 and 10 monosaccharides. Like lactose, sucrose or maltose.

- Polysaccharides: big polymers with more tan 10 monosaccharides. Like starch, glycogen, cellulose and chitin.

This molecules give diferent functions in the organisms, energy (starch, glicogen), and structural components (cellulose, ribose, chitin and desoxiribose).

Saccharides yield 4,2 Kcal/gr and are abundant in fruits, sweets, honey... They are a common source of energy in living organisms.

Our objectives in thes case were identify different sugars from its properties, differentiate mono and disaccharides and understand the relation between structure and some properties. 

MATERIALS:

- Test tube rack

- 10mL pipet

- Distilled water

- 5 test tubes

- 1 dropper

- 5 spatula

- Clock glass

- Lactose

- Maltose

- Glucose

- Sucrose

- Starch

- Lugol's iodine

PROCEDURE:

	GLUCOSE (G)	MALTOSE (M)	SUCROSE (SU)	LACTOSE (L)	STARCH (S)
Flavour	Sweet	Sweet	Sweet	No sweet	No sweet
Crystaline form	Crystalline	No crystalline	Crystalline	No crystalline	No crystalline
Colour	White	White	White	White	Cream
Solubility	Soluble	Soluble	Soluble	Soluble	Insoluble
Lugol's iodine	-	-	-	-	+

In the first part of the experiment we tested some phisycal properties of the saccharides: flavour, cristal structure and colour. How?

1. Flavour: Put small amunt of each saccharide in your hand and taste it! it can be sweet or not.
2. Crystals: Observe a small amount of each saccharide on clock glass under magnification.
3. Colour: White, transparent or creamy.

To test solubility:

1. Clean and dry 5 test tubes and label them "G, M, L, SU, S"
2. Put 5mL of water in each test tube.
3. With the aid of a spatula, put small amount of each saccharide inside the labelled test tube and test if they are soluble or insoluble.
4. Observe if each saccharide forms a misture called dissolution or coloidal suspension.

Lugol's iodine test:

Finally, add 2 drops of lugol's iodine to each test tube and test if the reaction is positive or negative. Lugol's is a solution of a elemental iodine (I) and potassium iodine (KI) in water that is use to test a saccharide. The reaction is positive when iodine reactss by turning from yellow to purple, dark-blue/black colour.

RESULTS AND CONCLUSIONS:

We can see that lactose is a disaccharide that don't have sweet flavour. The starch have a cream colour, it's insoluble and it's positive with the Lugol's iodine test because is the only polysaccharide of the experiment. The rest of the saccharides are disaccharides. 

QUESTIONS:

1.- Write the empirical formula of each saccharide that you have use. Show structures of the five saccharides. Classify each one in one group: mono, oligo or polysaccharide.

G: C6H12O6 - monosaccharide.

M: C12H22O11 - oligosaccharide.

SU: C12H22O11 - oligosaccharide.

L: C12H22O11 - oligosaccharide.

S: C6H10O5 - polisaccharide.

2.- Which of the monosaccharides are aldoses and which are ketoses?

ALDOSES: glucose, maltose, lactose, starch.

KETOSES: sucrose.

3.- Which bond links monosaccharides?

Oligosaccharides.

4.- Which saccharide/s is/are sweet? Is this property related to the structure of the molecule?

Glucose, maltose, sucrose

5.- Which saccharide/s is/are soluble? Is this property related to the structure of the molecule?

Starch, It's a big molecule, it have a high molecular weight and is relacionated with the structure.

6.- Which saccharide has reacted with Lugol's iodine solution?

Starch

7.- Which kind of foods contains starch?

Cereals, pasta, rice, legums...                             

8.- Calculate the energy from the nutrition facts labelfrom a cereal

• Calculate the energy that comes from the saccharides.

23gr saccharides

4,2 Kcal => 1g

23x4,2= 96.6Kcal

L4. pH

INTRODUCTION:

The pH is a quantitatively measure to degree of acidity and basicity in a solution. A solution with pH less than 7 are said to be acidic and solutions with a pH greater than 7 are basic or alkaline. Pure water (distilled water) has a pH close to 7, this solution is neutral. 

The equation of pH is this:

                            

pH= -log (H + )

For measuring the pH of a solution at the lab we have different pH indicators:

1. Universal indicator paper: It's an absorbent paper that has been impregnated with universal indicator. This method consists as from about pH 2 to pH 10. 
2. pH-meter: Is an electronic device used for measuring he exact pH af a solution. It consists of a glass electrode connected to an electronic meter that measures and displays de pH reading. 

We made this experiment because this week we studied the dissolutions and dispersions in biology class. The objectives were: measure different pH values of organic and inorganic solutions and prove different methods of measuring pH.

MATERIALS:

- Distilled water

- Milk

- Wine

- Lemon

- Tomato

- Coffee

- Carbonated beverage

- 10% NaOH solution

- 10% HCl solution

- NH3 solution

- Soap solution

- Universal indicator paper (strips)

- pH-meter

- Acetic Acid

- Tongs

- 8 x 100 mL Beakers

- 2 Clock glass

- 1 x 250 mL Beaker

- 5 tests tubes

- test tube rack

- 10 mL pipet

- Funnel

- Graduated Cylinder

PROCEDURE:

To measure de pH of different solutions we are going to put the different solutions in small beakers of 250 mL.

1. Squeeze the lemon and tomato in two clock glasses.
2. Take a pice of indicator paper and place one end of it into the soution. Leave for at least 20 seconds.
3. Remove the indicator paper and compare its colour with the appropriate colour chart. 
4. Repeat points 1 to 2 with as many others solutions as you are provided with.
5. Record your results.

How does concentration affect pH?

Squeeze the rest of the lemon inside a beaker and filter the solution with a funnel and cellulose paper. 

1. Prepare a test tub track with 5 tests tubes cleaned with distilled water. Mark the tubes with the labels: A, A1, A2, A3 and B.
2. Add 10 mL of lemon juice to tubes A and B.
3. Take the A tube and put 5mL of its lemon juice to test tube A1.
4. Take the A1 tube and put 2,5 mL of its lemon juice to tube A2.
5. Take the A2 tube and put 1,2 mL of its lemon juice to tube A3.
6. Add distilled water to each test tube until it has the same volume as test tube B (10 mL).
7. Calculate the concentration of each test tube with the formula you have below

                - Concentration (%)= ( Volume of Juice/ Total Volume) x 100

• Find the pH of the test tube:

TEST	VOLUME OF	TOTAL	CONCENTRATION	pH
B	10 mL	10 mL	100%	3,38
A1	2,5 mL	10 mL	25%	3,37
A2	1,25 mL	10 mL	12’5 %	3,33
A3
A	5 mL	10 mL	50%	3,35

RESULTS AND CONCLUSIONS:

We measure the pH of different substances, and we observed that if the pH was acid the color of the indicator, was red, pink, orange, yellow. LIke the HCl (pH=1) or the wine (ph=3) the tomato (pH=4) and vinegar (pH=2) and the milk that present a neutral- acid pH (pH=6-7) are acid too.

And if the substance present a basic pH, the color of the indicator was more blue and purple like the soap (pH=10), the bleach, that present an a high pH (pH=12-14) and NaOH (pH=14). 
And in a neutral pH is when this solution have pH of seven, the color is more green and this substances present the same number of H+ and oH-, like the distilled water. 

Then we do an other experiment with the lemon juice, with  tubs (A,A1, A2, A3, B) we do a different concentrations of lemon juice and distilled water and with the pH- meter we can calculate, measure the exact pH of the solution. We can see the results in the above table. With this results we can confirm that the lemon juice are acid. It shows a high concentration of hydrogen ions.

QUESTIONS:

1.- Which of the solutions gave an acid pH?

HCl, vinegar, wine, tomato, coffee and milk.

2.- Which of the solutions was alkaline?

Soap, bleach and NaOH.

3.- Which of the solutions were neutral? Did you expect this results? Explain.

Distilled water, because it has the same concentration of H+ and OH, and thats why we expect that water is with a pH neutral.  

                                                                                                  

4.- How does a pH of 3 differ from pH of 4 in terms of H+ concentrations?
10 times large value.

5.- In the second part of the experiment, you have compared the pH of the same product (lemon juice) in different concentrations. In this case explain:

• Which is the dependent variable? The pH.
• Which is the independent variable?The concentration of lemon juice.
• Which is the problem that we want to solve? The problem that we want to solve is equalize the concentrations.
• Which is the control of the experiment? The control is the test B.
• Write the results and conclusions of your experiment: The pH is very similar on the all concentrations, but the normal is that the pH depends of the protons concentrations. 

6.- Which pH do you think that gastric juices might have? Why? Do you think that intestinal pH has the same pH? Why?
They have an acid pH because its used to improve food digestion. I think that all intestinal pH are acid.

7.- Which pH do you think that blood might have? Why?
The blood have an neutral pH, because if is hypotonic the cell bloat, if is an hypertonic the cell crease. The pH causes it to keep the concentration, because we could die.

8.- What is acid rain? Which are the consequences in the ecosystems and how is its formation pattern? Is rain Barcelona acid or alkaline?

The acid rain is a phenomenon, that have a pH levels lower than normal. Is mainly due to the mixture of water with molecular products of sulfur and nitrogen in the combustion incurred, such as nitrous oxide NO2, NO3 nitric oxide or sulfur dioxide SO2 which when combined with the water generated nitrous acid HNO2, HNO3 nitric acid and sulfuric acid H2SO4. 
Acid rain can have harmful effects on plants, animals and buildings. Causes corrosion of metal structures such as bridges or erosions in stone buildings. It also burns the leaves of vegetation and ecosystems it affects mainly vegetables, such as forests, but can also affect lakes and reservoirs acidifying them.
In Barcelona this rain, is alkaline.