Chemistry

In this physics coursework, I have been asked to carry out research of my selection and to develop it. I have selected to research the life cycle of a star, and I would conduct this by gathering the necessary information in a form of a report which explains this in detail. I have chosen to explore this particular topic firstly because I am extremely fascinated in space and the universe and secondly because I do not know much about the life cycle of a star and I deem this will help extend my knowledge.

Firstly when carrying out this research before describing the life cycle of a star I need to be familiar of what a star is, and how it is formed

What is a star, and how does it form?

Stars are basically huge balls of hydrogen gas. Hydrogen is by far the most common element in the Universe, and stars form in clusters when large clouds of hydrogen, which naturally forms a hydrogen ‘molecule’ (H+H=H2) with another atom, collapse.

The hydrogen clouds collapses very slowly, although they can be speeded up by the effects of a passing star, or the shockwave from a distant supernova explosion. As the cloud collapses, it speeds up its rotation, and pulls more material into the centre, where a denser ball of gas, the ‘proto-star’ forms. The proto-star collapses under its own weight, and the collisions between hydrogen molecules inside it generate heat. Eventually the star becomes hot enough for the hydrogen molecules to split apart, and form atoms of hydrogen.

The star keeps on collapsing under its own weight, and getting even hotter in the core, until finally it is hot enough there (roughly 10 million degrees) for it to start generating energy, by nuclear fusion – combining hydrogen atoms to form a heavier element, helium. Energy is released from the core, and pushes its way out through the rest of the star, creating an outward pressure which stops the star’s collapse. When the energy emerges from the star, it is in the form of light, and the star has begun to shine.

A Star is formed from a cloud of gas, mostly hydrogen, and the dust that is initially spread over a huge volume, but which is pulled together by its own collective gravity. This gravitational collapse of the cloud creates a body of large density, and the loss of gravitational potential energy in the process is very large indeed. The result is that the original particles acquire high kinetic energy, so that the collisions between them are very violent. Atoms lose their electrons. Not only has that, collisions taken place in which electrical repulsion of nuclei is no longer strong enough to keep them apart. They can become close enough together for the strong nuclear force to take effect, so that they merge. Fusion takes place, with hydrogen as the principal key material. This begins the process of conversion of mass to energy, and much of the released energy takes the form of photons which begins to stream from the new star.

Every star then exists in a state of slowly evolving stability. On the one hand there is the trend for the material to continue to collapse under gravity. On the other hand there is a tendency for the violent thermal activity and the emission of radiation resulting from fusion to blow the material apart. The more bigger star in general, the greater is the gravitational pressure and so the higher rate of energy is released by fusion, therefore bigger stars use up their supply of fusing nuclei more quickly than do smaller stars, such that bigger stars have shorter lives.

The enormous luminous energy of the stars comes from nuclear fusion processes in their centres. Depending upon the age and mass of a star, the energy may come from proton fusion, helium fusion, or the carbon cycle. For brief periods near the end of the luminous lifetime of stars, heavier elements up to iron may fuse, but since iron is at the peak of the binding energy curve, the fusion of elements more massive than iron would soak up energy rather than deliver it. This links to the below graph:

Fusion in stars makes energy available to create radiation, consuming mass at an amazing rate. The sun, for example loses a mass of 4.5 million tonnes every second. Also, heavier nuclei are formed from smaller ones, so that the compression of a star changes. Concluding this, as the star dies the material dependant on its size is scattered in space.

The Hertzsprung – Russell Diagram

This simple graph shows ways in which to classify stars. Temperature is plotted on the x-axis. This is related to the colour as cooler stars are redder, hotter stars are bluer. Relative luminosity is plotted on the y-axis. Because of the very wide range of temperatures and stellar luminosities, logarithmic scales are used. The location of an individual star on such a graph lets us establish a loose system of classification. This graph aids us to find out what star has what temperature so we can easily classify it using the relative luminosity and temperature. Here is a diagram of the graph which shows the stars in their classified points showing their rough temperature and luminosity.

So how do the changes in the stars take place?

Very massive stars experience several stages in their cores.

o First hydrogen fuses into helium then helium to carbon creating larger nuclei. Such large stars in later life can have shells or layers with heavier nuclei towards their centres. It is not only the life expectancy of a star that depends on its mass, but also the way which it dies.

o Older stars have outer layers in which hydrogen is the fuel for fusion, while the inner layers helium is the fuel, and for massive stars there may be further layers beneath. Most stars, including the sun become red giants after the end of their equilibrium phase.

o This process is started by cooling in the inner core, resulting in reduced thermal pressure and radiation pressure and so causing gravitational collapse of the hydrogen shell. But the gravitational collapse provides energy for heating the shell, and so the rate of fusion in the shell increases. This makes the shell expand enormously.

o The outermost surface of the star becomes cooler, and its light becomes redder, but the larger surface area means that the stars luminosity increases.

o Meanwhile the gravitational collapse affects the core as well, and ultimately the process of fusion of helium in the core cause the outer shell to expand further and thin leaving the hot extremely dense core as a white dwarf.

o Slowly this cools and becomes a black dwarf.

o For the stars that are several times bigger then the sun, death may be even more dramatic. A core of carbon is created by fusion of helium, and once this core is sufficiently compressed then fusion of the carbon itself takes place. The rapid release of energy makes the star briefly as bright as a galaxy, as bright as 10 billion stars.

o The star explodes into a supernova and its material spreads back into the space around. In even larger stars, fusion of carbon can continue more steadily, producing still larger nuclides and ultimately creating iron nuclei. The iron nuclei also experience fusion, but these are different as they are energy consuming meaning they keep it in. The central core of the star collapses under gravity. This increases temperature but cannot now greatly increase the rate of fusion, so collapse continues. Outer layers also collapse around the core, compressing it further. It becomes denser then an atomic nucleus, protons and electrons join together to create neutrons.

o Meanwhile, the collapse of the outer layers heats these, increasing the rate of fusion so that suddenly the star explodes as a supernova. This spreads the material of these layers into space, leaving a small hot body behind a neutron star.

o Furthermore if this supernova is big enough, its gravity continues to pull the matter towards a single point with a huge gravitational field where not even light can escape from is known as the black hole.

Star pictures obtained from Internet http://www.enchantedlearning.com/subjects/astronomy

Here is an illustration of a star life cycle followed by the theory

How long a star lives for and how it dies…

How long a star lives and how it dies, depends entirely on how massive it is when it begins. A small star can sustain basic nuclear fusion for billions of years. Our sun, for example, probably can sustain reactions for some 10 billion years. Really big stars have to conduct nuclear fusion at an enormous rate to keep in hydrostatic equilibrium and quickly falter, sometimes as fast as 40,000 years.

If the star is about the same mass as the Sun, it will turn into a white dwarf star. If it is somewhat more massive, it may undergo a supernova explosion and leave behind a neutron star. But if the collapsing core of the star is very great at least three times the mass of the Sun nothing can stop the collapse. The star implodes to form an infinite gravitational warp in space, a hole. This is exemplified in a very simple diagram highlighting the consequence of each mass of the stars and what they will revolve into.

Normal stars such as the Sun are hot balls of gas millions of kilometres in diameter. The visible surfaces of stars are called the photospheres, and have temperatures ranging from a few thousand to a few tens of thousand degrees Celsius. The outermost layer of a star’s atmosphere is called the “corona”, which means “crown”. The gas in the coronas of stars has been heated to temperatures of millions of degrees Celsius.

Most radiation emitted by stellar coronas is in X-rays because of its high temperature. Studies of X-ray emission from the Sun and other stars are therefore primarily studies of the coronas of these stars. Although the X-radiation from the coronas accounts for only a fraction of a percent of the total energy radiated by the stars, stellar coronas provide us with a cosmic laboratory for finding out how hot gases are produced in nature and how magnetic fields interact with hot gases to produce flares, spectacular explosions that release as much energy as a million hydrogen bombs

The Orion Trapezium as observed. The colours represent energy; where blue and white indicate very high energies and therefore extreme temperatures. The size of the X-ray source in the image also reflects its brightness, i.e. more bright sources appear larger in size.

The Life Cycle of a star:

In Large Stars

In hot massive stars, the energy flowing out from the centre of the star is so intense that the outer layers are literally being blown away. Unlike a nova, these stars do not shed their outer layers explosively, but in a strong, steady stellar wind. Shock waves in this wind produce X-rays; from the intensity and distribution with energy of these X-rays, astronomers can estimate the temperature, velocity and density of this wind.

Medium sized Stars

In medium-sized stars, such as the Sun, the outer layers consist of a rolling, boiling disorder called convection. A familiar example of convection is a sea-breeze. The Sun warms the land more quickly than the water and the warm air rises and cools as it expands. It then sinks and pushes the cool air off the ocean inland to replace the air that has risen, producing a sea-breeze. In the same way, hot gas rises from the central regions of the Sun, cools at the surface and descends again.

From Red Giant To supernova

Once stars that are 5 times or more massive than our Sun reach the red giant phase, their core temperature increases as carbon atoms are formed from the fusion of helium atoms. Gravity continues to pull carbon atoms together as the temperature increases and additional fusion processes proceed, forming oxygen, nitrogen, and eventually iron.

As the shock encounters material in the star’s outer layers, the material is heated, fusing to form new elements and radioactive isotopes. While many of the more common elements are made through nuclear fusion in the cores of stars, it takes the unstable conditions of the supernova explosion to form many of the heavier elements. The shock wave propels this material out into space. The material that is exploded away from the star is now known as a supernova remnant.

The White Dwarf

A star experiences an energy crisis and its core collapses when the star’s basic, non-renewable energy source, hydrogen which is used up. A shell of hydrogen on the edge of the collapsed core will be compressed and heated. The nuclear fusion of the hydrogen in the shell will produce a new surge of power that will cause the outer layers of the star to expand until it has a diameter a hundred times its present value. This is called the ‘red giant’ phase of a star’s existence.

There are other possible conditions that allow astronomers to observe X-rays from a white dwarf. These opportunities occur when a white dwarf is capturing matter from a nearby companion star. As captured matter falls onto the surface of the white dwarf, it accelerates and gains energy. This energy goes into heating gas on or just above the surface of the white dwarf to temperatures of several million degrees. The hot gas glows brightly in X-rays. A careful analysis of this process can reveal the mass of the white dwarf, its rate of rotation and the rate at which matter is falling onto it. In some cases, the matter that gathers on the surface can become so hot and dense that nuclear reactions occur. When that happens, the white dwarf suddenly becomes 10,000 times brighter as the explosive outer layers are blown away in what is called a nova outburst. After a month or so, the excitement is over and the cycle begins anew.

The Supernova

Every 50 years or so, a massive star in our galaxy blows itself apart in a supernova explosion. Supernovas are one of the most violent events in the universe, and the force of the explosion generates a blinding flash of radiation, as well as shock waves analogous to sonic booms.

There are two types of supernovas:

o Type II, where a massive star explodes

o Type I, where a white dwarf collapses because it has pulled too much material from a nearby companion star onto itself.

The general picture for a Type II supernova is when the nuclear power source at the centre or core of a star is exhausted, the core collapses. In less than a second, a neutron star (or black hole, if the star is extremely massive) is formed. When matter crashes down on the neutron star, temperatures rise to billions of degrees Celsius. Within hours, a disastrous explosion occurs, and all but the central neutron star is blown away at speeds in excess of 50 million kilometres per hour.

A thermonuclear shock wave races through the now expanding stellar debris, fusing lighter elements into heavier ones and producing a brilliant visual outburst that can be as intense as the light of ten billion Suns. The matter thrown off by the explosion flows through the surrounding gas producing shock waves that create a shell of multimillion degrees gas and high energy particles called a supernova remnant. The supernova remnant will produce intense radio and X-radiation for thousands of years.

In several young supernova remnants the rapidly rotating neutron star at the centre of the explosion gives off pulsed radiation at X-ray and other wavelengths, and creates a magnetized bubble of high-energy particles whose radiation can dominate the appearance of the remnant for a thousand years or more.

Eventually, after rumbling across several thousand light years, the supernova remnant will disperse.

The Neutron Stars

The nucleus contains more than 99.9 percent of the mass of an atom, yet it has a diameter of only 1/100,000 that of the electron cloud. The electrons themselves take up little space, but the pattern of their orbit defines the size of the atom, which is therefore 99.9% open space. What we perceive as solid when we bump against a rock is really a disorder of electrons moving through empty space so fast that we can’t see or feel the emptiness. Such extreme forces occur in nature when the central part of a massive star collapses to form a neutron star. The atoms are crushed completely, and the electrons are jammed inside the protons to form a star composed almost entirely of neutrons.

The result is a tiny star that is like a gigantic nucleus and has no empty space. Neutron stars are strange and fascinating objects. They represent an extreme state of matter that physicists are eager to know more about. The intense gravitational field would pull your spacecraft to pieces before it reached the surface. The magnetic fields around neutron stars are also extremely strong. Magnetic forces squeeze the atoms into the shape of cigars. Even if a spacecraft carefully stayed a few thousand miles above the surface neutron star so as to avoid the problems of intense gravitational and magnetic fields, you would still face another potentially fatal hazard. If the neutron star is rotating rapidly, as most young neutron stars are, the strong magnetic fields combined with rapid rotation create an amazing generator that can produce electric potential differences of trillions of volts.

Such voltages, which are 30 million times greater than those of lightning bolts, create deadly blizzards of high-energy particles. If a neutron star is in a close orbit around a normal companion star, it can capture matter flowing away from that star. This captured matter will form a disk around the neutron star from which it will spiral down and fall, or accrete, onto the neutron star. The in falling matter will gain an enormous amount of energy as it accelerates. Much of this energy will be radiated away at X-ray energies. The magnetic field of the neutron star can funnel the matter toward the magnetic poles, so that the energy release is concentrated in a column, or spot of hot matter. As the neutron star rotates, the hot region moves into and out of view and produces X-ray pulses.

Black Holes

When a star runs out of nuclear fuel, it will collapse. If the core, or central region, of the star has a mass that is greater than three Suns, no known nuclear forces can prevent the core from forming a deep gravitational damage in space called a black hole. A black hole does not have a surface in the usual sense of the word. There is simply a region, or boundary, in space around a black hole beyond which we cannot see.

This boundary is called the event horizon. Anything that passes beyond the event horizon is doomed to be crushed as it descends ever deeper into the gravitational well of the black hole. No visible light, nor X-rays, nor any other form of electromagnetic radiation, or any particle, no matter how energetic, can escape. The radius of the event horizon (proportional to the mass) is very small, only 30 kilometres for a non-spinning black hole with the mass of 10 Suns.

The purpose of this experiment was to observe the process in which cells must partake in a respiration process called anaerobic fermentation and as the name suggests, oxygen is not required. This particular procedure, which Is catabolic meaning, it breaks down energy, can be present In to types of fermentation; alcohol In yeast or lactic acid in muscles. This Is a continued reaction from glycoside, where glucose Is broken down Into three carbon sugars.

The products of alcohol fermentation are ethanol and carbon dioxide and the products produced by lactic acid fermentation is lactate. As we observed the effects of yeast fermentation, It Is Imperative to know that yeast makes energy through fermentation. Yeast fermentation was combined with several different saccharine such as glucose, sucrose, starch, and fructose. Dolled water was also included In this experiment as another variable. The control was simply a vial of yeast and distilled water at room temperature.

Each vial was filled completely with the mixture (the solution was composed of individual saccharine and water) and then the gap was measure in 2 minute increments. The spectrometer was set at a 600 mm absorbency and each vial was measure, once again, in every two minute intervals. The purpose of this experiment was to better understand the logistics behind the fermentation process. In tube one, the succinctness was fumigated. The second tube differed in the fact that there was boiled water, which is not a suitable living indention for yeast, and therefore the enzyme was denatured.

There was no carbon dioxide produced when mixed with boiled water but without that variable’s presence, there was a greater amount of carbon emission. Tube three had an added inhibitor so therefore the rate of reaction was considered slow which can be observed in figure 1-1 . Adding the inhibitor meant that the enzyme was occupied and not in absorbency. Tube four, the final tube, had the most substrate included and due to this, the enzyme had a chance to bind to an activation site despite the inhibitor.

We did this by titrating Hydrochloric Acid into 10 com of Calcium Hydroxide. As we titrated HCI into the Calcium Hydroxide solution, the phenolphthalein’s pink shade given to the Calcium Hydroxide slowly became clearer until the complete lack of pink color in the solution of base and acid. My value of 1. 58 g/mol was close to the accepted value of . 59 g/mol with only a 0. 6 + 0. 007% error, which is almost identical to the theoretical value. Because of such a small percentage error, one can attribute the error to random error, and not systematic because of its small value.

Limitation/weakness How much did it affect my result Human error exists even outside uncertainty, where the HCI solution was possibly not correctly stirred along with the Calcium Hydroxide. Because of the fact that this is a titration experiment, the stirring and conglomeration of hydrochloric acid and calcium hydroxide needed to be constant in order for the reaction and correct mount of hydrochloric acid was used to neutralize the basic aspect of calcium hydroxide. Because of the inconsistent stirring, there must have been a small amount of error attributed to this limitation, though it may not look as much.

Possibility: The Calcium Hydroxide solid that did not dissolve into the water may have passed the filter into the actual solution used, causing discrepancies in the data. A second limitation/weakness that may have caused our error is because of the fact that we may not have been careful enough with the filtration of the saturated Calcium Hydroxide. Because the filters themselves may not have been perfect experiment might have taken a wrong turn with solid Calcium Hydroxide within the solution as well, needing a smidgen more Hydrochloric Acid to neutralize it.

Evaluation of quality of the result: Although there is limitations to our measuring, since we were only able to measure to 2 decimals as a volume, this experiment is still a valid way to figure out the concentration of Calcium Hydroxide with Just the amount and concentration of Hydrochloric Acid, and can be used by other scientists who would theoretically have no idea about how to go about the method of determining the concentration of Calcium Hydroxide. Limitation or weakness Improvement 1 . Human error exists even outside uncertainty, where the HCI solution was possibly not correctly stirred along with the Calcium Hydroxide.

To lower this error and improve this limitation, we could have the same person stir the solution/titration constantly, with more precision and concentration over this mixture, as to get the correct amount of HCI needed to neutralize the solution without any disagreements over what neutralized means or not. With two people concentrated over specific tasks, the uncertainty would also become much smaller as well due to less error attributed to our mistakes. 2. Possibility: The Calcium Hydroxide old that did not dissolve into the water may have passed the filter into the actual solution used, causing discrepancies in the data.

One of the biggest improvements we could make is that we could, this time around, not add such a large amount of Calcium Hydroxide powder into water blindly, as had been done before, so that we would be able to have more saturated water to work with, as opposed to what we had to work with at that moment when the experiment was done. By doing that, we would be able to reduce uncertainty that may have come along with the solid Calcium Hydroxide mixed in with the saturated solution.

University of Santo Tomas Faculty of Pharmacy Organic Chemistry Laboratory APPLICATION OF DIFFERENT KINDS OF TEST TO CLASSIFY HYROXY- AND CARBONYL-CONTAINING COMPOUNDS Jane Catherine SP. Villanueva, Edenn Claudine C. Villaraza, Lorenz Oliver C. Villegas and Cristel Bernice T. Wee Group 10 2G-Medical Technology Organic Chemistry Laboratory ABSTRACT Hydroxyl group refers to a functional group containing OH- when it is a substituent in an organic compound. It is also known as the characteristic functional group of alcohols and phenols.

On the other hand, carbonyl group refers to a divalent chemical unit consisting of a carbon and an oxygen atom connected by a double bond. It is known as the characteristic functional group of aldehydes and ketones instead. In this experiment, hydroxyl- or carbonyl- containing samples were given to the group for examination. The samples were analyzed through different tests namely the involvement of the solubility of alcohols in water, the Lucas Test, the Chromic Acid Test or also known as Jones Oxidation, the 2,4-Dinitrophenylhydrazone (2,4-DNP) Test, the Fehling’s Test, the Tollens’ Silver Mirror Test, and the Iodoform Test.

The solubility of alcohols in water test showed that the sample, benzyl alcohol was immiscible while ethanol was the most miscible from all the other compounds used. While in Lucas Test which was used to differentiated the primary, secondary, and tertiary alcohols had turned tert-butyl alcohol into a cloudy solution afterwards. In Chromic Acid Test which was a test for oxidizable compounds or any compounds that possess reducing property would yield to a blue green solution if it reacted positively. This was seen in all the sample used in this test except for acetone.

Whereas Dinitrophenylhydrazone (2,4-DNP) Test was preformed to test for aldehydes and ketones which would result to a yellow orange precipitate if it was positively reacted. All the compounds subjected to this test namely n-butyraldehyde, benzaldehyde and acetone gave a positive result. Fehling’s Test and Tollens’ Silver Mirror Test were used to tests for aldehydes. In Fehling Test, both the n-butyraldehyde and benzaldehyde gave a positive result which was a brick red precipitate but acetone gave a negative result which was only a blue solution.

While the Tollens’ Silver Mirror Test had shown that both n-butyraldehyde and benzaldehyde gave a positive result which was a silver mirror and then again acetone gave a negative result which was the absence of a silver mirror. Lastly Iodoform test was performed and was known as a test for methyl carbinol and methyl carbonyl groups. Both acetone and isopropyl alcohol resulted to a positive outcome in this test which was formation of yellow precipitate but n-butyraldehyde on the other hand yield to a negative result which was a yellow solution containing black precipitate. INTRODUCTION

In organic chemistry, classification of test was tests that categorize a substance into one of several classes. They were used to detect functional groups and other structural features. Alcohol were derivatives of hydrocarbons in which one or more of the hydrogen atoms have been replaced by a hydroxyl (-OH) functional group. Hydrocarbons are compounds which contain hydrogen (H) and carbon (C) only. The hydroxyl group imparts particular properties to the radical to which it is attached. [1] Figure 1. Alcohol Alcohols are classified into three categories: primary (1°), secondary (2°) and tertiary (3°).

This classification is based on the number of carbon-containing groups (R for an alkyl or an aromatic group) attached to the carbon bearing the hydroxyl group. If the carbon bearing the OH has one R group, the molecule is a primary alcohol. If two R groups are attached, it is then a secondary alcohol. If three R groups are attached, then the alcohol is tertiary[1][4] Figure 2. Three alcohol groups There are other molecules that contain an -OH group. Even though water (H2O) contains OH, it is not considered as an alcohol because alcohols were defined as organic compounds that have little or no ionization of the ydrogen. Other organic compounds that contain -OH groups but are not alcohols are phenol (C6H5OH) and acetic acid (CH3COOH). These compounds are not alcohols because they are acidic. The term alcohol, then, is another representation of a type of electronic structure in the molecules of substances. [3] [4] Phenols are aromatic compounds in which a hydroxide group is directly bonded to an aromatic ring system. They are very weak acids, and like alcohols, form ethers and esters. The main phenols are phenol itself, cresol, resorcinol, pyrogallol, and picric acid.

Phenol itself (C6H5OH), also known as carbolic acid, is a white, hygroscopic crystalline solid, isolable from coal tar, but made by acid hydrolysis of cumene hydroperoxide, or by fusion of sodium benzenesulfonate with sodium hydroxide. Formerly used as an antiseptic, phenol has more latterly been used to make bakelite and other resins, plastics, dyes, detergents, and drugs. [4] [15] The hydroxyl- containing compounds used in the experiment were ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, isopropyl alcohol, and benzyl alcohol.

Ethanol also known as ethyl alcohol is a clear, colorless liquid with a characteristic, agreeable odor. In dilute aqueous solution, it has a somewhat sweet flavor, but in more concentrated solutions it has a burning taste. Its low freezing point has made it useful as the fluid in thermometers for temperatures below –40°C, the freezing point of mercury, and for other low-temperature purposes, such as for antifreeze in automobile radiators. Ethanol is miscible in all proportions with water and with most organic solvents. It is useful as a solvent for many substances and in making perfumes, paints, lacquer, and explosives. 15] Figure 3. Structure of Ethanol n-butyl alcohol also known as n-butanol, 1-Butanol or 1-butyl alcohol is a four carbon straight chain alcohol. It is a volatile, clear liquid with a strong alcoholic odor, and is miscible with water. It is a highly refractive compound which corrodes some plastics, and rubbers. It is miscible with many organic solvents, and incompatible with strong oxidizers. It is also used as a direct solvent and as an intermediate in the manufacture of other organic chemicals. [7] Figure 4. Structure of n-butyl alcohol

Sec-butyl alcohol, a four carbon secondary alcohol, is a volatile, clear liquid with a strong alcoholic odor with a water solubility of 12. 5%. This substance is most hazardous when peroxide levels are concentrated by distillation or evaporation. It is a highly refractive compound which corrodes some plastics, and rubbers. It is miscible with many organic solvents, and incompatible with strong oxidizers. It is flammable strongly with a luminous flame. It is used as a direct solvent and as an intermediate in the manufacture of other organic chemicals. [8] Figure 5.

Structure of Sec-butyl alcohol Tert-butyl alcohol is a clear, noncorrosive liquid. It is miscible with water as well as most common organic solvents. The sterically hindered tertiary butyl group imparts stability compared to primary and secondary alcohols. As a result, the solubility and oxidative stability characteristics provide many industrial applications as a reaction and process solvent and chemical intermediate. It is used as a non-reactive solvent for chemical reactions, a non-surfactant compatibilizer for many solvent blends, and a non-corrosive solvent.

It is used in free radical polymerizations to dissolve monomers. TBA is a main raw material of tert-butyl functional group in organic synthesis. [9] Figure 6. Structure of Tert-butyl alcohol Isopropyl alcohol also known as propan-2-ol, 2-propanol is a common name for a chemical compound with the molecular formula C3H8O. It is a colorless, flammable chemical compound with a strong odor. It is the simplest example of a secondary alcohol, where the alcohol carbon is attached to two other carbons. Being a secondary alcohol, isopropyl alcohol can be oxidized to acetone, which is the corresponding ketone.

Isopropyl alcohol dissolves a wide range of non-polar compounds. It is also relatively non-toxic and evaporates quickly. Thus it is used widely as a solvent and as a cleaning fluid, especially for dissolving lipophilic contaminants such as oil. [10] Figure 7. Structure of Isopropyl alcohol Benzyl alcohol (C6H5CH2OH) is a colorless liquid with a mild pleasant aromatic odor. It is a useful solvent due to its polarity, low toxicity, and low vapor pressure. Benzyl alcohol is partially soluble in water (4 g/100 mL) and completely miscible in alcohols and diethyl ether.

Like most alcohols, it reacts with carboxylic acids to form esters. Benzyl alcohol is used as a general solvent for inks, paints, lacquers, and epoxy resin coatings. It is also a precursor to a variety of esters, used in the soap, perfume, and flavor industries. It is often added to intravenous medication solutions as a preservative due to its bacteriostatic and antipruritic properties. [15] Figure 8. Structure of Benzyl alcohol Carbonyl group is a divalent chemical unit consisting of a carbon and an oxygen atom connected by a double bond.

The group is a constituent of carboxylic acids, esters, anhydrides, acyl halides, amides, and quinones, and it is the characteristic functional group of aldehydes and ketones. Carboxylic acid and their derivatives, aldehydes, ketones, and quinones are also known collectively as carbonyl compounds. Aldehydes and ketones contain carbonyl groups attached to alkyl or aryl groups and a hydrogen atom or both. These groups have little effect on the electron distribution in the carbonyl group; thus, the properties of aldehydes and ketones are determined by the behavior of the carbonyl group.

In carboxylic acids and their derivatives, the carbonyl group is attached to one of the halogen atoms or to groups containing atoms such as oxygen, nitrogen, or sulfur. These atoms do affect the carbonyl group, forming a new functional group with distinctive properties. Figure 9. Carbonyl Group An aldehyde is an organic compound containing a terminal carbonyl group. This functional group, called an aldehyde group, consists of a carbon atom bonded to a hydrogen atom with a single covalent bond and an oxygen atom with a double bond.

Thus the chemical formula for an aldehyde functional group is -CH=O, and the general formula for an aldehyde is R-CH=O. The aldehyde group is occasionally called the formyl or methanoyl group. The word aldehyde is a combination of parts of the words alcohol and dehydrogenated, because the first aldehyde was prepared by removing two hydrogen atoms (dehydrogenation) from ethanol. Molecules that contain an aldehyde group can be converted to alcohols by the addition of two hydrogen atoms to the central carbon oxygen double bond (reduction).

Organic acids are the result of the introduction of one oxygen atom to the carbonyl group (oxidation). Aldehydes are very easy to detect by smell. Some are very fragrant, and others have a smell resembling that of rotten fruit. [15] On the other hand, Ketone features a carbonyl group (C=O) bonded to two other carbon atoms. They differ from aldehydes in that the carbonyl is placed between two carbons rather than at the end of a carbon skeleton. They are also distinct from other functional groups, such as carboxylic acids, esters and amides, which have a carbonyl group bonded to a hetero atom.

Ketone compounds have important physiological properties. They are found in several sugars and in compounds for medicinal use, including natural and synthetic steroid hormones. [15] The difference between aldehydes and ketones is in the groups that are attached to the carbonyl carbon atom. In the case of an aldehyde, there is always at least one H atom attached to the carbonyl carbon atom. An aldehyde has one R group attached. R stands for any other organic chain or group. In the case of ketones, there are no H atoms attached to the carbonyl carbon. The ketone has two R groups attached. [2] [15] Figure 10.

Structure of Aldehyde and Ketone Some of the carbonyl-containing compounds used in the experiment were benzaldehyde, n-butraldehyde, acetaldehyde, acetone and acetophenone. Benzaldehyde (C6H5CHO) also known as benzenecarbonal is a colorless liquid aldehyde with a characteristic almond odor. It boils at 180°C, is soluble in ethanol, but is insoluble in water. It is formed by partial oxidation of benzyl alcohol, and on oxidation forms benzoic acid. It is called oil of bitter almond, since it is formed when amygdalin, a glucoside present in the kernels of bitter almonds and in apricot pits, is hydrolyzed, e. . , by crushing the kernels or pits and boiling them in water; glucose and hydrogen cyanide (a poisonous gas) are also formed. It is also prepared by oxidation of toluene or benzyl chloride or by treating benzal chloride with an alkali. Benzaldehyde is used in the preparation of certain aniline dyes and of other products, including perfumes and flavorings. [13] Figure 11. Structure of Benzaldehyde Acetaldehyde (CH3CHO) also known as ethanol is a colorless liquid aldehyde, sometimes simply called aldehyde. It is soluble in water and ethanol.

Acetaldehyde is made commercially by the oxidation of ethylene with a palladium catalyst. It is used as a reducing agent (e. g. , for silvering mirrors), in the manufacture of synthetic resins and dyestuffs, and as a preservative. [11] Figure 12. Structure of Acetaldehyde n-butyraldehyde (CH3(CH2)2CHO) also known as butanal is an aldehyde derivative of butane. It is a colorless flammable liquid that smells like sweaty feet. It is miscible with most organic solvents. n-butyraldehyde is used as an intermediate in the manufacturing plasticizers, alcohols, solvents and polymers.

It is also used as an intermediate to make pharmaceuticals, agrochemicals, antioxidants, rubber accelerators, textile auxiliaries, perfumery and flavors. [12] Figure 13. Structure of N-butyraldehyde Acetone ((CH3)2CO) also known as propanone is colorless, mobile, flammable liquid with a characteristic sweetish smell is the simplest example of the ketones. Acetone is miscible with water and serves as an important solvent in its own right, typically as the solvent of choice for cleaning purposes in the laboratory. [6] Figure 14. Structure of Acetone Acetophenone (C6H5C(O)CH3) is the simplest aromatic ketone.

This colorless, viscous liquid is a precursor to useful resins and fragrances. It can be obtained by a variety of methods. In industry, acetophenone is recovered as a by-product of the oxidation of ethylbenzene, which mainly gives ethylbenzene hydroperoxide for use in the production of propylene oxide. [5] Figure 15. Structure of Acetophenone The hydroxyl- and carbonyl- containing compounds were analyzed by utilization of different tests such as testing the solubility of alcohols in water, Lucas Test, Chromic Acid Test (Jones Oxidation), 2,4-Dinitrophenylhydrazone Test, Fehling’s Test, Tollens’ Silver Mirror Test, and Iodoform Test.

Most organic compounds were not soluble in water with the exception of low molecular-weight amines and oxygen-containing compounds like alcohols, carboxylic acids, aldehydes, and ketones. Low molecular-weight compounds are generally limited to those with fewer than five carbon atoms. [14] Lucas Test often provides classification information for alcohols, as well as a probe for the existence of the hydroxyl group. Substrates that easily give rise to cationic character at the carbon bearing the hydroxyl group undergo this test readily; primary alcohols do not give a positive result.

Since the Lucas Test depends on the appearance of the alkyl chloride as a second liquid phase, it is normally applicable only to alcohols that are soluble in the reagent. This limits the test in general to monofunctional alcohols lower than hexyl and certain polyfunctional molecules. [4] Chromic Acid Test also called Jones Oxidation detects the presence of a hydroxyl substituent that is on a carbon bearing at least one hydrogen, and therefore oxidizable. It is detected by the appearance of Cr+3 ion. This test can be used to differentiate aldehydes and ketones.

A positive result would show green or blue-green solution. [4] 2,4-Dinitrophenylhydrazone Test can be used to qualitatively detect the carbonyl functionality of a ketone or aldehyde functional group. Ketones and Aldehydes would form yellow to orange precipitate after undergoing in this test. [4] Fehling’s Test and Tollens’ Silver Mirror Test are used to detect aldehydes. However, Fehling’s solution can only be used to test for aliphatic aldehydes, whereas Tollens’ reagent can be used to test for both aliphatic and aromatic aldehydes.

A positive result in Fehling’s Test would give a brick red precipitate while in Tollens’ Silver Mirror, it is the formation of silver mirror. [4] Iodoform Test is a test for methyl carbinol and methyl carbonyl group. A positive result would yield to yellow crystals or precipitate. Its mechanism occurs through a series of enolate anions which are iodinated. [4] The objectives of the experiment were to distinguish whether a compound was a hydroxyl- or carbonyl-containing, to differentiate the three types of alcohols, to differentiate aldehydes from ketones and to explain the mechanisms involved in the differentiating tests.

EXPERIMENTAL A. Compounds Tested * Ethanol * n-butyl alcohol * Sec-butyl alcohol * Tert-butyl alcohol * Benzyl alcohol * n- butyraldehyde * Benzaldehyde * Acetone * Acetophenone * Isopropyl alcohol * Acetaldehyde * Lucas reagent * Chromic acid reagent * 95% ethanol * Fehling’s A and B * Tollen’s reagent * 5% NaOCl solution * Iodoform test reagent * 2,4-dinitrophenylhydrazine B. Procedure 1. Testing the solubility of alcohols in water The samples involved in the experiment were ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, and benzyl alcohol.

Five test tubes were labeled with each of the alcohol samples. With the aid of a Pasteur pipette, 10 drops from each of the samples were taken then placed into the appropriate test tube. To the tube containing ethanol, 1-ml of water was then added drop wise to the tube containing alcohol and the mixture was shaken thoroughly after each addition. If cloudiness resulted, 0. 25-ml of water at a time was added continuously with vigorous shaking until a homogeneous dispersion results. The total volume of water added was noted. If cloudiness resulted after the addition of 2. -ml of water, the alcohol is said to be immiscible in water but if there was no cloudiness then it is miscible to water. The results were noted down. The same procedure was performed on the test tubes containing n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, and benzyl alcohol. 2. Using the Lucas Test This test was performed on n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol and isopropyl alcohol. Lucas reagent was prepared by dissolving 16 g of anhydrous zinc chloride in 10-ml of concentrated hydrochloric acid. The mixture was then allowed to cool.

The Lucas Reagent was already prepared beforehand. 50-mg or 2-3 drops of the sample was added to 1-ml of the reagent in a test tube and the mixture was shaken vigorously for a few seconds. The mixture was allowed to stand at room temperature. The rate of formation of the cloudy suspension or the formation of two layers was observed. 3. Using the Chromic Acid Test / Jones Oxidation This test was performed on n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzaldehyde and acetone. 1 drop of liquid or a small amount of the solid sample was dissolved in 1-ml of acetone in a small vial or test tube. drops of 10% aqueous Potassium chromate solution and 5 drops of 6M sulphuric acid were added into the mixture. 4. Using the 2,4-DNP Test This test was performed on acetone, n-butyraldehyde and benzaldehyde. The reagent was prepared by slowly adding a solution of 3 g of 2,4-dinitrophenylhydrazine in 15-ml of concentrated sulphuric acid, while stirring to a mixture of 20-ml of water and 70-ml of 95% ethanol. The solution was then stirred and filtered. This reagent was already prepared beforehand. A drop of a liquid sample was placed into a small sample. 5 drops of 95% ethanol was added and well shaken.

Afterwards, 3 drops of 2,4-DNP was added and if no yellow or orange precipitate formed, the solution was allowed to stand for at least 15 minutes. 5. Using the Fehling’s Test This test was performed on acetone, n-butyraldehyde, and benzaldehyde. Fehling’s reagent was prepared by mixing equal amounts of Fehling’s A and Fehling’s B. Fehling’s A was prepared by dissolving 7 g of hydrated copper (II) sulfate in 100-ml of water. Fehling’s B was prepared by mixing 35 g of Potassium sodium tartrate and 10 g of Sodium hydroxide in 100-ml water. Then, 1-ml of freshly prepared Fehling’s reagent was placed into each test tube. drops of the sample to be tested was added in to the tube. The tubes were then placed in a beaker of boiling water and changes within 10-15 minutes were observed. 6. Using the Tollens’ Silver Mirror Test This test was performed on benzaldehyde, acetone and n-butyraldehyde. The reagent was prepared by adding 2 drops of 5% Sodium hydroxide solution to 2-ml of 5% Silver nitrate solution and mixing thoroughly. Next, only enough 2% ammonium hydroxide (concentrated ammonium hydroxide is 28%) was added drop by drop and with stirring to dissolve the precipitate.

Adding excess ammonia will cause discrepancies on the result of the test. Then, four test tubes with 1-ml of freshly prepared Tollens’ reagent were prepared. Two drops each of the samples were then added. The mixture was shaken and allowed to stand for 10 minutes. If no reaction has occurred, the test tube was placed in a beaker of warm water (35-50 oC) for 5 minutes. Observations were recorded. It was noted that if Tollens’ reagent is left unused for a period of time, it may form explosive silver. This was avoided by neutralizing unused reagent with a little nitric acid and discarded afterwards. . Using the Iodoform Test This test was performed on acetone, n- butyraldehyde and isopropyl alcohol. 2 drops of each sample was placed into its own small vial or test tube. 20 drops of fresh chlorine bleach (5% Sodium hypochlorite) was slowly added while shaking to each test tube and then, mixed. The formation of a yellow participate was noted. RESULTS AND DISCUSSION 1. Solubility of Alcohols in Water In the experiment, five compounds were tested to determine the presence of the –OH, hydroxyl group through solubility of the sample in water.

The presence of an –OH group was indicated by the miscibility of the substance. This follows the general rule in solubility that “like dissolves like”. Meaning, a polar solute will dissolve in a polar solvent and a non polar solute will be insoluble in a polar solvent. [14] Going back to the experiment, it was observed that alcohol was soluble in water but as the number of carbon atoms in the carbon chain of the alcohol increased, the solubility of the alcohol sample decreased. It was also observed that branching of the compound increased its solubility in water.

Branching will increase solubility since more branching will reduce the size of the molecule and make it easier to solvate the molecules with the solvent. [14] The results of the experiment show that the solubility of alcohols in water depends on the balance between the strength of the hydrogen bonds formed between water and the hydroxyl group, and the strength of the Van der Waals forces between the hydrocarbon chains of the alcohol. Alcohol| Condensed Structural Formula| Amount of Water (in ml) needed to produce a homogeneous dispersion| Solubility in Water| Ethanol| CH3CH2OH| 0. ml| Most Miscible| n-butyl alcohol| CH3CH2CH2CH2OH| 2. 0 ml| Miscible| Sec-butyl alcohol| | 1. 4 ml| Miscible| Tert-butyl alcohol| | 0. 5 ml| Miscible| Benzyl alcohol| | More than 2. 0 ml| Immiscible| Table 1. Solubility of alcohols in water The table above showed that ethanol, n-butyl alcohol, sec-butyl alcohol, and tert-butyl alcohol were all miscible with water. Only benzyl alcohol had exhibited immiscibility with water. As stated, all alcohols were soluble in water except under C6. Hence, ethanol, n-butyl alcohol, sec-butyl alcohol, and tert-butyl alcohol are all miscible with water.

Ethanol has two carbon atoms, while the other three all have four carbons since they are all derivatives of the alcohol, butanol. Benzyl alcohol was immiscible with water because it is an aromatic alcohol. Ethanol was the most miscible alcohol followed by tert-butyl alcohol, sec-butyl alcohol, and n-butyl alcohol. Ethanol exhibited fastest solubility because it has only two carbon atoms as compared to the butanol derivatives having four carbon atoms. Tert-butyl alcohol was the most miscible among the butanol derivatives because it has the most branching substituents present. 2. Lucas Test

The four types of alcohols namely n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol and isopropyl alcohol were differentiated from each other by way of the Lucas Test. Lucas Test differentiates primary, secondary, and tertiary alcohols. Reagents used include anhydrous ZnCl2 and HCl. Positive result was based on its turbidity or alkyl chloride formation and its rate of the reaction. Tertiary alcohols formed the second layer in less than a minute. Secondary alcohols required 5-10 minutes before formation of second layer while primary alcohols were usually unreactive. Substance| Condensed Structural Formula| Reaction| -butyl alcohol| CH3CH2CH2CH2OH| Clear solution(+)| Sec-butyl alcohol| | Clear solution(+)| Tert-butyl alcohol| | Turbid (+++) /Cloudy solution and formation of two layers| IsopropylAlcohol| | ClearSolution(+)| Table 2. Lucas Test Based on Table 2, it was only tert-butyl alcohol which had immediately formed two layers or a cloudy solution; hence, it was known to be a tertiary alcohol. Sec-butyl alcohol and Isopropyl alcohol when subjected to Lucas test resulted to a clear solution although theoretically, a secondary alcohol dissolves to give a clear solution then form chlorides which would yield to a cloudy solution within five minutes. -butyl alcohol was considered as a primary alcohol. It was unreactive but eventually would react after long period of time. Generally, the order of reactivity of the alcohols toward Lucas reagent was 3°;2°;1° because the reaction rate was much faster when the carbocation intermediate was more stabilized by a greater number of electron donating alkyl group bonded to the positive carbon atom. This means that the greater the alkyl groups present in a compound, the faster its reaction would be with the Lucas solution. [1] Figure 16.

Reaction in Lucas Test 3. Chromic Acid Test (Jones Oxidation) This test was performed on n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzaldehyde and acetone. The chromic acid test classifies the three types of alcohols by oxidizing the alcohol. The test was also used to be able to distinguish aldehydes from ketones. Since primary and secondary alcohols were also oxidized by the chromic acid reagent, this test was not useful for distinguishing aldehydes unless a positive identification of a carbonyl group has been obtained from the 2,4-DNP test.

Chromic acid has an orange-red color due to the presence of Cr+6 ions, upon oxidation of the aldehyde, the chromium was reduced to Cr+3, which had a green color. A positive result was indicated by a green precipitate due to chromous sulfate, Cr? (SO? )?. [1] From the results, it was noted that the formation of an opaque blue-green suspension within 2-3 seconds, accompanied by disappearance of the orange color of the reagent, indicates a primary or secondary alcohol. A primary alcohol oxidizes readily, first to an aldehyde, then to a carboxylic acid.

These two oxidation steps made sense because the primary alcohol functional group has two C-H bonds that can be broken; secondary alcohols were oxidized to ketones, a secondary alcohol only has one C-H bond that can be broken, so it can only oxidize once, to a ketone; a tertiary alcohol has no C-H bond that can be broken, so it was not oxidized, no matter how strong the oxidizing agent was. During the oxidation, the orange-red color of the chromic acid changed to a blue-green solution. Figure 17. Oxidation of the three types of Alcohols The results also show that aldehydes gave the same result but reacted more slowly.

With aliphatic aldehydes, the solution turned cloudy in about 5 seconds, and the opaque blue-green suspension formed within 30 seconds; aromatic aldehydes required 30-90 seconds or longer before a suspension formed. The generation of some other dark color, particularly with the liquid remaining orange, was considered a negative test. It was concluded that alcohols and aldehydes are oxidized by chromic acid but ketones were not because they don’t have a hydrogen atom attached to their carbonyl group that can be used for oxidation. Figure 18. Oxidation of Aldehyde Substance| Condensed Structural Formula| Reaction| -butyl alcohol| CH3CH2CH2CH2OH| Blue green solution(+)| Sec-butyl alcohol| | Blue green solution(+)| Tert-butyl alcohol| | Blue green solution (+)| n-butyraldehyde| | Blue greenSolution (+)| Benzaldehyde| | Blue green solution(+)| Acetone| | Green solution(-)| Table 3. Reactions to the Chromic Acid Test It was observed that all the compounds tested gave a visible positive result, a blue green solution, except for acetone which had yielded to a green solution. 4. 2,4-dinitrophenylhydrazone Test This test was performed on acetone, n-butyraldehyde and benzaldehyde.

The 2,4-dinitrophenylhydrazone (2,4-DNP) test determined the presence of a carbonyl group in the sample compound. The test used an organic reactant, 2,4-dinitrophenylhydrazine, to distinguish the carbonyl compounds, aldehydes and ketones, from the non-carbonyl compounds, alcohols. The 2, 4-dinitrophenylhydrazine reagent was a translucent yellow solution. When this reagent was subjected in the presence of a carbonyl compound, a yellow colored precipitate would form while in the presence of an alcohol, the solution would remain translucent yellow with no precipitate formed.

The reaction of 2,4-DNP with an aldehyde or ketone was a condensation reaction. Under less acidic conditions, in this type of reaction, a nucleophile donates a pair of electrons toward the carbonyl carbon forming a single bond to it. [2] At the same time the double bond between the carbonyl carbon and oxygen becomes a single bond as one bonding pair of electrons in the double bond moves to become an unshared pair on the oxygen. The oxygen now has one bond to it and it holds three pairs of unshared electrons, so it has a negative charge.

Consequently, the oxygen picks up a proton from somewhere and becomes an -OH group. The proton from the acid attaches itself to one of the unshared pairs of electrons on the oxygen. The carbonyl group now has a +1 charge and is very inviting to even a weak nucleophile. So, the nucleophile attacks the carbonyl carbon forming a bond and the doubly bonded oxygen of the carbonyl becomes an -OH, as before. [1] Figure 19. Nucleophilic addition of 2,4-DNP to Acetone. As seen just below, this product is not usually the one that was isolated.

Rather this product undergoes an elimination reaction in which the -OH was removed from the carbon to which it is attached and the hydrogen was removed from the nitrogen immediately to the right, resulting in a double bond between the nitrogen and carbon and a molecule of water. The final product was known as a 2,4-dinitrophenylhydrazone. That is why this reaction was also considered as an elimination reaction. Figure 20. Elimination reaction of DNP Figure 21. Reaction of 2,4-DNP with a Carbonyl group Substance| Condensed Structural Formula| Reaction| n-butyraldehyde| | Yellow- orange precipitate(+)|

Benzaldehyde| | Yellow –orange precipitate(+)| Acetone| | Yellow –orange precipitate(+)| Table 4. Reactions to the 2,4- DNP Test As shown on table 4, it was observed that there was a formation of a yellow – orange precipitate in all the compounds used. This would then indicate a presence of either an aldehyde or a ketone. 5. Fehling’s Test Fehling’s test differentiated aldehydes and ketones. It was based upon the ability of the aldehyde group to reduce the Cu+2 ion of Cu(OH)? , a blue color, to the Cu?? ion of Cu? O, a dark red color, in the presence of a base.

Fehling’s solution contains copper (II) ions complex with tartrate ions in sodium hydroxide solution. Complexion of the copper (II) ions with tartrate ions prevents precipitation of copper (II) hydroxide. Aldehydes reduce the complex copper (II) ion to copper (I) oxide, changing the color of the solution to brick red or dark green. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. [2] In short it involved a redox reaction wherein aldehyde was oxidized to carboxylic acid and ketones did not undergo oxidation. Copper was reduced from Cu2+ to Cu+. Figure 22.

Oxidation of aldehyde to carboxylic acid through Fehling’s test It was a test for aldehydes. Reagents include CuSO4, NaOH. A positive result is the formation of brick red precipitate (Cu2O/cuprous oxide). This test was performed on acetone, n-butyraldehyde, and benzaldehyde. Substance| Condensed Structural Formula| Reaction| n-butyraldehyde| | Brick red precipitate (+)| Benzaldehyde| | Brick red precipitate(+)| Acetone| | Clear blue solution(-)| Table 5. Reactions to the Fehling’s Test As shown in Table 5, n-butyraldehyde and benzaldehyde exhibited positive result while acetone exhibited an absence of brick red precipitate.

It can be concluded that n-butyraldehyde and benzaldehyde were both aldehyde. 6. Tollens’ Silver Mirror Test Tollens’ silver mirror test was a test for aldehydes. Tollen’s reagent was an ammoniacal solution of silver ion prepared by dissolving silver oxide in ammonia. The preparation of the reagent is based on the formation of a silver diamine complex that is water soluble in basic solution. In this reaction, the aldehyde was oxidized to a carboxylic acid while the Ag+1was reduced to silver metal, which deposited as a thin film on the inner surface of the glass.

The generic reaction was as follows and was specific for aldehydes. [16] Figure 23. Oxidation of aldehyde with Tollen’s reagent Substance| Condensed Structural Formula| Reaction| n-butyraldehyde| | Silver Mirror(+)| Benzaldehyde| | Silver Mirror(+)| Acetone| | Clear grayish-black solution (-)| Table 6. Reactions to Tollen’s Silver Mirror Test This test was performed on benzaldehyde, acetone and n-butyraldehyde. Based on the results seen in table 6, it was concluded that the aldehyde samples produced a silver mirror on the inner surface of the test tube since aldehydes were easily oxidized.

The ketones sample, acetone, on the other hand, didn’t form this mirror image because of its inability to oxidize due to the lack of a hydrogen atom attached to its carbonyl group that could be used for oxidation. 7. Iodoform Test Iodoform test was a test for methyl carbinol, secondary alcohol with adjacent methyl group, and methyl carbonyl. Methyl ketones, but not other ketones, were oxidized by iodine in aqueous sodium hydroxide. The ketone was oxidized to a carboxylic acid which yellow iodoform/ precipitate would be formed. It was the yellow precipitate formed would be the basis of a positive result.

Acetaldehyde, but not other aldehydes, would yield to a positive result in this test owing to its structural similarity to methyl ketones. It was also true that ethanol would be oxidized to acetaldehyde and secondary alcohols that could be oxidized to methyl ketones given this test. [2] Figure 24. Oxidation of a methyl ketone Substance| Condensed Structural Formula| Reaction| N-butraldehyde| | Yellow solution with black precipitate (-)| Acetone| | Yellow precipitate| Isopropyl alcohol| | Yellow precipitate| Table 7. Reaction to the Iodoform Test

This test was performed on acetone, n- butyraldehyde and isopropyl alcohol. Based on table 7, the results indicate that the methyl ketones of isopropyl alcohol and acetone were oxidized by iodine to carboxylic acids because the compounds formed a yellow precipitate while n-bytraldehyde didn’t. It was then concluded that compounds with a methyl group next to the carbonyl group would give a positive result in the iodoform test, ethanol and secondary alcohols with the methyl group attached to the same carbon as the OH- group would also give a positive result.

During the experiment, the compounds acetaldehyde and acetophenone were not available. This was the reason why results of these compounds in different tests were not observed but based from different informations which were gathered from different sources. Acetophenone would give a positive result in the following test namely 2,4 DNP test and Iodoform test. While acetaldehyde would give a positive result in the following test namely Chromic Acid test, 2,4 DNP test, Fehling’s test and as well as Tollens’ Silver Mirror test. REFERENCES: From books: [1]Lehman, John W(2009).

Operational Organic Chemistry: A Problem-Solving Approach to the Laboratory Course. Upper Saddle River, New Jersey: Pearson Prentice. [2]Martin, Stephen F(2011). Organic Chemistry Laboratory Experiments: Miniscale and Microscale. Brooks/Cole Cengage Learning. [3]McMurry, John(2010). Foundations of Organic Chemistry, Philippine Edition. Cengage Learning. [4]Shriner, Ralph Lloyd (1980). Systematic Identification of Organic Compound: A Laboratory Manual (6th Ed. ). John Wiley ; Sons, Inc. New York: Van Hoffmann Press. From Websites: [5]Acetophenone. www. chemicalland21. om/industrialchem/solalc/ACETOPHENONE. htm 09/09/11 [6]Acetone. www. chemicalland21. com/industrialchem/solalc/ACETONE. htm 09/09/11 [7]n-butyl alcohol. www. chemicalland21. com/industrialchem/solalc/NBUTYLALCOHOL. htm 09/09/11 [8]Sec-butyl alcohol. www. chemicalland21. com/industrialchem/solalc/2-BUTANOL. htm 09/09/11 [9]Tert-butyl alcohol. www. chemicalland21. com/industrialchem/solalc/TERTBUTYL%20ALCOHOL 09/10/11 [10]Isopropyl alcohol. www. chemicalland21. com/petrochemical/ISOPROPANOL. htm 09/10/11 [11]Acetaldehyde. www. ntp. niehs. nih. gov/ntp/roc/eleventh/profiles/s001acet. df09/10/11 [12]n-butyraldehyde. www. chemicalland21. com/industrialchem/organic/N-BUTYRALDEHYDE. htm 09/10/11 [13]Benzaldehyde. www. chemicalland21. com/specialtychem/perchem/BENZALDEHYDE. htm 09/10/11 [14]Solubility of Things. www. solubilityofthings. com/water/alcohols 09/09/11 [15]Alcohols, Aldehydes and Ketones. www. ipfw. edu/chem/112/kimble/3-Alcohol%20Aldehyde%20Ketones. pdf 09/10/11 From scientific journals: [16]Ennis, J. L. and E. S. Shanley. “Silver Nitrides. ” Journal of Chemical Education (1991): 68, A6. “Silver Nitrides. ” Journal of Chemical Education (1991): 68, A6.

THE MOLECULAR BASIS OF INHERITANCE I. History A. Discovery of “transformation” – a change in genotype and phenotype due to the uptake of external DNA by a cell 1. Griffith 1920s did experiments with Streptococcus pneumoniae (p294 fig16. 2) a. took two strains of S. pneumoniae, one virulent, one not b. heat killed virulent strain, then mixed them with the living nonvirulent strain c. living nonvirulent strain became virulent d. nonvirulent strain took on virulent strain’s DNA ? became virulent e. see p294 fig16. 2 S strain = virulent, R = nonvirulent f. ventually Griffith’s work lead way to more studies on DNA being the carrier of genetic info. B. Proof that viral DNA and not viral protein contains genetic information to make more viral particles 1. Hershey and Chase 1950s p295 fig16. 4 a. knew that viruses could infect bacteria and make more viruses using the host cell’s replicating ‘machinery’ b. background: sulfur gets incorporated into virus’ protein/phosphorus into virus’ DNA c. took T4 (bacteriophage) and plated with a lawn of E. coli and radioactively labeled sulfur, result = T4 with radioactive labeled protein (DNA not labeled) d. took T4 and plated with E. oli and radioactively labeled phosphorus = T4 with radioactively labeled DNA (protein not labeled) e. background: when virus + bacteria is spun down, viral particles in supernatant and bacteria in pellet f. took T4 (S-labeled) infected new E. coli lawn, spun down, found S-radioactive labels in supernatant g. took T4 (P-labeled) infected new E. coli lawn, spun down, found P-radioactive labels in pellet h. result = it’s the DNA that’s injected into the host to make more virus (even plated these spun down pellet bacteria, and they lysed and released new virus C. Discovery of the structure of DNA 1.

Watson and Crick a. used an x-ray crystallography picture (p297) by Franklin to determine DNA as a double-helical structure b. review p298 – A pairs with T and G with C/ A and G are purines and C and T are pyrimidines/double hydrogen bonds between A and T, and triple between G and C II. DNA Replication A. 3 models of DNA replication p300 fig 16. 10 1. Conservative model – the parental helix splits, copies, then goes back together again to remain intact while a second entirely new copy is made 2. Semiconservative model – the parental helix splits, copies and remains a part of the two new helixes 3.

Dispersive model – the parental helix splits unevenly, copies and remains a part of the two new helixes but in pieces B. Experimental proof p300 fig16. 11 1. added radioactively labeled heavy nitrogen to replicating bacteria, then placed this culture into radioactively labeled light nitrogen (used to distinguish strands) 2. allowed bacteria to replicate again, results gave hybrid DNA strands (ruled out conservative model) (note: both hybrids half and half and totally mixed look the same, so semiconservative and dispersive models both upheld this time- see below) 3. llowed bacteria to replicate again, results gave hybrid strands and only light double strands (ruled out dispersive model since all should be mixed if this was right) C. Origins of replication p301 fig16. 12 1. origin of replication – site where DNA replication begins a. proteins recognize a specific sequence on the template DNA, open the dsDNA to make a bubble, and begin replication b. replication fork – location on DNA strand where new DNA strand is growing 1. prokaryotes plasmid (single circular dsDNA helix) have one origin of replication and replication occurs in both directions 2. ukaryotes have linear dsDNA have many origins and replication occurs in both directions D. Elongation of new DNA 1. DNA polymerase – enzyme that synthesizes the new DNA strand by adding nucleotides to the growing strand 2. DNA polymerase receives energy to do this by nucleotides being nucleoside triphosphate (CTP, GTP, ATP, TTP) since they lose Pii = exergonic reaction to supply energy E. DNA is antiparallel p302 1. carbon numbering – carbon attached to base is 1’, count clockwise, carbon attached to phosphate group is 3’, carbon attached to other phosphate group is 5’ 2. be able to find 5’ vs 3’ end . (p302 fig 16. 14) replication occurs 5’ ? 3’, so strand being made in this direction is called the leading strand and replication occurs toward the replication fork 4. lagging strand is replication that occurs 5’ ? 3’ but replication moves away from the replication fork a. lagging strand produces Okazaki fragments which must be connected with DNA ligase p303 fig 16. 15 F. Priming DNA synthesis (getting replication started) p303 fig16. 15 1. primer – existing RNA polynucleotide on the template DNA strand since DNA polymerase cannot just start adding new nucleotides on its own a. rimer is laid down by enzyme primase b. only one primer required for leading strand to begin synthesizing/new RNA primer required for each lagging strand beginning c. DNA polymerase eventually replaces RNA nucleotides with DNA ones and occurs before ligase connects any lagging DNA strands G. Other assisting proteins 1. helicase – enzyme that unwinds dsDNA at the replication form 2. single-strand binding proteins – hold apart template DNA while replication occurs **FINAL GOOD SUMMARY P304 fig 16. 16 III. DNA Proofreading and Repair

A. Mismatch repair 1. as DNA polymerase lays down nucleotides, if it notices a mismatched one to template, will remove and replace with correct one 2. Excision repair p305 fig16. 17: consists of nuclease – enzyme that can cut out damaged segments of a DNA strand, then new nucleotides are filled in based on what the other DNA strand sequence is by DNA polymerase and ligase IV. Replication of the ends of DNA strands p306 fig16. 18 A. DNA polymerase can only add nucleotides to a 3’ end (since it grows in a 5’ ? 3’ direction) B.

For lagging strand, there is no problem since it replaces RNA primer and joins DNA with ligase C. For leading strand, there is a problem, since the 3’ end of the template strand has a RNA primer, which cannot be replaced with DNA nucleotides (by DNA polymerase) since there is no 3’ end to start from (DNA polymerase cannot just add nucleotides opposite of the DNA template strand – must use a RNA primer) D. This results in successive replicated strands becoming shorter and shorter – the remedy? E. Telomeres – eukaryotic cells have short repetitive nucleotide sequences that do not code for anything 1. elomeres protect the cell from false alarms that there is DNA damage and cause the cell to die since losing these ends don’t mean anything (note that prokaryotes do not have this problem since their DNA is circular with no “end”) 2. but when telomeres are lost, are they replaced? Yes by telomerase – enzyme that works in conjunction with DNA polymerase to add length to telomeres a. p306 fig16. 19 have shortened “just made” DNA strand b. telomerase is associated with an RNA strand and DNA polymerase c. telomerase lines up the RNA strand with the 3’ DNA strand to serve as a template to have the 3’ end grow d. hen the RNA strand serves as a primer for new growth onto the 5’ strand, then the primer is removed e. result is an elongated DNA strand that was shorted during replication *telomerase is not present in most cells of multicellular organisms (like us) *DNA of older individuals tends to be shorter *telomerase is abundant in germ line cells – those that give rise to gametes *researchers find telomerase in cancer cells – makes sense since these cells replicate often and would have very short DNA (possible cancer therapy is to target their telomerase)

In many applications, Eastman methyl acetate can be an effective replacement for acetone and other fast-evaporating solvents. Strength?assay Methanol Water Acidity as acetic acid color, PC’s Some common cleaning applications include: universal cleaner LOW-VOCE and environmental cleaners Aerosol carburetor cleaners Paint gun cleaners Cleaners for printing inks Table 1 Sales specifications Property Eastman methyl acetate can be used alone or In easily blended formulations to optimize cleaning efficiency. Because methyl acetate is miscible with most organic aerospace, marine, and industrial.

Eastman methyl acetate methyl acetate, high purity 96. 0% mint. 2. 5% Max. 1. 5% Max. 0. 15% Max. 5 Max. 99. 5% mint. 0. 10% Max. 0. 05% Max. Cleaners for industrial wipes Regulatory and VOCE-exempt status Concerns for work place safety and the environment have led to the deselecting of chlorinated solvents in many applications. In the United States, methyl acetate was added to the list of compounds excluded from the definition of volatile organic compound (VOCE) on the basis that these compounds have been determined to eve negligible photochemical reactivity.

Methyl acetate is relatively nontoxic, nonrestrictive, and readily biodegradable, making it useful in environmentally friendly formulations. Fast evaporation rate Fast drying is often a key performance requirement in cleaning applications. A slow-drying solvent can impede the cleaning process, adding additional labor cost. In addition, slow-drying solvents can attract airborne contaminants and leave residues that negate the effectiveness of the cleaning processes. Using Eastman methyl acetate in cleaning applications (Continued)

Table 2 shows physical properties of methyl acetate versus other fast-evaporating solvents. Methyl acetate evaporates faster than MEEK and ethyl acetate, allowing its use as a replacement for those solvents in applications where VOCE reduction is required. Methyl acetate is similar to acetone in evaporation rate, VOCE exemption, and non-HAP status but offers a higher flash point and hydrophobic property. The hydrophobic nature of methyl acetate can be a key performance criteria in cleaning applications where moisture-related problems can damage or corrode parts, eating to defects or rejects.

Lab 5 DECOMPOSITION (Nov 2, 2011) Introduction: Decomposition is the breakdown of organic material into its smaller molecules and elements. (This term is generally considered as a biotic process but one may find it also used to describe an abiotic process, e. g. , due to weathering. ) The decomposing organisms may use the release of elements for nutrients and by breaking apart the carbon-carbon bonds in organic matter this can release energy for them.

These smaller molecules and nutrient elements may also become available for use by the primary producers (i. e. , plants and phototropic microorganisms). Decomposition is an important step in the food chain and contributes to the nutrient cycling within an ecosystem. Most of the organic matter in an ecosystem ultimately passes through the decomposer subsystem. Decomposition of organic matter is a major ecosystem process involving an array of different organisms.

The catabolism (breakdown of molecules into smaller units) of the organic compounds is mostly accomplished by bacteria and fungi. However if one considers decomposition as the disappearance or breakdown of organic litter then the soil fauna (invertebrates such as the springtails, mites, isopods, etc) must be included in this array of soil biota that contributes to the decomposition of organic matter. Wood decomposition is also influenced by the fungal species that break it down.

Some of these species form brown rot (where only cellulose and hemicellulose are broken down leaving lignin which is brown), while others form white rot where all three are broken down). The majority of fungi are white rotters, but brown rot fungi are ecologically important because they form long-lived nurse logs. Decomposition rates vary due to abiotic factors such as moisture level, temperature, and soil type. The rates also vary depending on the amount of initial breakdown caused by the prior consumers in the food chain.

The more broken down the organic matter (greater surface area exposed), the faster is the final decomposition. There are a variety of methods to determine decomposition rates. For example, 1) weight loss (a change in organic matter mass over time) – such as using litter bags or core sampling; 2) organic tissue or component substrate changes (e. g. , weight or concentration changes of cellulose or lignin); 3) microbial populations (fingerprinting the microbial populations present and their changes) and/or their activity (e. g. CO2 evolution using alkali traps [eg, soda lime, sodium hydroxide] or detection of CO2 in gaseous samples [e. g. , InfraRed Gas Analyzer-IRGA, gas chromatography-GC]. Objectives 1. Determine CO2 evolution as an indicator of decomposition and microbial populations from the hardwood, conifer and garden soils using a static soda lime trap. 2. Determine the effects of isopods on decomposition of vine maple leaves 3. Examine differences between brown and white rot in wood decay 4. Solve a problem set using conifer needle mass loss data from litterbags. . Soil CO2 evolution using the Soda Lime technique (a static-chamber method)  CO2 evolution will be determined from the soil surface beneath conifer trees (Douglas-fir and cedar), deciduous hardwood trees adjacent to Winkenwerder Hall, and a nearby garden soil on campus using the static trap soda lime technique. Soda lime gains weight when exposed to CO2. The main components of soda lime are : • Calcium hydroxide – Ca(OH)2 (about 75%) • Water – H2O (about 20%) • Sodium hydroxide – NaOH (about 3%) Potassium hydroxide – KOH (about 1%) The method is based on the adsorption of CO2 by soda lime that is measured by a weight gain. The following absorption reactions occur: 2NaOH+CO2[pic]Na2CO3+H2O Ca(OH)2+CO2[pic]CaCO3+H2O Procedure: 1. Obtain soda lime 2. Dry the soda lime in a clean beaker at 105 C in a drying oven to remove adsorbed moisture (212 Bloedel) 3. When dry (probably overnight or until it stops losing weight), weigh out approximately 10 g into a soil can (record to at least the nearest 0. 001g). 4.

A plastic container (16 cm diam) is used as a chamber to trap CO2 evolving from the soil. 5. At the field sites place the soil can with soda lime on the soil and then place the plastic container upside down over it and push its edges into the soil to form a seal around the beaker to trap CO2 from the soil respiration. 6. Also place a control (blank) sample of soda lime in a soil can in the field also under a plastic container, but one that has a bottom on it (aluminium foil) so that it does not allow CO2 evolving from the soil to be adsorbed.

This control (blank) is treated as all other samples except that it is not exposed to soil CO2 evolution. 7. Incubate for 24 hr (leave in situ so that CO2 evolution has been subjected to abiotic/biotic fluctuations occurring over the diurnal period). 8. After 24hr remove the soda lime from under the can and put the top on the soil can to keep CO2 exchanges from occurring. 9. Dry the soil can of soda lime (uncovered) in the drying oven at 105 C (overnight sufficient) and then reweigh. 10.

Three replicate samples are used for the conifer, hardwood and garden soils as well one blank at each site. 11. At each site record pH and temperature in the upper 5 cm of mineral soil. Make general observations about the amount of roots you see at each site Calculation: The difference in weights before and after incubation is an estimate of the grams of carbon dioxide evolved from the soil. Multiply this weight by a correction factor* of 1. 69 (due to 1 mole of water generated by each mole of CO2 absorbed by the lime) (Grogan 1998).

The units are g CO2 per ‘container area’ per 24hr. This is converted to g CO2 m-2 hr-1. S = (Wsl x 1. 69) / (Ac x T) where, S is CO2 evolution (g CO2 m-2 h-1), Wsl is the soda lime weight gain, 1. 69 is the C absorption rate of soda-lime, Ac is the chamber area (m2), and T is the sampling time in hours. Do the same calculation for the control (blank) and subtract that value from the sample calculation to derive the correct CO2 evolution from the soil. In Excel conduct an Analysis of Variance (ANOVA) to determine if there are significant differences (P