Principles of Chemistry - Essay Example

Topic – Insert & Affiliaton) Question The n-butane molecule is represented by the following: Scientists have suggested a number of ways that organic molecules may be represented. Among them are the ‘Sawhorse’, ‘Newman’, ‘Fischer’, ‘Haworth’, and ‘Natta’ projections. In the Sawhorse projection the molecule C-C bond is viewed at an oblique angle. In the Newman projection the C-C bond is viewed from one end. The closer C is represented by a dot and the C further away as a circle. This projection makes it easier to view the torsional angles and is thus the preferred projection. The Carbon atom closer to the viewer is called the ‘proximal’ and the one further the ‘distal’ atom. The two extreme conformations of Butane are the staggered anti and fully eclipsed. Their conformations are shown as sawhorse and Newman projections below: Staggered (Anti) Fully Eclipsed All staggered conformations have lower strains compared to eclipsed conformations. The staggered (anti) conformation is the most stable and therefore the most preferred. In this conformation there is no steric strain and no torsional strain. Due to the migration of electrons towards the more electronegative nucleus in a bond between two different elements the molecules become polarized. In larger molecules a number of poles form and this causes some parts of the molecule to attract and some to repel each other. In the butane molecule both ends of the molecule are rotating around the C-C bond and take different configurations (these are constantly changing. Torsional strain is the extra potential energy that a molecule possesses due to the presence of eclipsed bonds. It is so called due to the kind of force it exerts on the C-C bond – a twisting force. Steric strain is caused by the crowding together of groups – as explained above the groups are forced to approach each other till they touch, but because of their similar polarity they are under forces of repulsion also. This leads to the molecule possessing extra potential energy, this energy is called steric strain. The energy profile for rotation about the central C-C bond in Butane is shown below (Chapter 4). Sketch copied from the website. Question - 2 The reaction of buatan-1-ol with phosphorous bromide will be as under: C4H9OH + PBr3  C4H9Br + HOPBr2 The molecular weights of the reactants and the products are calculated as under: C4H9OH = (12x4+1x9+16+1) = 74 PBr3 = (31x4+80x3) = 271 C4H9Br = (12x4+1x9+80) = 137 HOPBr2 = (1+16+31+80x2) = 208 Since the reaction contains one mole of both the reactants and products, 74g of buatan-1-ol will react with 271g of phosphorous bromide to give 137g of 1-Bromobutane. a. The question states that one mole equivalent of phosphorous bromide was used, the mass of this chemical used was 271g. b. Using 74g of C4H9OH the yield of C4H9Br should be = 137g of C4H9Br 27g was actually used, therefore, the yield should be = 137x27/74 32.8g was the actual yield, thus the yield percent is = 32.8 x 74 x100 = 65.62%. 137x27 Question – 3 a. In 1870, Markonikov laid down a basic principle of chemistry that holds till today. The Markonikov Rule states that when a hydrogen halide (Hydrochloric acid, HCl being an example) reacts with an asymmetrically substituted alkene the hydrogen attaches itself to the carbon atom that has the larger number of its hydrogen atoms substituted. The halogen automatically attaches itself to the carbon atom having the least number of its hydrogens substituted. An example of such an addition reaction: Cl2HC=CH2Cl + HCl  Cl2H 2 C=CH2Cl2 Here the halide chlorine attached to the right hand side of the molecule that had fewer hydrogen atoms replaced. b. Enantiomers are two molecules of the same chemical that are the mirror image of each other, like our hands, they are similar but yet not the same and can not be superimposed on each other. Such molecules are called Enantiomers. They are not isomers and are not different from each other in any way except in their configuration. c. An Emperical formula gives the proportion of the constituent elements in the molecule but does not give the exact composition of the molecule. The empirical formula represents the mole fraction/ ratio of each element in the compound. For example the empirical formula may show that there is hydrogen and oxygen present in the mole ratio 1:1, and the empirical formula would be HO, the formula does not show that this is Hydrogen peroxide H2O2. d. When a bond is formed between two atoms sharing a pair of electrons (covalent bond) the shared electrons spend an equal amount of time close to the nuclei of the two atoms if they are the same for example a Carbon to Carbon bond, such a bond is termed non-polar . However, if the two atoms are different they exert an unequal pull on the electrons and therefore the electrons spend more time closer to the nucleus exerting the larger pull. Such a bond is called Polarised. For example oxygen is more electronegative that carbon and therefore the bond is polarised towards the oxygen atom in a carbon to oxygen bond, but when carbon and hydrogen form a bond the hydrogen is less electronegative than the carbon and therefore the bond is polarised towards the carbon and that end of the molecule displays as slightly negative charge. e. Atoms of some elements either alone, or in combination with others (functional group) attach to an organic molecule giving it distinct properties. The chemical properties are characteristic of the functional group, and except for reaction rates, are similar to other compounds having the same functional group. For example when one atom of Chlorine attaches to any of the alkenes it gives compounds like methyl chloride, ethyl chloride and so on. All these have the similar properties and are referred to as Chlorides (Halides, if we speak generally of all the Halogen elements); similarly a combination of O and H as a hydroxyl functional group (OH) will form Methyl Alcohol, Ethyl Alcohol and so on. The physical properties depend upon the length of the carbon chain with the longer carbon chain compounds being heavier, with higher melting and boiling points etc. A group of compounds, having similar properties due to the presence of a functional atom or group is termed as a Homologous series. In simpler terms we can say that a homologous series is a set of molecules with the same general formula and the same functional group. A homologous series can also be said to be a range of compounds that differ from each other only in the number of Carbon and Hydrogen atoms in the backbone chain. The alkanes (Methane, Ethane, Propane, Butane and so on) form a Homologous series. f. Carbon, like many of the first-row elements of the Periodic Table, has atomic orbitals that can hybridize. This is because the s-orbital and p-orbitals of carbons second electronic shell have very similar energies. As a result, carbon can adapt to form chemical bonds with different geometries (hybridization). Carbon is capable of forming many different compounds with the same empirical formula but displaying different properties due to this property. This is called hybridisation of carbon. Question – 4 A number of routes are available for the preparation of Alkanes. Some of these are: 1. The Wurtz reaction, In this two molecules of an alkyl halide react with elemental Alkali metal to form the corresponding alkane and Alkali metal halide. An example of this reaction is: 2CH3Br + 2Na  CH3 - CH3 + 2NaBr 2. The Grignard reagent method. A grignard reagent is (an organo-metallic reagent) is first formed and then this is reacted with water. CH3-Br + Mg  CH3MgBr (Grignard Reagent) This reaction takes place in the presence of anhydrous ethyl ether, CH3MgBr + H2O  CH4 + MgBrOH 3. Direct catalytic hydrogenation of the corresponding alkenes or alkynes will also yield the alkane. This is a direct addition of the hydrogen. CH3=CH3 + H2  CH3 -CH3 Finely divided Platinum catalyst and a high temperature are required to complete this reaction. Question – 5 The three Isomers of C2H4O, are Acetaldehyde, Ethylene Oxide and Vinyl Alcohol. The same elements combine in the same ratio, but due to the difference in their atomic structure they show the behaviour of three different functional groups, CHO - Aldehyde, O – Oxide and OH – Alcohol. Sketches of the structure of the three isomers are given below. Vinyl Alcohol Acetaldehyde Ethylene Oxide Note I am unable to sketch in the bond angles. Please draw freehand H-O-H is 105o and R-O-H is 109o References Website: Chapter 4.1: Conformations, accessed on 14 May 2006 from: http://www.chem.wwu.edu/lampman/1 Website: Hybridisation Carbon, accessed on 14 May 2006 from: http://invsee.asu.edu/nmodules/Carbonmod/hybrid.html Bibliography Book: Johnson, WA; ‘Invitation to Organic Chemistry’, Jones & Barlett, 1999 Read More