Radical Chlorination of 1-Chlorobutane. The radical chlorination of 1-chlorobutane was carried out using sulfuryl chloride and azoisobutyronitrile (AIBN). From the reaction there were for possible products which are as follows 1,1-dichlorobutane, 1,2- dichlorobutane, 1,3-dichlorobutane, and 1,4-dichlorobutane. The structures produced from the reaction are as follows; Attached to the four carbons in 1-chlorobutane are hydrogens that can react readily with chlorine, because of its electron withdrawing character. Chemical environment surrounding the carbons are different and therefore affecting the character of the hydrogens attached.
The fact that the energy needed to break the necessary bonds falls within the visible light spectrum is the basis on which the experiment is based. This brings up the issue of selectivity. For example, the bromine radical is more selective than the chlorine radical. This has to do with electronegativity. It is known that chlorine is more electronegative than bromine, and thus chlorine is more reactive, and less discriminatory as to what it will react with, thus making bromine more “selective”.
An acid is one which can employ a lone pair from another molecule in completing the stable group of one of its own atoms." The first two theories, Arrhenius and Brønsted-Lowry are to do with the protons; exchange between substances, the Lewis however looks at the electron arrangement of the substance. Acetic acid HC2H3O2 acetate Carbonic acid H2CO3 carbonate Hydrochloric acid HCl chloride Nitric acid HNO3 nitrate Phosphoric acid H3PO4 phosphate Sulphuric acid H2SO4 sulphate 2. Give 3 common properties of acids Turns blue litmus red Conducts an electric current Taste sour Reacts with metals to produces hydrogen gas Reacts with carbonate and hydroen-carbonates to form carbon dioxide gas Reacts with Metal oxides to produce a salt and water React with metal hydroxides to produce a salt plus water 3. Define a base.
John Simmers and Keena Dickey Reducing Benzil Using Sodium Borohydride Purpose: The Purpose of this lab this lab is to reduce benzil using sodium borohydride, and then identify the product that was produced by testing the products melting point, mixture melting point, thin layer chromatography and infrared spectroscopy. Theory: Reductions in organic chemistry results from the addition of hydrogen or lose of oxygen from an organic molecule. When NaBH4 is at room temperature it is only able to reduce aldehydes and ketones to form the corresponding alcohol. NaBH4 is a much easier reducing agent than LiAlH4 because it reacts very slowly with protic solvents at room temperature. However NaBH4 has its down falls and decomposes in the presence of acidic functional groups like carboxylic acids, which means that acid groups must be reduced with a base such as sodium hydroxide before NaBH4 will be able to reduce an aldehyde or ketone.
A. ethanoic acid B. ethyl ethanoate C. hexane D. hex-1-ene 8. A compound boils at –33oC. It also dissolves in water to give an alkaline solution. Which type of bonding is present within the compound? A. metallic B. covalent (polar) C. ionic D. covalent (non-polar) 9.
But HOH is a weaker base, and better leaving group. Adding a strong acid to the mixture allows protonation of the –OH group to give water as a leaving group. Once this protonation occurs, the mechanism that is followed depends on the nature of the R group (Williamson, Kenneth L.). In the presence of a strong acid, an alcohol can be dehydrated to form an alkene. The acid used in this experiment is 85% phosphoric acid and the alcohol is cyclohexanol.
However, the desired product is (-)-Isopinocampheol, in which the -OH group need to attach to the less substituted carbon instead. In order to form (-)-Isopinocampheol, we need to do hydroboration-oxidation reaction, in which (+)-α-Pinene will react with borane-tetrahydrofuran complex as hydroborating reagent to form an intermediate that have -OH group in anti-Markovnikov position. First of all, the carbon-carbon double bond in (+)-α-PInene will connect with H-BH group. Based on anti-Markovnikov rule, the H ion which has partial positive charge will connect to the most substituted carbon while -B which has partial negative charge will connect to the less substituted carbon. Since (+)-α-Pinene's double bond is trisubstituted and sterically hindered, only two compounds will reaction with borane to form dialkylborane as intermediate.
General mechanism The chain component is as per the following, utilizing the chlorination of methane as an issue illustration: 1. Start: Splitting or homolysis of a chlorine particle to structure two chlorine particles, started by ultraviolet radiation or daylight. A chlorine particle has an unpaired electron and goes about as an issue radical. Methane chlorination: launch 2. Chain engendering (two steps): a hydrogen particle is pulled off from methane leaving a 1˚ methyl radical.
CHE 111 Laboratory 3 Hydrates Introduction Hydrates Water molecules combine with the molecules of certain substances, forming loose chemical combinations called hydrates. An example of a hydrate is MgSO4•7H2O. This formula means 7 water molecules are loosely attached to a magnesium sulfate molecule. Other examples of hydrates are Na2SO4•10H2O and Ba(OH)2•8H2O. When the hydrate is heated, it easily loses water molecules attached and becomes an anhydrous salt.
This is done by a procedure called refluxing. Refluxing is the process of heating a product to the boiling point and re-condensing the vapor continuously. The procedure halogenation is the addition of a halogen to a π bond forming a halo alkane. In this synthetic reaction bromine was used in the process called bromination. The bromine is acting first like an electrophile, and then after bromine has broken the π bond, a carbocation has formed, and a bromide ion has been created, the bromide ion then acts as the nucleophile and forms a bond with the carbocation.