Bromination of Arenes This lab demonstrated the application of adding bromine to various arenes, hydrocarbons with alternating single bonds. This process, bromination, is a mechanism which treats hydrogen as a functional group. This being the case, the rate of reaction of certain arenes can be measured and compared to that of other arenes upon the addition of the bromine. The reaction occurs when the bromine radical generates from the halide diatomic molecule, using light energy. 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.
Hydrogens on the alkene have different reactivity which determines which isomer is favored. Hydrogens attached to carbons, and have a weak bond strength are most reactive therefore are favored during abstraction by chlorine. The chloro substituent has an effect on carbon reactivity due to the chlorine atom being an electron withdrawing substituent. The chlorine
Electron donating substituents attached to the diene, accelerate the reaction as do electron withdrawing groups on the dieneophile . The Alcohol acts as the diene and the anhydride as the dieneophile. The Diels–Alder reaction is a highly stereoselective reaction. In order for the reaction to occur the diene must be in the s-cis formation .The trans, trans-2,4-Hexadiene-1-ol can adopt the cis formation by rotating around a sigma bond while still maintaining Trans stereochemistry . The position of the diene to the dieneophile in the transition state determines the
LAH is highly reactive and will reduce acid chlorides, esters, carboxylic acids, amides, and nitriles as well as aldehydes and ketones. NaBH4 is less reactive and therefore more selective than LAH: it reduces aldehydes and ketones, but not carboxylic acids and amides, and only slowly reduces esters. NaBH4 is much safer to handle than is LAH, making it a good choice whenever the functional group to be reduced is an aldehyde or ketone. The purpose of the below methodology isto synthesize meso-hydrobenzoin from benzyl using the reducing agent sodium borohydride. Reaction and Mechanism: Formation of meso-hydrobenzoin | | Benzil | + | Sodium Borohydride | → | Hydrobenzoin | | C14H10O2 | | NaBH4 | | C14H14O2 | | | | | | | Molecular Wight | 210.23 g/mol | | 37.83 g/mol | | 214.26 g/mol | Observed Mass/Volume | 400 mg | | 150 mg | | 354.4 mg | Observed Moles | 1.902 mmol | | 3.96 mmol | | 1.65 mmol | Theoretical Yield | | | | | 407.52 mg | Percent Yield | | | | | 88.5 % | Theoretical Melting Point | 94-96 °C | | | | 137 °C | Observed Melting Point | | | | | 136.4-137.3 °C | Calculations: Observed Moles ofBenzil-0.2 g of Benzil x 1 mol210.23gmol=0.000951 mol→0.951 mmol x 2 reaction doubled=1.902 mmol Observed Moles ofSodium Borohydride-0.075 g of Sodium Borohydride x 1 mol37.83gmol=0.00198 mol→1.98 mmol x 2 reaction doubled=3.96 TY-0.000951 mol x 1
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.
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.
B) permanent dipoles of molecules containing covalent bonds between atoms of very different electronegativities. C) the hydrophobic effect. D) ion pairing between oppositely charged functional groups. Answer: A Page Ref: Section 5 28) The aggregation of nonpolar molecules or groups in water is thermodynamically due to the A) increased entropy of the nonpolar molecules when they associate. B) decreased enthalpy of the system.
We resulted that lead, silver, and copper are the strongest oxidizing agents, and that magnesium and zinc are the weak oxidizing agents. The strong oxidizing agent oxidized the weak oxidizing agent and in turn the strong oxidizing agent got reduced while oxidizing the weak agent. When a reaction occurred, the solid metal reduced the ion, and in turn made it the more reactive metal. In part two we used a solvent extraction technique to derive an activity series for the halogens. With the use of this technique we placed chlorine, bromine, and iodine into solutions containing chloride, bromide, and iodide.
The process of salting out was used to separate cyclohexanone from the aqueous solution to purify the product. Which are weak intermolecular forces (e.g., hydrogen bonds) between organic molecules or nonelectrolytes and water are easily disrupted by the hydration of electrolytes. -------NaOCl------- MW 74.392 Cyclohexanol Cyclohexanone Bp 161 ºC bp 157 ºC MW MW 100.16 MW 98.14 Ml/g 8.0 ml 7.4g Moles 0.0075 mol 0.0075 mol Oxidation of Cyclohexanol Mechanism of Oxidation Cyclohexanolis an irritant. Avoid contact with skin, eyes, and clothing. Acetic acid is a dehydrating agent, an irritant, and causes burns.
Synthesis of 1-Bromobutane I. Conclusion In this experiment we prepare 1-bromobutane from 1-butanol by using Sn2 reaction. By heating the primary alcohol with two reagents: NaBr and H2SO4, an aqueous solution containing the alko-halide and water will be produced. The overall equation for the reaction: H2SO4 + NaBr + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Br + H2O + NaHSO4 Questions: 3- By changing the source of the halide from NaBr to NaCl: H2SO4 + NaCl + CH3CH2CH2CH2OH -------> CH3CH2CH2CH2Cl + H2O + NaHSO4 2- In refluxing you gently heating the mixture without losing product to evaporation. If the mixture were boiled some of the solvent that contains some products will be gone with evaporation.