Since (+)-α-Pinene's double bond is trisubstituted and sterically hindered, only two compounds will reaction with borane to form dialkylborane as intermediate. After we get borane in the desired position, we will need to do oxidation process to replace borane by -OH group. Treat dialkylborane with NaOH and H202 to form alcohol and boric acid. Oxygen in hydrogen peroxide will attach to
Synthesis of Methyl Stearate Post-Lab Submitted by Matthew Sharma TA: Evan Determining the Limiting Reagent Methyl Oleate (MW= 296.49 g/mol) Amount used = 1.141g (1.141g)(1mol / 296.49g) = 0.003848 mol ∴ The limiting reagent for the reaction is methyl oleate because hydrogen is used in excess Theoretical Yield Methyl Stearate (0.003848 mol)(298.504g/mol product) = 1.1486 g methyl stearate Percent Yield Given the 1:1 stoichiometry of the reaction: 100% x (0.2551g product / 1.1486g theoretical yield) = 22.2% Conclusion A 22.2% yield of synthesized methyl stearate was obtained via the catalytic hydrogenation of methyl oleate in the presence of the catalyst 10% palladium on carbon. The product
(2 marks) 8 Show 2 possible products that could form when Compound H undergoes a halogenation reaction with iodine. Draw the structural isomers and name them. 2 marks 9a Is this halogenation reaction is an addition or substitution reaction? ½ mark 9b Under what conditions would this reaction occur? ½ mark 10 Outline the reaction pathway to produce propanoic acid from propane.
This will help determine the types of ions present in the water sample. Ions also absorb light differently according to the concentration of the ions in the solution. Using absorption spectroscopy the absorption rate is used to determine which ions are present in the solution and at what concentration. The equation m1v1=m2v2 will be need where m1 and v1 are the initial concentration and volume, and m2 and v2 and the final concentration and volume. Experimental: List of Chemicals 0.5 M NaCl solution 0.5 M LiCl solution 0.5 M KCl solution 0.5 M CaCl2 solution 0.5 M SrCl2 solution Fe/Cu solution containing 400 ppm Cu2+ and 20 ppm Fe3+ in SCN- solution 20 M iron(III) nitrate solution
The net reaction is: This reaction has been studied extensively and occurs for a wide variety of ketones. In general, the halogenations of a ketone can be represented as follow: The main evidence for any mechanism is provided by kinetic studies to determine an experimental rate law. Following the rate law of chemical kinetics, the differential rate equation for the reaction could be written as follow: Where k = rate constant; a, b,c are the orders of the reaction of S, I3-, and H+ respectively. I3- ion is the only coloured species in the reaction mixture, a spectrophotometer can is used to measure the change in its concentration, by applying the Beer-Lambert Law Where A= absorbance, ε= molar absorption coefficient, [I3-]= concentration and /= optical path length, that is, the distance travelled by the light through the solution. The ideal wavelength for the measurement of
With the use of this technique we placed chlorine, bromine, and iodine into solutions containing chloride, bromide, and iodide. In the reaction the free halogen (X2) oxidizes the other halide ion (Y-) and gets reduced by gaining electron(s). In table 3, chlorine was the strongest oxidizing agent and iodine was the weakest oxidizing agent. Since chlorine was the strongest oxidizing agent it will react more and the weak agent will react less. This explanation can be demonstrated in table 3 also because the results of the reactions demonstrates that chloride reacted more by the color of the product compared to the color of chloride in the mineral oil.
2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4. 1-butanol (bp=118 degrees C) 5. butanone (bp=80 degrees C) Give a better separation for the mixture to be distilled tetrahydofuran (bp=65 degrees C) because it is farthest from 100 degrees C Which alkyl halide would react fastest in a nucleophilic substitution using silver nitrate in ethanol (weak nucleophile, protic solvent)? 3-bromo-3-methylpentane (most
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. This experiment uses bromination, the specific name of halogenation with bromine. In order to for bromination to occur bromine must first be generated. This is done in situ through the oxidation of glacid acetic acid and puridinum bromide perbromide. Once generation is accomplished the available bromine can be brominated.
Introduction Nucelophilic substitutions are chemical reactions in which an electron rich nucleophile attacks the electron poor electrophile1. There are two classes of nucelophilic reactions – SN1, and SN2. The SN1 reaction is a 2 step, uni-molecular reaction, which is independent of the nucleophile. It requires a highly substituted electrophile since there is a formation of a carbocation in its rate determining step, good polar protic solvents which stabilize the carbocation and a good leaving group1. On the other hand, a SN2 reaction is a concerted, bimolecular reaction which has one slow, transition state1.
One means of classification depends on the way in which carbon atoms are connected. Chain aliphatic hydrocarbons are compounds consisting of carbons linked either in a single chain or in a branched chain. Cyclic hydrocarbons are aliphatic compounds that have carbon atoms linked in a closed polygon (also referred to as a ring). For example, hexane (single) and 2-methylpentane (branched) are chain aliphatic molecules, while cyclohexane is a cyclic aliphatic compound. Another means of classification depends on the type of bonding that exists between carbons.