In this experiment, the dark blue colour is visible because of the helical amylose and amylopectin reacting with iodine (Travers et al., 2002). The starch-iodide complex forms because of the transfer of charge between the starch and iodide ion and results in spacing between the energy levels. This allows the complex to absorb light at different wavelengths resulting in a dark blue colour (Travers et al., 2002). A blue colour would indicate a positive test while a yellow colour would show a negative test. The Benedict’s test is useful for reducing sugars.
First, the alkyl halide reacts with a strong Lewis Acid catalyst, usually aluminum chloride, to form a complex, which will then lose the halide to the Lewis acid to give the electrophilic acylium ion. The ion, stabilized by resonance, will react with the p-electrons from a double bond in benzene (acting as a nucleophile) and form the cyclohexadienyl cation intermediate and the tetrachloroaluminate anion. The anion then acts as a base to remove a proton from the ring and reform the initial Lewis acid. The ring regains its aromaticity and the product, an aromatic ring with an acyl substituent is fully synthesized.
This helps to indicate whether or not the reaction follows Markovnikov’s Rule, which states that the electrophile (E+) will add to the carbon involved in a double bond that produces the most stable carbocation. If the rule is followed, the reaction will proceed according to the mechanism in Figure 1. In the silver nitrate test, the alkyl bromide is added to AgNO3. The rate of precipitation with 2° should be faster than the solution with the 1° alkyl halide. In the sodium iodide test, the alkyl halide is added to sodium iodide in acetone.
The Bile Esculin agar test has its medium as selective and differential. Black medium shows a positive result for esculin hydrolysis. In the agar, Gram-positive cannot grow in the presence of bile while certain Gram-negative bacteria can hydrolyze esculin with bile present. MR-VP broth contains glucose and peptone. The enteric bacteria will oxidize glucose for ATP, but there are different fermentative pathways that allow glucose to be fermented.
The purpose of this experiment was to synthesize a Grignard reagent with 1-bromobutane and homogenized magnesium in anhydrous diethyl ether. This solution was refluxed in a flask connected to condenser and drying tube. As seen in the mechanism, maintaining a dry condition is important to avoid the Grignard reagent from attacking water, which will result in loss of the bromine. It is important to reduce the amount of moisture and water vapors to avoid destroying the Grignard reagent, which is essential to the synthesis of 2-methylhexanol. Hence, a calcium chloride and cotton were filled inside a drying tube.
In the second step, the addition of sodium borohydride reduced the imine into another derivative, which was yellowish lime color. The solution turned clear when acids and anhydrides was added, which indicated the precipitate were dissolved. However, after refluxing for a while, yellow precipitates begin to form near the top of the flask. It was assumed that the remaining starting material was concentrated from a decrease volume to reappeared in solution. Nevertheless, this may have been a sign of contamination that will negatively affect the entire reaction.
2-bromobutane would have been expected to react next, due to bromine being a better leaving group than chlorine, then 2-chlorobutane. Tert-butyl-chloride would be expected to never react in a SN2 reaction, as it is so unreactive under these conditions. For each of the molecules used in this experiment (except tert-butyl-chloride), the nucleophile, iodine, would attack the electrophilic carbon bonded to the leaving group, chloride or bromide. That leaving group would then take the
The average equivalent mass for the acid is 135g/mol H+. 3. The answer obtained in Question #2 is the equivalent mass of the acid rather than the molar mass because the acid could be polyprotic, which would mean the equivalent mass is different from the molar mass since it is depending on moles of H+ per molecule, and there could be multiple moles of H+ ions in one mole of a molecule. 4. The KHP and the acid samples must be dried, because there would still be extra water which would skew the molarity.
Water will act as initial solvent for caffeine extraction. This is due to water that slowly soluble with caffeine at ambient temperature but highly soluble when temperature is at 100°C. Then, methylene chloride is chosen as the extraction solvent, due to its miscibility with caffeine and immiscibility with water. As mentioned above, the immiscible pair is chose for the extraction part because to allow the aqueous and organic layers to be separated. Basically, the bottom layer is the aqueous layer while the upper layer is the organic compound.
When tin chloride is added to the solution, Sn2+ took away positive charged ions; Fe3+ to Fe2+ (2Fe3+(aq) + Sn2+(aq) → 2Fe2+(aq) + Sn4+(aq)). This took away Fe3+ from the solution, causing the equilibrium to change to the reactants to balance the concentration of Fe2+. The solution turned a lighter color to increased rate of the reverse reaction. When AgNO3 was added to the solution, the silver nitrate broke down into Ag+ and