2. Using the atom and bond library below, construct the following molecule. It may be
Of the alcohols tested 1-Butanol was found to contain the strongest intermolecular forces (IMF) of attraction, with Methanol containing the weakest. It was discovered through experimentation that Methanol induced the highest ?T of all alcohols tested, and that conversely 1-Butanol induced the lowest ?T. The atomic structure of all four alcohols is very similar, as starting with 1-Butanol a CH2 group is lost as you move from 1-Butanol to 1-Propanol to Ethanol and then again to Methanol. Each structure is fairly linear and contains an H-bond with Oxygen, so the real change is found in the loss of the CH2 group, this lowers the liquid’s Molecular Mass, thus lowering the London forces as you move from 1-Butanol through
The goal of this experiment was to synthesize an alkene (4-methylcyclohexene) from an alcohol (4-methylcyclohexanol) by dehydration. The reaction, consist of 4-methylcyclohexanol, phosphoric acid, and sulfuric acid, was refluxed at a given time frame. The product was isolated by distillation and purified by adding sodium chloride to help the extraction. The final product had a 125% yield and was characterized by the IR spectroscopy and chemical reaction. The alkene resulted in a colorless liquid after adding molecular bromine dissolved in dichloromethane.
~ ~ is the simplest, and most important of the Alkenes. It is one of the most important Petrochemicals
3. In Molecular Series I (Straight-Chain Alcohols): As you go through the group from methanol ethanol 1-propanol 1-butanol:
4. Write a structural diagram equation to represent the reaction between each alcohol and HCl(aq).
The goals in this lab were to have a reaction occur with 4-methylcyclohexanol and an acid catalyst to form our product of 4-methylcyclohexene via an E1 reaction. This reaction is accomplished by removing the –OH group on 4-methylcyclohexanol via dehydration and to have a double bond form via a loss of a hydrogen on a β-Carbon.
3. The IR spectrum of the starting material shows a medium/strong C-O bond at around 1500cm-1, also the starting material shows a strong C-H bond at around 3000cm-1 and another medium C-H bond at 2865cm-1 indicating an aldehyde group whereas the product does not. The IR spectrum of the product shows a two weak broad O-H peaks at around
2) Explain the trend in the solubility of the three alcohols in water. (In your discussions, bring out the theoretical concepts on which
have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To
The purpose of this lab is to understand the process of eliminating an alkyl halide to form an alkene. The experiment is carried out by first converting the alcohol, 2-methy-2-butanol, into the alkyl halide of 2-chloro-2-methylbutane that will then be put through dehydrohalogenation that favors elimination reaction (E2) to create a mixture of 2-methyl-2-butene and 2-methyl-1-butene. A fractional distillation will be taken to purify the mixture and an additional gas chromatography will be done to further analyze the mixture composition. A bromide test will be done to determine the product of an alkene in the experiment.
The results from the NMR of 1-propanol showed 3 different prominent peaks with the peak at 2.2 cm-1 being the acetone. Because 1-bromopropane has three non-equivalent hydrogens it was found to represent this set of NMR data. The other product, 2-bromopropane only had 2 different types of hydrogens and would have only had 2 peaks. Further analysis of the structure of 1-bromopropane showed that the hydrogens closest the bromine group were an indication of peak A in the graph. Because of the electronegativity of the bromine, this peak was located further downfield. There were 2 neighboring hydrogens so using the n+1 rule gave the 3 peaks. Going down peak B showed the next carbon which had 5 neighboring hydrogens thus giving 6 peaks. Finally, the carbon furthest away from the bromine was found at peak C. It had 2 neighboring hydrogens and provided 3 peaks.
With the increased temperature, the isopropanol-water solution contracts faster than the lower temperature due to the fact that molecules move faster with increased temperature. The hydrogen bonding between isopropanol and water is quite strong between the solvents and within themselves. As a result of the compounds’ properties they rearrange to assume a more efficient hydrogen bond between them, as the angles required for efficient hydrogen bonding are not present; the bond angles change due to the repulsive force of free electrons within the compound. The intermolecular forces that cause the molecules of ethanol, isopropanol, and water to rearrange and contract primarily include dipole-dipole interaction and hydrogen bonding. In terms of dipole-dipole interaction, the forces occur when polar molecules are