
Organic Chemistry I CHEM 2323 Packet #10 1. Draw the preferred product when 3-methylbut-1-ene reacts with acidic water. H+, H2O ~H H2O OH Indicate two different ketones that can be converted to the same product using a Grignard reaction. Need to look at the carbon attached to the alcohol. This carbon would be the carbonyl carbon of the ketone if a Grignard reaction was performed to form the alcohol. Attached to this carbon is either a methyl or ethyl group, which would necessarily be the Grignard reagent. CH MgBr O 3 OH CH3CH2MgBr O Indicate an ester that can be converted into the same product using a Grignard reaction. In a reaction between an ester and a Grignard reagent, two equivalents of the Grignard react. Therefore the Grignard must be a methyl magnesium bromide, since there are two methyl groups and only one ethyl group attached to the alcohol carbon. CH MgBr O 3 OH O Draw the preferred product obtained if the product from the first reaction is reacted with sulfuric acid. H SO OH 2 4 2. Consider the following three isomers of C4H10O1. OH OH OH acidity 1 2 3 (1 being most acidic) boiling point 1 2 3 (1 having highest b.p.) water solubility 3 2 1 (1 having greatest solubility) Rank the three isomers by acidity, boiling point and water solubility. Explain reasoning for each ranking scheme. Acidity: The stability of the resultant anion after deprotonation will determine the acidity. As the number of electron donating alkyl groups attached to the alcohol carbon increases, the stability of the anion decreases. Therefore the 1˚ alcohol is more acidic than a 2˚ alcohol and the 3˚ alcohol is the least acidic. Boiling Point: The boiling point is predicted by the amount of energy a molecule has in the liquid state versus the gaseous state. All three alcohols can hydrogen bond in the liquid state and as such these compounds will have a higher boiling point compared to compounds that cannot hydrogen bond. The linear 1˚ alcohol will also have a greater van der Waals force due to packing of the adjacent alkyl chains. This packing is disrupted with the 2˚ and 3˚ alcohols. Combining these forces the linear 1˚ alcohol will have the greatest boiling point while the 3˚ alcohol will have the least van der Waals forces and thus the lowest boiling point. Water Solubility: The solubility of these alcohols in water is dependent on the relative amount of hydrophilic and hydrophobic components of the molecules. The alcohol is a hydrophilic component due to the ability to hydrogen bond with the water. The alkyl chain is a hydrophobic component. The 1-butanol has the least water solubility due to the larger surface area of the hydrophobic component. The t-butyl alcohol has the greatest water solubility because the hydrophobic component has a lower surface area due to the more spherical nature of the alkyl chain. In addition compare 1-butanethiol with 1-butanol by acidity, boiling point and water solubility. Explain differences for each property. 1-Butanethiol replaces the oxygen atom with a sulfur atom. The sulfur is more acidic due to the greater polarizability of sulfur. The boiling point and water solubility will be less for 1- butanethiol due to the lower hydrogen-bonding ability of the sulfur versus oxygen atom. 3. Indicate a method to prepare the following products from the indicated starting materials. OH OH O OH OH H2SO4 KMnO4 O O NaBH4 OH O OH OH meso HO2C CO2H 1,2,3,4-butanetetraol fumaric acid 1) RCO3H CO2H CH2OH OH LAH 2) H+, H2O HO H HO H HO2C HO2C = CO2H CO2H HO H HO H OH CO2H CH2OH .
Details
-
File Typepdf
-
Upload Time-
-
Content LanguagesEnglish
-
Upload UserAnonymous/Not logged-in
-
File Pages3 Page
-
File Size-