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Organic Chemistry 102 1. Problems #4 Solutions Due Friday, Feb. 9 Give the proper name or structure for each of the following. Name and show any stereochemistry present in these. names: OH b) c) a) OH CH3 HO 2,4-diethyl-1-heptanol 1R,6S(cis)-6-methylcyclohex-3-en-1-ol structures: d) trans-cyclohexene glycol H exo-bicyclo[2.1.1]hex-2-en-5-ol OH e) pentamethylene glycol HO OH OH 2. a) For each set of experimental facts presented below, give reasonable explanations for the observed trends, using the relationships between structure and physical properties. CH3 CH3CH2CH2CH2OH CH3 CH2 CH CH3 Each of the alcohols can hydrogen bond with the water to help it OH dissolve. The more polar of the solubility infinitely soluble 8.3g/100 mL H2O 26.0g/100 mL H2O in water in H2O Explain three is 1-butanol, which would } each seem to indicate it should be the separately boiling 118°C 99.5°C 82.5°C one that is most soluble in water. point Therefore there must be another explanation. The intermolecular interactions between the 2° and 3° alcohols are less due to smaller surface areas and poorer intermolecular hydrogen bonding. This makes it easier for the water molecules to separate the alcohols. The other reason is that the 2° and 3° alcohols are more compact and disrupt the water molecules less. Boiling Point: This is just what one would expect based on intermolecular interactions. The 1° alcohol would have more effective H-bonding, as well as more effective London Dispersion forces (more surface area). We learned earlier that branching reduces boiling point due to a reduction in the dispersion forces. But also here, the hydrogen bonding is made less effective with an increase in the steric size of the alkyl group near the oxygen. H3C C CH3 OH b) CH3CH2CH2CH2OH CH3CH2CH2CH2CH2CH2OH CH3 CH2 H C H C CH2 CH3 OH OH solubility in water boiling point 8.3 g/100mL H2O 118°C 0.59 g/100mL H2O 157°C infinitely soluble in H2O 207°C } Explain each separately Solubility: the number of hydroxyl groups will increase the solubility in water due to increased Hbonding interactions. Butanol has fewer non-polar methylene groups than hexanol making it more soluble. Boiling point: both butanol and hexanol are primary alcohols but hexanol has a higher molecular weight, therefore increasing intermolecular London forces and the boiling point. The hexanediol, with the two OH groups, has extra intermolecular interactions, including additional H-bonding, which will raise the boiling point. 3. Provide two different syntheses, using structurally different starting materials, for the following compounds where a Grignard reagent is used in each method. 1. BrMgCH2CH2CH3/ether CH3CH2CH2 CH a) or H 2. H+/H2O CH3CH2CH2 CH CH2CH2CH3 O C OCH2CH3 1. 2 BrMgCH2CH2CH3/ether O 2. H+/H2O OH O 1. BrMg-CH(CH3)2/ether b) 2. H3O+ H O CH3 OH MgBr C H /ether 1. or 2. H3O+ CH3 H BrMg CH2CH3 c) H3C O C O H3C H3O+ 2. or H3C 4. C 1. CH2CH3 C OH CH2CH3 1. BrMgCH3 C 2. H3O+ O 1. BrMgCH2CH3 2. H3O+ Provide reagents and conditions to carry out the following transformations. If more than one method is possible, choose the one that is most commonly used for that transformation. H H2 C PCC/CH2Cl2 OH O a) OH O H2CrO4/H2SO4 b) " acetone O OH H2CrO4/H2SO4 c) acetone CH3 CH3 1. OsO4 d) OH 2. NaHSO3/H2O OH OH H2 C 1. LiAlH4/ether O e) O NaBH4/EtOH " f) O H NaBH4/EtOH g) HO OH 2. NaOH/H2O H O + enantiomer (1:1) O HO H2 C OH 5. Predict the major organic products expected in the following reactions. Pay particular attention to stereochemistry in your answers. When mixtures are expected, indicate the relative amounts of each. H3 C OH H3 C CH3 a) HBr H3C Br H3 C CH3 HCl/ZnCl2 b) OH H3C c) Cl H3C CH3 Br HBr H3C + H3 C OH H3C H3 C d) Br H3C CH3 PBr3 " H3 C Br H3C 1. Ts-Cl/py H3 C e) H3 C OH CH3 f) H C 3 H CH2 2. KO(t-Bu)/HO(t-Bu) H3C 1. Ts-Cl/py CH3 H3 C 2. NaI/acetone I OH H H3C H3 C g) CH3 H3C 1. BH3:THF 2. H2O2/NaOH CH3 H3C OH h) + enantiomer (1:1) H3 C OH H3 C CH3 1. Hg(OAc)2/THF/H2O " 2. NaBH4/NaOH