b The reaction shows that each carbon atom in the cyclohexane loses a bond to

B the reaction shows that each carbon atom in the

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b) The reaction shows that each carbon atom in the cyclohexane loses a bond to hydrogen to form benzene. Fewer bonds to hydrogen in the product indicates oxidation . catalyst + 3 H 2 15.48 a) reduced b) oxidized 15.49 Step 1 substitution Step 2 addition 15.50 a) The structures for chloroethane and methylethylamine are given below. The compound methylethylamine is more soluble due to its ability to form H-bonds with water. Recall that N-H, O-H, or F-H bonds are required for H-bonding. H C C Cl H C C N C H H H H H H H H H H H H
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15-19 b) The compound 1-butanol is able to H-bond with itself (shown below) because it contains covalent oxygen- hydrogen bonds in the molecule. Diethylether molecules contain no O-H covalent bonds and experience dipole- dipole interactions instead of H-bonding as intermolecular forces. Therefore, 1-butanol has a higher melting point because H-bonds are stronger intermolecular forces than dipole-dipole attractions. CH 3 CH 2 CH 2 CH 2 O H H O CH 2 CH 2 CH 2 CH 3 Hydrogen Bond c) Propylamine has a higher boiling point because it contains N-H bonds necessary for hydrogen bonding. Trimethylamine is a tertiary amine with no N-H bonds, and so its intermolecular forces are weaker. CH 3 CH 2 CH 2 N H H CH 3 N CH 3 CH 3 Propylamine Trimethylamine 15.51 Upper right result: R CH CH 3 OH Bottom result: R CH CH 3 Br 15.52 Addition, this is due to resonance stability of the aromatic ring. 15.53 The C=C bond is nonpolar while the C=O bond is polar, since oxygen is more electronegative than carbon. Both bonds react by addition. In the case of addition to a C=O bond, an electron-rich group will bond to the carbon and an electron-poor group will bond to the oxygen, resulting in one product. In the case of addition to an alkene, the carbons are identical, or nearly so, so there will be no preference for which carbon bonds to the electron-poor group and which bonds to the electron-rich group. This may lead to two isomeric products, depending on the structure of the alkene. When water is added to a double bond, the hydrogen is the electron-poor group and hydroxyl is the electron-rich group. For a compound with a carbonyl group, only one product results: CH 3 CH 2 C CH 3 O + H O H CH 3 CH 2 CH CH 3 OH
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15-20 However, when water adds to a C=C, two products result: CH 3 CH 2 CH CH 2 H O H + CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH CH 3 OH + In this reaction, very little of the second product forms. 15.54 R N H H C CH 3 O OR' 15.55 Alcohol: R - OH + H 2 O ' R - O - + H 3 O + Carboxylic acid: R C OH O + H 2 O R C O O R C O O + H 3 O + The resonance-stabilized carboxylate ion allows the transfer of the proton to water. The alkoxide ion cannot show any resonance stabilization. 15.56 Esters and acid anhydrides form through dehydration-condensation reactions. Dehydration indicates that the other product is water . In the case of ester formation, condensation refers to the combination of the carboxylic acid and the alcohol. The ester forms and water is the other product.
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