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Solutions to Problems 19 49. (a) (1) The negatively charged carbon is bonded to three other atoms and has a lone pair, similar to N in sp³. (2) Compare 38(d): Carbon will be sp² (double bond requires a p orbital). (3) Compare 38(e): Carbon will be sp (triple bond requires two p orbitals). (b) How is orbital energy related to ability to accommodate negative charge? Species containing electrons in lower energy orbitals are more stable than those with electrons in higher energy levels. Given the orbital energy order sp CH₂=CH⁻ (sp²) > CH₃CH₂⁻ (sp³). (c) From (b), HC=C⁻ is more stable than which is more stable than These are formed in the equilibria HC=CH + (Most favorable-gives most stable anion), CH₂=CH₂ + CH₂=CH⁻ (Less and CH₃CH₃ + CH₃CH₂⁻ (Least favorable-gives least stable anion). Thus, we have the following order of acidity: HC=CH > CH₂=CH₂ > CH₃CH₃. 50. e>c>d>a>b e After the cation, positive character on carbon is related to number of (polarized) bonds to electronegative atoms. 51. The letters in the structure below refer to the letters in the problem. In some cases, only representative bonds or atoms are labeled. H a i CH₃ a g d d g h f CH₃ f e f e g b e N CH₃ H HH 52. (a) H:C:C:C:C:C:H H:C:::N: H:C:C:C:C:C:H H N δ⁻ N δ⁻ C δ⁺ (b) and (c) N N :0 C C :OH (d) :N:::C: +H, δ⁺ The δ⁺ carbon being attacked by the cyanide ion already has an octet. So one electron pair of the two in the double bond is forced to move up to the oxygen atom to avoid violating the octet rule.