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270 Chapter 13 ALKYNES: THE CARBON-CARBON TRIPLE BOND in ethyne contributes to the acidity of the hydrogen (together with the enhanced stability of the conjugate base. the ethynyl anion. also a result of hybridization effects). It may seem paradoxical to you that the strongest C-H bond is the easiest one to deprotonate. Remember, however. that bond strength relates to homolytic cleavage (to and H·), whereas acidity refers to heterolytic cleavage (to and H⁺). 30. The bond strength should be greatest and the bond length smallest in again as a result of the sp (50% character) orbital at C2. 31. In analogy to alkynes, alkenes, and alkanes. the three compound types differ in the hybridization of the nitrogen atom. Acid strengths vary in the same way: > > 32. Stability order is cyclopentene > 1,4-pentadiene > 1-pentyne. Cyclopentene has the most σ bonds, which are generally stronger than π bonds. 1,4-Pentadiene and 1-pentyne both have two π bonds, but the alkyne is of higher energy. Notice (Section 13-2) that heats of hydrogenation for alkynes are 65-70 kcal mol⁻¹, or 32.5-35 kcal mol⁻¹ per bond, whereas the normal range for alkenes is 27-30 kcal mol⁻¹ (Section 11-10). 33. (a) 3-Heptyne > 1-heptyne (internal more stable than terminal) (b) Stability decreases from left to right. The first two, isomers of propynylcyclopentane. follow the order "internal more stable than terminal." The last, cyclooctyne, despite being internal, is less stable than either as a result of bond angle strain. Make a model: The alkyne carbons cannot have 180° bond angles. The compound has actually been made, but it doesn't hang around very long and has a strain energy in excess of 20 kcal mol⁻¹. 34. Degree of unsaturation is calculated for each compound. (a) = 12 + 2 = 14; degree of unsaturation = (14 10)/2 = 2 bonds or rings. NMR looks like an ethyl group: CH₃ (t, δ = 1.0) next to CH₂ (q, δ = 2.0). Because the molecule has 10 H's, there must be two equivalent ethyl groups, 2 CH₃CH₂-, adding up to C₄H₁₀. Two C atoms are all that are left to account for. To get two degrees of unsaturation, make a triple bond between them (2 bonds): 2 and (b) = 14 + 2 = 16; degree of unsaturation = (16 12)/2 = 2 bonds or rings. IR: terminal -C=CH. NMR: δ = 0.9 (t, 3 H) CH₃, next to CH₂ δ = 1.3 (m, 4 H) ? δ = 1.5 (quintet, 2H) CH₂, with CH₂ groups on both sides δ = 1.7 (t, small J, H) Aha! How about Split by (Compare Figure 13-5.) δ = 2.2 (m, 2H) Perhaps the CH₂ referred to here? So far, you have and or C₅H₈; you need C₂H₄ more. The simplest way is 1-heptyne.