北京化工大学：《有机化学》课程教学资源（双语习题与答案）Chapter 13 Alkynes（The Carbon-Carbon Triple Bond）.pdf_大学文库

13-3. Spectroscopy of the Alkynes The high-field position of the hydrogen resonance is surprising at first, but it is the logical result of the cylindrical symmetry of the triple bond, which allows electrons to rotate in a tight circle about its axis. These alkyne signals are often easy to spot because of their characteristic splitting patterns, which arise from long-range coupling to the neighboring nuclei across the triple bond (see, for instance, the alkynyl hydrogen triplet in Fig. 13-5). Other NMR properties of these compounds are pretty normal. Diagnostic bands for CqC and qCOH in the IR spectra of terminal alkynes complement the NMR data, especially when the latter become more complex as a result of overlapping signals. Take note, however, of the weakness or even absence of an IR band for internal CqC triple bonds. 13-4. Preparation of Alkynes by Double Elimination The elimination route to alkynes involves removal of two moles of hydrogen halide by strong base from a dihaloalkane. The most common kind of sequence involves forming the dihaloalkane by addition of halogen to an alkene double bond. Because you already know how to make alkenes, you now have access to alkynes in the following way: 13-5. Preparation of Alkynes from Alkynyl Anions The other major alkyne preparation is based on the easy accessibility of nucleophilic carbanions from terminal alkynes (Section 13-2). So, a wide variety of internal alkynes may be made from any terminal alkyne, via the scheme where “E” is the electrophilic carbon in a primary haloalkane, a strained cyclic ether, or a carbonyl compound. 13-6, 13-7, and 13-8. Reactions of Alkynes Just as you saw with alkenes, alkynes are subject to a variety of addition reactions. These can occur in two stages: a single addition to make an alkene, and then a second addition to give an alkane derivative. The mechanisms, stereochemistry and regiochemistry, are essentially analogous to those you’ve already seen. So, you can look back to the reactions of the previous chapter as points of reference. Only an occasional detail or two will be different. When you choose to stop the addition reaction at the alkene stage, there is often the possibility of picking reagents and conditions to allow for specific formation of either the trans or cis alkene product. This flexibility further contributes to the usefulness of alkynes in synthesis. A special note should be made concerning hydration reactions of alkynes. Remember that hydration of alkenes leads to alcohols. Alkynes can be hydrated, too. Markovnikov addition is achieved with an aqueous Strong base Any electrophile “E” R Nucleophile R C C H C C R C C E Addition, e.g., Br2 Double elimination, e.g., NaNH2 CH CH CHBr CHBr Alkene 1,2-Dihaloalkane C C Alkyne Halogenation, e.g., Br2, hv Elimination, e.g., KOC(CH3)3 CH2 CH2 CH2 CHBr Alkane Haloalkane 248 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 248

acidic Hg(II) catalyst. Anti-Markovnikov addition occurs via a modified hydroboration–oxidation sequence. Both initially give vinylic alcohols (or enols) as products, but these are kinetically and thermodynamically unstable and isomerize to carbonyl compounds in a reaction called tautomerism. The latter is thermodynamically favorable because of the very strong carbon–oxygen double bond that is formed. It is kinetically rapid because the enol OOH bond is acidic and readily deprotonated, allowing the proton eventually to find its way to the nearby carbon. More details concerning the process will be upcoming when carbonyl compounds are discussed. Note that the hydration of alkynes is a new synthesis of aldehydes and ketones. 13-9. Alkenyl Halides Although the carbon–halogen bonds in alkenyl halides do not participate in the usual substitution reactions, they may be converted into carbon–metal bonds, thus allowing formation of lithium and Grignard reagents that can be used in the synthesis of allylic alcohols by exposure to ketones and aldehydes. The alkenyl halide carbon–halogen bond is also reactive in a variety of transformations involving transition metals. The Heck reaction utilizes this property to enable the coupling of the halogen-bearing carbon of an alkenyl halide to one of the double-bonded carbons of a second alkene, giving rise to dienes as products. We won’t be exploring this kind of chemistry in much depth, partly because it falls outside the scope of mainstream mechanistic organic chemistry. However, take note of the fact that such processes now make up an important part of the so-called synthetic toolbox of the pharmaceutical and medicinal chemist: Reactions such as these have made the syntheses of many types of compounds of therapeutic value much easier than was the case before their discovery and development. Solutions to Problems 21. (a) (b) (c) 22. (a) 3-Chloro-3-methyl-1-butyne (b) 2-Methyl-3-butyn-2-ol (c) 4-Propyl-5-hexyn-1-ol (d) trans-3-Penten-1-yne (e) E-5-Methyl-4-(1-methylbutyl)-4-hepten-2-yne (f) cis-1-Ethenyl-2-ethynylcyclopentane 23. Bond strengths: ethyne ethene ethane. Ethyne COH bond uses the sp orbital from carbon, which overlaps best with the 1s orbital on hydrogen. The high (50%) s character strongly attracts the bonding electrons to the carbon nucleus. This effect shifts these electrons closer to carbon, thereby enhancing the bond polarity, which follows the same order: COH greatest in ethyne. In turn, the greater bond polarity HO Br Cl O H C C H O C C Solutions to Problems • 249 1559T_ch13_247-258 11/03/05 20:12 Page 249

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 COH bond is the easiest one to deprotonate. Remember, however, that bond strength relates to homolytic cleavage (to C and H), whereas acidity refers to heterolytic cleavage (to C and H). 24. The bond strength should be greatest and the bond length smallest in propyne, again as a result of the sp (50% s character) orbital at C2. 25. 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: CH3CqNH CH3CHPNH2 CH3CH2NH2 26. Stability order is cyclopentene 1,4-pentadiene 1-pentyne. Cyclopentene has the most bonds, which are generally stronger than bonds. 1,4-Pentadiene and 1-pentyne both have two bonds, but the alkyne is of higher energy. Notice (Section 13-2) that heats of hydrogenation for alkynes are 65–70 kcal mol1 , or 32.5–35 kcal mol1 per bond, whereas the normal range for alkenes is 27–30 kcal mol1 (Section 11-7). 27. (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 mol1 . 28. Degrees of unsaturation are calculated for each compound. (a) Hsat 12 2 14; degrees of unsaturation (14 10)/2 2 bonds or rings. NMR looks like an ethyl group: CH3 (t, 1.0) next to CH2 (q, 2.0). Because the molecule has 10 H’s, there must be two equivalent ethyl groups, 2 CH3CH2O, 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 CH3CH2O and OCqCO F CH3CH2OCqCOCH2CH3 (b) Hsat 14 2 16; degrees of unsaturation (16 12)/2 2 bonds or rings. IR: terminal OCqCH. NMR: 0.9 (t, 3 H) F CH3, next to CH2 1.3 (m, 4 H) F ? 1.5 (quintet, 2H) F CH2, with CH2 groups on both sides 1.7 (t, small J, H) F Aha! How about So far, you have CH3OCH2O and OCH2OCqCH, or C5H8; you need C2H4 more. The simplest way is CH3CH2CH2CH2CH2CqCH, 1-heptyne. HCC Split by CH2 (Compare Figure 13-5.) 2.2 (m, 2H) Perhaps the CH2 referred to here? 250 • Chapter 13 ALKYNES: THE CARBON–CARBON TRIPLE BOND 1559T_ch13_247-258 11/03/05 20:12 Page 250

(c) The molecular formula is C5H8O. How? Briefly: Carbon, 71.4% of 84 60; 60/12 (atomic mass of C) 5 Hydrogen, 9.6% of 84 8; 8/1 (atomic mass of H) 8 Oxygen, 19% (the remainder) of 84 16; 16/16 (atomic mass of O) 1 Double check using exact masses from Table 11-5: 5(12.00000) 8(1.00783) 15.9949 84.05754 Hsat 10 2 12; degrees of unsaturation (12 8)/2 2 bonds or rings. IR: OCqCO stretch shows at 2100 cm1 , broad band from 3200–3500 cm1 suggest OOOH. NMR, focus on the signals with the simplest splitting patterns first: 1.8 (broad s, 1 H) F OH, broad singlet gives it away 3.7 (t, 2 H) F CH2, next OH (chemical shift tells you this), and also next to another CH2 (triplet splitting tells you that) 1.9 (t, 1 H) F CqCH (narrowness of splitting is typical), “long-range” coupled to a CH2 on the other side of the triple bond. Let’s see what we know so far. We have figured out that the molecule contains the two pieces HOOCH2OCH2O and OCH2OCqCH. Add them up and you get C5H8O: that’s all there is in the formula, so just put them together: HOOCH2OCH2OCH2OCqCH. The two middle CH2 groups are responsible for the two signal sets that we didn’t bother to try to interpret because they were more complicated. Try to figure out on your own why they look the way they do. 29. qCOH of terminal alkyne has ˜COH 3300 cm1 . (a) DOCqCCH2CH2CH2CH2CH2CqCOD (b) CqCOD (˜COD) (c) Before reaction, m1 is H (mass 1) and m2 is C9H11 (mass 119). Rewrite the Hooke’s law equation as ˜ 2 k2 f(m1 m2)/m1m2. So (3300)2 k2 f(120/119), or k2 f 1.1 107 . Because k and f are assumed to be constant, use this value for k2 f to predict ˜ 2 for the product. Now m1 is D (mass 2), so ˜ 2 (1.1 107 )(122/240) 5.6 106 and predicted ˜COD 2366 cm1 . The discrepancy of about 10% is typical and due to changes in k and f. 30. (a) CH3CH2CH(CH3)CqCH (b) CH3OCH2CH2CH2CqCCH3 (after aqueous work-up) (c) (d) The reverse of the meso compound, this gives 31. (a) trans-3-octene, via two sequential one-electron reductions as described in Section 13-6: (b) Upon sodium/liquid ammonia reduction of a triple bond, the trans stereochemistry of the double bond that normally results is determined in the first two steps of the mechanism. Addition of one electron gives the alkyne radical anion, in which the two substituents on the original alkyne carbons Z Cl C H C R R E Cl C H C R R ) CH3 (R CH3CHCH2 Rotate Meso Cl C H R Cl C H R H Cl R C Cl C H R OCH3 (anti) Solutions to Problems • 251 1559T_ch13_247-258 11/03/05 20:26 Page 251