ICARE-CNRS and OSUC, 1C Avenue de la Recherche Scientifique, 45071 cedex 02 Orléans, France.
J Phys Chem A. 2012 Jun 21;116(24):6113-26. doi: 10.1021/jp211355d. Epub 2012 Feb 29.
The reactions of three unsaturated alcohols (linalool, 6-methyl-5-hepten-2-ol, and 3-methyl-1-penten-3-ol) with ozone and OH radicals have been studied using simulation chambers at T ∼ 296 K and P ∼ 760 Torr. The rate coefficient values (in cm(3) molecule(-1) s(-1)) determined for the three compounds are linalool, k(O3) = (4.1 ± 1.0) × 10(-16) and k(OH) = (1.7 ± 0.3) × 10(-10); 6-methyl-5-hepten-2-ol, k(O3) = (3.8 ± 1.2) × 10(-16) and k(OH) = (1.0 ± 0.3) × 10(-10); and 3-methyl-1-penten-3-ol, k(O3) = (5.2 ± 0.6) × 10(-18) and k(OH) = (6.2 ± 1.8) × 10(-11). From the kinetic data it is estimated that, for the reaction of O(3) with linalool, attack at the R-CH═C(CH(3))(2) group represents around (93 ± 52)% (k(6-methyl-5-hepten-2-ol)/k(linalool)) of the overall reaction, with reaction at the R-CH═CH(2) group accounting for about (1.3 ± 0.5)% (k(3-methyl-1-penten-3-ol)/k(linalool)). In a similar manner it has been calculated that for the reaction of OH radicals with linalool, attack of the OH radical at the R-CH═C(CH(3))(2) group represents around (59 ± 18)% (k(6-methyl-5-hepten-2-ol)/k(linalool)) of the total reaction, while addition of OH to the R-CH═CH(2) group is estimated to be around (36 ± 6)% (k(3-methyl-1-penten-3-ol)/k(linalool)). Analysis of the products from the reaction of O(3) with linalool confirmed that addition to the R-CH═C(CH(3))(2) group is the predominant reaction pathway. The presence of formaldehyde and hydroxyacetone in the reaction products together with compelling evidence for the generation of OH radicals in the system indicates that the hydroperoxide channel is important in the loss of the biradical [(CH(3))(2)COO]* formed in the reaction of O(3) with linalool. Studies on the reactions of O(3) with the unsaturated alcohols showed that the yields of secondary organic aerosols (SOAs) are higher in the absence of OH scavengers compared to the yields in their presence. However, even under low-NO(X) concentrations, the reactions of OH radicals with 3-methyl-1-penten-3-ol and 6-methyl-5-hepten-2-ol will make only a minor contribution to SOA formation under atmospheric conditions. Relatively high yields of SOAs were observed in the reactions of OH with linalool, although the initial concentrations of reactants were quite high. The importance of linalool in the formation of SOAs in the atmosphere requires further investigation. The impact following releases of these unsaturated alcohols into the atmosphere are discussed.
三种不饱和醇（芳樟醇、6-甲基-5-庚烯-2-醇和 3-甲基-1-戊烯-3-醇）与臭氧和 OH 自由基的反应在 T ∼ 296 K 和 P ∼ 760 Torr 的模拟室中进行了研究。三种化合物的速率系数值（以 cm(3)分子(-1) s(-1)表示）分别为芳樟醇，k(O3) = (4.1 ± 1.0) × 10(-16)和 k(OH) = (1.7 ± 0.3) × 10(-10)；6-甲基-5-庚烯-2-醇，k(O3) = (3.8 ± 1.2) × 10(-16)和 k(OH) = (1.0 ± 0.3) × 10(-10)；3-甲基-1-戊烯-3-醇，k(O3) = (5.2 ± 0.6) × 10(-18)和 k(OH) = (6.2 ± 1.8) × 10(-11)。从动力学数据估计，对于 O(3)与芳樟醇的反应，R-CH═C(CH(3))(2)基团的攻击约占总反应的（93 ± 52）%（k(6-甲基-5-庚烯-2-醇)/k(芳樟醇）），而 R-CH═CH(2)基团的反应约占（1.3 ± 0.5）%（k(3-甲基-1-戊烯-3-醇)/k(芳樟醇））。类似地，已经计算出对于 OH 自由基与芳樟醇的反应，OH 自由基对 R-CH═C(CH(3))(2)基团的攻击约占总反应的（59 ± 18）%（k(6-甲基-5-庚烯-2-醇)/k(芳樟醇）），而 OH 自由基对 R-CH═CH(2)基团的加成约占（36 ± 6）%（k(3-甲基-1-戊烯-3-醇)/k(芳樟醇））。对 O(3)与芳樟醇反应产物的分析证实，对 R-CH═C(CH(3))(2)基团的加成是主要的反应途径。反应产物中存在甲醛和羟基丙酮，以及系统中生成 OH 自由基的有力证据表明，在 O(3)与芳樟醇的反应中，氢过氧化物通道在形成的双自由基[(CH(3))(2)COO]*的损失中很重要。对 O(3)与不饱和醇反应的研究表明，与存在 OH 清除剂相比，在不存在 OH 清除剂的情况下，二次有机气溶胶（SOA）的产率更高。然而，即使在低-NO(X)浓度下，OH 自由基与 3-甲基-1-戊烯-3-醇和 6-甲基-5-庚烯-2-醇的反应对大气条件下 SOA 的形成也只有很小的贡献。在 OH 与芳樟醇的反应中观察到相对较高的 SOA 产率，尽管反应物的初始浓度相当高。芳樟醇在大气中形成 SOA 的重要性需要进一步研究。讨论了这些不饱和醇释放到大气中后的影响。
