## 4 Discussion
### Reaction Mechanism to Form Gaseous Products  
The great variety of products detected by three complementary techniques allows to propose a reaction mechanism. What seems clear is that HCl, CO, 3-chloro-2-butanone, and acetaldehyde are the products for which the yields are higher. There are two reaction pathways, depicted in Figures 8 and 9, that can explain the presence of these reaction products: the addition of Cl to the double bond in the C-3 position (Figure 8) and the H-abstraction from the aldehyde group (Figure 9). In both figures, the species in bold are the ones identified in this study. Figure 8: Proposed mechanism for the Cl reaction of _trans_-2-methyl-2-butenal through for the Cl-addition to C-3. The addition of Cl to C-3 of T2M2B can explain the formation of the observed products: 3-chloro-butanone, methylglyoxal, acetaldehyde, acetyl chloride, methanol, and formaldehyde. The initial step of this pathway leads to the formation of a chlorinated alkyl radical (CH\({}_{3}\)CH(Cl)C(CH\({}_{3}\))C(O)H) that, following the general reactivity trend, reacts with O\({}_{2}\) and a peroxy radical to generate the alkoxy radical CH\({}_{3}\)CH(Cl)CO(CH\({}_{3}\))C(O)H. This alkoxy radical can react with O\({}_{2}\), as described by [PERSON] et al. [29] for carbonyl containing peroxy radicals, to form 3-chloro-2-butanone, CH\({}_{3}\)CH(Cl)C(O)CH\({}_{3}\), or it can decompose to generate methylglyoxal (CH\({}_{3}\)C(O)C(O)H)) and the CH\({}_{2}\)CHCl radical that continues to react with O\({}_{2}\), RO\({}_{2}\), and decompose to generate acetaldehyde (CH\({}_{3}\)C(O)H) and acetyl chloride (HC(O)Cl), and, through further reactions of the methyl radical, methanol (CH\({}_{3}\)OH) and formaldehyde (HC(O)H). However, the decomposition route of the CH\({}_{3}\)CH\({}_{2}\)Cl radical seems to be almost negligible, given the low yield of methyl glyoxal, (1.17 \(\pm\) 0.08)%. The C-3 addition mechanism was expected, according to previous studies for unsaturated aldehydes [9], to be inhibited because of the conjugation between the double bond and the carbonyl group that reduces the partial negative charge on that carbon atom. However, the presence of the methyl group in this molecule seems to activate this channel, since one of the most abundant products observed, 3-chloro-2-butanone (CH\({}_{3}\)CH(Cl)C(O)CH\({}_{3}\)) can only be explained by this reaction pathway. The observed yield for this product indicates that this pathway represents around 53% of the overall reaction. The H-abstraction from the -C(O)H group can explain the presence of acetaldehyde, HCl, CO, 2-methyl-2-butenoic acid, butanone, butadiene, and ketene. The first step leads to the formation of HCl, a major product, and a carbonyl radical that can decompose or react with O\({}_{2}\) followed by the reaction with RO\({}_{2}\) or HO\({}_{2}\) radicals. The reaction with HO\({}_{2}\) can generate the 2-methyl-2-butenoic acid, CH\({}_{3}\)CH=C(CH\({}_{3}\))C(O)OH, one of the minor products of this reaction, the decomposition generates the CH\({}_{3}\)CH=CCH\({}_{3}\) radical, and the RO\({}_{2}\) reaction leads to the formation of the CH\({}_{3}\)CH=C(CH\({}_{3}\))C(O)O radical that can decompose to generate CO\({}_{2}\) and the unsaturated alkyl radical CH\({}_{3}\)CH=CCH\({}_{3}\), as described in various previous studies for unsaturated aldehydes [30; 31; 32]. The CH\({}_{3}\)CH=CCH\({}_{3}\) radical reacts with O\({}_{2}\) to form acetaldehyde, CH\({}_{3}\)C(O)H, and the acetyl radical, CH\({}_{3}\)C(O), or the  
Figure 9: Proposed mechanism for the Cl reaction of _trans_-2-methyl-2-butenal through the initial H-abstraction from the aldehyde group. CH\({}_{3}\)CH=C(CH\({}_{3}\))OO radical. The acetyl radical can react with Cl atoms and can generate HCl and ketene [33], another product observed at high concentration. The CH\({}_{3}\)CH=C(CH\({}_{3}\))OO radical can react either with HO\({}_{2}\) or RO\({}_{2}\) radicals.