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The gas‐phase pyrolysis of alkyl nitrites VI. t ‐Amyl nitrite
Author(s) -
Batt L.,
Islam T. S. A.,
Rattray G. N.
Publication year - 1978
Publication title -
international journal of chemical kinetics
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.341
H-Index - 68
eISSN - 1097-4601
pISSN - 0538-8066
DOI - 10.1002/kin.550100904
Subject(s) - chemistry , isobutane , acetone , nitrite , reaction rate constant , nitric oxide , nitric acid , decomposition , inorganic chemistry , carbon monoxide , medicinal chemistry , kinetics , organic chemistry , catalysis , nitrate , physics , quantum mechanics
The rate of decomposition of tert ‐amyl nitrite ( t ‐AmONO) has been studied in the absence (120°–155°C) and presence (160°–190°C) of nitric oxide. In the absence of nitric oxide for low concentrations of tert ‐amyl nitrite (∼10 −4 M ) and small extents of reaction (∼1%), the first‐order homogeneous rates of acetone formation are a direct measure of reaction (1) since k 3a ≫ k 2 (NO):The rate of acetone formation is unaffected by the addition of large amounts of carbon tetrafluoride or isobutane (∼1 atm) but is completely suppressed by large amounts of nitric oxide (1 atm 120°–155°C). The rate of reaction (1) is given by k 1 = 10 16.3±0.1 10 −40.3±0.1 /θ sec −1 . Since ( E 1 + RT ) and Δ H ° 1 are identical, both may be equated with D ( t ‐AmO – NO) = 40.9 ± 0.1 kcal/mol and E 2 = 0 ± 0.1 kcal/mol. The thermochemistry leads to the result that Δ H ° f ( t ‐AmO) = −26.6 ± 1 kcal/mol. From Δ S ° 1 and A 1 , k 2 is calculated to be 10 10.5 ± 0.2 M −1 ·sec −1 . Although the addition of nitric oxide completely suppresses acetone formation at lower temperatures, it reappears at higher temperatures. This is a result of reaction (3a) now competing with reaction (2), thus allowing k 3a to be determined. The rate constant for reaction (3a) is given by k 3a = 10 14.7 ± 0.2 10 −14.3 ± 1 /θ sec −1 . There are two possible routes for the decomposition of the tert ‐amyloxyl radical:The dominating process is (3a). From the result at 160°C that k 3a / k 3b = 80, we arrive at the result k 3b = 10 15.0–18.7 /θ sec −1 . In addition to the products accounted for by the radical split (1), methyl‐2‐but‐1‐ene and methyl‐2‐but‐2‐ene are produced as a result of the six‐centre elimination of nitrous acid (5):The ratio k 5a / k 5b was 0.35. Unlike tert ‐butyl where the rates of the two paths were comparable [(l) and (5)], here the total rate of the elimination process was only 0.5% that of the radical split (1). The reason for this is not clear.