A number of HCFCs were recently delisted by EPA. (See Federal Register
"FINAL RULE: Exclusion of 16 Compounds as VOCs, 8-27-97" on the EPA's
The scientific information justifying delisting "...consisted mainly of
the rate constant for the reaction of the compound with the hydroxyl
(OH) radical. This rate constant (kOH value) is commonly used as one
measure of the photochemical reactivity of compounds. The petitioner
compared the rate constants with that of ethane which has already been
listed as photochemically negligibly reactive (ethane is the compound
with the highest kOH value which is currently regarded as negligibly
reactive)." (FR 8-27-97)
The kOH value for ethane is 2.4 times ten to the minus thirteen
centimeters cubed per molecule per second (2.4 E-13 cm3/molecule-sec).
The kOH value for 2-propanol (IPA, CAS# 67-63-0) is 5.07 E-12
cm3/molecule-sec. I looked up these values after I saw the EPA rule and
asked the same question you are now asking. The kOH values are from the
Environmental Science Center (ESC) of Syracuse Research Corporation,
which has an on-line database (http://esc.syrres.com/~ESC/database.htm).
Since the kOH value for IPA is twice the kOH value of ethane, it would
not meet the criteria used to delist the HCFCs on 8/27/97.
Of course, other measures of ozone formation potential may show that IPA
is OK. I have heard that Dr. W.T. Carter published typical values for
grams of ozone formed per gram of chemical released to the atmosphere.
Ethane had a value of 0.25 g-O3/g-ethane. (Journal of Air and Waste
Management, July 1994, p. 881, Table 3.)
A contact at EPA, who may be able to help you is Bill Johnson,
919-541-5245. (He referred me to the above article, which I have not
had a chance to look at yet.)
Keep us posted on what you find.