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Previous by Date: Subject: Re: [DCHAS-L] Benzaldehyde Date: Tuesday, February 18, 2014 7:55:08 PM Author: Samantha D <samantha.dannenberg**At_Symbol_Here**GMAIL.COM> |
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It is most likely benzoic acid or perbenzoic acid. Oxidation of benzaldehydes that have been freshly distilled and hence are free of inhibitors is extremely fast. I have seen crystals at the end of a 3-hour undergrad. lab in a bottle that was fresh at the beginning.
If it forms a peroxide, then this would almost certainly not occur at the aromatic ring, but by removal of H radical from the aldehyde. This yields PhC=O radical and HOO radical, which would re-combine to form the peracid. PhC=O also likely undergoes loss of carbon monoxide, giving Ph radical that would end up as phenyl peroxide. The peroxy compounds would likely also photo-dissociate into carboxy radicals and alkoxy radicals, and so on. I would also speculate that one molecule of per-acid oxidizes another of aldehyde, thereby giving two benzoic acids. So overall it is unlikely that much peracid would build up.
The other key safety point is that aryl per-acids such as per-benzoic acid or meta-chloro derivative, mCPBA, are much more stable than alkyl peroxides such as those from ether. They likely can be handled by cautious addition of a reducing agent. Try iodide, and watch the iodine form!
If the bottle is large, however, or if you are not a chemist by training, I would urge more caution.
Paul
On Tue, 18 Feb 2014 23:59:19 +0000 Paul Harrison If you have an accommodation need for a planned meeting, please e-mail me directly and I will do my best to make appropriate arrangements. Should you require any materials sent via this e-mail address in an alternate/accessible format, please let me know
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Wayne Phan
> I just did a lab safety inspection and found a bottle of Benzaldehyde with crystal formation. The bottle is dated back to 5-25-1989. I am not sure if this is peroxide or just polymer being formed. What are your thoughts? If it is peroxide, what is the proper protocol for handling it?
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> Thanks,
>
> Wayne
>
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