At the very bottom of the page you will find what I have posted in italics above. Has this been touched upon recently? I have utfse and found nothing, basically looking for anything in regards to this amazingly easy **sounding** route to the styrene.
https://www.thevespiary.org/rhodium/Rhodium/chemistry/2cb-new.txt (https://www.thevespiary.org/rhodium/Rhodium/chemistry/2cb-new.txt)
Do NOT use 2 liters of 31% HCl! That's WAY too much!
You scaled up the above synthesis x10, but used 200 mL 10.5 M NaOH instead of 20 mL?
0.2 L * 10.5 mol/L = 2.1 moles (you added 2.1 moles NaOH instead of 0.21 moles, i.e. ~1.9 moles too much).
You were supposed to use 2000 mL 4% HCl (4% HCl = ~1.3 M), which is equal to 2.0 L * 1.3 mol/L = 2.6 moles HCl
To neutralize the excess NaOH (1.9 moles) plus adding the originally required amount (2.6 moles), you need to use 1.9 + 2.6 = 4.5 moles. How much 31% HCl does that correspond to? Look here:
31% HCl = 10.0 M (moles/L)
4.5 moles HCl
------------- = 0.45 L
10.0 moles/L
Thus, you need 450 mL 31% HCl.
Either you dilute that to the originally suggested volume (2 L) by adding 1550 mL water, or you dilute it to the originally suggested concentration (4%) by adding 3 L water, so that you end up with ~3.5 L 4% HCl. The latter dilution was calculated like this:
0.45 L * 31%
------------ = ~3.5 L
4%
External heat (some heat is evolved upon mixing, but not enough to make it to 60°C). The solution added to the acid, as we want a rapid change in pH. See Post 320904 (https://www.thevespiary.org/talk/index.php?topic=8824.msg32090400#msg32090400)
(Rhodium: "2,5-Dimethoxynitrostyrene w/ NaOH catalyst", Methods Discourse)