Talk:Nile red

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Within the structure of Nile Blue hydrochloride, the C=NH2+ group is an iminium group, or imine without the HCl. Imine and iminium are featuring sp2 hybridized C and N, in contrast to an amine, which features sp3 hybridized C and N. Thus I revised "amine" to "iminium". — Preceding unsigned comment added by Biochemputer (talkcontribs) 21:20, 22 December 2013 (UTC)[reply]

Is it really 2-naphthol for the preparation of nile red? Shouldn't it be 1-naphthol? — Preceding unsigned comment added by 141.35.40.137 (talk) 10:44, 24 January 2014 (UTC)[reply]

Project[edit]

Making sure this is going to nile red..this Chrystal Augustyniak. Can not talk now but you can you reply if this is indeed nile red. I can contact you sometime tomorrow with further information.. Cauggie (talk) 03:37, 26 May 2016 (UTC)[reply]

Please, consider revising the structure in the reaction scheme. The reaction consumes naphthalenediols and not naphthol, as shown.

Adding a headnote[edit]

I don't know how to do it but there should be a head note sitting that this is about the chemical not the YouTuber ChaseTOM4YT (talk) 23:51, 22 October 2023 (UTC)[reply]

@ChaseTOM4YT I'm not sure if there is a way to do that as he doesn't have a Wikipedia article. —Panamitsu (talk) 01:11, 23 October 2023 (UTC)[reply]
do YouTubers have Wikipedia pages? ChaseTOM4YT (talk) 01:17, 23 October 2023 (UTC)[reply]
@ChaseTOM4YT Yes, YouTubers have Wikipedia pages, for example, Stampy. Nigel just doesn't have one, for whatever reason that may be. —Panamitsu (talk) 01:18, 23 October 2023 (UTC)[reply]
Oh ok might as well make one for him ChaseTOM4YT (talk) 01:19, 23 October 2023 (UTC)[reply]
@ChaseTOM4YT Hey I recently wanted to make a page about him,but I need to get MOAR info about him. DoraTheMora (talk) 18:30, 28 March 2024 (UTC)[reply]
go to the youtuber wiki and there's a whole page about him. 2601:247:4582:2A0:7CD2:7AAE:C49E:4FD5 (talk) 23:40, 28 March 2024 (UTC)[reply]
@2601:247:4582:2A0:7CD2:7AAE:C49E:4FD5 yep ok DoraTheMora (talk) 10:28, 29 March 2024 (UTC)[reply]