April 18, 2024, 12:16:39 AM
Forum Rules: Read This Before Posting


Topic: how to work out a reaction ??  (Read 1265 times)

0 Members and 1 Guest are viewing this topic.

Offline Pickard88

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
how to work out a reaction ??
« on: November 04, 2020, 09:16:26 AM »
hi, i was just wanting a bit of advice. i know i cannot expect to learn the intricacies of chemistry from a few posts on a forum but if anyone has any advice i would really appreciate it.
how to you formulate a reaction? say i had a precursor that i wanted to oxidize, and i have an oxidizing agent, how would i work out how much of said oxidizing agent to use?
as i said i don't expect to become an instant chemist and it may take years of university before getting to a point where it is possible to know how to go about it but thought you never know it may be simpler than i thought haha often synthesis are posted as the basics giving the precursor/reagents used and the basic process but no details in terms of quantities ect...
thankyou :)

Offline wildfyr

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 1771
  • Mole Snacks: +203/-10
Re: how to work out a reaction ??
« Reply #1 on: November 04, 2020, 09:59:46 AM »
Typically, we use molar equivalencies to work that problem out. 1 mol of oxidizing agent to 1 mol of precursor.

There are exceptions and more complicated systems, but that is the most general explanation.

Offline Pickard88

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: how to work out a reaction ??
« Reply #2 on: November 04, 2020, 10:02:51 AM »
ok wow that is more simle than i expected lol, does that just apply to oxidizing or is it the same for reductions/other reactions?

Offline wildfyr

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 1771
  • Mole Snacks: +203/-10
Re: how to work out a reaction ??
« Reply #3 on: November 04, 2020, 10:38:44 AM »
It depends a lot on your reagents. A single mole of lithium aluminum hydride contains 4 hydrides, so it can reduce something 4 times (potentially, as each hydride reacts from the core, the remaining ones are less reactive). However sodium hydride contains only 1 hydride, so it reacts 1 to 1.

Its really about moles of "reactive moieties" rather than mols of reagents per se. Putrecine has 2 equivalent reactive groups, so to acetylate both of them you would need 1 mol of putrecine and 2 moles of acetylating agent (like acetyl chloride).

Offline kriggy

  • Chemist
  • Sr. Member
  • *
  • Posts: 1520
  • Mole Snacks: +136/-16
Re: how to work out a reaction ??
« Reply #4 on: November 05, 2020, 07:38:22 AM »
ok wow that is more simle than i expected lol, does that just apply to oxidizing or is it the same for reductions/other reactions?

More or less yes but there are other considerations as well.
For example, reducing the nitro group to amino group needs 6 electrons, therefore when using 2 electron donor like ZnCl2 you need at least 3 molar equivalents of it.
WHen reducing esters to alcohols by hydride reagents, the reaction goes stepwise: at first single hydride is added to the ester to yield aldehyde which then accepts another hydride and becomes an alcohol.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27652
  • Mole Snacks: +1800/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: how to work out a reaction ??
« Reply #5 on: November 05, 2020, 08:13:44 AM »
electron donor like ZnCl2

You've lost me. Not Zn?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Pickard88

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: how to work out a reaction ??
« Reply #6 on: November 06, 2020, 11:04:57 AM »
youve lost me there as well i must say haha, how do you know which moieties of a reagent are reactive?

Offline hollytara

  • Chemist
  • Full Member
  • *
  • Posts: 317
  • Mole Snacks: +39/-0
Re: how to work out a reaction ??
« Reply #7 on: November 07, 2020, 05:13:48 PM »
This is where the literature is extremely useful.

Sponsored Links