Billzene
Thanks so much Billzene! Legend 🙂

So just to clarify - we could say H2SO4 for alkene hydration... ??? Also - could we ignore it's state for the oxidation of alcohols as well?
Alcohol → Aldehydes → Carboxylic Acid (Reagents: Cr2O72-(aq) or MnO4-; Catalyst: H2SO4 (l))

I think I'm just going to forget H3PO4 existed..... 😆

      God
      Technically you can state H2SO4 for alkene hydration, according to the 2020 examiner’s report. I was taught H3PO4, but since VCAA accepts any strong inorganic acid except HCl, you can assume H2SO4 is fair game

      For oxidation of alcohols, I think the state of the reagent is required. In VCAA exams, they don’t want you to write H2SO4 next to the oxidant (just H+). I usually write H+(aq)/MnO4-(aq) just cuz it’s easier to write (in uni chem you also learn that MnO4- is a better oxidant than dichromate, as well as being safer cuz Cr(VI) compounds are very carcinogenic).

          Hi everyone!
          Just a quick question:
          why is core charge always positive?

          Thank you!

            PizzaMaster

            Since core charge is given by number of protons in nucleus - number of non-valence e-s, and the number of non-valence e-s is always less than the number of protons in nucleus, core charge will always be positive

              Can someone help me with this chemistry question?

              There are a number of structural isomers for the molecular formula C4H10O. (Mass Spec Graph) Identify the fragment at 29 m/z.

              The two possible answers are [C2H5]+ or [COH]+

              My question is.... shouldn't the [COH] have a charge of +2? The molecule has lost two hydrogens...? not just 1 like in C2H5.....

              Or does one electron drift off with the hydrogen, and the other one just happen to stay with the molecule?

              Thanks!
              -G

                God For the purposes of VCE chem, you should assume that all fragments have a +1 charge.

                "Or does one electron drift off with the hydrogen, and the other one just happen to stay with the molecule?"

                This is called a radical, since they're uncharged, they won't be detected by the mass spec machine since it can only "see" charged fragments

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                    Struggling a bit with states for organic reaction pathways, always messing up gaseous and liquid states. Any tips?

                      robinsonada

                      H2SO4 should be (l) when it's catalysing an esterification because it's concentrated

                      Oxidising reagents are written as H+(aq)/Cr2O72- (aq) or H+(aq)/MnO4- (aq)

                      H2O will be (g) in alkene hydration since it's conducted at 300˚C

                      All reactants for esterification are (l) and not (aq) since water can hydrolyse the ester product back into the reactants

                        Hi everyone,
                        just quickly,
                        does a high melting point also mean that an element will have a high boiling point too?
                        thank you!

                          PizzaMaster
                          Most of the time - i'd say yes. High melting point means high intermolecular forces - which shouldn't disappear when it becomes a liquid. (Although, knowing chemistry...... )

                            Can someone help with finding the splitting patterns in butanal?

                            My question is.... for the hydrogen bonded to the C=O group, should it be a triplet or a singlet?

                            If a triplet, shouldn't that mean the CH2 group (also bonded to the C=O) should be a quartet?

                            Thanks! 🙂

                              God
                              The carbonyl hydrogen should be a triplet. The H environment on the alpha carbon has complex splitting so I don’t think VCAA will ask you that (it will be a quartet by the n+1 rule, but the reality is that the aldehyde H and the CH2 are in very different chemical environments that split the alpha carbon’s H environment by different amounts, so the observed pattern has a lot of overlapping peaks and isn’t a quartet. You’d call this a multiplet)

                                Im currently doing U2 Aos1 but i don't get ions and charges at all

                                  When you are oxidising a primary alcohol, how do you prevent the reaction from going all the way to a carboxylic acid? It will first make an aldehyde, but how do I prevent the aldehyde from becoming oxidised?

                                  Alcohol ---> Aldehyde ----> Carboxylic acid MnO4-(aq) / H+(aq)

                                    Will
                                    In uni chem you learn that you wouldn't use acidified dichromate (Jones reagent) or acidified permanganate if you want an aldehyde final product because they're strong oxidising agents that will oxidise any aldehyde formed into a carboxylic acid. You'd use Swern's reagent or Ley's reagent (TPAP) if you want to oxidise a 1˚ alcohol to an aldehyde because they're weaker oxidants that don't over-oxidise the alcohol to a carboxylic acid

                                      Will

                                      To reduce carboxylic acid production when oxidising alcohols:

                                      • Limit the heat
                                      • Limit the oxidising agent
                                      • Remove the aldehydes as it is produced.

                                          God
                                          I know it's what your teacher may have taught you (I was taught that mild oxidation under 60˚C produces aldehydes in high school too), but this one of those really shameless lies taught in VCE chem. Acidified permanganate or dichromate is aqueous, and in aqueous solutions, aldehydes react with H2O to form geminal diols (an alcohol that has 2 hydroxyl groups attached to the same carbon). Hence you can't actually make a pure sample of aldehyde with the oxidising reagents you learned about in VCE, only its geminal diol (hydrate) form which has very different chemical properties from aldehydes