VCE Chem Questions Thread

God

Billzene
Ahhhh... Ok. Cheers!

robinsonada

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

Billzene

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

PizzaMaster

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

God

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...... )

God

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! 🙂

Billzene

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)

God

Cheers!

pel0010

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

ArtyDReams

pel0010

What are you getting stuck on specifically?

Will

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)

Billzene

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

God

Will

To reduce carboxylic acid production when oxidising alcohols:

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

Billzene

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

Will

I have trouble understanding hydrogen/carbon groups environments.
I can identifying them, I just seem to get tripped up on whether different hydrogens are part of the same group.
My common mistake is believing that two carbon environments are part of the same environment, when they are in reality different.

For carbon/hydrogen sections to be a part of the same environment does the molecule HAVE to be symmetrical?

Billzene

Will

If there is a H-environment identical to another in the molecule, then yes it's symmetrical. One hack I learned in uni for identifying whether or not 2 given environments are chemically equivalent is to pretend to cut off the C atom the H-environment is attached to from the rest of the molecule. You repeat it for the other environment (with a separate structure) and then you compare if the rest of the molecule is identical in both cases. If it is, then these 2 environments are equivalent (will post a visual example)

Billzene

There done, the squiggly line is just the convention we use in uni organic chem to represent a removed or arbitrary fragment.
3-pentanone example

Will

Billzene

Thank you so much!
I really appreciate it!

God

Hi Billzene!

Quick question:
For the substitution reaction: Haloalkane + H₂O(g) → Alcohol: NaOH / OH-

Is water a gas? Or a liquid?

Cheers!

Btw: Absolutely love that method for h-environments! Maybe you should be called god, lol....

Billzene

God
H₂O isn't involved in the reaction converting haloalkanes to alcohols cuz it's uncharged and therefore a weak nucleophile. Unless you have a suitable catalyst, this reaction won't happen at a significant rate, that's why VCAA expects NaOH or KOH or OH-. You need OH- since converting a primary haloalkane into a primary alcohol is what we call an SN2 reaction, which requires a strong (charged) nucleophile. You may think OH- is the catalyst cuz it deprotonates water, but this isn't actually the case. You use up 1 particle of OH- to deprotonate 1 molecule H₂O (making 1 particle of OH-) which then reacts with a haloalkane molecule (so you end up using 2 particles of OH- while only producing 1, for a net use of 1 OH- particle). Since catalysts by definition aren't consumed, OH- in this case is a reactant, not a catalyst

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