Chemistry 3/4 2013 Thread

[lzxnl]

I don't know this guy's AN name or anything...I haven't even seen him...


Mod edit: Again, please do not use real names here.

[Kayte]

How do you determine the number double bonds in a fatty acid?

Such as something like C17H29COOH?

Thanks!

[KevinooBz]

How do you determine the number double bonds in a fatty acid?

Such as something like C17H29COOH?

Thanks!

If you're finding the number of carbon double bonds you can use the rule 2n+1, where n is the number of carbons,  to test if the molecule is saturated. 2(17)+1=35. So you know it's unsaturated since the number of hydrogens is actually 29. Every time you have a carbon double bond you lose the space for two hydrogen bonds. To lose 6 hydrogen bonds, there would need to be 3 carbon double bonds.

[xenial]

Okay, a few questions:

What proportion of the exam will these 'essay style questions' make up? Is there any specification about what they'll be about - is it just experiment design, like the last question of the sample exam? Or are worded questions slowly taking over the entire assessment itself? Maybe I should revise analytical techniques more.. if they ask me to explain things like IR and Mass spec in detail them I'm screwed haha.

(Also, on that tangent about spec - how many marks did you lose for like 47ish in 2011? I'd love to say I could aim for a 50.. but I I made stupid errors on my first and last sacs and I think I'm rank 3/4 now..)

[lzxnl]

How do you determine the number double bonds in a fatty acid?

Such as something like C17H29COOH?

Thanks!

I personally prefer a 'first principles' approach to this question as opposed to rote-learning formulas that are easily confused. This is just my perspective.

So, C17H29COOH? From the perspective of carbon-carbon double bonds, you can replace the COOH group with another hydrogen. Think about it. The number of C=C bonds does not change as you've replaced a -CH2-COOH with -CH3
That leaves us with C17H30. If you think about what straight-chain alkanes are, they consist of two terminal CH3 groups and then everything in between is CH2. So really, if we have a chain of length n, the molecular formula of your alkane is CnH(2n+2); two hydrogens per carbon in the CH2 section and then the two extra terminal hydrogens.
So compare your C17H30 with CnH(2n+2). n=17 obviously, but our alkane has six more hydrogens. I trust you can see that removing two hydrogens from neighboring carbons adds one to the C=C count. Thus six less hydrogens than alkane => three C=C bonds.

Okay, a few questions:

What proportion of the exam will these 'essay style questions' make up? Is there any specification about what they'll be about - is it just experiment design, like the last question of the sample exam? Or are worded questions slowly taking over the entire assessment itself? Maybe I should revise analytical techniques more.. if they ask me to explain things like IR and Mass spec in detail them I'm screwed haha.

(Also, on that tangent about spec - how many marks did you lose for like 47ish in 2011? I'd love to say I could aim for a 50.. but I I made stupid errors on my first and last sacs and I think I'm rank 3/4 now..)

These longer questions, from experience, seem to be on analytical and experimental techniques, such as chromatography, NMR etc. I would certainly revise explanations of how these techniques work; they're bound to ask at least one in detail I think. I mean, only those who can explain these things well show that they know what goes on in the techniques.

[clıppy]

Just another question for cells again.
What are the differences (if any) between electrolytic cells and a recharging secondary cell?

[saba.ay]

I don't think there is any difference. A secondary cell is an electrolytic cell when it is connected to a power supply and being recharged.

[EliStone]

Hey, I'd doing a design prac next week with the following description, I was just wondering if anyone has any tips or has done this prac before...
The design practical will investigate the relationship between something, we get told on the day

The reaction being used in this investigation is between potassium iodide solution and hydrogen peroxide solution as described by the following equation:
Reaction 1   2I−  + H2O2 +  2H+  → I2 + 2H2O   
   20 mL samples of this are used    for each trial. Immediately prior to reaction, 10mL of 2 molL−1 sulfuric acid and 2 mL of    starch solution are added. The other reactant is a hydrogen peroxide solution made by    diluting 5 mL of 20 volume H2O2 up to a total 30 mL with distilled water.
   The S2O3= reduces the I2 as it forms in reaction 1 according to the following equation:
      Reaction 2   2S2O3=  + I2 → S4O6=  + 2I− 
Reaction 2 continues until all of the S2O3= is used up. The slight excess of I2 which now    forms reacts with the starch to indicate a completion point for Reaction 1

Any help would be appreciated,
Regards Eli 

[barydos]

So we know that Group 1 metals react vigorously with water, but what is the reaction pattern that they follow?
Group1Metal + H2O -->  ? + ?
e.g. Na_(s) + H₂O_(l) --> ? + ?
Thanks

[BasicAcid]

So we know that Group 1 metals react vigorously with water, but what is the reaction pattern that they follow?
Group1Metal + H2O -->  ? + ?
e.g. Na_(s) + H₂O_(l) --> ? + ?
Thanks

NaOH(aq) + H2(g)

The rapid production of H2 is the reason why they react violently and explode in water I believe

[lzxnl]

No...that's the effect, not the cause.

The actual cause is the weak attraction alkali metals have for their lone valence electrons, leading to a low activation energy and hence a fast and vigorous reaction.

Note that activation energy is not the same as reducing strength; lithium reacts very slowly with water and in a rather boring matter because its valence electrons is held more strongly due to its proximity to the nucleus, leading to a higher activation energy, but it is more "reactive" than sodium.

[barydos]

Thanks guys.

Also I just want to clarify with the systematic naming of organic compounds.
Is this what we do?
- (highest priority) functional group goes to suffix
    * highest priority (smallest number on carbon atom): carboxylic acid > alkanol > amine ?
- any other functional groups are alphabetically ordered in prefix (secondary functional groups like Chloro, Bromo, and the alkyl groups like methyl/ethyl are always prefixed?)

Additionally, are chloro/bromo actually called secondary functional groups (or am I making this up?), and are methyl/ethyl also ordered alphabetically in addition to the chloro/bromo stuff? Oh and are the chloro/bromo and methyl/ethyl equal in priority?

And double bonds... are they written before the suffix functional group? But-2-en-3-ol (also is it ene or en)

And finally is the following called: 3-chloro-2-methylbut-4-en-2-ol?


Sorry for all the questions! Help is greatly appreciated

[lzxnl]

Thanks guys.

Also I just want to clarify with the systematic naming of organic compounds.
Is this what we do?
- (highest priority) functional group goes to suffix
    * highest priority (smallest number on carbon atom): carboxylic acid > alkanol > amine ?
- any other functional groups are alphabetically ordered in prefix (secondary functional groups like Chloro, Bromo, and the alkyl groups like methyl/ethyl are always prefixed?)

All of these are correct

Additionally, are chloro/bromo actually called secondary functional groups (or am I making this up?), and are methyl/ethyl also ordered alphabetically in addition to the chloro/bromo stuff? Oh and are the chloro/bromo and methyl/ethyl equal in priority?

Secondary functional groups are functional groups bonded to a carbon which is bonded to two other carbons. So a secondary amine would be something like CH3CH(NH2)CH3

And double bonds... are they written before the suffix functional group? But-2-en-3-ol (also is it ene or en)

That is what I'd call it. Unless of course, you have a methyl or halogen group.

[quote
And finally is the following called: 3-chloro-2-methylbut-4-en-2-ol?


Sorry for all the questions! Help is greatly appreciated
[/quote]

It's 3-chloro-2-methylbut-3-en-2-ol. Number designates where the double bond begins.

[random_person]

Question 10 - VCAA 2008 Sample Exam

Some Gaseous PCl₅ is placed in an empty container.
PCl₅ PCl₃ + Cl₂        K_eq = 2.5 M
When equilibrium is reached, the mass of the gas mixture, compared to the initial mass of PCl₅, is

A   Halved.
B   Unchanged.
C   One and a half times greater.
D   Doubled.

Can someone please tell me the answer?

[barydos]

All of these are correctSecondary functional groups are functional groups bonded to a carbon which is bonded to two other carbons. So a secondary amine would be something like CH3CH(NH2)CH3That is what I'd call it. Unless of course, you have a methyl or halogen group.

It's 3-chloro-2-methylbut-3-en-2-ol. Number designates where the double bond begins.

You're a legend thank you very much!