Author Topic: ed_saifa's chem questions =) (Read 5778 times) Tweet Share

ed_saifa · « **on:** April 27, 2009, 01:47:32 pm »

Help would be awesome. Thanks

[IMG]http://img8.imageshack.us/img8/7357/questionq.th.jpg[/img]

Mao · « **Reply #1 on:** April 27, 2009, 09:02:36 pm »

Haven't done this, but I think the =O breaks to become -O-, an -OH donates the H to -O-, and ether link is formed.

so for (a), it is C₆H₁₁(OH)O(CH₂)₂OH

ed_saifa · « **Reply #2 on:** April 27, 2009, 09:24:31 pm »

Thanks Mao.

What are the products when Benzamide is involved in a reaction in a basic hydrolysis?

Mao · « **Reply #3 on:** April 27, 2009, 09:44:12 pm »

Wait, I take that back. Turns out I know nothing about this.

The thing I have drawn is the 'hemiacetal', the intermediate between the ketone and the acetal.

for (a), the hemiacetal reacts again with the diol, which then becomes

C₆H₁₀(O(CH₂)₂OH)₂

[attached drawings]

ed_saifa · « **Reply #4 on:** May 05, 2009, 11:48:06 pm »

I've got a question about aromaticity.
Furan is aromatic yet I can only count 5 pi bonds. Where is the 6th pi bond?
Thanks

Mao · « **Reply #5 on:** May 05, 2009, 11:58:35 pm »

I thought aromaticity is just delocalisation of electrons over the ring. For furan, a lone pair from the oxygen delocalises.

Wikipedia has pictures.

ed_saifa · « **Reply #6 on:** May 06, 2009, 12:05:01 am »

I dunno. I don't get pi bonds. Is delocalised electrons similar to pi bonds?

Collin Li · « **Reply #7 on:** May 09, 2009, 07:27:00 pm »

In acid, the carbonyl groups tautomerise into enols (wow it's all coming back to me, gross!).

Enols are double bonds and OH groups (en and ol). The electron-rich double bonds can attack electrophilic carbons. (Carbons next to oxygens?)

Follow the mechanistic arrows of where the electrons go and you should be fine.

ed_saifa · « **Reply #8 on:** May 29, 2009, 11:41:02 pm »

Hydrogen reacts with iodine to form hydrogen iodide according to the following reaction
H2(g) + I2(g) ↔ 2HI(g)
Calculate the equilibrium concentration of HI after 1 mole of H2 and 1 mole of I2 are placed in a 1
litre vessel, given that KC for the reaction is 55.3 at 25 ºC.

Help with this would be appreciated thanks!

Collin Li · « **Reply #9 on:** May 30, 2009, 05:36:01 pm »

$55.3 = \frac{\mbox{[HI]}^2}{\mbox{[H}_2\mbox{][I}_2\mbox{]}}$

Those are referring to the concentrations at equilibrium, of course.

Since $n = cV$ , and $V = 1$ , then $n = C$ and hence the equilibrium concentrations can be denoted as:

$\mbox{[H}_2\mbox{]} = 1 - x$
$\mbox{[I}_2\mbox{]} = 1 - x$
$\mbox{[HI]} = 2x$

Our job is to find $x$ :

$\implies 55.3 = \frac{4x^2}{(1-x)^2}$

$\implies 55.3\times(1-x)^2 = 4x^2$

$\implies \sqrt{55.3}\times(1-x) = \pm 2x$

$\implies \sqrt{55.3} = x\left(\sqrt{55.3} \pm 2\right)$

$\therefore x = \frac{\sqrt{55.3}}{\sqrt{55.3} \pm 2}$

Since the concentrations must be zero or positive, then $1-x\geq 0 \implies x\leq 1$

$\implies x = \frac{\sqrt{55.3}}{\sqrt{55.3} + 2} \approx 0.7881$

So, $\mbox{[HI]} = 2x \approx 1.5761$

Collin Li · « **Reply #10 on:** May 31, 2009, 09:45:01 am »

Note that the key difference between this type of problem and low $K_a$ problems is that we assumed that the $-x$ for the equilibrium reactant concentrations were negligble instead.

This approximation is justified because for low $\mbox{K}_a$ , the reaction does not proceed forward very much (favours reactants moreso), so it makes sense to assume little change to the reactant concentrations.

Hence, we would have had a much simpler result (if we applied this approximation to the above question):
$\implies K_c = \frac{4x^2}{1\times1} = 4x^2$

$\therefore \mbox{[HI]} = \sqrt{55.3} \approx 7.4364$ (a bad approximation in this case)

But you probably already realised this!

Collin Li · « **Reply #11 on:** May 31, 2009, 09:54:22 am »

Quote from: ed_saifa on May 06, 2009, 12:05:01 am

I dunno. I don't get pi bonds. Is delocalised electrons similar to pi bonds?

Pi bonds are able to delocalise because they are "secondary" to sigma bonds, and hence these electrons are "freer" to move (and delocalise across the structure). Hence, we can only move around the electrons in the "second bond" of a double bond - not possible for a single bond, otherwise we'd be breaking the framework of the molecule!

ed_saifa · « **Reply #12 on:** May 31, 2009, 12:43:20 pm »

Thanks HEAPS Coblin!

ed_saifa · « **Reply #13 on:** August 22, 2009, 11:09:17 am »

Hi guys, I got question on kinetics. Cheers

[IMG]http://img38.imageshack.us/img38/3/kinetics.th.jpg[/img]

I tried finding an equation for it using a graphics but that didn't work.
I also tried using $log{[A]}=log{[A_0]} -\frac{kt}{2.303}$

Mao · « **Reply #14 on:** August 23, 2009, 12:49:19 am »

Have you tried modelling that as a second order reaction?

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