Some Ochem/Destroyer Help!

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woox

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Hey guys a few more questions on Ochem help from various sources:

Question 1 said:
So this one, the question is: Which of the compounds shown below will show only one resonance peak in its H NMR spectrum?

1f8jug.png

The answer is all of these structures but I dont understand why the nitrile (far right), has only 1 peak? Won't it show 1 and 4 peaks?

Question 2 said:
The heat of formation reaction for solid CaCO3 is:

Answer: Ca(s) + C(graphite) + 3/2 O2 (g) -> CaCO3 (s)

Can someone explain what to do for these types of questions? I don't know what its asking and don't recall Chad covering it.

Question 3 said:
How many chiral centers are in compound below?

2n3a4w.png

The answer is 8 and I marked the ones I thought are centers - my question is are these the right chiral centers? (the one marked in green is the one I am not sure about).

Thnx :)

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Question #1, the carbon attaching to N in methyl cyanide doesn't have a H, so nothing will show on the H-NMR. That only leaves you with 1 peak (a singlet).

Question #2, the heat of formation is the same thing as the enthalpy of formation of things. You need to consider each atom's own enthalpy (separately).

Question #3, yes the one in green is a chiral center.
 
Question #1, the carbon attaching to N in methyl cyanide doesn't have a H, so nothing will show on the H-NMR. That only leaves you with 1 peak (a singlet).

The carbon is also attached to a CH3 though, wont that show 4 peaks?
 
The carbon is also attached to a CH3 though, wont that show 4 peaks?

Remember, it's "H"-NMR, not C-NMR! So for any carbon to show a peak, it has to have a proton attached to it.
 
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Hey, have to make it a quick reply: 1. The first has 4 equivalent H. Since they are all equivalent, they don't split each other. The rest don't have adjacent H, non-equivalent H to split. So only one peak is shown for all of them. 2. Heats of formations is a convenient way to calculate the enthalpy change for complex reactions. Basically, you take the elemental forms (most stable form in regards for allotropes) and you form compounds from these elemental forms. This always releases energy, for instance, creating water 2C (graphite) + O2 -> CO2 (heat of formation = -X) As you can see, we took the elemental forms as reactants (graphite being the more stable allotrope of carbon) and formed a compund that released heat. The nice thing about this takes advantage of Hess' Law. You can take the reactants of a more complex reaction, decompose them into their elemental forms (the opposite of forming these compounds, and thus you will need to reverse the sign on the heats of formation). This i the energy needed to "invest" or input into the systemto break them apart. Then you can calcualte the formation of the products from the elements you just formed in the previous step. This of course, releases energy. Do any manipulation to the equation to balance everything and then add up these two heats of formations. This will give you the overall heat of reaction. Just to note, reactions don't acutually proceed like this, but Hess' Law says don't worry about it, the numbers work out to be the same regardless what path(s) you take. 3. One of your red dots is wrong. The green dot looks fine.

EDIT: sorry, for some reason all of the newline characters were removed, so its one continuous paragraph. Must have been my NoScript add-on. Sorry
 
Yeah, one of the red dots is on a C double-bond (the cyclic with the Br on it). It should be to the one adjacent to it.
 
Well what about the neighbor carbon for CH3-CN?

The CH3 has a C carbon neighbor, it has 0 H atoms, so n+1 peaks = 1.

The C in CN has 1 carbon neighbor with 3 hydrogen, so it should be n+1 = 4?

This is why the first 2 structures in image work because in the benzene, 4 same H, the only one diffrent is the carbon which is attached to Br, and that has NO hydrogens so n+1 = 1 peak.

The CH3-O-CH3, has no adjacent hydrogens, so again only 1 peak.
 
Well what about the neighbor carbon for CH3-CN?

The CH3 has a C carbon neighbor, it has 0 H atoms, so n+1 peaks = 1.

The C in CN has 1 carbon neighbor with 3 hydrogen, so it should be n+1 = 4?

This is why the first 2 structures in image work because in the benzene, 4 same H, the only one diffrent is the carbon which is attached to Br, and that has NO hydrogens so n+1 = 1 peak.

The CH3-O-CH3, has no adjacent hydrogens, so again only 1 peak.

I understand your confusion, but if a carbon does not have hydrogen attached to it, how would it show any peaks on the H-NMR?! Yes, the carbon with 3 hydrogens has a neighboring carbon, which is attached to N. And it will show a peak, BECAUSE it has 3 hydrogens attached to it. But the other carbon doesn't have a hydrogen attached to it (only a Nitrogen and a Carbon), so it won't show any peaks. Just know that if a carbon doesn't have a hydrogen attached to it, it won't show a peak! It's the Hydrogen you're analyzing not the carbon!
 
Yes, you seem to be confusing H-NMR and C-NMR. Maybe even mixing the two together at the same time. Btw, has anyone ever gotten a C-NMR on their DAT?
 
Yes, you seem to be confusing H-NMR and C-NMR. Maybe even mixing the two together at the same time. Btw, has anyone ever gotten a C-NMR on their DAT?

I think I did! It was 2 years ago though, but I remember having 2 graphs, and I think 1 was HNMR and 1 was CNMR.
 
Hmm so what would be an example of 2 or 3 peaks for H NMR so I can compare?

Anything with an ethyl group should give you 2 peaks:

CH3CH2:

1st carbon (with 3H's) = triplet
2nd carbon (with 2H's) = quartet

But then again, this is just an example. There are so many other ways you can get more than just 1 peak. Any carbon that has a H attached to it must have a peak of some sort.
 
I think one of the chiral centers is incorrect. There is one marked in red that is attached directly to a double bond. The Carbon directly above the green dot should be a chiral center instead.
 
I think one of the chiral centers is incorrect. There is one marked in red that is attached directly to a double bond. The Carbon directly above the green dot should be a chiral center instead.
+1 I noticed that too.

Woox, Chad covers heat of formation in one of his thermodynamics vids. That problem can be solved by definition of deltaH, which says something about the compound getting formed from their elemental forms (I'm too lazy to look it up in my big pile of notes lol).
 
+1 I noticed that too.

Woox, Chad covers heat of formation in one of his thermodynamics vids. That problem can be solved by definition of deltaH, which says something about the compound getting formed from their elemental forms (I'm too lazy to look it up in my big pile of notes lol).

I actually did this section today. Its 6.4 Enthalpies of Formation (Delta Hf)

Woox- I didn't understand that question either until Chad went over some examples. For Delta H of Formation Reactions you want to show how you created 1 mol of products from its reactants in STANDARD STATE. Some things to remember:

Diatomics: (N,H,F,O,I,Cl,Br) always expressed as N2, H2, F2 etc and must be expressed this way in these types of questions
All Metals are SOLID at standard state except for Br2 and Mercury
Carbon is graphite at standard state

For 1 mol of HNO3 ---> (1/2)H2 + (1/2) N2 + (3/2)O2
For 1 mol CO ---> C(graphite) + (1/2)O2

Thanks, Chad lol
 
I actually did this section today. Its 6.4 Enthalpies of Formation (Delta Hf)

Woox- I didn't understand that question either until Chad went over some examples. For Delta H of Formation Reactions you want to show how you created 1 mol of products from its reactants in STANDARD STATE. Some things to remember:

Diatomics: (N,H,F,O,I,Cl,Br) always expressed as N2, H2, F2 etc and must be expressed this way in these types of questions
All Metals are SOLID at standard state except for Br2 and Mercury
Carbon is graphite at standard state

For 1 mol of HNO3 ---> (1/2)H2 + (1/2) N2 + (3/2)O2
For 1 mol CO ---> C(graphite) + (1/2)O2

Thanks, Chad lol
+1
Br2 and mercury are liquids
 
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