Nuclear Magnetic Resonance

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I am having trouble with figuring out the number of unique carbons and unique hydrogens. What exactly does it mean when we say unique? If we say that a molecule has 5 unique carbons are we saying that each carbon has different bonds in relation to one another?

For n-butyl acetate do we say that there are six carbons because each carbon has unique bonds? TBR identifies the number of unique carbons within a molecule and then identifies which of those unique carbons has hydrogens. Will identifying the number of unique carbons and then identifying which of those has hydrogens ALWAYS give you the number of unique hydrogens?

The alkoxy group on C1 and the methyl group on C4 are unique carbons so as are the the carbons that they are attached to. I know this but I do not understand it. If it goes by the unique bonds that carbons forms then it would go as follows:
C2 C-H C-C C=C
C3 C-H C-C C=C
C5 C-H C-C C=C
C6 C-H C-C C=C

Carbons 2, 3, 5, and 6 all share the same type of bonds yet TBR says that there are 6 unique bonds, of which only four contain hydrogens. These examples fall under the subheading "Symmetry and NMR Signals" and I am having trouble relating symmetry to n-butyl acetate and para-methylanisole.

Seeing symmetry in these to molecules is something it is easy to do. One exhibiting mirror symmetry while the other exhibiting point symmetry. My last questions pertains to benzene. Does it have 1 unique carbon and 1 unique hydrogen?

 

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I am having trouble with figuring out the number of unique carbons and unique hydrogens. What exactly does it mean when we say unique? If we say that a molecule has 5 unique carbons are we saying that each carbon has different bonds in relation to one another?

When we say "unique" we're referring to the atom's electronic environment; personally I prefer to avoid the word unique because it's inherently relative/implies a context so I prefer to think of "types" of a given atom. Now, when I say "electronic environment" I mean the set of bonding groups that the atom has. A discrete "type" of carbon (if that's the atom we're concerned with) has a particular set of bonding groups. There are two things important about this set: 1) sets are equivalent only when every single component of the set is identical and 2) the bonding groups are the components of the set and they extend as far as possible. Let's take ethylene (H3C-CH3) as an example. When we examine the first (either) carbon's bonding groups, we see four bonding groups:

1) bond to H
2) bond to H
3) bond to H
4) bond to CH3

Note that the 4th group is not simply "bond to C" because of the 2nd key principle of the "type" classification: the bonding groups extend as far as possible. When we follow the path along each of those bonding groups we've identified, we run out of bonds to continue down.

For n-butyl acetate do we say that there are six carbons because each carbon has unique bonds?

For n-butyl acetate I would say there are six "types" of carbons because there are six carbons with sets of bonding groups where each of those sets is different from the others.

TBR identifies the number of unique carbons within a molecule and then identifies which of those unique carbons has hydrogens. Will identifying the number of unique carbons and then identifying which of those has hydrogens ALWAYS give you the number of unique hydrogens?

No, simply because for 1H-NMR hydrogen atoms attached to atoms other than carbon will show up (e.g. alcohols).

The alkoxy group on C1 and the methyl group on C4 are unique carbons so as are the the carbons that they are attached to. I know this but I do not understand it. If it goes by the unique bonds that carbons forms then it would go as follows:
C2 C-H C-C C=C
C3 C-H C-C C=C
C5 C-H C-C C=C
C6 C-H C-C C=C

Carbons 2, 3, 5, and 6 all share the same type of bonds yet TBR says that there are 6 unique bonds, of which only four contain hydrogens. These examples fall under the subheading "Symmetry and NMR Signals" and I am having trouble relating symmetry to n-butyl acetate and para-methylanisole.

Rewrite the sets of bonding groups for carbons 2 and 6 of para-methylanisole keeping in mind what I said above about the bonding groups going out as far as possible along bonds (this will be time-consuming but hopefully drive the point home.) You should find that you come back around to the same carbons 2 and 6 after completing the ring and that the bonding groups you've written are identical. From a symmetry perspective, the plane runs along the para-substitution axis; carbons 2 and 6 and 3 and 5 are pairs of equivalent carbons.

Seeing symmetry in these to molecules is something it is easy to do. One exhibiting mirror symmetry while the other exhibiting point symmetry. My last questions pertains to benzene. Does it have 1 unique carbon and 1 unique hydrogen?

On the topic of benzene, this is why I don't like using "unique" to describe them; to me, saying benzene (a six-carbon compound) has one "unique" carbon implies that the other 5 are both different from the one "unique" one and also the same amongst themselves (if that makes no sense just forget I said anything.) Yes, for benzene all six carbons are equivalent, and so all six hydrogens are equivalent.

Edit: to clarify, for p-methylanisole carbons 2 and 6 are equivalent carbons, and carbons 3 and 5 are equivalent carbons. Each of those pairs represents a "type" of carbon.

 

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When we say "unique" we're referring to the atom's electronic environment; personally I prefer to avoid the word unique because it's inherently relative/implies a context so I prefer to think of "types" of a given atom. Now, when I say "electronic environment" I mean the set of bonding groups that the atom has. A discrete "type" of carbon (if that's the atom we're concerned with) has a particular set of bonding groups. There are two things important about this set: 1) sets are equivalent only when every single component of the set is identical and 2) the bonding groups are the components of the set and they extend as far as possible.

If we were to look at the C1 carbon for n-butyl acetate:

Is this what you mean by the bonding groups extending as far as possible? If so, I had trouble applying this to cyclic compounds.

Rewrite the sets of bonding groups for carbons 2 and 6 of para-methylanisole keeping in mind what I said above about the bonding groups going out as far as possible along bonds (this will be time-consuming but hopefully drive the point home.)

This was a little confusing. I still don't understand why C6 and C2 are equivalent and why C3 and C5 are equivalent. If every single component of a bonding set must be equivalent then how are C2 and C6 equivalent? It is the double bond that is confusing me. I understand that both are equidistant from the C1 and C4 carbons and you encounter the same groups if you go clockwise for C6 and counter-clockwise for C2. Running counter-clockwise for C2 you encounter a double bond where as you encounter a single bond for C6 running in the clockwise direction. Do you consider double bonds to not be a component of the bonding set?

 

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If we were to look at the C1 carbon for n-butyl acetate:

Is this what you mean by the bonding groups extending as far as possible? If so, I had trouble applying this to cyclic compounds.

You've got the idea but you actually skipped some atoms there. In the format I would use, the bonding groups would be:

1) bond to H
2) bond to H
3) bond to H
4) bond to COO(CH2)3CH3

I'll write them out for you for p-methylanisole below.

This was a little confusing. I still don't understand why C6 and C2 are equivalent and why C3 and C5 are equivalent. If every single component of a bonding set must be equivalent then how are C2 and C6 equivalent? It is the double bond that is confusing me. I understand that both are equidistant from the C1 and C4 carbons and you encounter the same groups if you go clockwise for C6 and counter-clockwise for C2. Running counter-clockwise for C2 you encounter a double bond where as you encounter a single bond for C6 running in the clockwise direction. Do you consider double bonds to not be a component of the bonding set?

Keep in mind there are no discrete double bonds; the resonance hybrid most accurately represents the structure and the cyclic bonds all have bond order of 1.5.

The component groups for Carbon 2 in p-methylanisole are:

1) bond to H
2 - going counterclockwise) 1.5 bond to C(OCH3)CHCH [at this point I will arbitrarily choose to stop as the next bond to follow leads to C4, and the next bonding group will lead there as well and we can follow it with the next group]
3 - going clockwise) 1.5 bond to CHCCH3

We have successfully written the bonding groups and included all atoms in the molecule. Let's do the same for Carbon 6:

1) bond to H
2 - going counterclockwise) 1.5 bond to CHCCH3 [again arbitrarily choosing to stop because I've hit C4 and followed through its ring substituent, I'll get the rest of the molecule going the other direction around the ring in the next bonding group]
3 - going clockwise) 1.5 bond to C(OCH3)CHCH [and again now I've hit every atom in the molecule]

Compare the bonding groups in the sets belonging to Carbons 2 and 6:

C2 Group 1 = C6 Group 1
C2 Group 2 = C6 Group 3
C2 Group 3 = C6 Group 2

All bonding groups are identical; carbons have, equivalent electronic environments.