Quick NMR Question (From TBR Orgo)

[Josh138]

---

Members don't see this ad.

I just had trouble with the following question #15 found on pg 150 of the TBR orgo book 1:

Which compound shows a 2H quartet in its 1H NMR spectrum?
A. 2-chloropentane
B. 3-chloropentane
C. 2,2-dichloropentane
D. 3,3-dichloropentane

The answer key says the answer is D except I am not sure why that is the answer and why A is wrong.

I thought A was right because I thought the 2H referred to the hydrogens attached to the 3rd carbon in the molecule 2-chloropentane. That 3rd carbon is also attached to the 2nd carbon (which is bonded to 1 hydrogen) and to the 4th carbon (which is attached to 2 hydrogens). Thus, 3 hydrogens are attached to the adjacent carbons which makes it a quartet doesnt it? How would the NMR look differently if it didn't?

I thought D was wrong because the 2nd and 4th carbon would be the same to each other which would make it 4H instead of 2H.

Thanks in advance!

 

[Jepstein30]

I just had trouble with the following question #15 found on pg 150 of the TBR orgo book 1:

Which compound shows a 2H quartet in its 1H NMR spectrum?
A. 2-chloropentane
B. 3-chloropentane
C. 2,2-dichloropentane
D. 3,3-dichloropentane

The answer key says the answer is D except I am not sure why that is the answer and why A is wrong.

I thought A was right because I thought the 2H referred to the hydrogens attached to the 3rd carbon in the molecule 2-chloropentane. That 3rd carbon is also attached to the 2nd carbon (which is bonded to 1 hydrogen) and to the 4th carbon (which is attached to 2 hydrogens). Thus, 3 hydrogens are attached to the adjacent carbons which makes it a quartet doesnt it? How would the NMR look differently if it didn't?

I thought D was wrong because the 2nd and 4th carbon would be the same to each other which would make it 4H instead of 2H.

Thanks in advance!

The integral (ex. 2H) is a relative quantity. What is 2H and 1H in NMR may actually be 4 hydrogens and 2 hydrogens in the molecule. By itself, its not really that useful. You should be looking for a quartet here which means you need a hydrogen next to an atom with three hydrogens on it (and no hydrogens attached to other neighbors).

A doesn't have a quartet at all since the hydrogen on the 2nd carbon is split into a quartet of triplets (or whatever) from the hydrogens to its left and right. H on carbon 4 is split into a quartet of triplets as well but would be 2H. Hydrogens are split differently by each 'type' of hydrogen. That middle hydrogen isn't a quartet because it is split into a doublet from the H on C2 and a triplet from the 2H on C3 = doublet of triplets.

D is the answer because the hydrogens on carbons 2 and 4 are split only by the 3 H on the neighboring carbon -> quartet. We have 4H total of these and 6H of another 'type' of H. Since the integral is a relative value, thats simplified to 2H and 3H... so D has a quartet with an integral 2H.

 

[Deadlifts]

The integral (ex. 2H) is a relative quantity. What is 2H and 1H in NMR may actually be 4 hydrogens and 2 hydrogens in the molecule. By itself, its not really that useful. You should be looking for a quartet here which means you need a hydrogen next to an atom with three hydrogens on it (and no hydrogens attached to other neighbors).

A doesn't have a quartet at all since the hydrogen on the 2nd carbon is split into a quartet of triplets (or whatever) from the hydrogens to its left and right. H on carbon 4 is split into a quartet of triplets as well but would be 2H. Hydrogens are split differently by each 'type' of hydrogen. That middle hydrogen isn't a quartet because it is split into a doublet from the H on C2 and a triplet from the 2H on C3 = doublet of triplets.

D is the answer because the hydrogens on carbons 2 and 4 are split only by the 3 H on the neighboring carbon -> quartet. We have 4H total of these and 6H of another 'type' of H. Since the integral is a relative value, thats simplified to 2H and 3H... so D has a quartet with an integral 2H.

Well shoot. My professor and maybe textbook (Bruice) are explaining integration and neighbors improperly then if this is the case.

You have to differentiate multiplicities if the neighboring electrons aren't chemically equivalent?

Looking at 2-chloropentane: H3C-CClH-CH2-CH2-CH3, why isn't Carbon-3 a 2H quartet (n=3)? You'd see two separate signals for a 2H doublet and a 2H triplet due to Carbon-3?

 

[Jepstein30]

Well shoot. My professor and maybe textbook (Bruice) are explaining integration and neighbors improperly then if this is the case.

You have to differentiate multiplicities if the neighboring electrons aren't chemically equivalent?

Looking at 2-chloropentane: H3C-CClH-CH2-CH2-CH3, why isn't Carbon-3 a 2H quartet (n=3)? You'd see two separate signals for a 2H doublet and a 2H triplet due to Carbon-3?

Yea, if you want I can get into the nitty gritty of NMR to explain it. If you think about why nearby hydrogens produce splitting, then you'll see why chemical equivalency matters.

Heads up, if you ever aren't sure about something NMR wise, you can always google for the NMR spectrum and analyze it.. this online database is very helpful: http://sdbs.riodb.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng and you can even search by molecule or look at the IR/CNMR, etc.

I can't seem to find 2-chloropentane on there (may not have it) but 2-chlorobutane is similar: http://sdbs.riodb.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

If you look at the NMR page, it even identifies which hydrogens are responsible for the various peaks. If you look at hydrogen group B, it should be split into a doublet quartet. Hard to tell from the picture because the peaks are so clumped together so hit "peak data". You can work out which of those peaks belong to each group.. because you know that group D is definitely a simple triplet at 1.71. Leaves us with 8 peaks for group B which fits the doublet quartet (if it was just splitting like you described, it would be a simple quartet).

Sorry if thats confusing to follow but you can use that site to look up simple structures to help you understand.

Also, when you're on "peak data" take a look at the integrals. They are all rather large because again, they are relative numbers. You'll notice that (although not as cleanly as made up data) if you divide each one by the smallest integral, they simplify down towards the actual numbers of hydrogen in each group! Remember that you are adding the integrals together for each splitting..