Help with NMR

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rcwha

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Hi guys,

I was wondering if somebody could help me with the nmr problems that are usually on the DATs. Problems that ask number of peaks or how many singlets and triplets etc.

For example, can somebody walk me through this problem:

The h'nmr of m-xylene will show:

The answer is 4 peaks with 2 singlets and 1 doublet and 1 triplet....

Also, does a singlet have one or two adjacent hydrogens?


Thanks

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I believe m-xylene or 1,3-dimethylbenzene should have 3 peaks: one singlet, one doublet, and one triplet. EDIT: Sorry this is wrong :)

Here's a real NMR of that compound: http://www.chem.ucla.edu/cgi-bin/webspectra.cgi?Problem=bp22&Type=H yeah 1 s, 1 d, 1 t... Why is there a singlet missiong from this NMR?

To determine whether a peak is a singlet, doublet, etc. you take the number of adjacent hydrogens and add one. Therefore for a singlet n + 1 = 1, n has to be 0... that hydrogen must have no adjacent hydrogens like the smart man above me said :)

EDIT: Whoooops forgot about the hydrogens on the methyl groups! Yeah, the answer was right... 4 peaks: two singlets, one doublet, one triplet
 
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SHould not it be two singlets and 2 doublets?
We have four types of Hydrogen here.

SOrry, I was wrong. 2 singlets, one doublet, and one triplet is correct, and I get it. Made a mistake!
 
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I still don't understand how you are approaching the question. Like, which is the singlet, doublet and triplet...
I believe m-xylene or 1,3-dimethylbenzene should have 3 peaks: one singlet, one doublet, and one triplet. EDIT: Sorry this is wrong :)

Here's a real NMR of that compound: http://www.chem.ucla.edu/cgi-bin/webspectra.cgi?Problem=bp22&Type=H yeah 1 s, 1 d, 1 t... Why is there a singlet missiong from this NMR?

To determine whether a peak is a singlet, doublet, etc. you take the number of adjacent hydrogens and add one. Therefore for a singlet n + 1 = 1, n has to be 0... that hydrogen must have no adjacent hydrogens like the smart man above me said :)

EDIT: Whoooops forgot about the hydrogens on the methyl groups! Yeah, the answer was right... 4 peaks: two singlets, one doublet, one triplet
 
I hope this picture loads. The image is M-Xylene or 1,3-dimethlybeneze. If you notice there should be 4 different peaks because there are 4 different hydrogens. The hydrogens dimethyl groups will see a large singlet. The carbon #2, between the methyl groups will also see a singlet, however which will be located further upstream or downstream of the previous singlet. The carbon 4 and 6 ( the carbons on the left of the methyl groups are gonna see a doublet (n+1, where n=1) of the C5's hydrogen. Carbon #5 will see a triplet (n+1, where n=2) due to one hydrogen on C4 and C6.

It make sense to me, but its hard to explain it online. Let me know if I can clarify.
 

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SHould not it be two singlets and 2 doublets?
We have four types of Hydrogen here.


Wrong. The methyl group H's produce singlets(a). The proton between the two methyl groups on the ring is a singlet(b). The two protons in the other direction of the methyl groups are doublets since they are adjacent to another inequivalent hydrogen(c). The last proton remaning, para to the hydrogen between the two methyl groups, is a triplet because it is next to 2 hydrogens (1 on either side which are equiv to each other) (d).

Therefore, ( I will put the relative heights in parentheses)
a- singlet (2x)
b- singlet (x)
c- doublets (2x)
d- triplets (x)

Hope this clears it up.
 
Wrong. The methyl group H's produce singlets(a). The proton between the two methyl groups on the ring is a singlet(b). The two protons in the other direction of the methyl groups are doublets since they are adjacent to another inequivalent hydrogen(c). The last proton remaning, para to the hydrogen between the two methyl groups, is a triplet because it is next to 2 hydrogens (1 on either side which are equiv to each other) (d).

Therefore, ( I will put the relative heights in parentheses)
a- singlet (2x)
b- singlet (x)
c- doublets (2x)
d- triplets (x)

Hope this clears it up.

I agree with maygayver, but I think for 'a', there should be one singlet, as the Hydrogens on both methyl groups are equivalent.
also there should be one doublet only, again because the hydrogens on C4 and C6 are equivalent.
 
Makes sense! thank you so much for your explanation. Now im having trouble with 1, 4 dimethlybenzene. The answer is 2 singlets (one from each methly group).. But I'm getting 1 doublet (one of the hyrogens on the benzene) and 1 singlet (from the methyl group). The right answer is counting both the methyl groups, but i thought those methyl groups were equivalent!
Please explain,
Thanks


I hope this picture loads. The image is M-Xylene or 1,3-dimethlybeneze. If you notice there should be 4 different peaks because there are 4 different hydrogens. The hydrogens dimethyl groups will see a large singlet. The carbon #2, between the methyl groups will also see a singlet, however which will be located further upstream or downstream of the previous singlet. The carbon 4 and 6 ( the carbons on the left of the methyl groups are gonna see a doublet (n+1, where n=1) of the C5's hydrogen. Carbon #5 will see a triplet (n+1, where n=2) due to one hydrogen on C4 and C6.

It make sense to me, but its hard to explain it online. Let me know if I can clarify.
 
Makes sense! thank you so much for your explanation. Now im having trouble with 1, 4 dimethlybenzene. The answer is 2 singlets (one from each methly group).. But I'm getting 1 doublet (one of the hyrogens on the benzene) and 1 singlet (from the methyl group). The right answer is counting both the methyl groups, but i thought those methyl groups were equivalent!
Please explain,
Thanks
That's weird, I am getting 1 singlet, 1 doublet, & 1 sixtet.
 
Makes sense! thank you so much for your explanation. Now im having trouble with 1, 4 dimethlybenzene. The answer is 2 singlets (one from each methly group).. But I'm getting 1 doublet (one of the hyrogens on the benzene) and 1 singlet (from the methyl group). The right answer is counting both the methyl groups, but i thought those methyl groups were equivalent!
Please explain,
Thanks

RCWHA, I get two singlets too, and that is right. How many TYPES of H are there in this compound? Two! One type is the one on CH3 [both CH3s are equivalent]. The other type is any of the 4 Hydrogens on the ring [all these four are equivalent]. When you use the (n+1) rule in HNMR, you never add the neibhouring hydrogen that is equivalent to the one you are looking at. SO we will have a singlet for the H's on CH3, and a singlet for any of the 4 equivalent H on the ring.
By the way, did you add me on yahoo?
 
Ohhh so the key thing is that you don't add the neighboring hydrogens that are equivalent!
No, i didn't add you on yahoo..but I did click on your member name to send you a personal message bout this question (but decided to just post it on the board)

Thanks for the help!
RCWHA, I get two singlets too, and that is right. How many TYPES of H are there in this compound? Two! One type is the one on CH3 [both CH3s are equivalent]. The other type is any of the 4 Hydrogens on the ring [all these four are equivalent]. When you use the (n+1) rule in HNMR, you never add the neibhouring hydrogen that is equivalent to the one you are looking at. SO we will have a singlet for the H's on CH3, and a singlet for any of the 4 equivalent H on the ring.
By the way, did you add me on yahoo?
 
Ohhh so the key thing is that you don't add the neighboring hydrogens that are equivalent!
No, i didn't add you on yahoo..but I did click on your member name to send you a personal message bout this question (but decided to just post it on the board)

Thanks for the help!

Yeah, let me illustrate it this way.
Let's say we have Hydrogen type A, type B, and type C on three different carbons. let's say B is in the middle, and A and C are both B's neighbours.we wanna figure out B's singnal now. If A and C are equivalent Hydrogens, but different that B, you still count 2, and do 2 + 1 = triplet.
If any of A or C are equivalent to B, then you don't count it.
Let's say A is equivalent to B, but C is different. Then we only count C, and have 1 + 1 = 2.
If A, B and C are all equivalent, then you have don't count either A or C. 0+1 = singlet.
I hope this helps.
 
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Yeah, let me illustrate it this way.
Let's say we have Hydrogen type A, type B, and type C on three different carbons. let's say B is in the middle, and A and C are both B's neighbours.we wanna figure out B's singnal now. If A and C are equivalent Hydrogens, but different that B, you still count 2, and do 2 + 1 = triplet.
If any of A or C are equivalent to B, then you don't count it.
Let's say A is equivalent to B, but C is different. Then we only count C, and have 1 + 1 = 2.
If A, B and C are all equivalent, then you have don't count either A or C. 0+1 = singlet.
I hope this helps.
Hi this makes so much more sense. Ok now I get it why there are 2 singlets. But what about the carbon carrying the methyl group, wouldn't it get sixtet? since it has 5 neighbouring hydrogens?
 
none of the carbons have 5 neighboring hydrogens.

the methyl group you are talking bout has zero neighboring hydrogens so it is a singlet.

CH3- C=CH

the middle carbon has zero hydrogens

I think im explaining it right! anybody else have an input on this?

Hi this makes so much more sense. Ok now I get it why there are 2 singlets. But what about the carbon carrying the methyl group, wouldn't it get sixtet? since it has 5 neighbouring hydrogens?
 
none of the carbons have 5 neighboring hydrogens.

the methyl group you are talking bout has zero neighboring hydrogens so it is a singlet.

CH3- C=CH

the middle carbon has zero hydrogens

I think im explaining it right! anybody else have an input on this?
The carbon bearing the methyl group itself doesnt have any hydrogens, but it has 5 neighbouring hydrogens....so it should give 5+ 1 = sixtet peak.
 
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What about, for example, isobutane? Would there be one doublet and one dectet??
 
I think it would be 1 doublet & 1 quartet...right harrygt?

There are two types of hydrogen in isobutane. the one on the tertiary carbon has 9 neighboring hydrogens, so it will give a dectet. The other type of H is any of the ones on any of the three methyl groupd. Those H have only one neighboring H [the one on the tertiary carbon], so that would give a doublet.
I hope I'm not making anymore mistakes.
 
this stuff is starting to click...anybody have more!

There are two types of hydrogen in isobutane. the one on the tertiary carbon has 9 neighboring hydrogens, so it will give a dectet. The other type of H is any of the ones on any of the three methyl groupd. Those H have only one neighboring H [the one on the tertiary carbon], so that would give a doublet.
I hope I'm not making anymore mistakes.
 
Try to find the HNMR of 1,3,5-trimethylbenzene. Yeah, it IS starting to CLICK! Good job by all of you guys.
 
There are two types of hydrogen in isobutane. the one on the tertiary carbon has 9 neighboring hydrogens, so it will give a dectet. The other type of H is any of the ones on any of the three methyl groupd. Those H have only one neighboring H [the one on the tertiary carbon], so that would give a doublet.
I hope I'm not making anymore mistakes.
yea you are right! I got confused again. I thought since those 9 hydrogens are similar to each other, they won't give a dectet. They would give a quartet instead. But I forgot that they each give their own individual peak eventhough they are similar to each other. You are suppose to compare them to the center carbon, not to each other. duh! Sorry for the confusion!
 
yea you are right! I got confused again. I thought since those 9 hydrogens are similar to each other, they won't give a dectet. They would give a quartet instead. But I forgot that they each give their own individual peak eventhough they are similar to each other. You are suppose to compare them to the center carbon, not to each other. duh! Sorry for the confusion!

Yup, its like B at the middle, and 9 A around it. The A Hydrogens are similar, but that still counts. You would only not count a hydrogen, if it was equivalent to the hydrogen at the center.

YEEEEEEEEEy! 2 singlets for 1,3,5-trimethylbenzne.
 
Yup, its like B at the middle, and 9 A around it. The A Hydrogens are similar, but that still counts. You would only not count a hydrogen, if it was equivalent to the hydrogen at the center.

YEEEEEEEEEy! 2 singlets for 1,3,5-trimethylbenzne.

pheww!! finally :cool:
 
got em', now i don't have to be scared when i get these problems on a practice test....Got more? lol
 
Sweet, got the same thing as harry

Next questoin: What would the H NMR of cyclopropane look like? [carefult! No regrets will be accepted!]:D

This thread has gone so far, and it is actually becoming more and more interesting. Keep it up.
 
hmmm, all the hydrogens are equivalent..so one singlet....hope im right :)

Next questoin: What would the H NMR of cyclopropane look like? [carefult! No regrets will be accepted!]:D

This thread has gone so far, and it is actually becoming more and more interesting. Keep it up.
 
For cyclopropane, shouldn't it be one quintet?

For chlorocyclopropane, would it be one quintet and one one quartet?


Try this: 1-chloro-3-(2-propoxy)-benzene ... should be a challenge :)

Answer: highlight to read (and doublcheck cuz I may have gotten it wrong)
1 septet, 1 triplet, 3 doublets, 1 singlet
 
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Also, What was the final say on chlorocyclohexane from the other board?

I'm getting 3 different hydrogens with 1 trip and 2 quints...
 
Also, What was the final say on chlorocyclohexane from the other board?

I'm getting 3 different hydrogens with 1 trip and 2 quints...

Chlorocyclohexane should have 4 peaks/4 nonequivilant hydrogens. Let's start with C1 (the C with the Cl on it) That carbon has one hydrogen on it, and the number of adjacent hydrogens is 4, so that peak is a quintet. The carbons on C2 and C6 are equivilant, and they have 3 adjacent hydrogens, so that peak is a quartet. The C3 and C5 H's are equiv. and they have 4 adj. H's, so that peak is a quintet. the C4 H's has 4 adj. H's so that is also a quintet.

So the total is 1 quartet, 3 quintets.
 
All the answers to the questions:


Cyclopropane: One singlet

chlorocyclopropane: One quintet, one doublet

chlorocyclohexane: three quintets, one quartet.
 
For cyclopropane, shouldn't it be one quintet?

For chlorocyclopropane, would it be one quintet and one one quartet?


Try this: 1-chloro-3-(2-propoxy)-benzene ... should be a challenge :)

Answer: highlight to read (and doublcheck cuz I may have gotten it wrong)
1 septet, 1 triplet, 3 doublets, 1 singlet

Andy, regarding cyclopropane, refer to the 14th post on this thread, which is one of my posts. All Hydrogens are equivalent on cyclopropane, so you don't count the equivalent hydrogens. The same applys to chlorocyclopropae. Read that post for sure.
Your answer to 1-chloro-3-(2-propoxy)- benzene is correct!:thumbup:
 
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There are two types of H out there. Be careful. Ouch

Wow, talk bout rookie mistake..I forgot to draw out the hydrogen with the chlorine! I did the same thing with chlorocyclohexane...I guess i was in the habit of drawing single hydrogens for benzenes :(

Better make the mistakes now than this fri!
 
HarryGT, I get it now...thanks! I'm almost sure my prof didn't talk about that when we (very quickly) went over HNMR


I was bored at work so I made this little walkthrough of how to solve the HNMR of 1-chloro-3-(2-propoxy)-benze...
View attachment Orgo_NMR.PDF
 
HarryGT, I get it now...thanks! I'm almost sure my prof didn't talk about that when we (very quickly) went over HNMR


I was bored at work so I made this little walkthrough of how to solve the HNMR of 1-chloro-3-(2-propoxy)-benze...
View attachment 10521

Lol, that PDF was pretty neat dude, I mean with all those colors and step by step solution. I liked it!
 
Wow, talk bout rookie mistake..I forgot to draw out the hydrogen with the chlorine! I did the same thing with chlorocyclohexane...I guess i was in the habit of drawing single hydrogens for benzenes :(

Better make the mistakes now than this fri!

:laugh:Lol, yeah man. I already guessed that you have not counted the hydrogen attached to the C1. I'm sure you won't make this mistake on the real thing if you get one of these [I doubt though, has there ever been any of these type of problems on DAT?]. I mean it is still enjoyable to do them, but do they show up on the real thing?
 
HarryGT, I get it now...thanks! I'm almost sure my prof didn't talk about that when we (very quickly) went over HNMR


I was bored at work so I made this little walkthrough of how to solve the HNMR of 1-chloro-3-(2-propoxy)-benze...
View attachment 10521

Since the adjacent hydrogens, B and F are non equivalent, the A hydrogens will be split into a doublet and each peak of the doublet will be split into a doublet.
 
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