Hydrogen Bonding Question

[JakeMUSC]

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Hey guys. Please help with the answer and reasoning to this question:

Which of the following compounds is not capable of hydrogen bonding:
A. C4 H11 N
B. C4 H9 N O
C. C4 H7 F3
D. C4 H8 O2
E. C4 H10 O

???? All I know is that H bonds occur with F, O, N....hmmmm.....im guessing it has to do with the free valence electrons or something?!?

 

[luder98]

Hey guys. Please help with the answer and reasoning to this question:

Which of the following compounds is not capable of hydrogen bonding:
A. C4 H11 N
B. C4 H9 N O
C. C4 H7 F3
D. C4 H8 O2
E. C4 H10 O

???? All I know is that H bonds occur with F, O, N....hmmmm.....im guessing it has to do with the free valence electrons or something?!?

E: Ether can not make H bonds.

 

[JakeMUSC]

ok. Thanks Luder, So if this question comes up again, i just need to write out the structure and know that ethers dont H bond. Are there any other class of compounds that even with FON cant hydrogen bond?

 

[luder98]

ok. Thanks Luder, So if this question comes up again, i just need to write out the structure and know that ethers dont H bond. Are there any other class of compounds that even with FON cant hydrogen bond?

Make sure you have H to bond with the FON. In the case of ethers, they don't have H's (R1-O-R2).

Drawing the structures always works, but it's better (and faster) if you can visualize it in your head. This requires your ability to recognize different functinal groups from molecular formulas. That way you can do it in less than a min. Drawing out may frustrate you if they intentionally throw in some "complicated" compounds to trick you. Good luck.

 

[jdcinza13]

but keep in mind that there can be many different structures for the molecular formulas listed above. Just don't get caught up drawing them all. Typically, there will be something that stands out automatically (such as the ether, or no FON).

 

[coodoo]

Hey guys. Please help with the answer and reasoning to this question:

Which of the following compounds is not capable of hydrogen bonding:
A. C4 H11 N
B. C4 H9 N O
C. C4 H7 F3
D. C4 H8 O2
E. C4 H10 O

???? All I know is that H bonds occur with F, O, N....hmmmm.....im guessing it has to do with the free valence electrons or something?!?

But can't E be writtien as Ch3Ch2Ch2Ch2-OH? that can make H-bonds. I think the answer is C because F making more than one bond makes the entire molecule extremely unstable (R-F-H) <---this would give F a +1 charge.

 

[DrTacoElf]

But can't E be writtien as Ch3Ch2Ch2Ch2-OH? that can make H-bonds. I think the answer is C because F making more than one bond makes the entire molecule extremely unstable (R-F-H) <---this would give F a +1 charge.

I agree with your first point but that molecule can also be written as an ether which will not hydrogen bond -- that can not be said about any other of the above molecules. About choice c, a hydrogen bond isn't a covalent bond. Therefore you can't use it to assess formal charge. Flourine will be partially negative and attract partially positive hydrogens in adjacent molecules.

 

[drat]

But can't E be writtien as Ch3Ch2Ch2Ch2-OH? that can make H-bonds. I think the answer is C because F making more than one bond makes the entire molecule extremely unstable (R-F-H) <---this would give F a +1 charge.

A. seems to be an amine -- can H bond
B. seems to be a ketone with an amine -- both functionalities can H bond
C. seems to be a fluorinated hydrocarbon (CH3-CH2-CH2-CF3)...F doesn't form two covalent bonds like R-F-H . . . unless the poster meant R-F- - -H, to symbolize an H bond (fluorinated hydrocarbons can H bond as well)
D. seems to be a carboxylic acid -- can H bond
E. is ambiguous. I'm thinking the the orginal question had the structures drawn out...if an alcohol -- it can H bond. but as previous posters said, if it's an ether, it cannot.

 

[luder98]

But can't E be writtien as Ch3Ch2Ch2Ch2-OH? that can make H-bonds. I think the answer is C because F making more than one bond makes the entire molecule extremely unstable (R-F-H) <---this would give F a +1 charge.

I read it somewhere, I could be wrong. But it says... "F has the strongest electronegativity. It will only possess -1". BTW, if R-F-H existed, I think F would carry 0 charge not +1.
Plus, high electronegative atoms with positive charge are not stable. You can reason that E can be written as ROH, but that is the best choice. This reminds me a question on one of my bio tests: "Science is like doing multiple choice tests. The least incorrective answer is the correct one for now."

 

[JakeMUSC]

I wrote that question, word for word. Thats exactly the way the practice test had it, no lines, no nothing, and that specific order.

 

[ToothFairy423]

I read it somewhere, I could be wrong. But it says... "F has the strongest electronegativity. It will only possess -1". BTW, if R-F-H existed, I think F would carry 0 charge not +1.
Plus, high electronegative atoms with positive charge are not stable. You can reason that E can be written as ROH, but that is the best choice. This reminds me a question on one of my bio tests: "Science is like doing multiple choice tests. The least incorrective answer is the correct one for now."

Hi guys,
so, this is my first post on this website! Its an awesome website and I read through it everyday, but never registered until now hehe. I was reading through this post, and found it interesting that everyone has different opinions about this question. So here are my thoughts 🙂

Formal charge is the number of valence e - 1/2 bonding - bonding.
So if we had the scenario where F had two bonds, such as C-F-C. The formal charge would be 7-6, which gives us a FC of +1.

This would be highly unstable (as mentioned above) so the only realistic possibility would be CH3CH2CH2CF3. This molecule cannot H bond because, in order for it to H bond, the H would need to be bonded to F, O, or N. But here it is bonded to C, which is not electronegative enough to give it a high partial positive charge. In fact, the difference in electronegativities between C and H is so small, that its close to being considered a Non polar bond (so I just learned/read today!). Ok hopefully this all makes sense and I dont sound too dumb 🙂.

 

[luder98]

FC = valence -1/2bonding - non-bonding. In this case, I believe, it's 7-1/2(4)-5 = 0.

 

[ToothFairy423]

FC = valence -1/2bonding - non-bonding. In this case, I believe, it's 7-1/2(4)-5 = 0.

F will have 8 e total around it

so if it is bonded to two atoms, there will be 4 bonding, and 4 non-bonding, not five non bonding. C-F-H. there are 4 lone pairs of e here (non-bonding) and 4 bonding e here (2 bonds). Still disagree or agree??

 

[ToothFairy423]

F will have 8 e total around it

so if it is bonded to two atoms, there will be 4 bonding, and 4 non-bonding, not five non bonding. C-F-H. there are 4 lone pairs of e here (non-bonding) and 4 bonding e here (2 bonds). Still disagree or agree??

oops. not 4 lone pairs of e, but 2 lone pairs of e, and 2 bonds.

think of it this way, if Flourine had a Formal charge of 0 when it is bonding to two atoms, then this would be the most stable, and we would alwyas prefer to bond flourine to two atoms, rather than to bond it to one.

 

[phosphorylation]

Hi guys,
so, this is my first post on this website! Its an awesome website and I read through it everyday, but never registered until now hehe. I was reading through this post, and found it interesting that everyone has different opinions about this question. So here are my thoughts 🙂

Formal charge is the number of valence e - 1/2 bonding - bonding.
So if we had the scenario where F had two bonds, such as C-F-C. The formal charge would be 7-6, which gives us a FC of +1.

This would be highly unstable (as mentioned above) so the only realistic possibility would be CH3CH2CH2CF3. This molecule cannot H bond because, in order for it to H bond, the H would need to be bonded to F, O, or N. But here it is bonded to C, which is not electronegative enough to give it a high partial positive charge. In fact, the difference in electronegativities between C and H is so small, that its close to being considered a Non polar bond (so I just learned/read today!). Ok hopefully this all makes sense and I dont sound too dumb 🙂.

toothfairy, welcome to SDN, its great to add people to our family. I think you are right actually. The whole ether think has bugged me all day today when I saw this post earlier, I tried drawing it, then envisioning it, and I just couldn't see the H bonding with anything that'll make it a hydrogen bond. I tried drawing CH3CH2CH2CF3, and it totally makes sense.

On the other hand, a lot of things make sense to me, and they turn out wrong in the end.

once again though, welcome to SDN, its great to have you here...

 

[ToothFairy423]

toothfairy, welcome to SDN, its great to add people to our family. I think you are right actually. The whole ether think has bugged me all day today when I saw this post earlier, I tried drawing it, then envisioning it, and I just couldn't see the H bonding with anything that'll make it a hydrogen bond. I tried drawing CH3CH2CH2CF3, and it totally makes sense.

On the other hand, a lot of things make sense to me, and they turn out wrong in the end.

once again though, welcome to SDN, its great to have you here...

Thanks!! i taking the DAT in three weeks, my first time. wish me luck!!

 

[luder98]

we argue on a what-if case. so many impossible possibility. F has 7 e- on its outer most shell. it only needs one electron to fill up its octet. thus, it doesnt need to bond with two atoms. plus, it is highly electronegative, i'm almost 100% sure that it will only possess -1.

 

[phosphorylation]

Thanks!! i taking the DAT in three weeks, my first time. wish me luck!!

i wish you the best of luck...

 

[drat]

we argue on a what-if case. so many impossible possibility. F has 7 e- on its outer most shell. it only needs one electron to fill up its octet. thus, it doesnt need to bond with two atoms. plus, it is highly electronegative, i'm almost 100% sure that it will only possess -1.

Correct -- F doesn't bond with H in this example.
The structure is CH3-CH2-CH2-CF3