Sciencemadness Discussion Board - Homework Questions-

Homework Questions-

Tricia - 27-10-2014 at 09:44

So I've been given an NMR for a compound that has carbon, hydrogen, and oxygen in it. It has a molecular weight of 114 g/mol. My NMR is showing a triplet at the 1.0 ppm range with 3 hydrogens detected from it, a multiplet around 1.5 ppm detecting 2.02 hydrogens, and another triplet around 2.5 ppm detecting 1.98 hydrogens. Based upon the NMR alone, it looks like 7 hydrogens are attached to carbons.

There is 73.6% carbon and 12.3% hydrogen. I took the difference of this and got 61.3% oxygen. When calculating the empirical formula I got 6.1 mols for carbon, 12.3 mols for hydrogen, and 3.8 mols for oxygen, and so when I calculate the formula I got this: C1.6H3.2O1.

Now this is where I start running into difficulties with my calculations. When I added up the number of carbons, hydrogens, and oxygens in the empirical formula I came back with 38.4 grams. When I divide that from 114 g/mol I got 2.96 mols. I thought I needed to multiply this back into the empirical formula but the numbers ended up wrong. I keep getting a molecular formula that doesn't look right to me. My most recent attempt yielded this formula: C5H9O3.

Can someone help me with what I'm doing wrong? Thanks!

[Edited on 31-10-2014 by Bert]

DraconicAcid - 27-10-2014 at 09:54

Quote: Originally posted by Tricia

There is 73.6% carbon and 12.3% hydrogen. I took the difference of this and got 61.3% oxygen.

Here's your problem. What should these percentages add up to?

Tricia - 27-10-2014 at 10:02

Quote: Originally posted by DraconicAcid

Quote: Originally posted by Tricia

There is 73.6% carbon and 12.3% hydrogen. I took the difference of this and got 61.3% oxygen.

Here's your problem. What should these percentages add up to?

Oh my God, you're absolutely right! I just realized having the 61.3% would put it WAY over 100! No wonder I wasn't coming back with the right answer! Thank you!

Okay, after adding 73.6 and 12.3 together I got 85.9. When I subtracted that from 100 I got 14.1 - which should hopefully be the correct concentration for oxygen! Thanks again!

Predict the Major Product for the Following Reaction

Tricia - 28-10-2014 at 02:21

I'm out of tries with this one homework question so I just want to make sure I have the correct reaction drawn before I submit this answer. I tried to follow Markovnikov's rule and attach the Cl to the sp2 carbon with the most hydrogens and I attached the CH3CH2O- to the one with the least hydrogens. Am I pretty close? Thanks!

Praxichys - 29-10-2014 at 05:19

You are correct. The alcohol is nucleophilic towards the chloronium intermediate and generates an ether favoring the more substituted area. Just make sure they are in the anti position.

The mechanism is given at the bottom of page 3 here.

Tricia - 29-10-2014 at 07:26

Thank you!

More Homework help

Tricia - 30-10-2014 at 02:21

I'm starting to realize that I'm not very good at ozonolysis. I think I've got the hang of it with my online homework, but with the following two questions I'm literally on my last try with these. I've drawn out in both what I believe to be the correct starting formula. Would anyone be able to tell me if I'm correct? If I'm not, could you please point out to me what I'm doing wrong?

CuReUS - 30-10-2014 at 03:24

this is brilliant ,i thought that the alcohol was only a solvent in the reaction

,so the normal halogenation should be done only in non nucleophilic solvent to get dihalogenated products.but can haloamines also be formed this way or will the bromine just attack the amine to form a quaternary ammonium salt?

also ,if you study the reaction mechanism ,in the normal dihalogenation a carbocation is formed whereas the cohalogenation proceeds via a bridge intermediate(no carbocation formed) so there is no chance of a rearrangement even if a more substituted carbon is available .

but if only dihalogenation is done ,is there any chance of a ring expansion

CuReUS - 30-10-2014 at 03:57

ok ,so your first compound is a dicarboxylic acid (butane dioic acid) whereas your second compound is an aldehyde derivative with an oxo group(the aldehyde is given more priority to the keto while naming the compound)

now you must first understand why its different although the same ozone has been used
that's because in the first conversion ,oxidative ozonolysis is used (that's why there is H₂O₂ ) which will oxidise the intermediate ozonide to acid

the second conversion uses reductive ozoloysis(that's why Zn/acetic acid has been used .the Zn supplies electrons to the ozonide intermediate and the solvent donates the protons(H⁺ as it is an acid) to protonate the ozonide,resulting in the formation of the aldehyde and the keto groups and Zn takes away one of the O of the ozonide to form Zinc oxide(ZnO)

now the best way to get back the starting compound which has been ozonolysed is to just join the two carbons with the double bond O with a double bond (first rub away the two O and then just join the two carbon atoms with a double bond)

so for example if your ozonolysis product was 2 molecules of formaldehyde(H₂C=O) then by following the above method you can easily tell that that the starting product was ethene((H₂C=CH₂)

now back to your first compound,by following the method you see that by joining the first and the last carbon atoms with a double bond after removing the two O(ignore the -OH ,they are there because of the peroxide) ,you get a ring compound(which should look like a square or rectangle with a double between the two carbon atoms you just joined ) this is called cyclobutene

doing the same thing with the second compound(join the rightmost carbon and the 5th carbon with a double bond you get another ring compound ,but a part of you final product i.e the two carbons(-C₂H₅ group)just to the left of the 5th carbon is outside the ring, forming 1- ethyl cyclopentene (the numbering is important as you have to specify the location of the ethyl group(Et) )

[Edited on 30-10-2014 by CuReUS]

[Edited on 30-10-2014 by CuReUS]

Tricia - 30-10-2014 at 12:53

Now I get it! Thank you!

Chemosynthesis - 30-10-2014 at 15:32

Just for future reference, you may consider making a homework thread in Beginnings, since that section is devoted more towards homework (even in o-chem, as seen in the textbook thread in that subforum: https://www.sciencemadness.org/whisper/viewthread.php?tid=66...).

Bert - 30-10-2014 at 17:00

Quote: Originally posted by Chemosynthesis

Just for future reference, you may consider making a homework thread in Beginnings, since that section is devoted more towards homework (even in o-chem, as seen in the textbook thread in that subforum: https://www.sciencemadness.org/whisper/viewthread.php?tid=66...).

Yes- Homework questions should be in "beginnings", and a single thread seems best for now...

I'll help you out by collecting all these homework threads...

Stereochemical structures

Tricia - 5-11-2014 at 09:23

Okay, this one has me really confused. The question asks me to draw the structures that are part of the fumarase-catalyzed reaction. I got the product correct but I keep getting the reactant incorrect. The structure drawn in the picture below that is marked with a red X is the incorrect answer. I thought this was the structure for fumarate, but it was incorrect. I also thought this was trans but it still coming up wrong. Do I have the overall structure correct but have the placement of oxygens wrong? Or am I completely off the mark?

Thanks!

DJF90 - 5-11-2014 at 09:32

The structure is correct but the software you're using doesn't appear to like your wonky bond angles on the left hand side.

Tricia - 5-11-2014 at 09:51

You're correct. Once I redrew the structure and straightened out that one double-bonded oxygen, then it said my answer was correct. Thank you!

Stereochemistry problems

Tricia - 5-11-2014 at 10:35

Some of my online homework problems are really puzzling me. After a previous question of mine posted basically led to the revelation that my structure was technically correct, but for some reason didn't look correct in the program, I'm wondering if I'm yielding the same issue with other problems. The two structures below are ones I thought I had correct but I honestly can't tell if I'm completely wrong or if I'm drawing the stereochemistry incorrectly. The first one in the list I've drawn the Br attaching to the alkene with different wedges and dashes attached at different points and it's still incorrect. With the second one I know that the BH2 that's supposed to attach to one of the double bonds forms an OH. Is this just a matter of me not following syn/anti addition correctly with these two problems?

Dr.Bob - 5-11-2014 at 11:22

In the left example, I think there are more products possible, so try to figure out what they might be.

UnintentionalChaos - 5-11-2014 at 11:46

I think the issue in the left example is that when you draw a chiral tetrahedral carbon, two of the bonds can be in-plane (normal lines), one has to be dashed and one has to be a wedge.You have 3 bonds in-plane for both carbons, although your bromines are correct. Dr.Bob also has a point in that there is another possible product.

For the right example, you seem to have added an extra methyl onto the carbon with the OH. BH3 addition to a double bond is a syn process, and the oxidative cleavage to the alcohol preserves the stereochemistry. You've drawn the product of anti-addition. Additionally, there are two possible products since the borane can attack the double bond from either side.

Draw the structure of the alcohol(s) formed in the following reaction sequence.

Tricia - 6-11-2014 at 02:30

I posted this in the incorrect thread so it got deleted before I could see if anyone had responded to it. I've been stumped about this one for awhile though, and right now I want to check and make sure I have the setup for these correct but simply have the placement of the atoms wrong. Below is my most recent attempt at this problem and by the hint I think I'm on the right track, but I'm on my last try with this problem so I want to make sure I get this right. So, am I really close and just have some placements wrong? Or am I completely off? Thanks!

DJF90 - 6-11-2014 at 05:06

Your regioselectivity is correct. But, you've got the stereochemistry wrong. Boron and hydrogen add to the same face, and the oxidation proceeds with retention of configuration at boron, so you end up with H and OH on the same face; this is not what you've drawn.

Whether you need to draw both diastereomers when you've ticked the "racemic" box on the right hand side is to be determined.

CuReUS - 6-11-2014 at 08:06

tricia ,please post all your questions in this thread
http://www.sciencemadness.org/talk/viewthread.php?tid=41534#...

i think your question has been moved to the above thread,not deleted

[Edited on 6-11-2014 by CuReUS]

Finding Two Stereoisomers

Tricia - 6-11-2014 at 13:43

This is the tricky thing with online chemistry homework in that I don't know if a structure I draw is wrong because of stereochemistry, or if it's wrong because the whole thing is wrong. So I've got this problem below, and I thought with the excess H2O that there would be an OH- to attach to the structure. Does this mean that there is an HBr leftover on the side for bother stereoisomers? I forgot to draw the HBr's in the problem, but am I at least on the right path?

Chemosynthesis - 6-11-2014 at 16:42

You have the right idea; this is producing a vicinal haloalçohol and the addition is anti... however, you mistakenly drew only one Markovnikov product. You will get a racemic mixture of Markovnikov stereoisomers as your main product.

You will produce HBr, but it's generally of little significance.

[Edited on 7-11-2014 by Chemosynthesis]

Tricia - 7-11-2014 at 07:16

I'm ready to pull my hair out over all these problems. It would be better if examples of these excess problems were in my textbook, but they're not! I can't even find anything on the internet relating to this! So after solving the trans portion (thank you for your help Chemosynthesis) I thought it would be a cakewalk to solve the cis version of the same problem. But I keep getting it wrong. And again, I don't know if it's because my stereochemistry is incorrect or because my structures are completely wrong. I have the image below of my recent try with this. Do I have the stereochemistry wrong or is this wrong because I drew it wrong?

Chemosynthesis - 7-11-2014 at 09:20

Your stereo chemistry would be correct, but you placed the bromine and hydroxyl group incorrectly, so your regiochemistry is mistaken.

When performing a Markovnikov addition, the most electronegative group adds to the most substituted or stable carbon. Oxygen is more electronegative than bromine, so alcohols or ethers form on the most substituted carbon here.
Some confusion may come from acid addition; If it were HBr addition, bromine is more electronegative than hydrogen, so the bromine would be on the more substituted carbon. If you have elemental bromine in an inert, aprotic solvent, the dihalide is equally electronegative and so it doesn't matter.

Edit: if you don't want to just memorize this, think about the mechanism that is occurring. In organic chemistry, you will see a pattern of nucleophiles attacking electrophiles. The way we draw electron arrows is supposed to convey this, because your negative charge is what you start the arrow from.

When you look at your products here, the double bond is a source of electron density, and so if there is a reaction, it will attack the least electronegative group. In the case of HBr, this is hydrogen. In your example, it is bromine. Now you are stuck with a positive charge in your resonance structures. Consider where it is most stable. Your remaining nucleophile (excess water with the lone pairs on oxygen, solvating your reactant) logically attacks this and adds to that carbon.

[Edited on 7-11-2014 by Chemosynthesis]

Tricia - 7-11-2014 at 09:57

Quote: Originally posted by Chemosynthesis

Your stereo chemistry would be correct, but you placed the bromine and hydroxyl group incorrectly, so your regiochemistry is mistaken.

When performing a Markovnikov addition, the most electronegative group adds to the most substituted or stable carbon. Oxygen is more electronegative than bromine, so alcohols or ethers form on the most substituted carbon here.
Some confusion may come from acid addition; If it were HBr addition, bromine is more electronegative than hydrogen, so the bromine would be on the more substituted carbon. If you have elemental bromine in an inert, aprotic solvent, the dihalide is equally electronegative and so it doesn't matter.

Edit: if you don't want to just memorize this, think about the mechanism that is occurring. In organic chemistry, you will see a pattern of nucleophiles attacking electrophiles. The way we draw electron arrows is supposed to convey this, because your negative charge is what you start the arrow from.

When you look at your products here, the double bond is a source of electron density, and so if there is a reaction, it will attack the least electronegative group. In the case of HBr, this is hydrogen. In your example, it is bromine. Now you are stuck with a positive charge in your resonance structures. Consider where it is most stable. Your remaining nucleophile (-OH) logically attacks this and adds to that carbon.

[Edited on 7-11-2014 by Chemosynthesis]

I swapped the Br and the OH- onto opposite carbons, but it still came back wrong. Since the regiochemistry was incorrect I tried to rearrange the molecules but it still came back wrong. Here's my most recent attempt. Like I said, the way the online software is set up I don't know if I'm drawing the molecule completely wrong or if I've misplaced certain molecules.

Tricia - 7-11-2014 at 10:40

Never mind, I finally got the answer. Not only were my structures incorrect, but I had the incorrect molecules selected for the dashes and wedges. It's a relief to finally get this figured out! Now I know what to do for future problems like this!

chemrox - 7-11-2014 at 13:48

Quote: Originally posted by Praxichys

The mechanism is given at the bottom of page 3 here.

Thanks for posting the text. Would mind posting the title/author?
Thanks,
CRX

Tricia - 11-11-2014 at 05:47

Okay, this one has me really puzzled. I thought that in order to get from acetylene to trans-2-pentene I needed to first convert the acetylene into propyne, and then convert that into 1-butyne in order to achieve the trans-2-pentene. But I don't know if I have the mechanisms for each reaction wrong or if the products for each reaction is wrong. Below is an image of my latest attempt with this problem.

Chemosynthesis - 11-11-2014 at 06:11

Your step 2 product is missing the methyl substituent from the previous acetylide substitution. I don't see anything else that looks out of place.
Acetylene to acetylide, substitution. Repeat with different alkyl halide on remaining terminus. Dissolving metal trans reduction pdt.

Tricia - 11-11-2014 at 08:19

According to Markovnikov's rule, the hydrogen should add to the carbon with the most hydrogrens, right? When doing the problem below, this is what I thought but I'm still getting this wrong. Are the chlorine and the hydrogen supposed to be attached to the same carbon?

Chemosynthesis - 11-11-2014 at 08:58

Look at the mechanism. If you ever have a carbocation intermediate, you have to watch out for methyl or hydride shifts.

Tricia - 12-11-2014 at 12:21

I'm working on some word problems, and the question asks me what what major product forms from the acid catalyzed hydration of 1-phenylcyclohexene. I drew what I think is the mechanism below, leading to the product 1-phenylcyclohexanol. Am I close?

Update: I just realized that my drawing in the context of this question doesn't make any sense. For the acid catalyzed hydration to occur then the reagents reacting with the reactant has to be H2SO4 and H2O and not what I listed below. So this drawing is completely wrong anyways.

But would the product of the hydration be similar to the product shown below?

[Edited on 12-11-2014 by Tricia]

[Edited on 12-11-2014 by Tricia]

Tricia - 12-11-2014 at 17:31

I was able to solve the above question (I needed to add a hydrogen to the ring) but now I have another question. I need to properly convey in my report as to why the NMR I have reads a syn addition product that follows Anti-Markovnikov's rule. It's a hydroboration reaction involving 1-phenylcyclohexene with an end product of trans-2-cyclohexanol. Our professor went over the NMR in class and indicated that the NMR has six peaks near 4 ppm, indicating that it was Anti-Markovnikov. There is also a very high peak near 8 ppm, indicating that it's syn addition.

So the issue I'm having is that I need to explain why it's Anti-Markovnikov because of the six peaks and why it's syn addition because of the high peaks. All I have is my professor's say-so on this, and when I tried to do some research into this topic anything close to what I was looking for asked me to pay for an answer. I'm not asking for an answer to this, but if somebody could provide me links that can help me with this (and so I can develop a works cited page other than by what my professor said) I would greatly appreciate it.

Chemosynthesis - 14-11-2014 at 19:49

You want to look into peak integration and how the chemically equivalent hydrogens (H-NMR/P-NMR assumed) are indistinguishable.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/...

http://www.chem.ucla.edu/harding/ec_tutorials/tutorial49.pdf

http://users.wfu.edu/ylwong/chem/nmr/h1/integration.html

Tricia - 18-11-2014 at 08:17

It's funny how ones that look simple cause me the most trouble. I'm really stuck on sorting these out, and I honestly don't know which is right and which is wrong. Can someone clue me in if I'm on the right track? Thanks!

gdflp - 18-11-2014 at 08:26

Try drawing out the condensed structures. The radicals in the primary and allylic bins are correct, but the secondary and tertiary bins are mismatched. There should be 2 primary radicals, 2 secondary radicals, 2 tertiary radicals, and 1 allylic radical.

[Edited on 11-18-2014 by gdflp]

Magpie - 18-11-2014 at 19:13

Count the number of non-hydrogen groups bonded to the radical carbon. If there is one group this = primary, if two groups = secondary, and if three groups = tertiary. As already said drawing out the radical carbon and the attached groups & hydrogens will make this clear.

Carbon is tetravalent so there will be 3 attached groups/hydrogens and then the lone electron. But you know that already.

Tricia - 21-11-2014 at 09:44

I'm not sure that this question is looking for when it asks for the neutral organic product. I've tried looking this up in my notes, textbook, and online and I haven't found anything useful. Here's my most recent attempt. Am I close or am I off the mark? Thanks!

Chemosynthesis - 21-11-2014 at 11:06

Remember the Friedel Crafts mechanism; you first have to activate a reagent/catalyst. In this case (acylation), you go through an acylium cation. There is no chlorination.

The reason it asks for the neutral product is that acylation gives a charged complex of ketone and Lewis acid. This is broken during workup.

Tricia - 1-12-2014 at 08:02

For this problem, the hint stated it was an E2 Elimination reaction. I thought this meant that the Br would be eliminated entirely from the benzene ring, but that was wrong. Does the methyl attached to the oxygen get eliminated or am I looking at this the wrong way?

Tricia - 1-12-2014 at 08:06

I'm out of tries with this problem. Can anyone tell me if I'm close to the correct answer? Thanks!

forgottenpassword - 1-12-2014 at 08:39

Step 2: Br2, hv gives benzylic bromination.
Step 3: KOtBu gives styrene.
Step 4: mCPBA gives styrene oxide.

Tricia - 1-12-2014 at 08:47

Quote: Originally posted by forgottenpassword

Step 2: Br2, hv gives benzylic bromination.
Step 3: KOtBu gives styrene.
Step 4: mCPBA gives styrene oxide.

KOtBu isn't an available choice.

smaerd - 1-12-2014 at 10:01

KOtBu is not a choice but, (CH3)3CO - is.

Something tells me there's another way with-in the choices but I don't really have time to explore it.

Rather then just giving the answers it might be best to sort of hint at solutions, or explain why a given approach is not appropriate. Then again if you're 'out of tries' it is what it is.

forgottenpassword - 1-12-2014 at 10:15

Quote: Originally posted by smaerd

Rather then just giving the answers it might be best to sort of hint at solutions

Hence I wrote KOtBu!

This one's absolutely abysmal: http://www.sciencemadness.org/talk/files.php?pid=367964&...
Perhaps you should re-read the chapter in the book on Friedel Crafts acylation (and alkylation). You obviously haven't understood it at all -- it is quite central to your further studies that you do.
In this one: http://www.sciencemadness.org/talk/files.php?pid=366544&... you haven't grasped it at all either. There's little point in helping you get the correct answers because you will struggle with slightly different examples. You've obviously just started with the basics so take this opportunity to understand it properly; or else you will struggle tremendously later on. Luckily this stuff is the easiest to learn. You will just have to spend an hour or so reading a chapter in a book. Don't bother trying the questions until you've read over it again, because it would only be guess work.

[Edited on 1-12-2014 by forgottenpassword]

Darkstar - 2-12-2014 at 17:01

Quote: Originally posted by smaerd

Something tells me there's another way with-in the choices but I don't really have time to explore it.

What about :

1) CH₃CH₂Cl, AlCl₃ to give ethyl benzene
2) Br₂, hv to brominate at benzylic position
3) (CH₃)₃CO- to give styrene
4) Br₂, H₂O to give halohydrin
5) (CH₃)₃CO- to give epoxide via intramolecular substitution

Not sure if you're allowed to use the same reagent twice, though. Either way, forgottenpassword's route is the better and more direct one.