Does anyone know what chemical could have MW=70 (ORGANIC)?

Question

I am doing an assignment in chemistry on unknown organic chemicals.

Through various test like:- solubilty in H2O, Solubilty in NaHCO3, 2, 4 DNP and Lucas test - which has shown that the chemical is a secondary alcohol.

The mass spectrum which I was given had peaks at 31, 42, 55 (which I have found the answers too) - but I am unable to find what MW of 70 could be.

Can any off you good people help - you will acknowledgement in the bibliography. PLEASE!!!

Answer 1

Well, first find out what it could be. You have a great deal of information in the MW of 70. Now you want to find out what possible combinations of atoms could there be to make up your molecule.

Divide 70 by 12, and you get 5 remainder 10. This means you have a maximum of 5 carbons in the molecule, as carbon atoms weigh 12 g/mole. The other 10 could be hydrogens. So your compound could have a formula of C5H10.

Alternatively, there could be oxygen. Start swapping one carbon (mass 12) for four hydrogens to get one oxygen. We can go from C5H10 to C4H6O. Or you could have C3H2O2.

Remember that the number of hydrogens cannot exceed 2n+2 where n is the number of carbons.

So, our likely compounds are C5H10, C4H6O, and C3H2O2. Find out how many double bonds/rings/triple bonds are in the system using the Degrees of Unsaturation Equation, which is d = (2*C+2-H-X)/2, where X is a halogen, H is the number of hydrogens, and C is the number of carbons.

For C5H10, using the equation we find d = 2(5)+2-10, or 2, which is divded by 2 to give 1. That means C5H10 has one double bond or one ring.

For C4H6O, d = 2(4)+2-6, or 4/2 = 2. Two double bonds, one ring and a double bond, two rings, or one triple bond.

For C3H2O2, d = 2(3)+2-2, or 6/2 = 3, which could be a combination of things.

For C5H10, we know it's a hydrocarbon, so it's not going to be soluble in water. It has no OH's, so it's not going to react with base. DNP reacts with ketones, which we don't have, and since there are no alcohols, a Lucas test will give nothing.

C4H6O could have one double bond and one ring, two double bonds, two rings, or a triple bond. If the O is part of an alcohol, it should react with the sodium bicarbonate, and that it should be, to some extent, soluble in water.

If it's a ring and double bond, look for the double bond O to react with DNP.

If it's cyclobutenol, look for the Lucas test to give a positive.

I'm reasonably sure it can't be C3H2O2. If could be an acid (which would react with base), it would be soluble in water, but it would be a triple bond in conjugation with an acid, and that would effectively decompose on its own.

Now let's look at the MS. Notice that the loss from 70 to 55 is a loss of 15, which means a loss of CH3. On the other end, 31 is a CH2-OH (alpha cleavage), and you should see a corresponding 39 peak for half. The only thing that would fit that would be 2-butyne-1-ol.

So, as you're looking at a conjugated primary alcohol, try the base test to see if it deprotonates, the Lucas test to see if it is a secondary alcohol. If it's not, then it's a primary alcohol and it's 2-butyne-1-ol.

Answer 2

Well, you didn't give much data and I can only guess whether you made a mistake, or any other possibilities, but I kind of doubt your analysis. You could have 31 if you have primary alcohol. I can't come up with a reasonable 42 and 55 to go with it and mw 70. What makes you think the mw is 70?

Okay, you have 3-methyl-1-butanol, search in the SDBS database for the MF. The MW is 88, but not surprisingly, it will readily lose water to give a peak at 70. it also has fragments at 55, 42 and 31.

I don't even know if I am right on this, but I want to complain about the purpose of this exercise? If you can't rationally analyze the MS, what is the purpose of a MS? Further, mass spectra did not become popular until chemical ionization and either GC or HPLC interfaces became popular. That was in part because the data was of limited value. I say that because unless you have a known compound, it is very difficult to determine an unknown structure from a fragmentation pattern. Let me state that again. If you didn't know the originating structure and you were to perform a chemical reaction on it in which the results are even more nebulous than they were before, of what value is that exercise? If you couldn't figure this out from the MS, I am not surprised. I couldn't either (nor did the other answers).

Can you get an IR? Compare it with the IR in the SDBS database. The peaks must match peak for peak. How about an NMR? We are in the 21st century or is your class stuck in 1930?

While I am complaining about your instruction, let me go further. I don't doubt you did the Lucas test. What I do doubt however is whether the Lucas test (HCl/ZnCl2) is a good diagnostic for a primary, secondary, or tertiary alcohol? I am asserting that you looked at the results of the Lucas test and concluded you had a secondary alcohol based upon the test. It isn't the test that is at fault. No one doing real chemistry would use a Lucas test to determine a structure. Look at your results. Does that tell you anything? Too bad it doesn't tell the people teaching organic chemistry something. (For those people who don't use the Lucas test and use IRs, NMRs, and other data, ignore what I said.)

Answer 3

70 - 55 = 15 = CH3

So if you know what fragment 55 represented, just add on a CH3 group (which may not necessarily have been intact in the parent; it could have formed by rearrangement in the spectrometer).

Alternatively, you know it is a secondary alcohol so you know it contains the group -CHOH- , mass 30. The rest adds up to 40, which could be C3H4, or much less probably C2O, and nothing else if CH, H,O are the only elements.

So something like cyclobutenyl alcohol?? (That would be an enol; wouldn't it convert to cyclobutanone?)

Are you SURE that 70 is the parent ion?

Answer 4

OH
l
CH3 -C - CH2CH3
l
H

Mwt = 74