C4H6O2 is the same as C4H6 (oxygens can be ignored) which if fully saturated would be C4H10 (from CnH2n+2). (also known as degrees of unsaturation or DBE).
There's a doublet with an integration of 6, and a multiplet with an integration of 1. Once you have a few clues from the NMR, start drawing structures! So even though fluorine is a poor leaving group, the negative carbon will cause it to eliminate to form benzyne. Grignards are also nucleophilic, and so react with carbonyls (which are electrophiles). Every proton on cyclohexane appears identical, but remember than cyclohexane is usually in a chair form, so there are really two types of protons: axial and equatorial.
Draw some C6H12O and C12H18O structures and elminate those that don't fit the data, then learn and repeat. (also known as DBE or degrees of unsaturation). The sharp IR peak at 1,700 cm-1 tells you this molecule contains a carbonyl (C=O).
I can't stress how important it is to just draw something!
So there must be at least one oxygen (and 1 IHD.). 6 C's would be 72, so let's try 5 C's. The trick to these retrosynthesis problems is to determine where the connections or "cuts" were made.
The formula is C10H14O. The problem told us it was a heterolytic cleavage, in which both electrons form the bond go towards the positive charge.
Note that the chlorine radical doesn't give an MS peak because it is neutral.
The 1H and 13C NMR spectra of a compound with chemical formula C10H14O are shown below.
In most ungraduate organic chemistry courses, being able to draw an alpha cleavage is much more important than a being able to draw a McLafferty rearrangement (which tends to only show up on bonus problems).
The proton NMR of cyclohexane gives only one peak when the NMR is run at room temperature. the NMR shows a doublet with an integration of 6 and a multiplet with an integration of 1. The difference between C10H22 and C10H14 is 8 hydrogens, which corresponds to 4 IHD. In most ungraduate organic chemistry courses, being able to draw an alpha cleavage is much more important than a being able to draw a McLafferty rearrangement (which tends to only show up on bonus problems).
The mass spec of 4-nonanone shows peaks at m/z = 58, 71, 86, 99. Here are the steps I would take to solve this problem: Yes.
The mass spec of 4-nonanone shows peaks at m/z = 58, 71, 86, 99. (draw out the spectrum you would expect to see). They're probably both methyls because they both have integrations of 3. They're both doublets with integration of 2 which points to a para-substitution pattern. Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol. Compound A has molecular formula C6H12O and shows a sharp peak at 1,710 cm-1 in its IR spectrum. Let's go through the steps you should take to solve any NMR structure elucidation problem. (also known as DBE or degrees of unsaturation). MS ionization will knock off an electron from the heteroatom (atom that's not C or H), in this case, the oxygen, leaving behind a positively charge compound. Grignards behave as though they are carbanions (negatively charged carbons), and so are very basic. So let's go with it.
The 1H and 13C NMR spectra of a compound with chemical formula C4H6O2 are shown below.
C12H18O is the smae as C12H18 which should be C12H26 if fully saturated. The 13C NMR peak at ~210 ppm indicates a carbonyl (specifically an aldehyde or ketone).
3. Are there any hints? Let's go through another way to make benzyne. That's the same as C10H14 (oxygens don't change IHD count). From the IR we know this molecule must have a carbonyl group. Draw a few benzene candidate structures with formula C10H14O, but then eliminate structures that don't fit the data.
Use curved arrows to show the heterolytic cleavage that accounts for this fragment. (n=2, so n + 1 = 3). Heteroatoms (atoms that are not C or H) are always the most likely atoms to lose an electron during mass spec, and this will be no exception. Use hooks to show the alpha cleavages that result in these two fragments. Some things we know form the NMR spectra: Draw a few structures based on these clues, and eventually you will come to the correct structure.
So protic solvents such as water or ethanol aren't suitable for Grignard reactions; the Grignard reagent will react with the alcohol in an acid-base reaction. Determine the structure of this compound. This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.
Determine the structure of this compound. Let's use steps similar those outlined in problem 662 to solve this NMR structure elucidation problem.
So first, let's make a guess of this compound's molecular formula.
The mass spec of methyl ethyl ether shows peaks at m/z = 45 and 59. The benzene ring "uses" 6 carbons which leaves 2 carbons for the other peaks.
Yes. Show how each alcohol can be prepared from a combination of a carbonyl and a Grignard reagent. Some textbooks use hooks instead, but the results are the same. 1O is 16.
C's hydrogens are on a carbon adjacent to a carbonyl (C=O), so its chemical shift will be around ~2 ppm.
Mass spec NMR problems require us to have done these two steps by now.
Compound B's 1H NMR spectrum is shown below.
Just starting drawing out structures with the proper formula and IHD count! A also is a CH3, so it will have an integration of 3.
For b), adding phenyl Grignard to cyclopentanone will do the job. B is also adjacent to oxygen, so it will be close to ~4 ppm, but a little further downfield than A because it's more substituted than A. So it must be a ketone. And then elminiate those that do not fit the data (too many signals, wrong multiplicity/integrations etc.).
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