The more substituted the carbocation, the most stable it is. This is because of the inductive effect- adjacent carbon atoms donate some of their electron density to neighboring carbocations (which are electron deficient), making them closer to neutral and more stable.

So the 3° carbocation is the most stable.

For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.

For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.

For c), there are two different types of beta hydrogens, and so two different alkenes are possible.

The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.

Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:

1. The alcohol will deprotonate to form a neutral alcohol (right), or
2. The alcohol will leave as a neutral molecule to form a carbocation.

Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards- don't draw structures you have drawn before.

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. This is the molecular ion (M⁺).

Oxygen usually undergoes homolytic cleavage- the bond splits and each atom gets one electron. Since only one electorn is involed, we use hooks instead of the usual curved arrows.

If your professor is into mass spec cleavage mechanisms, I suggest you practice this mechanism, called an alpha cleavage (see image below).

There are four types of protons (hydrogens) in the compound below, so we would expect to see four peaks (A, B, C, and D).

A is not adjacent to any other types of protons, so its multiplicity is 1, also called singlet (s). ( n=0, so n + 1 = 0 + 1 = 1). A also is a CH3, so it will have an integration of 3. Finally, A's proton are on a carbon adjacent to an oxygen, so its chemical shit will be about ~4 ppm.

B is a CH2 so its integration is 2. It's adjacent to a 2 hydrogens (a CH₂ ) so its multiplicty is 3, also called a triplet (t). (n=2, so n + 1 = 3). 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.

C is also a CH2 so its integration is 2. It's also adjacent to 2 hydrogens so it's a triplet (n+1 rule). C's hydrogens are on a carbon adjacent to a carbonyl (C=O), so its chemical shift will be around ~2 ppm.

D is an aldehyde proton, which is always a singlet with near ~10 ppm. It's only one proton, so its integration is 1.

Note: notice that the aldehyde proton doesn't couple to another other protons! So even through C is surrounded by a CH₂ on its left and an aldehyde proton on its right, when determining multiplicity we only count the CH₂, and not the aldehyde proton.

Let's go through the steps you should take to solve any NMR structure elucidation problem.

1. Are there any hints?

Yes. A broad IR peak around 3,300 cm^-1 tells you this compound contains an alcohol.

2. How many IHD are there?

(indices of hydrogen deficiency are also called degrees of unsaturation or double bond equivalents, depending on the textbook.)

The formula is C₁₀H₁₄O. That's the same as C₁₀H₁₄ (oxygens don't change IHD count). A fully saturated compound has formula C_nH_2n+2, so a C₁₀ molecule should have 22 hydrogens (2 x 10 + 2). The difference between C₁₀H₂₂ and C₁₀H₁₄ is 8 hydrogens, which corresponds to 4 IHD.

3. Draw some C₁₀H₁₄O structures with 4 IHD and eliminate, learn, repeat.

This isn't shown on the image below, but if you don't have an idea of the structure of the molecule, just start drawing structures! Never look at a blank page- just start drawing structures with the correct number of IHD (in this case, 4 IHD).

The peaks around ~7 ppm on the ¹H NMR tell you its probably aromatic (also, benzene has 4 IHD) so that's a good place to start. Draw a few benzene candidate structures with formula C₁₀H₁₄O, but then eliminate structures that don't fit the data. How? I like to go through this check list:

1. Eliminate by number of signals: Do the candidate structures you drew give the proper number of ¹H and ¹³C NMR signals? If not, eliminate!
2. Eliminate by multiplicity: Do your structures have the correct ¹H NMR multiplicities? If not, eliminate!
3. Eliminate by integration: Do your structures have the correct ¹H NMR integrations?If not, eliminate!
4. Eliminate by chemical shifts: Would the structures you drew have chemical shifts that are about the same as the chemical shifts in the ¹H spectrum given in the problem?

That's the order in which I usually eliminate candidate structures- fastest method (determining the number of expected NMR signals) to slowest method (predicting chemical shifts).

• Some things we know so far about this molecule:
• Form the IR we know It contains an alcohol.
• The gigantic singlet with integration of 9 screams tert-butyl
• The total integration of the aromatic region (~7 ppm) is 4, which means this molecule is disubstituted benzene. They're both doublets with integration of 2 which points to a para-substitution pattern.

As you elminate incorrect structures you will learn what fits the data, and be able to draw better candidate structures. Then you can repeat this process. Once you have a structure that passes these four problems, you probably have the correct structure.

Notice how we didn't even really need the ¹³C NMR to answer this problem. The ¹H NMR is usually enough.

I can't stress how important it is to just draw something! Even if you have no idea what the answer might be, don't even leave a blank page for an NMR problem. Just starting drawing out structures with the proper formula and IHD count! 

DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its ¹H NMR should only show one methyl signal (singlet, 6H).

So why is that the real life the ¹H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).

Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.

For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.

But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.

In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).

Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the ¹H NMR.

This is similar to problem 665- a very difficult NMR problem because we're not given the unknown's molecular formula, only its molecular mass (122 from the molecular ion peak on the mass spec).

So the first thing we have to do is guess the molecular formula based on this mass. Here's my thought process:

• There are signals in the aromatic region on the ¹H NMR (~7 ppm). So we probably have at least 6 carbon's with at least 4 IHD's (a benzene ring has 4 IHD).
• There are two other signals on the ¹H NMR, so there are probably at least 2 more carbons, for 8 carbons in total.
• 8 carbons with 4 IHD would give a formula of C₈H₁₀ (C₈H₁₈ from C_nH_2n+2 minus 8 H's for the 4 IHDs). which only gives a mass of 106. But we need a mass of 122- we're 16 mass units short. That's exactly one oxygen!
• Based on this, C₈H₁₀O is a good guess for the formula.

Now we can follow the steps laid out in problem 662:

1. Are there any hints? & 2. How many IHD are there?

Mass spec NMR problems require us to have done these two steps by now.

3. Draw some C₈H₁₀O structures with 4 IHD and eliminate, learn, repeat.

Some things we know form the NMR spectra:

• Because there are only four protons in the aromatic region (total integration of 4), the benzene ring is disubstituted. The two peaks in the aromatic region (~7 ppm) are doublets with integrations of 2, so this ring is probably para substituted.
• The benzene ring "uses" 6 carbons which leaves 2 carbons for the other peaks. They're probably both methyls because they both have integrations of 3. They're also both singlets, so there's not adjacent to any other carbons that have a hydrogen. The peak that's further downfield (¹H ~3.8 ppm) is probably near the oxygen.

Draw a few structures based on these clues, and eventually you will come to the correct structure.