A has a chair form that avoids having any substituents in the axial position.

B's most stable chair form has one axial substituent.

So A is more stable than B. Therefore, B is higher eneregy and has a higher heat of combustion than A.

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.

Predicting S_N1/S_N2/E1/E2 competition reactions tends to drive students crazy, but it's not so bad once you notice the general pattern:

• basic conditions (positive and negative charges) tend to go S_N2 or E2 (no carbocation)
• neutral or acidic conditions tend to go S_N1 or E1 (carbocation is formed).

That's how you determine a SN1/E1 reaction from an SN2/E2 reaction. But how to decide between substitution or elimination? General things to watch for are bulk, nucleophilicity, and heat:

• If you see heat (or Δ), the reaction will go elimination.
• If you see a big, bulky compound, the reaction will go elmination.
• If you see a strong base, the reaction will go elimination. Strong base is anything stronger than RO^-.

The exception: if everything is primary, it will probably go S_N2.

These rules probably seem confusing, so let's go through these eight examples and see how they apply.

a) NaCN is charged! (Na⁺ and CN^-), so it's S_N2 or E2. CN is not a strong base, so it's S_N2.

b) KOtBu (potassium tert-butoxide) is charged, so it's S_N2 or E2. ^-OtBu is a strong base, so if anything is more bulky than 1º it will go E2. ^-OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. ^-OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go S_N2.

e) Methanol (MeOH) is neutral so probably E1 or S_N1. Methanol is a weak base and there's no bulk, so S_N1. In general water and alcohol do a mixture of S_N1 and E1 with alkyl halides (mostly S_N1).

f) H₂SO₄ is acidic so probably E1 or S_N1. Can't be S_N1 though because there is no nucleophile in H₂SO₄. (HSO₄^- is a very weak nucleophile). An alcohol with H₂SO₄ or H₃PO₄ is a dehydration reaction- E1.

g) H₂SO₄ is acidic so probably E1 or S_N1. In this case we have a nucleophile- Cl^-, so it will go S_N1.

h) Amines are neutral but they don't so S_N1/E1- they tend to go S_N2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.

The 2° carbocation formed immediately undergoes a 1,2-methyl shift (a rearrangement) to form the more stable 3° carbocation, so the product is the 3° alkyl chloride instead of the 2º alkyl chloride, which would have formed in the absence of rearrangement.

To predict the stereochemistry of the alkene product from an E2 reaction, we have to rotate the leaving group (Br) and the beta-proton into anti-coplanar geometry. It's easiest to put the H and Br in the plane of the page.

After we do this, we see that both the ethyl and phenyl groups are wedges, and the two methyls are dashes. So after the E2 reaction, the resulting alkene will have its two methyl groups cis to each other.

This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.

Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.

Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).

1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction).  Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.

When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.

The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.

^-OtBu is a big, bulky base, and so will do an E2 reaction with the alkyl halide. The elimination product can be either a cis or trans alkene, but trans is usually the major product (cis alkenes have steric strain).

The alkene then reacts with Br₂ to form 2,3-dibromobutane, which can undergo two consecutive E2 reactions with ^-OtBu to form an alkene and then the alkyne (2-butyne).

Finally, the alkyne reacts with the first equivalent of Br₂ to form the trans 2,3-dibromo-2-butene, which reacts with a second equivalent of bromine to form the tetrabromo alkane.

Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.

1. Are there any hints?

Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.

Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.

2. How many IHD are there?

Compound A: C₅H₁₂O = C₅H₁₂ should be C₅H₁₂ (C_nH_2n+2) so 0 IHD.

Compound B: C₅H₁₀O₂ = C₅H₁₀ should be C₅H₁₂. Missing 2H, so 1 IHD.

Compound C: C₆H₁₂O₂ = C₆H₁₂ should be C₆H₁₄. Missing 2H, so 1 IHD.

These IHD counts fit our assumptions from part 1).

3. Draw some structures and eliminate, learn, repeat.

Some clues from the NMR:

• The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
• The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
• We know from before we have one IHD, and it's probably an ester.

So start drawing structures and eliminate those that don't fit the data!

a) Adding to alpha position, so this must have come from the combination of an enolate and an electrophile.

b) Adding to the beta position, so this must have come from a Michael addition (also called a conjugate addition).

To carry out a Michael addition, the first step is to convert propanal to an alpha, beta unsaturated carbonyl.

To do this, perform an alpha bromination with (Br₂/H₃O⁺) and then do an E2 elimination (tBuOK/heat).

Then add a Gilman reagent to the alpha,beta unsaturated aldehyde to add a carbon-carbon bond at the beta position.

Epoxides can open up in two different ways.

To add a nucleophile to the less substituted side of an epoxide, use basic conditions. This is done in #2 below.

To add a nucleophile to the more substituted side of an epoxide, use acidic conditions. This is done is #1 below.

Why do the conditions matter? Epoxides have two electrophilic carbons. Normally nucleophiles will preferentially attack the less substituted carbon, as they do in S_N2 reactions. Recall that S_N2 reactions usually happen with strong nucleophiles- that is, negative charges (basic conditions).

When an epoxide reacts under acidic conditions, the transition state has carbocation character, and so it's sort of like an S_N1 reaction. That is, instead of less substituted carbons being favored due to less steric bulk, more substituted carbons are favored do to a more stable carbocation. So acidic conditions cause an epoxide to open up on the more substituted side.

The trick to these retrosynthesis problems is to determine where the connections or "cuts" were made.

The carbonyl carbon becomes an alcohol after a Grignard reaction, so that's where the "cut" must be.

Note that there can often be more than one correct answer to these types of problems.

For a), adding propyl Grignard to acetone or methyl Grignard to 2-pentanone will result in the product.

We can also use two equivalents of methyl Grignard with 4-carbon ester, such as ethyl butanoate. Esters contains a build in leaving group (^-OR) and so react twice with Grignards.

For b), adding phenyl Grignard to cyclopentanone will do the job.

This is just an S_N2 reaction. 

The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).

Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.

Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).

Hydride reagents such as LiAlH₄ and NaBH₄ behave like hydride nucleophiles (H^-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.

Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.

Also note that you can't stop the reaction halfway at the aldehyde- LiAlH₄ will take an ester all the way down to an alcohol.

Note that I drew the McLafferty rearrangements using arrows (2 electrons moving at once). Some textbooks use hooks instead, but the results are the same.

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).

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 a difficult problem because we are never given the compound's molecular formula, only its mass (the molecular ion is the mass of the compound).

So first, let's make a guess of this compound's molecular formula.

Do we have any clues about what atoms are present in this molecule? Yes. The 13C NMR peak at ~210 ppm indicates a carbonyl (specifically an aldehyde or ketone). So there must be at least one oxygen (and 1 IHD.)

So let's do some math:

1O is 16. Mass is 86. 86-16 - 70. That's the total mass of C's and H's we have to work with.

How many C's can we "fit" in 70? 6 C's would be 72, so let's try 5 C's.

5 C's is 60, so we need 10 H's to make 70.

So a formula of C₅H₁₀O has the correct mass. It also has the correct number of IHD (1). So let's go with it.

Now we follow the series of steps laid out in problem 662:

1. Are there any hints?

We sort of did this step already, but yes- this molecule contains an aldehyde or ketone. Since we don't see a singlet at ~10 ppm on the 1H NMR, it can't be an aldehyde. So it must be a ketone.

2. How many IHD are there? (also known as DBE or degrees of unsaturation).

C₅H₁₀O is the same as C₅H₁₀ (ignore oxygen). C5 would be C₅H₁₂ if fully saturated (because of C_nH_2n+2), so this molecule is missing 2 hydrogens, which corresponds to 1 IHD.

3. Draw some C₅H₁₀O structures with 1 IHD and eliminate, learn, repeat.

Some things we know from the NMR spectra:

• There's a doublet with an integration of 6, and a multiplet with an integration of 1. This screams isopropyl group.
• That singlet with an integration of 3 is probably a methyl group.
• As mentioned above, since we don't see the aldehyde proton anywhere, the carbonyl must be a ketone.

Now draw a few candidate structures based on these clues and eliminate those structures that don't fit the data. If anything doesn't fit, you must elminate the entire structure.

These (no molecular formula, only MS data) are the hardest kinds of NMR problems you'll get in sophomore organic chemistry, so don't worry if you find it challenging- you're supposed to. 

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!