Show how each amine can be prepared from a carbonyl and an amine via reductive amination.
Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.
Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H2, NaBH4, NaBH3(CN), NaBH(OAc)3, etc.). For practice on forming imine and enamines, see problem 712).
MendelSet practice problem # 754 submitted by Matt on July 27, 2011.
For each amine below, show all Hofmann elmination products.
If more than one product is formed, predict which one will be the major product.
The Hoffman elimination first turns an amine into a better leaving group by methylating it with excess methly iodide, and then eliminates the methylated amine using base (such as -OH fomed from Ag2O in H2O). The products are alkenes.
This reaction is similar to an E2 elimination in that a beta proton is ripped off, and the alkene forms between alpha and beta. One notable difference from the E2 is that the less substituted product is favored, so Hoffman products are usually anti-Zaitsev.
a) There are three different types of beta protons, so three possible products can form. Product A is the least substituted alkene, so it's the major product.
b) There is only one possible product. The Hoffman elimination can't be done on aryl hydrogens (benzene rings).
MendelSet practice problem # 756 submitted by Matt on July 27, 2011.
The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.
Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.
When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).
Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.
MendelSet practice problem # 755 submitted by Matt on July 27, 2011.
Imidazole (shown below) has two nitrogen atoms, N-1 and N-3. Which nitrogen is more basic?
To answer this problem, draw the product after each nitrogen protonates, and compare their stabilities. Explain your reasoning.
Imidazole is aromatic. When N-3 protonates, the product is still aromatic.
But when N-1 protonates, the product is no longer aromatic (and therefore significantly less stable).
Because of this, N-3 is much more basic than N-1. Another way of thinking of this is that the lone pair on N-1 is involved in the aromatic circuit, and so is not available to pick up a proton.
MendelSet practice problem # 584 submitted by Matt on July 9, 2011.
Rank the amines A through D below in order of decreasing basicity (1 = most basic). Explain your reasoning.
sp2 carbons are more electronegative (electron withdrawing) than sp3 carbons due to increased s-character. Electron withdrawing groups (EWG) make acids more acidic and bases less basic.
For this reason, piperidine (compound A) is the more basic than pyridine (compound B):
(more basic) A > B (less basic)
Nitro (-NO2) is an EWG, so 4-nitropyridine (compound C) will be less basic than compound A.
Dimethylamine is an electron donating group (EDG), so it adds electron density to pyridine and increases basicity. So dimethylamino pyridine (DMAP, compound D) will be more basic than compound A:
(more basic) D > B > C (less basic)
So how does compound A compare with compound D? There is no way of predicting this based on EWG/EDG rules alone. After all, what should be more basic: piperidine, which has sp3 carbons (increases base strength), or DMAP, which has sp2 carbons (decreases base strength) but also an EDG (increases base strength) ?
There's no way to tell. But it turns out that Piperidine is a little more basic than DMAP. So the overal order is:
(more basic) A > D > B > C (least basic)
MendelSet practice problem # 752 submitted by Matt on July 27, 2011.
Propose a synthesis to accomplish each transformation. The only carbon sources allowed are alkenes and NaCN.
a) The starting alkene has 4 carbons, and the product has 5 carbons and a nitrogen. So you have to add 1C and 1N. This screams "add cyanide!"
Based on the location of the alcohol on the product, it looks like the way to go is to add NaCN to an epoxide. This puts the -CN and the -OH in the right positions.
b) The best way to prepare 2º and 3º amines is via reductive amination. Working backwards, the product can be prepared from a 4-carbon aldehyde (butyraldehyde) and a 3-carbon amine (propyl amine).
Performing hydroboration on the butene results in the anti-Markovnikov alcohol, which can be oxidized to the aldehyde using PCC.
To prepare propyl amine from an alkene, add HBr/peroxides to make the bromide (anti-Markovnikov), and then add NaN3 follow by H2/Pd.
MendelSet practice problem # 757 submitted by Matt on July 27, 2011.
Pyrrole undergoes eletrophilic aromatic substitution at C-2. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. What do you think? Why C-2 and not C-3?
Electrophilic substitution at C-2 leads to a carbocation intermediate with three resonance forms, while substitution at C-3 leads to a carbocation intermediate with only two resonance forms.
The C-2 intermediate has more resonance forms than the C-3 intermediate, and so is more stable. Therefore, EAS occurs at C-2.
MendelSet practice problem # 591 submitted by Matt on July 9, 2011.
a) Rationalize the relative stabilities of the cation species below.
b) Pyridine undergoes eletrophilic substitution at C-3. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. Consider part a) in your explanation.
a) The species on the left are much more stable than the species on the right because they have complete octets.
Nitrogen is less electronegative than oxygen, so a positive nitrogen (with a full octect) is more stable than a positive oxygen; oxygen is electronegative and so doesn't "like" being positive (have low electron density).
The carbocation and the "nitrocation" are both very unstable because they only have 6 valence electrons. Because nitrogen is more electronegative than carbon, the nitrocation is even less stable than the carbocation, because its worse at handling a positive charge.
b) The cation intermediates that result from substitution at C-3 only contain carbocations, while the intermdiates from C-2 or C-4 substitution contain a "nitrocation," which is less stable than a regular carbocation, and so are not favored. So pyridine undergoes EAS reactions at C-3.
One more note: For substitution at C-2 and C-4, it's tempting to draw a resonance form where the positive charge is on the nitrogen, which is flanked by two double bonds. But this cyclic compound is too strained to exist, and so doesn't contribute to this analysis.
MendelSet practice problem # 593 submitted by Matt on July 9, 2011.