Reactions (predict the product, show the starting material..)

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).

The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.

For the left route:

1. Br₂/FeBr₃ adds a bromine to benzene via EAS.
2. Mg/ether converts bromobenzene to phenyl Grignard (a nucleophile).
3. Phenyl Grignard attacks the aldehyde.
4. A protic workup gets rid of all negative charges (and quenches the Grignard reagent).

For the right route:

1. Benzene undergoes Friedel-Crafts acylation (an EAS reaction) to form the ketone.
2. The ketone is then reduced to an alcohol using NaBH₄.

Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!

Since chair forms can be cumbersome to deal with, always convert them to plain "overhead view" structures before doing the problem.

When in doubt, use numbering.

Chair forms can be difficult to draw, so it's a good diea to draw the "overhead view" first.

When approaching Diels-Alder problems, it's always good to user numbering: the product is always a cyclohexene compound.

In b), the trans dienophile leads to a trans product.

In c), the dienophile is an alkyne, and so becomes an alkene in the product.

^-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.

NaNH₂ is a very strong base (the pK_a of ammonia, its conjugate acid, is about 35) and will easily depronate a terminal alkyne (pK ~ 25), producing a negatively charged alkyne carbon (an acetylide).

The acetylide is a strong nucleophile and will undergo an S_N2 reaction with the 1ºalkyl halide. (acetylide is also a strong base, so with 2º or bulkier alkyl halides, it will go E2 instead).

Finally, Lindlar's catalyst will reduce the alkyne to a cis 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.

These are substitution reactions. In each case we're replacing the -OH with either -CN or -I, so the nucleophiles will be NaCN or NaI (S_N2 conditions).

Hydroxide (HO^-) is a poor leaving group, so the first step in each of these reactions will be to convert the alcohol into a better leaving group. Two good leaving groups are Br^- and SO₃R^-, but which one to use?

Every time an S_N2 reation takes place the wedge on the alpha carbon becomes a dash (and vice-versa).

For a), the wedge remains a wedge, so we have to do two S_N2 reactions (wedge to dash to wedge again). So we use PBr₃ to turn the OH into a Br. (Bromination of an alcohol with PBr₃ is an S_N2 reaction and so inverts stereochemistry).

For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr₃ to make the OH a better leaving group, we use RSO₂Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.