Organic Chemistry

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.

The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N₃^-). Stable molecules make good leaving groups, so the best leaving group of the three is SO₃R^-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN₃.

Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.

But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).

Hydroxide (HO^-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO₃R^-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.

In general, the more resonance forms a molecule has, the more stable it is.

Arrows in organic chemistry always go from regions of high electron density to regions of low electron density. Most of the time this means arrows start from negative charges and go towards positive charges.

Because bromine is electronegative, the carbon directly bonded to it (also known as the alpha carbon) has a partial positive charge, and can be attacked by a nucleophile such as azide (N₃^-). 

Because this is an S_N2 reaction, no carbocation is formed; as the nucleophile attacks the alpha carbon, the leaving group (Br^-) leaves.

The starting material (2-aminobutane) is a racemic; it has equal amounts of R and S enantiomers.

Enantiomers have the same the same chemical and physical properties, so we can't separate them using common lab techniques, such as recrystallization, chromatography, etc. But diastereomers can have different properties, and we can exploit this fact to separate otherwise inseparable compounds.

Adding an optically pure acid to the amine produces a diastereomeric mixture of salts which can be separated. Adding a base (NaOH) "breaks" the salt and allows us to isolate the pure R or S amine.

We are able to separate the (R,R) and (R,S) salts because they are diastereomers and so have different chemical and physical properties.

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.

The reaciton of an alkene with Br₂ is an anti-addition, and hydrogenation (H₂ or D₂) is a syn addition. From this we can figure our the relative stereochemistry of each product.

Each starting material is achiral, and therefore not optically active, so the products cannot be optically active. 

There are three ways for products not to be optically active:

• products can be achiral
• products can be a racemic mixture
• products can be meso

In both a) and b) each product has a sterocenter, so the products can't be achiral.

In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.

In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.

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).

There are several types of chirality. In undergraduate organic chemistry, most chiral molecules exhibit point chirality- they have at least one sterocenter and don't have a plane of symmetry. Molecules a) and b) both have stereocenters, but they also both have planes of symmetry, so neither is chiral (they are both meso compounds).

Molecule can also be chiral about an axis. The classic example of this is allenes- molecules with two consecutive double bonds. Compound c) has a plane of symmetry so it can't be chiral. It might be hard to see, but compound d) is in fact chiral- it's mirror images are non-super impossible. It has a "chirality axis." Just like a screw can be right-handed or left-handed, so can molecule d). 

This is two separate problems. Many textbooks describe how to mentally convert " zig-zag" (wedge/dash) structures to Fischer projections, but I've never met a student who can do this without making mistakes. So you should convert the name ((2R,3S) 2-bromo-3-chlorobutane) to a zig-zag structure, and then convert the name to a Fischer projection. Never try to convert a zig-zag structure directly to a Fischer projection.

To draw the zig-zag structure, first draw a structure with each halogen (the highest priority substituents) as a wedge, see the R/S configuration we drew, and then adjust as necessary. Why wedges? Because that puts the hydrogen as a dash, so R/S is easy to assign. With two wedges, the structure is (2R, 3R). So the 2R position is fine, and we just switch the wedge at C-3 to a dash, and the structure is correct (2R,3S).

To draw the Fischer projection, we do something similar- arbitaritly draw a structure, check R/S, and then adjust as necessary. In the structure below, I drew (arbitarily) 2S,3S, but we need 2R,3S, so I just switch C-2 and we have the correct structure.