Textbook: Brown 6th Ed. (2011)

Chapter 3: Stereochemistry and Chirality

Practice Problems (No matching mendel sets were found.)

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Individual Problems

Individual Problems

Problem # 525

On the molecule below, mark each stereocenter with an asterisk. (Note: in some textbooks, stereocenters are referred to as stereogenic centers, chirality centers, or asymmetric centers).

Solution

To be a stereocenter you need to:

be a carbon
be sp³ hybridized (all single bonds)
have four different substituents

Are there other types of stereocenters? Of course! ( sp³ sulfur atoms, for example). But in undergraduate organic chemistry, 99% of all sterocenters you come across will be sp³ hybridized carbon atoms.

So in the molecule below, there are four stereocenters.

Note that the sp²(alkene) carbons can't be sterocenters.

Problem Details

MendelSet practice problem # 525 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): stereocenters

Problem # 526

Assign R or S configuration for each molecule below.

a) is straightforward. I've started you off in b).

Solution

It's easiest to assign R or S configuration when the lowest priority substituent is "in the back" or "behind" the molecule, that is, a dash. Most of the time the lowest priority substituent will be a hydrogen atom.

So a) is straightforward. Because hydrogen is already a dash, we can ignore it and see in which direction the other three substituents decrease in priority (according to Cahn–Ingold–Prelog priority rules). Since they decrease in a counter-clockwise way, a) has S absolute configuration.

b) is a little harder to assign. If hydrogen were a wedge we would be able to just take the opposite of whatever answer we get. But in this case, H is neither a wedge nor a dash. So we use a trick: if we swtich any two pairs of substituents, the absolute configuration of the molecule remains the same. So we switch two pairs so that the hydrogen becomes a dash. On the resulting molecule, the priorities 1-3 decrease in a clockwise manner, so this molecule has an R absolute configuration.

Problem Details

MendelSet practice problem # 526 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): R and S configuration

Problem # 527

Draw the structure of (2R,3S) 2-bromo-3-chlorobutane using wedges and dashes. Also draw a Fischer projection.

Solution

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.

Problem Details

MendelSet practice problem # 527 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): Fischer projections

Problem # 528

Indicate which of the molecules below are chiral (if any).

Solution

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).

Problem Details

MendelSet practice problem # 528 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity, Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

Question Type: Concept (explain why this is so..)

Keyword(s): allenes, symmetry, chirality

Problem # 319

For a molecule to undergo an E2 reaction, the leaving group and the beta-proton must be in an anti-coplanar conformation (one atom straight up, the other straight down). Based on this, which compound undergoes E2 reaction with KOtBu faster? Why?

Solution

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.

Problem Details

MendelSet practice problem # 319 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): cyclohexane chair forms, E2 stereochemistry

Problem # 529

Indicate the major organic product of the reaction below. Include stereochemistry.

Solution

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).

Problem Details

MendelSet practice problem # 529 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule, Stereochemistry, Chirality, and Optical Activity, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*

Question Type: Concept (explain why this is so..)

Keyword(s): alkene addition, stereoselectivity

Problem # 530

Let's work through anti and syn additions to alkenes.

Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).

Solution

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.

Problem Details

MendelSet practice problem # 530 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule, Stereochemistry, Chirality, and Optical Activity, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*

Question Type: Concept (explain why this is so..)

Keyword(s): stereoselectivity

Problem # 531

E2 elimination reactions require anti-coplanar geometry. (note: some textbooks call this anti-periplanar).

Let's work through an E2 reaction, and rotate the molecule eblow into an anti-coplanar geometry to predict the product of this E2 reaction.

Solution

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.

Problem Details

MendelSet practice problem # 531 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): E2 stereochemistry

Problem # 532

Let's work through a chiral resolution. Write out the structure of the indicated compound in each box. Include stereochemistry.

Why is it possible to separate the (R,R) and (R,S) salts?

Solution

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.

Problem Details

MendelSet practice problem # 532 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): chiral resolution

Problem # 744

After a sample of optically pure (S)-2-ethyl-cyclohexanone is dissolved in an aqueous solution for several hours, a significant loss of optical activity is observed. Explain.

Solution

Carbonyls are in equilibrium with their enol forms. This process is called keto-enol tautomerization. (See problems 738 and 739). So when this compound is placed in water a small amount of it will constantly be converted to its enol form, and then back again to the carbonyl form.

But the enol form is achiral! (the alpha carbon is sp² instead of sp³as in the carbonyl form).

So when the achiral enol goes back to its carbonyl form, half will become (S) and the other half will become (R). The stereocenter becomes "scrambled." The process of an optically pure compound becoming a racemic mixture is called racemization.