Chapter 2: Isomers

Question 1

Get a fucking pen and paper, and draw the mother fucking isomer flow chart, damnit.

Answer 1

☺️

Question 2

Structural Isomers

Answer 2

Structural isomers (also called constitutional isomer) share their molecular formula, meaning that their molecular weights must be the same. Aside from this similarity, structural isomers are widely varied, with different chemical and physical properties.

Question 3

List examples of physical properties of molecules

Answer 3

Melting point
Boiling point
Solubility
Odor
Color
Density

Question 4

List the chemical properties of a molecule

Answer 4

Polarity
Electronegativity
Acidity/Basicity
Solubility
Oxidation state
Ionization energy
Chemical bond type

Question 5

Stereoisomers

Answer 5

Unlike structural isomers, stereoisomers share the same atomic connectivity. Stereoisomers differ in how these atoms are arranged in space, and all Iser that are not structural isomers fall under this category. The largest distinction within this class is between conformational and configurational isomers.

Question 6

Conformational Isomers

Answer 6

Of all of the isomers, confirmational isomers are the most similar. Conformational isomers are, in fact, the same molecule, only at different points in their natural rotation around single sigma bonds.

Question 7

Name the straight chain confirmations

Answer 7

In order of most stable to least stable (or least amount of potential energy to most amount of potential energy.

Anti/Staggered - 180°
Gauche - 60°
Eclipsed - 120°
Totally Eclipsed - 0°

For butane, the most stable confirmation occurs when the two terminal methyl groups are orientated 180° away from each other. Because there is no overlap of atoms along the line of sight, besides C2 and C3, the molecule said to be in a staggered confirmation. Specifically, it is called the anti-confirmation because the two largest groups are antiperiplanar (in the same plane, but on opposite sides).

Question 8

List the types of ring strain

Answer 8

Angle strain

Torsional strain

Nonbonded strain (van der Waal repulsion)

Question 9

Angle Strain

Answer 9

Angle strain results when bond angles deviate from their ideal values by being scratched or compressed.

Question 10

Torsional Strain

Answer 10

Torsional strain results when cyclical compounds must assume confirmations that have eclipsed or gauche interactions.

Question 11

Nonbonded Strain

Answer 11

van der Waal repulsion

Torsional strain results when nonadjacent atoms or groups compete for the same space. Nonbonded strain is the dominant source of steric stream in the flagpole interactions of the cyclohexane boat confirmation.

Question 12

Name the nonplanar conformations of cyclobutane, cyclopentane, and cyclohexane.

Answer 12

Question 13

Cyclohexanes
(5 key points)

Answer 13

The most stable confirmation of cyclohexane is the cheer confirmation, which minimizes all three types of strain.

The hydrogen atoms that are perpendicular to the plane of the ring are called axial, and those parallel are called equatorial.

The axial–equatorial alternate around the ring; if the wedge on C1 is an axial group, the dash on C2 will be an axial group, and the wedge on C3 will be an axial group, and so on.

Cyclohexanes can undergo a chair flip in which one chair form is converted to the other. in this process, the cyclohexane molecule briefly passes through a fourth confirmation called the half-chair confirmation. After the chair, flip, all axial groups become equatorial, and all equatorial groups become axial, but all dashes remain dashes and wedges remain wedges.

This half-chair conversion can be slow to a bulky group is attached to the ring; tert-butyl groups are classic examples of book groups on the MCAT. For substituted rings, the bulky group favors the equatorial position to reduce nonbonded strain (flagpole interactions) with axial groups in the molecule.

Question 14

Configurational Isomers
(3 key points)

Answer 14

Unlike conformational isomers that interconverted by simple bond rotation, configurational isomers can only change from one form to another by breaking and reforming covalent bonds.

The two categories of configurational isomers are enantiomers and diastereomers.

Both enantiomers and diastereomers can also be considered optical isomers because the different special arrangement of groups in these molecules affects the rotation of plain-polarized light.

Question 15

Chirality
(3 key points)

Answer 15

An object is considered chiral if its mirror image cannot be superimposed on the original object; this implies that the molecule lacks an internal plain of symmetry. Example: left or right hand.

Achiral objects have mirror images that can be superimposed; for example, a fork is identical to its mirror image and is there achiral.

on the MCAT, you will often see this constant tested when there is a carbon atom of four different substituents. This carbon will be an asymmetrical core of optical activity and is know as a chiral center.

Question 16

Enantiomers

Answer 16

Enantiomers have the same connectivity, but opposite configurations at every chiral center in the molecule.

Enantiomers have identical physical and chemical properties with two notable exceptions: optical activity and reactions in chiral environments.

Question 17

Optical activity

Answer 17

The compound is optically active if it has the ability to rotate plain polarized light.

One enantiomers will rotate plane-polarized at the same magnitude but in opposite directions of its mirror image.

Right Rotation: dextrorotatory (d-), (+)
Left Rotation: levorotatory (l-), (–)

Rotation can only be determined by experiment, and the amount of rotation depends on the concentration of the molecule and the length of the tube that light passes through.

Question 18

What are the standard conditions used to test and compare optical activity?

Answer 18

1 gram/mL concentration

1 dm (10 cm) for length

Question 19

What is the equation used to convert a rotation measured at different concentrations and tube lengths to a standard SPECIFIC ROTATION?

Answer 19

Question 20

What happens when (+) and (–) enantiomers are present in equal concentrations?

Answer 20

They for a RACEMIC MIXURE, and their optical rotation cancel each other out. No optical activity is observed.

Question 21

How do you separate a racemic mixture into two constituent isomers?
(4 key points)

Answer 21

The answer lies in the relationship between enantiomers and diastereomers.

Reacting two enantiomers with a single enantiomers of a single compound will lead to two diastereomers.
(+) and (–) + (+) = (+)(+) and (+)(–)

Diastereomers have different physical properties, and these differences enable one to separate these products by common laboratory techniques: crystallization, filtration, distillation, and others.

Once separated, these diastereomers can be reacted to regenerate the original enantiomers.

Question 22

Diastereomers

Answer 22

Diastereomers are non-mirror-image isomers, have two or more stereogenic centers, and differ at some, but not all, of these centers.

The term diastereomers encompass any stereoisomer that is not an enantiomers.

For molecules with n chiral chiral centers, there are 2^n possible stereoisomers.

Question 23

How many stereoisomers are there for a molecule with 2 chiral centers? How many enantiomers and diastereomers does this molecule have?

Answer 23

4 stereoisomers with 2 enantiomer combinations and 4 diastereomer combinations.

Question 24

How do the chemical and physical properties differ in diastereomers?

Answer 24

Diastereomers have different chemical properties, but they might behave similarly in a particular reaction because they have the same functional groups.

Because they have different arrangements in space, they will consistently have different physical properties.

Diastereomers will also rotate plane-polarized light; however, knowing the specific rotation of one gives no indication of the specific rotation of another.

Question 25

Cis-Trans Isomers
(3 key points)

Answer 25

Specific subtype of diastereomers in which substituents differ in their position around an immovable bond, such as a double bond, or a ring structure, such as a cycloalkane in which the rotation of bonds is greatly restricted.

In simple compounds with only one subunit on either side of the bond, we use the term cis and trans. If two subunits are on the same side, the molecule is considered cis. If they are on opposite sides, they are considered trans.

For more complicated compounds with polysubstituted double bonds, (E)/(Z) nomenclature is used instead.

Question 26

Meso Compounds

Answer 26

A molecule with chiral centers that has an internal plane of symmetry.

D- and L-tartaric acid are both optically active, but meso-tartaric acid has a plane of symmetry and is not optically active. This means that even though meso-tartaric acid has two chiral carbon atoms, the molecule as a whole does not display optical activity. Meso compounds are essentially the molecular equivalent of a racemic mixture.

Question 27

Answer 27

Question 28

Answer 28

Question 29

Answer 29

Question 30

Configuration

Answer 30

The configuration of a stereoisomer refers to the spatial arrangement of atoms or groups in the molecule.

Question 31

Relative configuration

Answer 31

The relative configuration of a chiral molecule is its configuration in relation to another chiral molecule. We can use the relative configuration to determine whether molecules are enantiomers, diastereomers, or the same molecule.

Question 32

Absolute confirmation

Answer 32

The absolute confirmation of a chiral molecule describes exact spatial arrangement of these atoms or groups, independent of other molecules.

Question 33

(E) and (Z) Forms

Answer 33

(E) and (Z) nomenclature is used for compounds with polysubstituted double bonds.

To determine the (E)/(Z) designation, start by identifying the highest-priority substituent attached to each double bonded carbon.

The alkene is named (Z) if the two highest priority substituents on each carbon are on the same side of the double bond.
(“z”ame side)

The alkene is named (E) if the two highest priority subunits on each double bonded carbon are on opposite sides.
(“e”pposite)

Question 34

(R) and (S) Forms and the sequence to determine absolute configuration

Answer 34

(R) and (S) nomenclature is used for chiral (stereogenic) centers in molecules.

Step 1: Assign Priority

Step 2 (Classics Version): Arrange in Space
Step 2 (Modified Version): Invert the Stereochemistry

Step 3: Draw a Circle

Step 4: Write the Name

Question 35

How do you assign priority to determine the (E)/(Z) designation?

Answer 35

Use the Cahn-Ingold-Prelog

Priority is assigned based on the atom bonded to the double bonded carbons: the higher, the atomic number, the higher the priority.

If the atomic numbers are equal, priority is determined by the next atoms outward again, whichever group contains the atom with the highest atomic number is given top priority.

If a tie remains, the atoms in this group are compared one by one descending atomic order until the tie is broken.

Question 36

Step 1 in determining (R)/(S) forms

Answer 36

Assign Priority

Using the Cahn-Ingold-Prelog priority rules, assigned priority to the fourth substituents, looking only at the atoms directly attached to the chiral center.

If the atomic numbers are equal, priority is determined by the combination of atoms attached to these atoms; if there is a double bond, it is counted as two individual bonds to that atom.

If a tie is encountered, work outward from the stereo center until tie is broken.

Question 37

Step 2 in determining (R)/(S) forms

Answer 37

Arrange in Space

Orient the molecule in three dimensional space so that the atom with the lowest priority (usually a hydrogen atom hydrogen atom) is at the back of the molecule.

Question 38

Step 3 in determining (R)/(S) forms

Answer 38

Draw a circle

Draw a circle based on priority.
Left = S
Right = R

Question 39

Step 4 in determining (R)/(S) forms

Answer 39

Write the name

Question 40

Fischer Projection

Answer 40

One way to represent three dimensional molecules is by a Fischer projection. In this system, horizontal lines, and Condons, the project out from the plane of the page (wedges), where vertical lands indicate bonds going into the plane of the page (dashes). The point of intersection of the lines represents a carbon atom.