Molecular Structure

Question 1

Lewis Dot Structures

Answer 1

1. Find the total number of valence electrons for all atoms in the molecule.
2. Use one pair of electons to form one bond between each atom.
3. Arrange the remaining electrons around the atoms to satisfy the duet rule for hydrogen and the octet rule for other atoms.

Exceptions: Sometimes atoms within a molecule break the octet rule. Boron and Beryllium do not contain a full octet. Atoms from the third period or higher may have an expanded octet due to vacant d orbitals available for hybridization.

Question 2

Formal Charge

Answer 2

Valence# - (Cont. bond # + Lone e^-s)

Question 3

Hydrocarbons

Answer 3

Alkane

• Acyclic C_<em>n</em>H_2n+2
• Cyclic C_nH_2n

Alkene

Alkyne

Question 4

Alcohol

Answer 4

Question 5

Ether

Answer 5

Question 6

Amine

Answer 6

Question 7

Carboxylic Acid

Answer 7

Question 8

Aldehyde

Answer 8

Question 9

Ketone

Answer 9

Question 10

Ester

Answer 10

Question 11

Amide

Answer 11

Question 12

Prefixes for Organic Nomeclature

Answer 12

Prefix Number of Carbons

Meth- 1

Eth- 2

Prop- 3

But- 4

Pent- 5

Hex- 6

Hept- 7

Oct- 8

Non- 9

Dec- 10

Question 13

Common Substituents Group

Answer 13

Question 14

Nomenclature Rules

Answer 14

• The longest carbon chain with the most substituents determines the base names.
• The end carbon closest to a carbon with a substituent is always the first carbon.
• Any substituent is given the same number as its carbon.
• If the same substituent is used more that once, use the prefixes di-, tri-, tetra, and so on.
• Order the substituents alphabetically.
• The ending of the molecules name is based on the most important functional group in the molecule based on IUPAC priority:
  
  • Carboxylic acids
  • Aldehydes
  • Ketones
  • Alcohols
  • Alkynes
  • Alkenes
  • Alkanes

Question 15

Newman Projections

Answer 15

Question 16

Hybridization and Molecular Shape

Answer 16

In order o figure out the type of hyrid orbital formed by an atom on the MCAT, simply count the number of sigma bonds and lone pairs of electons on that atom. Mtch this to the sum of the superscripts in a hybrid name.

*Remember, one orbital in the s subshell must be first.

s=1

p=3

d=5

Question 17

Chirality

Answer 17

A carbon is chiral when it is bonded to four different substituents.

In order to determine the configuration of a given molecule, number the atoms attached to the chiral center carbon with 1 going to the atom with the largest atomic weight.

R is clockwise

S is counterclockwise

Question 18

Optically Inactive

Answer 18

Optically inactive compounds may be compounds without a chiral center or molecules with internal mirror planes (meso compounds).

Question 19

Optically Active

Answer 19

Chiral molecules can be separated from their mirror images by chemicals. The result is a compound containing molecules with no mirror images in the compound. When plane-polarized light is projected through such a compound, the orientation of its electric field is rotated. Clockwise rotation is + or d. Counerclockwise is - or l.

Question 20

Stereoisomers

Answer 20

Two molecules with the same molecular formula and the same bond-to-bond connectivity are mirror images of each other, but not the same compound. There are two types:

• Enantiomers
• Diastereomers

Question 21

Enantiomers

Answer 21

Enantiomers have the same molecular formula, have the same bond-to-bond connectivity, are mirror images of each other, but are not the same molecule. They have opposite absolute configurations at each chiral carbon (mirror images).

Equal concentration mixtures of enantiomers is call a racemic mixture. They are optically inactive.

Question 22

Diastereomers

Answer 22

Diastereomers have the same molecular formula, have he same bond-to-bond connectivity, are NOT mirror images to each other, and are NOT the same compound.

• Cis-isomers have similar substituents on opposing carbons on the same-side.
  
  • cis molecule have a dipole moment
  • due to dipole moment, cis molecules have stronger intermolecular forces leading to higher boiling points.
  • due to their lower symmetry, cis molecules do not form crystals readily, thus having lower melting points.
  • Steric hindrance produces higher energy levels resulting in higher heat of combustion.
• Trans-isomers have similar substituents on opposing carbons on the opposite-side.
  
  • trans molecule do not have a dipole moment.
  • readily form crystals, thus having a higher melting point.

Question 23

Tri- and Tetra- Substituted Diasteromers

Answer 23

Question 24

Isomer Summary

Answer 24

Isomers are different compounds that have the same molecular formula.

Constitutional Isomers, When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways.