The most convenient representation of an organic molecule is a very simplified representation called a *skeleton structure*. In it, most carbon and hydrogen atoms are omitted. All C-C bonds are represented by line segments, with a C atom assumed at each end unless another atom symbol is shown. Double bonds are represented by two parallel lines, triple bonds by three lines. The hydrogen atoms are assumed to be present in order to fill each C to its usual bonding pattern of four bonds, if no other atoms are shown.

The molecule C_{3}H_{5}Cl could be written out as CH_{2}CHCH_{2}Cl or represented by the Lewis structure at left. It is time-consuming to write out all of the H atoms, so the more compact skeleton structure below is much quicker and (to the experienced eye) conveys all the same information.

- Confirm that adding the H atoms to each vertex of the skeleton structure to make a total of four bonds for each C gives the Lewis structure shown.
- Convert the skeleton structure of this molecule, a flavor ingredient in Juicy Fruit gum, to a full Lewis dot structure. Show that its formula is C7H12O2.

- Draw out the skeleton structures for these molecules:
- CH3CHClCH3
- CH3CH(OH)CH3
- CH3COCH3
- CH3CH2CH2CH(OCH2CH3)CH2CH=CHCH2CH=CH2.

a. b. c. d.

Formulas of Hydrocarbons

The simplest category of organic molecules is the hydrocarbons, those containing only C and H. Consider a series of molecules, CH4, C2H6, C3H8, C4H10 with increasing length of carbon chains.

- Draw out their skeleton structures (except CH4, which has no skeleton structure!)

What pattern do you see in the structure series? What is the pattern in the formula?

Add one more line segment to the zig-zag; add one more C and 2 more H.

- How many C and H atoms would you expect for the fifth in the series? The eighth? Draw their skeletons.

and

- Can you come up with a general rule predicting the number of hydrogen atoms for a hydrocarbon with
*n*carbon atoms?

It should be CnH(2n+2)

- Does this rule apply to branched hydrocarbons such as CH3CH2C(CH3)3 or CH3CH2CH(CH3)CH2CH(CH2CH3)CH2CH3?

It does for those; in fact, since you can “build” a molecule by inserting a new –CH2- unit in between an existing C-C or C-H bond it is always true (unless there are rings or double bonds)

The term *unsaturation* generally means the existence of double or triple bonds in a molecule. Each double bond replaces two H atoms on neighboring carbons with a second bond. –CH2–CH2– becomes –CH=CH– and the second bond is called a π bond. Thus, for each π bond in an organic molecule the formula decreases by two H atoms. A ring structure could also be interpreted as taking H atoms off of non-adjacent carbon atoms and connecting them with a single bond. This has the same effect on the formula. Therefore, ring structures and double bonded structures could be *isomers*.

- How many C and H atoms are there in C5H10

- Draw a skeleton and a full structure for a 5-carbon ring and count the H atoms. Is it an isomer of the above molecule? Yes

- What is the general rule for the formula of a hydrocarbon with a double bond or a ring?

There will be 2 less H for each double bond or ring as compared to the CnH(2n+2) rule prediction for the same # of C.

How many H atoms would there be in a 5-carbon molecule with *two* double bonds? With a triple bond? With both a ring and a double bond? Are these all isomers? Draw a skeleton structure for each.

They are all isomers of C5H8 or or

The term “degree of unsaturation” (DU) means the number of H2 molecules that a particular organic structure has *fewer* than the maximum number. Compare the formulas for five-carbon molecules you found for questions 5, 8, and 11. The number of hydrogen atoms decreases by two (or by one H2) for every ring or double bond.

You should have found earlier that a simple hydrocarbon fits the formula C*n*H*(2n+2)* whether it is linear or branched. A ring or a double bond will decrease that by two H, making C*n*H*2n*. This would have a DU of one since it is one H2 less than fully saturated.

A triple bond would change –CH2–CH2– to –CºC– with four fewer H atoms (or two H2 molecules) for a DU of 2 from the two π bonds.

**Rule**: For a hydrocarbon C*n*H*x*, the DU is calculated as (2*n*+2-*x*)/2 since (2*n*+2) would be the fully saturated, non-ring hydrocarbon.

The degree of unsaturation is a quick way of predicting the formula by looking at the number of double bonds and rings present. Use DU to predict the formula of each of the six-carbon hydrocarbons below and then verify by careful counting.

Complete the table:

Structure |
# rings |
# π bonds |
DU |
Formula |

a |
0 |
2 |
2 |
C6H10 |

b |
1 |
0 |
1 |
C6H12 |

c |
1 |
1 |
2 |
C6H10 |

d |
0 |
0 |
0 |
C6H14 |

e |
0 |
1 |
1 |
C6H12 |

f |
0 |
0 |
0 |
C6H14 |

g |
1 |
3 |
4 |
C6H6 |

Find three isomer pairs: _a_ and _c__, __b_ and _e__, _d__ and _f_

Draw another isomer with a DU of two.

Which structure has the highest DU? C6H6

What are the formulas of hydrocarbons with 7 C and a DU of 0, 1, or 2?

Draw *two* skeleton structures *each* for a 7-carbon molecule with DU of 1 and 2.

C7H16 for DU = 0; no ring or π bond

C7H14 for DU = 1; one ring or one π bond or

C7H12 for DU = 2; has 2 rings or 1 ring & 1π or 2 π

A *heteroatom* is any atom other than C or H. Different heteroatoms in the formula have different effects on the DU.

**Halogens**. Since halogens (F, Cl, Br, I) almost always form one bond to C, they should be considered as replacing an H. Thus, __add the number of halogens__ to the number of hydrogens when finding DU.

Example: Find the DU of C3H5Cl, pictured on page 1.

**Oxygen** can form either single bonds to C and H, or a C=O double bond. If it forms a C-O-H or C-O-C structure, it does not affect the formula. If it forms a C=O, the C is “missing” two H atoms and the double bond adds one to the DU. Thus, __you can ignore O atoms__ in determining the DU. They either have no effect or the C=O π bond shows up because of the “missing” H atoms that might otherwise be present.

Example: find the DU of isopentenyl acetate, page 1, and verify that this agrees with the given formula.

It has 2 π bonds and no ring, so DU = 2. The formula is C7H12, which is 4H less than the expected C7H1 for a saturated hydrocarbon.

What is the DU for C4H7BrO ?

A halogen counts the same as an H, and O doesn’t count, so this is like C4H8. That’s one DU short of the expected for saturated.

Draw at least *two* isomers of C4H7BrO as skeleton structures.

or