Chapter 3

Table 3.1 lists functional groups. Useful as a reference - no need to memorize it.
Classifications of hydrocarbons:
saturated: contains single bonds only (no π π pi\piπ bonds)
unsaturated: contains π π pi\piπ bonds
cyclic: contains one or more rings
acyclic: contains no ring
aromatic: contains a benzene ring or analogous structure
Cc1ccc(C(C)C)cc1

aliphatic: not aromatic (can be saturated or unsaturated)
IUPAC classification
alkanes: saturated. If acyclic, its formula is C n H 2 n + 2 C n H 2 n + 2 C_(n)H_(2n+2)\mathrm{C}_{n} \mathrm{H}_{2 n+2}CnH2n+2
CC1=CCC(=C(C)C)C=C1

alkenes: contains ≥ 1 ≥ 1 >= 1\geq 1≥1 double bond
alkynes: contains ≥ 1 ≥ 1 >= 1\geq 1≥1 triple bond
Isomers: molecules with the same empirical formula but different structure.
Constitutional isomers differ in their bond connectivity.
CCCCC
Linear
CCC(C)C
Branched
CC(C)(C)C

CC(C)(C)C
IUPAC nomenclature. (In Table 3.3, memorize names for C 1 − C 13 C 1 − C 13 C_(1)-C_(13)\mathrm{C}_{1}-\mathrm{C}_{13}C1−C13 )
C 5 H 12 C 5 H 12 C_(5)H_(12)\mathrm{C}_{5} \mathrm{H}_{12}C5H12 constitutional isomers
Alkane Alkyl group
CH 4 CH 4 CH_(4)\mathrm{CH}_{4}CH4 methane CH 3 CH 3 CH_(3)\mathrm{CH}_{3}CH3 methyl Me
C 2 H 6 C 2 H 6 C_(2)H_(6)\mathrm{C}_{2} \mathrm{H}_{6}C2H6 ethane C 2 H 5 C 2 H 5 C_(2)H_(5)\mathrm{C}_{2} \mathrm{H}_{5}C2H5 ethyl Et
C 3 H 8 C 3 H 8 C_(3)H_(8)\mathrm{C}_{3} \mathrm{H}_{8}C3H8 propane C 3 H 7 C 3 H 7 C_(3)H_(7)\mathrm{C}_{3} \mathrm{H}_{7}C3H7 propyl Pr
C 4 H 10 C 4 H 10 C_(4)H_(10)\mathrm{C}_{4} \mathrm{H}_{10}C4H10 butane C 4 H 9 C 4 H 9 C_(4)H_(9)\mathrm{C}_{4} \mathrm{H}_{9}C4H9 butyl Bu
Alkane Alkyl group CH_(4) methane CH_(3) methyl Me C_(2)H_(6) ethane C_(2)H_(5) ethyl Et C_(3)H_(8) propane C_(3)H_(7) propyl Pr C_(4)H_(10) butane C_(4)H_(9) butyl Bu| Alkane | | Alkyl group | | | | :--- | :--- | :--- | :--- | :--- | | $\mathrm{CH}_{4}$ | methane | $\mathrm{CH}_{3}$ | methyl | Me | | $\mathrm{C}_{2} \mathrm{H}_{6}$ | ethane | $\mathrm{C}_{2} \mathrm{H}_{5}$ | ethyl | Et | | $\mathrm{C}_{3} \mathrm{H}_{8}$ | propane | $\mathrm{C}_{3} \mathrm{H}_{7}$ | propyl | Pr | | $\mathrm{C}_{4} \mathrm{H}_{10}$ | butane | $\mathrm{C}_{4} \mathrm{H}_{9}$ | butyl | Bu |
Simple alkyl groups: C 3 H 7 C 3 H 7 C_(3)H_(7)\mathrm{C}_{3} \mathrm{H}_{7}C3H7 and C 4 H 9 C 4 H 9 C_(4)H_(9)\mathrm{C}_{4} \mathrm{H}_{9}C4H9
Two isomers of C 4 H 10 C 4 H 10 C_(4)H_(10)\mathrm{C}_{4} \mathrm{H}_{10}C4H10 :
CCCCC

CC(C)C

C 4 H 9 C 4 H 9 C_(4)H_(9)\mathrm{C}_{4} \mathrm{H}_{9}C4H9 groups derived from each C 4 H 10 C 4 H 10 C_(4)H_(10)\mathrm{C}_{4} \mathrm{H}_{10}C4H10 isomer:

Degree of alkyl substitution on C ( R R RRR is any alkyl group)

1 ∘ , 2 ∘ , 3 ∘ 1 ∘ , 2 ∘ , 3 ∘ 1^(@),2^(@),3^(@)1^{\circ}, 2^{\circ}, 3^{\circ}1∘,2∘,3∘ can be applied to C atoms or H atoms
a primary alkyl chloride

IUPAC names

CCCC(C)CCC(C)CC
  1. Find the longest chain and name it.
  2. Number the atoms in the chain, beginning at the end nearest a branch point.
  3. Name each substituent & give it a number.
  4. Write the name as a single word, with substituents in alphabetical order.
    CCCC(C)C(C)CCC
4,7-dimethylnonane
CCCC(C)CCC(C)CC
3,6-dimethylnonane
WRONG because the 1st substituent does not have the lowest possible number
CORRECT because the 1 st substituent does have the lowest possible number
CCC(CCCC(C)C)C[C@@H](Cl)CCSHCl

8-chloro-6-ethyl-2-methyldecane

In numbering the chain: if there are multiple substituents, number the chain in such a way that the first substituent has the lowest number. If a decision is not possible, then number the chain to minimize the number on the second subsituent, and so on, until a decision is reached.
CCC(CC)CC(C)C(C)CC
6-ethyl-3,4-dimethyloctane (not 3-ethyl-5,6-dimethyloctane)

Only as a last resort, number according to alphabetical order
CC(F)CC(C)CC(C)ClFCl
2-chloro-6-fluoro-4-methylheptane (based on the alphabet)
CC(Cl)CCC(C)C(C)FClF
6-chloro-2-fluoro-3-methylheptane (based on position of 2nd sub.)
CCCCC(C(CC)CCC)C(C)(C)CC
5-(1,1-dimethylpropyl)-4-ethylnonane
Name a branched substituent based on its longest chain, counting from the point of attachment to the main chain as carbon 1. Place the substituent name in parentheses. Use the first letter within the parentheses for alphebatizing, even if it starts with di, tri, etc.
IUPAC allows common names for simple branched substituents (isopropyl, tert-butyl, etc.).
CCCCC(CCC)C(C)C
4-isopropylheptane or 4-(1-methylethyl)heptane
Prefixes that are ignored in alphabetizing:
di, tri, tetra, ...
sec-, tert-
(exception: branched substituents - see example above)
Prefixes that are included in alphabetizing: iso, neo, cyclo

CONFORMATIONS OF ALKANES


Useful approximation: the 12 kJ / mol 12 kJ / mol 12kJ//mol12 \mathrm{~kJ} / \mathrm{mol}12 kJ/mol barrier is considered to be the sum of 3 H − H 3 H − H 3H-H3 \mathrm{H}-\mathrm{H}3H−H eclipsing interactions. Then one H − H H − H H-H\mathrm{H}-\mathrm{H}H−H eclipsing interaction contributes 4 kJ / mol 4 kJ / mol 4kJ//mol4 \mathrm{~kJ} / \mathrm{mol}4 kJ/mol to the barrier. The approximation is that we assume an H − H H − H H-H\mathrm{H}-\mathrm{H}H−H eclipsing interaction contributes 4 kJ / mol 4 kJ / mol 4kJ//mol4 \mathrm{~kJ} / \mathrm{mol}4 kJ/mol in any molecule.

Why is the staggered geometry lower in energy than eclipsed?

Two models:
  1. Steric interaction model: as two groups approach each other and their electron densities begin to overlap, electron-electron repulsion sharply raises the energy. The larger the group, the greater the steric repulsion.
  2. Hyperconjugation model (called "torsional strain" by McMurry): the staggered geometry allows greater electron delocalization than the eclipsed, via interaction of filled with empty σ σ sigma\sigmaσ orbitals.
For ethane, the concensus is that both effects are important. For molecules with interacting groups larger than H (Me, Et, iPr, t t ttt-Bu, etc.), steric interactions are dominant.
Hyperconjugation, or delocalization of σ σ sigma\sigmaσ orbitals, occurs throughout organic chemistry and is discussed briefly below. The description below is not in McMurry and will not appear on any exam or quiz.
As ethane rotates, an "electron delocalization window" alternately opens (staggered) and closes (eclipsed). In the staggered geometry, bonding and antibonding σ C − H σ C − H sigmaC-H\sigma \mathrm{C}-\mathrm{H}σC−H orbitals ( σ CH σ CH sigma_(CH)\sigma_{\mathrm{CH}}σCH and σ CH σ CH sigma_(CH)\sigma_{\mathrm{CH}}σCH ) on adjacent carbons interact to give a partial π π pi\piπ bond between C atoms. Compared to a σ CH σ CH sigma_(CH)\sigma_{\mathrm{CH}}σCH orbital, this gives a larger space in which electrons can move, encompassing 2 C atoms instead of one. By the Uncertainty Principle, delocalization lowers the energy of staggered ethane. This is not available to eclipsed ethane because of weak net overlap of σ CH σ CH sigma_(CH)\sigma_{\mathrm{CH}}σCH and σ ∗ CH σ ∗ CH sigma^(**)_(CH)\sigma{ }^{*}{ }_{\mathrm{CH}}σ∗CH orbitals of adjacent eclipsed C-H bonds.

Staggered

CC
Strong π π pi\piπ-type bonding overlap
Electrons are delocalized. Energy is lowered.

Eclipsed

CCC
Poor overlap (bonding & antibonding largely cancel)
Electrons are not delocalized. Energy is not lowered.

PROPANE

The energy plot is identical to ethane's, with a barrier of 14 kJ / mol 14 kJ / mol 14kJ//mol14 \mathrm{~kJ} / \mathrm{mol}14 kJ/mol instead of 12 .
Eclipsed
Staggered
Propane eclipsing interactions:
2 H − H + 1 Me − H 2 H − H + 1 Me − H 2H-H+1Me-H2 \mathrm{H}-\mathrm{H}+1 \mathrm{Me}-\mathrm{H}2H−H+1Me−H
Compute the value of Me-H eclipsing: Me − H = 14 − 2 ∗ 4 Me − H = 14 − 2 ∗ 4 Me-H=14-2**4\mathrm{Me}-\mathrm{H}=14-2 * 4Me−H=14−2∗4
= 6 kJ / mol = 6 kJ / mol =6kJ//mol=6 \mathrm{~kJ} / \mathrm{mol}=6 kJ/mol
How to think about rotations of two or more C-C bonds
(This section is not in McMurry, so it will not be on an exam or quiz.)
  1. In McMurry and in this course, we use Newman projections to consider each rotation separately. In propane, butane, etc. an alkyl group attached to the C − C C − C C-C\mathrm{C}-\mathrm{C}C−C bond is treated as a structureless lump that affects the C-C rotation barrier by steric interactions with nearby substituents.
  2. Here is a more realistic treatment that illustrates how chemists think about molecules like propane. It will not be on an exam because it is not in McMurry, and you can safely skip the rest of this page.
There are two C-C rotations, each with a barrier of 14 kJ / mol 14 kJ / mol 14kJ//mol14 \mathrm{~kJ} / \mathrm{mol}14 kJ/mol. Each rotation angle can have any value, and this leads to an energy surface, called an egg-carton surface. At the red peaks, the C 1 C 2 C 1 C 2 C_(1)C_(2)\mathrm{C}_{1} \mathrm{C}_{2}C1C2 and C 2 C 3 C 2 C 3 C_(2)C_(3)\mathrm{C}_{2} \mathrm{C}_{3}C2C3 rotations are both eclipsed, and at the blue minima they are both staggered. At other grid points one is staggered and the other eclipsed. The 1D curve above is recovered by picking any grid line and following it from 0 to 360 ∘ 360 ∘ 360^(@)360^{\circ}360∘ with the other angle constant. For example, the energy along the C 1 C 2 C 1 C 2 C_(1)C_(2)\mathrm{C}_{1} \mathrm{C}_{2}C1C2 axis line ( C 2 C 3 = 0 C 2 C 3 = 0 C_(2)C_(3)=0\mathrm{C}_{2} \mathrm{C}_{3}=0C2C3=0 ) oscillates between 14 and 28 , and the energy along the adjacent grid line with C 2 C 3 = 60 C 2 C 3 = 60 C_(2)C_(3)=60\mathrm{C}_{2} \mathrm{C}_{3}=60C2C3=60 varies between 0 and 14 .

Molecular motion on an egg-carton surface

The diagonal red dotted line illustrates a path in which two CH 3 CH 3 CH_(3)\mathrm{CH}_{3}CH3 groups rotate together (cooperative motion). The path shown maintains a constant energy of 14 kJ / mol 14 kJ / mol 14kJ//mol14 \mathrm{~kJ} / \mathrm{mol}14 kJ/mol as the CH 3 CH 3 CH_(3)\mathrm{CH}_{3}CH3 groups rotate together with a constant 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ difference. This and similar paths imply that conformational change does not have to occur by hops between adjacent minima, but can also involve long-distance hops between nonadjacent minima. This example illustrates the conceptual advantage of energy surfaces over energy curves.
Egg-carton type energy surfaces (2D periodic surfaces) are useful for describing solid surfaces like metals, graphene, etc. They have been used in studying catalysis and material science.
Butane eclipsing interactions
± 120 ∘ H − H + 2 Me − H = 4 + 2 ∗ 6 = 16 estimated 16 observed! ± 120 ∘ H − H + 2 Me − H = 4 + 2 ∗ 6 = 16  estimated  16  observed!  {:[+-120^(@)quadH-H+2Me-H=4+2**6=16" estimated "],[16" observed! "]:}\begin{aligned} \pm 120^{\circ} \quad \mathrm{H}-\mathrm{H}+2 \mathrm{Me}-\mathrm{H}=4+2 * 6= & 16 \text { estimated } \\ & 16 \text { observed! } \end{aligned}±120∘H−H+2Me−H=4+2∗6=16 estimated 16 observed! 
0 ∘ 0 ∘ 0^(@)quad0^{\circ} \quad0∘ Compute Me-Me eclipsing interaction:
2 H − H + Me − Me = 19 Me − Me = 19 − 2 ∗ 4 = 11 kJ / mol 2 H − H + Me − Me = 19 Me − Me = 19 − 2 ∗ 4 = 11 kJ / mol {:[2H-H+Me-Me=19],[Me-Me=19-2**4=11kJ//mol]:}\begin{aligned} & 2 \mathrm{H}-\mathrm{H}+\mathrm{Me}-\mathrm{Me}=19 \\ & \mathrm{Me}-\mathrm{Me}=19-2 * 4=11 \mathrm{~kJ} / \mathrm{mol} \end{aligned}2H−H+Me−Me=19Me−Me=19−2∗4=11 kJ/mol
Summary of eclipsing interactions ( kJ / mol kJ / mol kJ//mol\mathrm{kJ} / \mathrm{mol}kJ/mol ):
H − H H − H H-H\mathrm{H}-\mathrm{H}H−H 4
Me − H Me − H Me-H\mathrm{Me}-\mathrm{H}Me−H 6
Me − Me Me − Me Me-Me\mathrm{Me}-\mathrm{Me}Me−Me 11
H-H 4 Me-H 6 Me-Me 11| $\mathrm{H}-\mathrm{H}$ | 4 | | :--- | ---: | | $\mathrm{Me}-\mathrm{H}$ | 6 | | $\mathrm{Me}-\mathrm{Me}$ | 11 |
Me-Me gauche interaction: 3.8
Practice problem:
Show a qualitative energy diagram for rotation about the C 2 − C 3 C 2 − C 3 C_(2)-C_(3)\mathrm{C}_{2}-\mathrm{C}_{3}C2−C3 bond of 2-methylpentane
  1. Put C 2 C 2 C_(2)\mathrm{C}_{2}C2 in front
  2. Draw 3 eclipsed and 3 staggered Newman projections, label them A - F, specify dihedral angles. 0 ∘ 0 ∘ 0^(@)0^{\circ}0∘ should be eclipsed.
  3. Qualitatively assign highest and lowest energy eclipsed; same for staggered. Guiding principle: energy is lowest when the largest groups are separated.
  4. Any eclipsed structure is higher in energy than any staggered.
  5. Arrange A − F A − F A-FA-FA−F on energy diagram so that the angle increases monotonically.
A
0 ∘ 0 ∘ 0^(@)0^{\circ}0∘
High
B
120 ∘ 120 ∘ 120^(@)120^{\circ}120∘
High
C
240 ∘ 240 ∘ 240^(@)240^{\circ}240∘
Low
D
60 ∘ 60 ∘ 60^(@)60^{\circ}60∘
High
CCC1C2CC3CC(C)C(C2)C3C1(C)C
E
180 ∘ 180 ∘ 180^(@)180^{\circ}180∘
Low
CCC1C2CC3CC1C(C3C)C2C