Naming double bond-containing substituents IUPAC: alkenyl - - PowerPoint PPT Presentation

Naming double bond-containing substituents IUPAC: alkenyl 1-(2-butenyl)-2-methyl-1-cyclohexene

Naming double bond-containing substituents IUPAC: alkenyl 1-(2-butenyl)-2-methyl-1-cyclohexene Some common double bond-containing substituents: vinyl (ethenyl) 1-methyl-4-vinyl-1-cyclohexene

Naming double bond-containing substituents IUPAC: alkenyl 1-(2-butenyl)-2-methyl-1-cyclohexene Some common double bond-containing substituents: vinyl (ethenyl) 1-methyl-4-vinyl-1-cyclohexene allyl (2-propenyl) 1-allyl-4-vinyl-1-cyclohexene

Naming double bond-containing substituents IUPAC: alkenyl 1-(2-butenyl)-2-methyl-1-cyclohexene Some common double bond-containing substituents: vinyl (ethenyl) 1-methyl-4-vinyl-1-cyclohexene allyl (2-propenyl) 1-allyl-4-vinyl-1-cyclohexene phenyl 4,4-dimethyl-1-phenyl-1-cycloheptene

Naming double bond-containing substituents IUPAC: alkenyl 1-(2-butenyl)-2-methyl-1-cyclohexene Some common double bond-containing substituents: vinyl (ethenyl) 1-methyl-4-vinyl-1-cyclohexene allyl (2-propenyl) 1-allyl-4-vinyl-1-cyclohexene phenyl 4,4-dimethyl-1-phenyl-1-cycloheptene benzyl (phenylmethyl) (R)-1-benzyl-4-methyl-4-phenyl-1-cycloheptene

Cl

When the parent chain contains multiple C=C, use a multiplicative prefix before “ene”: (S)-6-chloro-4-methyl-2,4-heptadiene

H H H H H H H H

Alenes are classified according to their degree of substitution of the C=C functional group monosubstituted disubstituted terminally disubstituted trisubstituted tetrasubstituted

C H3 CH3 C H2 CH2 CH3 + . CH3 . :CH2 + :CH2

Csp2

• Csp2
• H1s

H1s H1s H1s

How strong is a C=C ? Is it twice the strength of a C–C? HDBE, kcal/mol 83 1 σ 146 1 σ + 1 π

• HBDE (1 σ + 1 π) - HBDE (1 σ ) = HBDE (1 π)

146

• 83 = 63

cis-alkenes and trans-alkenes Consider all the isomers of the alkene C4H8: are these conformational isomers?

• No! They are isomers that differ by the relative positions of the substituent methyl

groups in space.

• They cannot be converted into each other without breaking a bond.
• Therefore, they are diastereoisomers.
• Why does rotation about the C=C break a bond?

C C C H3 CH3 H H C CH3 H C C H3 H C C H CH3 H CH3

π bond no π bond π bond reforms These diastereoisomers are named just like disubstituted cycloalkanes:

• Identical groups on the same side of the C=C (that is, with a 0 dihedral angle)

are cis-alkenes

• Identical groups on opposite sides of the C=C (that is, with a 180 dihedral

angle) are trans-alkenes

O OH H H Cl OH CH3 Br

trans-2-butene cis-2-butene trans-cinnamic acid ?

The E,Z system of naming diastereomeric alkenes:

• Use the Cahn-Ingold-Prelog “Priority Rules”
• 1. For each C of the C=C, rank the priorities of the 2 atoms attached, according to

atomic number.

• 2. If the 2 atoms attached are identical, rank the atoms attached to those 2 atoms.

Priority of the entire substituent is determined by the first point of difference.

• 3. If identical priority numbers are on the same side of the C=C, the alkene is a

(Z)-alkene.

• 4. If identical priority numbers are on opposite sides of the C=C the alkene is a

(E)-alkene.

C X Y Z C X X C Y Z

For comprison purposes, identical atoms must be compared in the same hybridization state:

• The “Phantom atom” Rules
• 1. For naming purposes, a π bonded atom is considered to replicated by a phantom

σ bond to a phantom atom.

• 2. Rule 1 is applied to both atoms of the π bond.
• 3. Rules 1 and 2 apply to each π bond of multiply bonded atoms.
• 4. All other things being equal, phantom atoms rank below real atoms of the same

atomic number.

Cl Cl C4 C3 C C C C Cl Cl H H H H H H H 1 2 C C 2 1

An example: What is the IUPAC name of ?

• 1. The longest chain containing both C=C bonds is 5:

pentadiene

• 2. Number the chain to give the C=C bonds the lowest locants:

1,3-pentadiene

• 3. Add substituents:

1,5-dichloro-3-ethyl-4-methyl-1,3-pentadiene

• 4. Add stereodesignator for the C1=C2 bond:

(1E)-1,5-dichloro-3-ethyl-4-methyl-1,3-pentadiene

• 5. Add stereodesignator for the C3=C4 bond:

(1E,3E)-1,5-dichloro-3-ethyl-4-methyl-1,3-pentadiene

Another example using the phantom atom rule: Is this an E or a Z alkene?

C C O C O H H H H H

Another example using the phantom atom rule: Is this an E or a Z alkene? It’s an E alkene.

1 2 1 2

Thermodynamic stabilities of isomeric alkenes: Consider C4H8 + 6 O2 4 CO2 + 4 H2O H°rxn = H°rxn kcal/mol CO2 + H2O

The substitution - elimination continuum

what factors influence the balance between substitution and elimination for a given class of haloakanes?

Primary haloalkanes:

• SN2 for all Z-
• Increased steric hindrance in ‡ increases the proportion of E2

P

• branching in haloalkane

P

strong, bulky Lewis bases

Secondary haloalkanes:

• SN2 for all Z- less basic than HO-
• E2 for all Z- more basic than HO-
• Increased steric hindrance in ‡ increases the proportion of E2

P

• branching in haloalkane

P

strong, bulky Lewis bases

• SN1 or E1 when Z: is neutral and the solvent

CA of Nu: pKa H-CN 9.2 H-OH 15.7 H-CH2CH3 50

Tertiary haloalkanes:

• E2 for all Z-
• SN1 or E1 for most Z: and the anions of strong CA

Generally, regardless of the identity of Z: or Z- :

• SN is favored at lower temperatures
• E is favored at higher temperatures