1) <u>Stereo-selective (or enantioselective)</u> reactions form predominately or exclusively one enantiomer.
2) Epoxidation is the addition of a single oxygen atom to an alkene to form an epoxide.
3) <u>Hydrogenation (or reduction)</u> of an alkene forms an alkane by addition of H₂.
4) <u>Dihydroxylation</u> is the addition of two hydroxy groups to a double forming, a 1,2-diol or glycol.
5) <u>oxidative</u> cleavage of an alkene breaks both the σ and π bonds of the double bond to form two carbonyl groups.
6) <u>Regioselective</u> reactions form predominately or exclusively one constitutional isomer.
7) <u>Syn</u> dihydroxylation results when an alkene is treated KMnO4 or OsO4, where each reagent adds two oxygen atoms to the same side of the double bond.
A pi bond is a bond formed by the overlap of orbitals in a side-by-side fashion with the electron density concentrated above and below the plane of the nuclei of the bonding atoms. ... Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule.
Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.
