When (R)-3-bromo-2,3-dimethylpentane is treated with sodium hydroxide, four different alkenes are formed. Draw all four products, and rank them in terms of stability.

Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable

Explanation:

Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each -carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.

The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).

So, decreasing order of stability of alkenes from most stable to least stable:

2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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