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.

Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-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

codiepienagoya codiepienagoya · Answer 2 · 2021-12-16T09:28:55+01:00

Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.

In the first, sluggish phase, a two-degree cation with a positive charge is produced at carbon 3, and then it undergoes rearrangement to a 3-degree carbocation at carbon 2 either from the side.
So because carbocation only has 6 electrons in its outermost shell, it is sp2 hybridized and thus planar in structure.
In step two then occurs, as well as the attack of the -OH group, as the -OH (hydroxyl) group can strike from either side (top or bottom), leading to the formation of a racemic mixture of 2,4 dimethyl pentane-2-ol.
This quantity of alpha-hydrogens in an alkyl group can be used to determine its stability.
The greater the number of alpha-hydrogens inside an alkene, the overall larger the number of hyperconjugated structures, and thus the greater the stability.
Due to the obvious symmetrical structure of a trans-isomer, whenever the amount of alpha-hydrogens is the same, the trans-alkene isomer is much more stable than that of the cis-alkene isomer.

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