Identify the conformation of 2-bromo-3-methylpentane that will result in formation of trans-3-methylpent-2-ene by an E2 reaction.
A I
B II
C III
D IV
The correct answer and explanation is:
The correct answer is C (III).
In an E2 (bimolecular elimination) reaction, the elimination of a leaving group (such as a bromine atom) and a hydrogen atom from adjacent carbon atoms occurs in a single, concerted step. The product of the E2 reaction depends on the antiperiplanar (trans-diaxial) geometry of the leaving group and the proton that is abstracted.
Explanation:
- Conformation of 2-bromo-3-methylpentane: For 2-bromo-3-methylpentane to undergo an E2 elimination, the bromine (leaving group) on carbon 2 and the hydrogen on carbon 3 must be antiperiplanar. This means that both the leaving group and the hydrogen should be in opposite directions, ideally positioned in a trans-diaxial arrangement in a cyclic or planar conformation.
- Trans-3-methylpent-2-ene is the product, which indicates the need for the elimination of a hydrogen atom from the β-carbon (carbon 3), along with the bromine from the α-carbon (carbon 2). The elimination leads to the double bond between carbon 2 and carbon 3, with a methyl group on carbon 3.
- Conformation III: In this conformation, the bromine on carbon 2 and the hydrogen on carbon 3 are aligned in an antiperiplanar configuration. This allows the E2 reaction to proceed in a single step, resulting in the formation of the product trans-3-methylpent-2-ene.
In summary, the correct conformation is one where the bromine on carbon 2 and the hydrogen on carbon 3 are antiperiplanar, which is conformation III (C). This arrangement enables the E2 elimination to occur efficiently and gives the desired trans product.