Draw major product of substitution reaction of (2R,3S)2-bromo-3-methylpentane with CH3CH2O- nucleophile?
The major product of the substitution reaction of (2R,3S)-2-bromo-3-methylpentane with CH3CH2O- nucleophile is shown below:
CH3CH2O- + (2R,3S)-2-bromo-3-methylpentane
CH3CH2CH2CH2CH2CH2Br + CH3O-
To determine the major product of the substitution reaction between (2R,3S)-2-bromo-3-methylpentane and CH3CH2O- nucleophile, we need to consider the stereochemistry of the starting material and the potential reaction pathway.
Step 1: Identify the stereochemistry of the starting material, (2R,3S)-2-bromo-3-methylpentane.
The (2R,3S) configuration indicates that the molecule has a bromine atom (Br) attached to the second carbon (C2) of the pentane chain and a methyl group (CH3) attached to the third carbon (C3) of the pentane chain.
Step 2: Write the reaction mechanism for the substitution reaction with CH3CH2O- nucleophile.
In this case, the CH3CH2O- nucleophile will attack the carbon adjacent to the bromine atom (C2) and displace the bromine, forming a new bond to the carbon. The bromine atom will leave as a bromide ion (Br-).
Step 3: Determine the major product.
In the reaction, the CH3CH2O- nucleophile will replace the bromine atom, resulting in the formation of an ether. Since the starting material is optically active (due to the presence of stereocenters), the substitution can occur on either side of the molecule (R or S configuration).
The major product will be formed when the nucleophile attacks the carbocation formed after bromine leaves, leading to an inversion of configuration, resulting in an (S) configuration for the product.
So, the major product of the substitution reaction between (2R,3S)-2-bromo-3-methylpentane and CH3CH2O- nucleophile will be (S)-2-ethoxy-3-methylpentane.
To determine the major product of the substitution reaction between (2R,3S)-2-bromo-3-methylpentane and CH3CH2O- (ethoxide ion) as the nucleophile, we need to consider the stereochemistry and the regiochemistry of the reaction.
First, let's understand the reaction mechanism. This is an SN2 (substitution nucleophilic bimolecular) reaction. In this reaction, the nucleophile (CH3CH2O-) attacks the substrate (2-bromo-3-methylpentane) from the backside, leading to the displacement of the leaving group (the bromine atom).
Since the substrate is a chiral molecule, it has two possible enantiomeric configurations: (2R,3S) and (2S,3R). However, the reaction occurs at the chiral center (the carbon attached to the bromine atom), resulting in an inversion of the configuration.
In the (2R,3S) isomer of 2-bromo-3-methylpentane, the nucleophile will attack the carbon attached to the bromine from the backside, leading to the inversion of configuration at this carbon. The result is the formation of (2S,3S)-3-methylpentane, where the methyl group and the ethyl group are on the same side of the molecule.
Therefore, the major product of the substitution reaction between (2R,3S)-2-bromo-3-methylpentane and CH3CH2O- is (2S,3S)-3-methylpentane, where the methyl and ethyl groups are on the same side.
To draw the structure of (2S,3S)-3-methylpentane, start with a pentane chain, add the methyl group on the third carbon, and place the ethyl group on the second carbon (opposite to the methyl group). Finally, make sure the bromine (Br) atom is replaced by the ethoxide (CH3CH2O-) nucleophile.
Here's the structure of the major product:
CH3CH3
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CH3-C-CH-CH3
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CH2-OCH2CH3