Draw major product of substitution reaction of (2R,3S)2-bromo-3-methylpentane with CH3CH2O- nucleophile

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 reaction mechanism.

Let's break down the process step by step:

Step 1: Determine the nucleophilic site
In this case, the nucleophile, CH3CH2O-, will attack the carbon atom attached to the bromine (C-2) since it is an alkyl halide and a good leaving group.

Step 2: Determine the stereochemistry
The compound (2R,3S)-2-bromo-3-methylpentane tells us that the carbon chain has two chiral centers, specifically at C-2 and C-3. To determine the stereochemistry of the product, we need to consider the inversion or retention of the configuration at C-2 during the reaction.

Step 3: Understand the reaction mechanism
The reaction between an alkyl halide and a nucleophile typically follows an Sn2 (substitution nucleophilic bimolecular) mechanism, where the nucleophile attacks the electrophilic carbon while the leaving group is simultaneously kicked out. This mechanism leads to inversion of stereochemistry at the reaction center.

In this case, since we have a methyl group (CH3) attached to C-2, the inversion of configuration will occur, resulting in an optical antipode product.

Therefore, the major product of the substitution reaction between (2R,3S)-2-bromo-3-methylpentane and CH3CH2O- will be the (2S,3R)-2-ethoxy-3-methylpentane.