Question

Please answer as many as you can and please provide a full explaination.

Draw the substitution products for the following reaction, assuming that the reaction is carried out under SN2 conditions. If the products can exist as stereoisomers, show what stereoisomers are formed: (3R,4R)-3-bromo-4-methylhexane + CH₃O? Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Show the appropriate stereochemistry by choosing the dashed or wedged buttons and then clicking a bond on the canvas.

Answer 1

In an SN2 reaction of (3R,4R)-3-bromo-4-methylhexane with CH₃O⁻, the configuration at the carbon bonded to the leaving group (bromine) will invert, resulting in (3S,4R)-3-methoxy-4-methylhexane as the substitution product.

Step-by-step explanation:

The question asks for the substitution products under SN2 conditions for the reaction between (3R,4R)-3-bromo-4-methylhexane and CH₃O⁻. Under SN2 conditions, the attacking nucleophile (in this case, the methoxide ion CH₃O⁻) will approach the carbon atom that is bound to the leaving group from the opposite side, leading to an inversion of configuration at that carbon. Therefore, the resulting substitution product will have a configuration opposite to the starting material. If we consider the starting material as having (3R,4R) configuration, the product will have (3S,4R) configuration, because the methyl group at position 4 does not change its configuration.

Since the resulting product can exist as stereoisomers, we show both the original configuration and the inverted configuration at the carbon where the nucleophilic substitution occurs. The product of the reaction will be (3S,4R)-3-methoxy-4-methylhexane, which exhibits the change in stereochemistry due to the SN2 mechanism.