Only a substitution product is obtained when the compound 1-bromo-2,6-dimethylcyclohexane is treated with sodium methoxide. Draw the substitution product and explain why an elimination product is not obtained.
To determine the substitution product when 1-bromo-2,6-dimethylcyclohexane is treated with sodium methoxide, we first need to draw the structure of the starting compound:
Br
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CH3-C-CH(CH3)-CH2-CH2-CH3
Now, when sodium methoxide (NaOCH3) reacts with 1-bromo-2,6-dimethylcyclohexane, a substitution reaction occurs. The sodium methoxide replaces the bromine atom. This results in the formation of methanol (CH3OH) and the substitution product, which is 1-methoxy-2,6-dimethylcyclohexane.
CH3-O-CH3
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CH3-C-CH(CH3)-CH2-CH2-CH3
The reason an elimination product is not obtained is that sodium methoxide is a strong base as well as a nucleophile. In this case, it acts primarily as a nucleophile rather than a base. It attacks the electrophilic carbon atom (the one bearing the bromine) and replaces it with the methoxy group (CH3O-). Since an elimination reaction requires a strong base to remove a proton, the methoxide ions do not possess sufficient basicity to initiate an E1 or E2 elimination. Therefore, only a substitution product is obtained in this reaction.
To determine the substitution product formed when 1-bromo-2,6-dimethylcyclohexane is treated with sodium methoxide, we need to analyze the reaction mechanism and consider the factors that favor substitution over elimination.
First, let's draw the structure of 1-bromo-2,6-dimethylcyclohexane:
Br H
\ /
C - C - C - C - C
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CH3 CH3
When sodium methoxide (NaOCH3) is added, it acts as a nucleophile and replaces the bromine atom via a substitution reaction. The oxygen atom of the methoxide attacks the bromine atom, resulting in the formation of the substitution product.
To determine the exact product, let's identify the carbon atom where substitution occurs. In this case, there are two possible positions: carbon 1 and carbon 2.
If the substitution occurs at carbon 1, we get:
Br H
\ /
O - C - C - C - C
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CH3 CH3
If the substitution occurs at carbon 2, we get:
Br H
\ /
C - O - C - C - C
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CH3 CH3
Both of these substitution products are possible, depending on which carbon atom undergoes nucleophilic substitution. Drawn above are the potential substitution products that can be formed.
Now, let's consider why an elimination product is not obtained in this reaction. Elimination reactions typically occur under conditions that favor the formation of a double bond. However, in this case, the reaction is performed with sodium methoxide in a methanol solvent, conditions that favor a substitution reaction over elimination.
To favor elimination, commonly used reagents such as strong bases (e.g., sodium hydroxide - NaOH) or heating at high temperature are required. In this case, the presence of sodium methoxide as a nucleophile suggests that substitution is the preferred pathway rather than elimination.
In summary, when 1-bromo-2,6-dimethylcyclohexane is treated with sodium methoxide, a substitution product is obtained due to the nucleophilic attack of the methoxide ion on the bromine atom. The specific substitution product depends on whether the nucleophilic attack occurs at carbon 1 or carbon 2. An elimination product is not obtained because the reaction conditions and reagents favor substitution over elimination.