Formation of Substitution Product from 1-Bromo-2,6-Dimethylcyclohexane

In summary, when the compound 1-bromo-2,6-dimethylcyclohexane is treated with sodium methoxide, only a substitution product is obtained. This is because sodium methoxide is a strong base and a good nucleophile, favoring SN2 reactions. The most stable chair conformer of the reactant also shows why an elimination product is not obtained. Additionally, the products of an SN2 reaction can result in a change in stereochemistry.
  • #1
chiyakotiten
3
0
Only a substitution product is obtained when the compound below is treated with sodium methoxide. Draw the substitution product and explain why an elimination product is not obtained

1zlvg9h.jpg


(If the image doesn't show up, the compound is 1-bromo-2,6-dimethylcyclohexane)

Homework Equations


- The compound is a secondary structure
- It is going to be an SN2 reaction(?)

The Attempt at a Solution


2qlvdj6.jpg


Is it because sodium methoxide is a weak base, but a good nucleophile which favors SN2 reactions?
 
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  • #2
Sodium methoxide is a strong base which favours SN2.
Your product for SN2 is wrong. The nucleophile attacks from the rear end of the leaving group.
Regarding elimination -
Hint: What is the requirement for elimination? Is the hydrogen easily available for abstraction? Does it require high energy transition state?
 
  • #3
Edit: Sodium methoxide is not a strong base...just stable.
 
  • #4
Sodium Methoxide is stable because its a salt, as soon as you have a polar solvent involved you get sodium cation plus a methoxide anion. Alkoxides are fairly strong bases. pKa of alcohols are ~15.5-16.

To the OP. For a hint at why elimination won't happen, draw the most stable chair conformer of the reactant and you will see the answer.

And as far as the products of your substitution reaction go, remember what happens to stereochemistry as you go through SN1 or SN2 mechanisms.
 
  • #5


Yes, that is correct. Sodium methoxide (NaOCH3) is a strong nucleophile due to the presence of the negatively charged oxygen atom, but it is a weak base because it is not a good proton acceptor. In this case, the strong nucleophilic attack by sodium methoxide on the 1-bromo-2,6-dimethylcyclohexane molecule results in the substitution of the bromine atom with a methoxy group (-OCH3), forming the substitution product 1-methoxy-2,6-dimethylcyclohexane. This is because the SN2 mechanism involves a simultaneous bond-breaking and bond-forming process, which is favored by a strong nucleophile and a weak base. On the other hand, an elimination reaction, which involves the removal of a proton and a leaving group to form a double bond, would require a strong base. Since sodium methoxide is a weak base, it is not able to carry out an elimination reaction in this case.
 

1. What is the mechanism for the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane?

The mechanism for the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane is an SN2 (bimolecular nucleophilic substitution) reaction. This means that the substitution occurs in a single step and involves the attack of a nucleophile on the carbon atom bonded to the bromine, resulting in the displacement of the bromine and the formation of a new carbon-nucleophile bond.

2. What is the role of the solvent in the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane?

The solvent plays a crucial role in the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane. It serves as a medium for the reaction to take place and also helps to stabilize the reaction intermediates. In this case, polar aprotic solvents like acetone or DMF are preferred as they do not react with the nucleophile or the reactant, allowing the reaction to proceed smoothly.

3. How does the steric hindrance of the reactant affect the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane?

The steric hindrance of the reactant, which is determined by the size and bulkiness of the substituents, can have a significant impact on the formation of substitution product from 1-Bromo-2,6-Dimethylcyclohexane. In this case, the presence of two methyl groups on the cyclohexane ring creates steric hindrance, making it difficult for the nucleophile to attack the carbon atom bonded to the bromine. This leads to a slower reaction rate and may also result in the formation of unwanted side products.

4. What are the factors that affect the selectivity of the substitution product from 1-Bromo-2,6-Dimethylcyclohexane?

The selectivity of the substitution product from 1-Bromo-2,6-Dimethylcyclohexane can be affected by various factors, such as the nature of the nucleophile, the strength of the nucleophile, the solvent used, and the steric hindrance of the reactant. For example, a stronger nucleophile will have a higher tendency to attack the bromine, leading to a more selective reaction. Similarly, using a less bulky solvent can also improve the selectivity of the reaction.

5. How can the formation of unwanted side products be minimized during the reaction of 1-Bromo-2,6-Dimethylcyclohexane with a nucleophile?

To minimize the formation of unwanted side products, several strategies can be employed, such as using a less bulky solvent, using a more reactive and selective nucleophile, and optimizing the reaction conditions such as temperature and reaction time. Additionally, performing the reaction under controlled conditions and using a purification technique, such as column chromatography, can also help to minimize the formation of side products.

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