Conversion of tertiary haloalkane

[somecelxis]

Homework Statement

can someone teach me how do you convert this reactant to get the product?

I consider this as free radical substituition reaction and in the presence of UV , then the H atom in the 2 CH3 will also be substituted... i want ony the H atom on the carbon bonded to C to be substituted only.

Homework Equations

The Attempt at a Solution

 

[Yanick]

Why would you assume that the rate of addition of bromine to the terminal carbons will be so high as to make free radical bromination not useful for this conversion? Think about the mechanism and selectivity of free radical bromination. What type of carbons are favored to form radicals in these mechanisms? What type of carbon is the carbon of interest and what type are the terminal carbons?

Sometimes it is better to do a one step lower yielding synthesis than it is to go through several steps which may individually appear better but all that handling and loss etc makes the synthesis less efficient and more time/energy consuming etc.

 

[somecelxis]

Why would you assume that the rate of addition of bromine to the terminal carbons will be so high as to make free radical bromination not useful for this conversion? Think about the mechanism and selectivity of free radical bromination. What type of carbons are favored to form radicals in these mechanisms? What type of carbon is the carbon of interest and what type are the terminal carbons?

Sometimes it is better to do a one step lower yielding synthesis than it is to go through several steps which may individually appear better but all that handling and loss etc makes the synthesis less efficient and more time/energy consuming etc.

i was told that The H on the side chain is a "Benzylic hydrogen atom" it is very reactive and even chlorine will react mostly by attacking at that position. Bromine, as shown, is much more selective.
does it mean benzylic hydrogen atom more reactive than aliphatic hydrogen atom... why is it so?

 

[Yanick]

A majority of introductory O-Chem is explained by charge/electron density as influenced by the Inductive Effect and Resonance Effects. Although I'm not exactly sure why radicals become more stable when they are "spread out," I have just come to accept this fact.

http://www.chemgapedia.de/vsengine/.../radi_brom_benzyl/radi_brom_benzyl.vscml.html

For the record, I googled this in about 30 seconds (I certainly hope you took initiative yourself and haven't been waiting for someone to google for you).

 

[DeeNos]

Hi there,
Firstly, I would like to clarify that there are different methods to convert tertiary haloalkanes, depending on the specific reactant and product. Without knowing the specific molecules involved, it is difficult for me to provide a detailed answer. However, I can provide some general information and steps that may help you in your conversion process.

1. Understand the structure and properties of the reactant and product: Tertiary haloalkanes are compounds that have a halogen atom (such as chlorine, bromine or iodine) attached to a tertiary carbon atom. This means that the carbon atom is bonded to three other carbon atoms. The product, on the other hand, may have a different functional group attached to the tertiary carbon atom.

2. Identify the type of reaction: As you mentioned, this conversion can be considered as a free radical substitution reaction. This type of reaction typically involves the replacement of a halogen atom with another atom or group.

3. Determine the conditions for the reaction: In order for the reaction to occur, specific conditions must be met. In this case, the presence of UV light is required. Other conditions, such as temperature and solvent, may also play a role in the reaction.

4. Choose a suitable reagent: Depending on the specific reactant and product, different reagents may be used to achieve the desired conversion. For example, if you want to replace the halogen atom with a hydrogen atom, you can use a reducing agent such as zinc and hydrochloric acid.

5. Consider the mechanism of the reaction: It is important to understand the mechanism of the reaction in order to predict the products and determine the suitability of the reagents. In the case of free radical substitution, the reaction typically proceeds through three steps: initiation, propagation, and termination.

6. Perform the reaction in the laboratory: Once you have identified the appropriate reagents and conditions, you can carry out the reaction in the laboratory. It is important to follow proper safety protocols and use appropriate equipment.

I hope this information helps you in your conversion process. If you have any further questions, please let me know. Good luck with your experiment!