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How to prepare Me3CH using LiAlH4?Can anyone share me some important reagents and their machanism of reaction?

How to prepare Me3CH using LiAlH4?Can anyone share me some important reagents and their machanism of reaction?

Grade:11

1 Answers

Rajdeep
231 Points
5 years ago
HELLO THERE!
 
There’s something you must know in order to answer this question. That is, LiAlH4 (lithium Aluminium hydride) is a reducing agent that can only give it’s hydrogen atom to substitute an atom attached to primary or secondary Carbon atom, and it does not have any effect on Tertiary Carbon atom.
 
So, if you take Me3CX as the first reactant (where X is any halogen), the reaction won’t take place, as the carbon atom to which the halogen is attached is tertiary, and LiAlH4 cannot give it’s Hydrogen to substitute the X to form Me3CH. So, we need to find another way. That’s simple.
 
Just take the compound which has X substituted to either of the primary Carbon atoms.
Like:
 
X-Me-CH(Me2)
 
LiAlH4 will be effective on this compound as the halogen is attached to a primary carbon atom.
 
Reaction:
Cl-Me-CH-(Me2) -------(LiAlH4)-------> H-Me-CH(Me2) = Me3CH (we have got our product).
 
 
Some important reagents are:
LAH (Lithium Aluminium Hydride)
 
Grignard Reagent (R-Mg-X)
 
TEL (Tetra-ethyl Lead)
 
Benzyl Peroxide (used in Anti-Markovnikoff Reactions, and also to generate free radicals in reactions that proceed by free radical mechanism).
 
Post your questions regarding the use of the reagents if you don’t understand, we’ll answer it.
HAPPY POSTING!

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