Question

The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways.
As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides.
What are the products from these reactions?

Answer 1

The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways. As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides. The products from these reactions is I is 2- methylcyclohexanone; II is a mixture of cyclohexanone and 2- methylpropene.

Explanation:-

• Lithium enolates are formed from the reaction of ketones or esters and LDA at low temperatures. Pre-formed lithium enolates react cleanly with aliphatic aldehydes.
• Enolate formation is essentially just an acid-base reaction. The position of the equilibrium is controlled by a variety of factors: solvent, base, cation, temperature.

Answer 2

Answer:

The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways.

As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides.

Explanation:

I is 2-methylcyclohexanone; II is a mixture of cyclohexanone and 2-methylpropene

C7H12O is the formula for 2-Methylcyclohexanone

Isobutylene (or 2-methylpropene) is a hydrocarbon with the formula (CH3)2C=CH2. It is a four-carbon branched alkene (olefin), one of the four isomers of butylene.

C6H10Ois the formula for cyclohexanone.

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