Alkylation of acetyl anions They say only primary halides are used, so ?

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SUMMARY

The discussion centers on the alkylation of acetyl anions using terminal alkynes and primary halides, specifically addressing the reactions involving sodium amide (NaNH2) and various alkyl bromides. The participants clarify that CH3CH2CH2Br is a primary halide, while (CH3)2CHBr is a secondary halide, which cannot be used in the alkylation process. The solutions manual confirms that only primary halides are suitable for these reactions, leading to the formation of specific alkynes through defined pathways.

PREREQUISITES
  • Understanding of terminal alkynes and their reactivity
  • Knowledge of alkyl halides and their classifications (primary, secondary, tertiary)
  • Familiarity with sodium amide (NaNH2) as a strong base in organic synthesis
  • Basic principles of organic chemistry reaction mechanisms
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  • Study the mechanism of alkylation reactions involving terminal alkynes
  • Learn about the classification and reactivity of alkyl halides in organic synthesis
  • Explore the use of sodium amide (NaNH2) in various organic reactions
  • Investigate alternative methods for synthesizing alkynes using different reagents
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Organic chemistry students, researchers in synthetic organic chemistry, and anyone involved in the study of alkylation reactions and alkyne synthesis will benefit from this discussion.

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"Remember that the alkyne must be a terminal alkyne and the halide must be primary. More than one combination of terminal alkyne and halide may be possible."
-This is what it says in the solutions manual of my Ochem book.

​
In the first practice problem, they asked what you'd add to form
CH3CH2CH2C(triple bond)CCH3.

So the answer is CH3CH2CH2C(triple bond)C with 1. NaNH2 and 2. CH3Br
~OR~ HC(triple bond)CCH3 with 1. NaNH2 and 2. CH3CH2CH2Br

But I thought they said only primary halides can be used...?

Encyclopedia Britannica defines an alkyl halide as this: "In a primary alkyl halide, the carbon that bears the halogen is directly bonded to one other carbon, in a secondary alkyl halide to two, and in a tertiary alkyl halide to three."

primaryhalide.gif


But how come we used a secondary halide in the 2nd possible reaction?
CH3CH2CH2Br is a secondary halide, right?

In the second practice problem, it asks how we can form (CH3)2CHC(triple bond)CCH2CH3

So the answer is only one possible reaction: (CH3)2CHC(triple bond)CH with 1. NaNH2 and 2. CH3CH2Br

But why can't we also have another possible reaction using HC(triple bond)CCH2CH3 with
1. NaNH2 and 2. (CH3)2CHBr ?

Thanks so much for your help! :)
 
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Lo.Lee.Ta. said:
"Remember that the alkyne must be a terminal alkyne and the halide must be primary. More than one combination of terminal alkyne and halide may be possible."
-This is what it says in the solutions manual of my Ochem book.

​
In the first practice problem, they asked what you'd add to form
CH3CH2CH2C(triple bond)CCH3.

So the answer is CH3CH2CH2C(triple bond)C with 1. NaNH2 and 2. CH3Br
~OR~ HC(triple bond)CCH3 with 1. NaNH2 and 2. CH3CH2CH2Br

But I thought they said only primary halides can be used...?

Encyclopedia Britannica defines an alkyl halide as this: "In a primary alkyl halide, the carbon that bears the halogen is directly bonded to one other carbon, in a secondary alkyl halide to two, and in a tertiary alkyl halide to three."

primaryhalide.gif


But how come we used a secondary halide in the 2nd possible reaction?
CH3CH2CH2Br is a secondary halide, right?

No. CH3CH2CH2Br is a primary halide. CH3Br isn't though.

In the second practice problem, it asks how we can form (CH3)2CHC(triple bond)CCH2CH3

So the answer is only one possible reaction: (CH3)2CHC(triple bond)CH with 1. NaNH2 and 2. CH3CH2Br

But why can't we also have another possible reaction using HC(triple bond)CCH2CH3 with
1. NaNH2 and 2. (CH3)2CHBr ?

Thanks so much for your help! :)

Because (CH3)2CHBr is a secondary halide.
 

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