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

[Lo.Lee.Ta.]

"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.

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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."

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! :)

 

[chemisttree]

"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."

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.