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Author Topic: Alkylation of acetyl anions... They say only primary halides are used, so...?  (Read 566 times)

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Lo.Lee.Ta.

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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≡CCH3.
 
So the answer is CH3CH2CH2C≡C with 1. NaNH2 and 2. CH3Br
 ~OR~ HC≡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."
 
http://i1261.photobucket.com/albums/ii596/lolita56381/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≡CCH2CH3
 
So the answer is only one possible reaction: (CH3)2CHC≡CH with 1. NaNH2 and 2. CH3CH2Br
 
But why can't we also have another possible reaction using HC≡CCH2CH3 with
1. NaNH2 and 2. (CH3)2CHBr ?
 
Thanks so much for your *delete me* :)
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discodermolide

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""But how come we used a secondary halide in the 2nd possible reaction?
 CH3CH2CH2Br is a secondary halide, right?""
No it is a primary alkyl halide.

What do you thing will happen to the alkyl halide in the second reaction? That is (CH3)2CHBr. What type of alkyl halide is it?
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Lo.Lee.Ta.

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Oh, duh! Of course this is a primary alkyl halide! I don't know what I was thinking.

Every carbon on CH3CH2CH2Br is only bonded to one other carbon, so it is definitely a primary halide.

But if I had (CH3)3CBr, that would be an example of a tertiary halide, right?

And (CH3)2CH-Br would be a secondary halide.

That's why we can't have a second possible reaction for the 2nd problem! Because the halide, (CH3)2CHBr, is a secondary halide, and that's not allowed!

Got it!
Thank you, discodermolide!
You are the only one who ever helps me!  :)
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