Organic......NaNH2/NH3

[Toothguy80]

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In destroyer, this turns CH2BrCH2Br into ethylene.
In Kaplan, it says that 1-propyne will react more readily with NaNH2/NH3(l).

But I thought that this turns things into triple bonds, so how come a triple bond react with this more readily,....it can't turn a triple bond into a 6 bond carbon....lol

this is the question:

Which of the following would NaNH2 most readily react with?
1,2-dichloroethane
cis-1,2-dichloroethane
trans-2-butene
1-propyne
2-methyl-3-hexyne

 

[TimeforDAT]

what page of kaplan?

 

[Toothguy80]

Page?
I dont know, it's the last test, the "day before the test" Exam

 

[Jaylee777]

heres my take...

NaNH2 is a very strong base. It reacts with the dihalide to form the alkyne (terminal) at a high temp of 150 degrees celsius (KOH forms the internal alkyne at 200 celsius). Without the high temp, it is unlikely (will not readily) or not possible for it to react with the dihalide to form the triple bond. That rules out the dihalide answers. I think the other two are there to trip you up if you happen to mistake the reagent for Na/Nh3 which is a reducing agent that produces a trans product on a triple bond. Therefore, acting as a strong base it deprotonates the terminal Hydrogen on 1-propyne very readily. hope this helps

 

[Shunwei]

In destroyer, this turns CH2BrCH2Br into ethylene.
In Kaplan, it says that 1-propyne will react more readily with NaNH2/NH3(l).

But I thought that this turns things into triple bonds, so how come a triple bond react with this more readily,....it can't turn a triple bond into a 6 bond carbon....lol

this is the question:

Which of the following would NaNH2 most readily react with?
1,2-dichloroethane
cis-1,2-dichloroethane
trans-2-butene
1-propyne
2-methyl-3-hexyne

You also have to know that the higher s-content a hybridized orbital has, the more acidic the bond will be. Therefore, in a triple bond, the C-H bond is only sp-hybridized, and so will be quite acidic (relatively speaking) to the its more saturated counterparts.

 

[jdent]

Im still confused on this reaction. Where does it say that it needs high temperature. Someone please offer insight!

 

[Simiam]

It is D isn't it?

 

[winnie bear]

heres my take...

NaNH2 is a very strong base. It reacts with the dihalide to form the alkyne (terminal) at a high temp of 150 degrees celsius (KOH forms the internal alkyne at 200 celsius). Without the high temp, it is unlikely (will not readily) or not possible for it to react with the dihalide to form the triple bond. That rules out the dihalide answers. I think the other two are there to trip you up if you happen to mistake the reagent for Na/Nh3 which is a reducing agent that produces a trans product on a triple bond. Therefore, acting as a strong base it deprotonates the terminal Hydrogen on 1-propyne very readily. hope this helps

Thanks for posting your explanation, albeit 3 years ago😳). I had the exact same Q. I'm sure in the future someone else will google about this Destroyer/Kaplan contradiction on that problem!

Thanks!