Does anyone know what the third step does in option d?

[510586]

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The oh - and heat turn the Br thing into a triple bond correct? What does h3o+ then do?

 

[510586]

H3O+ usually protonates at the final step.

What is it protonating in this case?

 

[510586]

The terminal alkyne in the final step.

Isn't that already protonated like in this ?http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Images2/elimrx5.gif

 

[510586]

Your mechanism involves an internal alkyne below and a terminal above, which shows the final step at equilibrium. Your specific question deals with a terminal alkyne so you need that aqueous workup.

Oh so wouldn't that result in 1-hexyne then?

 

[DAT Destroyer]

View attachment 193490 The oh - and heat turn the Br thing into a triple bond correct? What does h3o+ then do?

Acid in answer D is used to neutralize the base to stop the reaction, to end with a terminal alkyne. If base is in access, the reaction may keep on going to make an internal alkyne.

First Br2 and CCl4 makes a viscinal halide, Then Excess NaNH2 and NH3 make a triple bond. Water just washes off the sodium at the end and protonates the Carbanion.

Acid and heat is used to make a double bond from a secondary or tertiary halide.

Hope this helps.

 

[510586]

Acid in answer D is used to neutralize the base to stop the reaction, to end with a terminal alkyne. If base is in access, the reaction may keep on going to make an internal alkyne.

First Br2 and CCl4 makes a viscinal halide, Then Excess NaNH2 and NH3 make a triple bond. Water just washes off the sodium at the end and protonates the Carbanion.

Acid and heat is used to make a double bond from a secondary or tertiary halide.

Hope this helps.

So if it ends in a terminal alkyne, would D also be correct?

 

[DAT Destroyer]

So if it ends in a terminal alkyne, would D also be correct?

NaNH2 and NH3 is a better choice to make an alkyne from the vicinal halide.
If one uses base and heat, due to the high temperature a triple bond may change positions, thus lowering the yield.
However, in NaNH2 and NH3, temperature is lower and yield is higher.

 

[510586]

Alright thank you.

 

[510586]

NaNH2 and NH3 is a better choice to make an alkyne from the vicinal halide.
If one uses base and heat, due to the high temperature a triple bond may change positions, thus lowering the yield.
However, in NaNH2 and NH3, temperature is lower and yield is higher.

So technically more than one would work. Hopefully the DAt makes it a bit more clear cut. But the question definitely taught me the oh thing