[a confused chiral girl]

Hi~

I am unsure how this reaction could give a product that is again a alkyne. I thought it would be an alkane.

In the next post, I drew what I got after the first reaction with NaNH2, NH3  and then I drew the mechanism with CH3CH2CH2Br according to how I think it would go. then I end up with an alkane, and that is none of the choices..can someone please show me the correct mechanism?

thank you!!

[a confused chiral girl]

According to my limited knowledge.... :-[I thought it would produce an alkane since something with "Br" produces any alkane, ..like HBr. so this is my approach aftr giving it many tried.

[PRIYA1022]

Hi,
  For your first reaction,
The hydrogen in a terminal alkyne is acidic, So a strong base like NaNH2 would deprototonate and give an anion, which would do a nucleophilic substitution on the alkyl halide( Sn2). This is a good way to synthesize higher alkynes from lower terminal alkynes.

[a confused chiral girl]

thank you for your reply, Priya~

I don't quite understand. Do you mean that the product I drew after reaction with NaNH2 is incorrect? i have not learnt the mechanism of alkynes yet, but I know that NaNH3 would produce a trans alkene from alkyne...

so I drew a trans alkene and tried to draw the mechanism of that with ch3CH2CH2Br..

[english]

This is just an archetypical nucleophilic substitution reaction.  The Br of the alkyl bromide is replaced and a new sigma bond is formed from a lone pair nucleophile produced from the reaction between the alkyne and NH₂^-.

The alkyne's conjugate base is very basic and readily displaces good leaving groups.

You are not reducing the alkyne to an alkene in this reaction (removal of a pi bond).


Amide ion is typically added from a salt in aqueous ammonia (ammonia auto-ionizes to amide ion, much like water to H⁺ and HO^-).  NH₂^- (amide) is a lone pair nucleophile.

Since the alkyl bromide is attached to a primary carbon, the formation of a carbocation isn't probable.  The C atom attached to Br is electrophilic because of the significant dipole moment in that sigma bond.  The C atom is a sigma bond electrophile.  

The C atom is electrophilic because C is relatively low in energy as bonding electrons are withdrawn through inductive forces from Br; in fact, when attached to a C atom, Br is commonly referred to as "bromide," or Br^-, because of the greater occupancy of the bonding electrons around bromine relative to C.  Thus, Br is almost like an ion.  This is also due to the transition state of the step in the reaction sequence that features the removal of Br as a nucleofuge, or leaving group, and the formation of the new sigma bond between C and the lone pair nucleophile, in which the C—Br bond is partially broken and bonding electrons are "sucked" towards Br.

Alkynes can be reduced to alkenes using ammonia, but a alkali metal (such as Na or Li) must be present.  These are called dissolving metal reductions.

[PRIYA1022]

Yes alkynes do undergo Birch Reduction( Na/liq NH3) to give trans alkenes...
But when you have a terminal alkene, the hydrogen attached to the triple bond is acidic and NaNH2 being a strong base results in an acid-base reaction...The salt you get here is sodium alkynide..
 Birch reduction.: The mechanism is a single electron transfer mechanism. requires an alkali metal. and the follow up requires some proton source.. usually an alcohol or NH3 itself..

[PRIYA1022]

The alkyne's conjugate base is very basic and readily displaces good leaving groups.
 As posted by g_english, this is the key to the substitution reaction.