Question

Given, $C{{H}_{3}}\left( Br \right)+CH\equiv {{C}^{-}}N{{a}^{+}}\to ?$
Also, do the same for F, Cl and I as the substitute.

Answer 1

The above reaction by the appearance seems to be the alkylation of acetylides with alkyl halides via substitution. The end product to this reaction will be substituted alkyne.

Complete Answer:
Let us study about the alkylation of acetylides with alkyl halides.
Acetylide anions are strong bases and nucleophiles. They perform $S{{N}_{2}}$ reaction with alkyl halides. Thus, they easily displace halides through substitution resulting in product as substituted alkyne. This is typically a substitution reaction as we are forming and breaking bonds on the same carbon.
This reaction is most suitable with methyl or primary halides. Secondary and tertiary halides will not usually support this type of reaction as elimination is more favourable than substitution in these halides.
General reaction-
$R-C\equiv {{C}^{-}}N{{a}^{+}}\xrightarrow{R'-X}R-C\equiv C-R'+NaX$
Now, seeing towards the given illustration;
1. When alkyl halide is methyl fluoride-
Double displacement reaction takes place which is also known as metathesis.
Reactants –Monosodium acetylide and Methyl fluoride.
Products –Methylacetylene and Sodium fluoride.
Reaction –
$CH\equiv {{C}^{-}}N{{a}^{+}}+C{{H}_{3}}F\to CH\equiv C-C{{H}_{3}}+NaF$
2. When alkyl halide is methyl chloride-
Double displacement reaction takes place which is also known as metathesis.
Reactants –Monosodium acetylide and Methyl chloride.
Products –Methylacetylene and Sodium chloride.
Reaction –
$CH\equiv {{C}^{-}}N{{a}^{+}}+C{{H}_{3}}Cl\to CH\equiv C-C{{H}_{3}}+NaCl$
3. When alkyl halide is methyl bromide-
Double displacement reaction takes place which is also known as metathesis.
Reactants –Monosodium acetylide and Methyl bromide.
Products –Methylacetylene and Sodium bromide.
Reaction –
$CH\equiv {{C}^{-}}N{{a}^{+}}+C{{H}_{3}}Br\to CH\equiv C-C{{H}_{3}}+NaBr$
4. When alkyl halide is methyl iodide-
Double displacement reaction takes place which is also known as metathesis.
Reactants –Monosodium acetylide and Methyl iodide.
Products –Methylacetylene and Sodium iodide.
Reaction –
$CH\equiv {{C}^{-}}N{{a}^{+}}+C{{H}_{3}}I\to CH\equiv C-C{{H}_{3}}+NaI$

Note:
Do note that this reaction is considerable for only primary or methyl substituted halides and not for secondary or tertiary halides.