Question

Statement: When $2 - $ butyne reacts with triethylamine followed by $B{r_2}$ then meso-$2,3 - $ dibromobutane is produced.
If the statement is true enter $1$, else enter $0$.

Answer 1

Hint: Meso compounds: Those compounds which have a superimposable mirror image. They are the compounds which have at least two chiral centres and centre plane of symmetry. If the plane of symmetry is not there then the compound will not be a meso compound.

Complete step by step solution:
First let us talk about chiral centres and then the plane of symmetry.
Chiral centres: These are those centres which have different atoms or groups of atoms around the central atom. For example: Chiral carbon is that carbon to which all the four groups attached to the carbon are different.
Plane of symmetry: It is defined as the plane which is when cut by the centre gives two identical compounds i.e. it is defined as the imaginary plane which bisects the molecule into two halves which are mirror images of each other.
Meso compounds: Those compounds which have a superimposable mirror image. They are the compounds which have at least two chiral centres and centre plane of symmetry. If the plane of symmetry is not there then the compound will not be a meso compound.
Enantiomers: These are those compounds in which all the carbon atoms or chiral centres have different geometries.
Now when $2 - $ butyne reacts with sodium in liquid ammonia followed by $B{r_2}$ then meso-$2,3 - $ dibromobutane is produced. The reaction is as:

${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}} \equiv {\text{CC}}{{\text{H}}_{\text{3}}}\xrightarrow{{Na}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = CHC}}{{\text{H}}_{\text{3}}}\xrightarrow{{B{r_2}}}{\text{meso 2,3 - dibromobutane}}{\text{.}}$

When $2 - $ butyne reacts with sodium in liquid ammonia then first reduction of $2 - $ butyne to $2 - $butene takes place and after that when $2 - $butene reacts with $B{r_2}$ then meso-$2,3 - $ dibromobutane is produced.
And in the question we are given with the reaction of $2 - $ butyne reacts with triethylamine followed by $B{r_2}$ then no meso-$2,3 - $ dibromobutane is produced. Because when $2 - $ butyne reacts with triethylamine then reduction of $2 - $ butyne is not possible and bromine will not be able to convert it into meso-$2,3 - $ dibromobutane.

Hence, the answer is false. So, enter $0$.

Note: Racemic mixture: It is defined as the mixture which contains equal amounts of opposite enantiomers. And hence they are not optically active (i.e. they do not rotate the plane of plane polarised light).