Question

$CH _{3} C \equiv CCH _{3} \xrightarrow{H 2 / P t} A \xrightarrow{D_{2} / P t} B$ The compounds $A$ and $B$, respectively are

• A. cis-butene-2 and rac-2, 3-dideuterobutane
• B. trans-butene-2 and rac-2, 3-dideuterobutane
• C. cis-butene-2 and meso-2, 3-dideuterobutane
• D. trans-butene-2 and meso-2, 3-dideuterobutane

Answer 1

Answer: (C)

cis-butene-2 and meso-2, 3-dideuterobutane

Answer 2

. Alkynes are catalytically hydrogenated to produce cis-alkene, which is then further hydrogenated in the presence of platinum to produce meso-2,3-dideuterobutane. The conversion of is the net response of C≡C to C-C with the addition of two molecules of H2 and D2

** Alkynes** may be hydrogenated by simply adding hydrogen atoms across their triple bonds. Alkane results from the formation of a carbon-carbon double bond, while alkene results from the formation of a carbon-carbon single bond.

A bond will be lost if hydrogen atoms are added across a double bond.

The result of hydrogenating the provided alkyne is

Furthermore, the double bond in cis-2-butene is broken down into a single bond and an alkane is produced when deuterium is added in the presence of a platinum catalyst.

Meso compounds are substances that contain a centre of symmetry and are optically active, meaning they exhibit a non-superimposable mirror image.

A racemic mixture is a collection of substances with both a (R) and a (S) structure.

The compound produced cannot be a racemic combination because it displays optical activity.

The resultant product is a meso compound in this case because it possesses a plane of symmetry and a non-superimposable mirror image that makes it optically active

So, the correct answer is cis-butene-2 and meso-2,3-dideuterobutane.