Question

The trans alkenes are formed by the reduction of alkynes with

• A. ${\mathrm{NaBH}}_4$
• B. $\mathrm{Na}/\mathrm{liq}.{\mathrm{NH}}_3$
• C. $\mathrm{Sn}–\mathrm{HCl}$
• D. ${\mathrm{H}}_2-\mathrm{Pd}/\mathrm{C},{\mathrm{BaSO}}_4$

Answer 1

The correct option is B

$\mathrm{Na}/\mathrm{liq}.{\mathrm{NH}}_3$

The explanation for the correct option:

B: $\mathrm{Na}/\mathrm{liq}.{\mathrm{NH}}_3$

Trans-Alkenes:

• The isomer is a cis alkene if two groups (connected to the double bond's carbons) are on the same side of the double bond.
• The isomer is a trans alkene if the two groups are on different sides of the double bond.

The explanation for Incorrect options:

A: ${\mathrm{NaBH}}_4$

• The main use of ${\mathrm{NaBH}}_4$ is the synthesis of sodium dithionite ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_4$, a bleaching agent for wood pulp, from ${\mathrm{SO}}_2$.
• With the use of this substance, aldehydes and ketones may be changed into equivalent alcohols.

C: $\mathrm{Sn}–\mathrm{HCl}$

• It can be used as a source of electrons i.e. a reducing agent and protons.

D: ${\mathrm{H}}_2-\mathrm{Pd}/\mathrm{C},{\mathrm{BaSO}}_4$

• Without further reduction into alkanes, it is utilized to hydrogenate alkynes to cis alkenes.
• When the triple bond is hydrogenated over Lindlar's catalyst, $\mathrm{Pd}/{\mathrm{BaSO}}_4$, it mostly produces cis alkene.

Therefore, Option B $\mathrm{Na}/\mathrm{liq}.{\mathrm{NH}}_3$ is the correct option.