Question

Which of the sequences below is the best synthesis of (E)-3-hexene?





A. A
B. B
C. C
D. D

Answer 1

Hint: Like alkene compounds, catalytic hydrogenation occurs in the case of alkyne compounds to cis-alkenes and trans-alkenes depending upon reaction conditions. In alkynes, anti-addition of hydrogen occurs in the presence of lithium or sodium metal dissolved in $N{{H}_{3}}$, resulting in a trans (E)-alkene.

Complete step by step solution:
For the synthesis of (E) alkene, metals like sodium, and lithium metal in liquid ammonia solution are used. In this case, an anti-addition of alkyne takes place and is reduced to trans alkene.

In the case of (a), alkyl lithium acts as a nucleophile, attacks carbonyl carbon, and gives an alcohol compound. Then after bromination, a bulky base${{(C{{H}_{3}})}_{3}}C{{O}^{-}}$ abstracts ortho H with respect to leaving group, and two products are formed. Here (E)-2-Hexene is the major product (Hofmann product) due to the presence of
 
For b, ethyne anion attacks ethyl bromide by the ${{S}_{{{N}^{2}}}}$reaction. Again a ${{S}_{{{N}^{2}}}}$ reaction follows in the next step and finally (E) 3-Hexene is formed in presence of sodium in liquid ammonia.

Like a, the same procedure follows for c and here (E) 3-Methyl-hex-3-ene is formed as a major product (Saytzeff’s product).

Like b, in (d) the same procedure follows in every step of the mechanism. Here the major product is (E) Hept-3-ene.

Thus, Option (B) is correct.

Note: Reduction of alkyne occurs also in presence of metal catalysts like Pt or Pd, but it gives alkane as the final product. It is impossible to stop the reduction of alkyne at equivalent addition and differentiate the required alkene. But by adding Pd/$CaC{{O}_{3}}$ mixture with $Pb{{(OAc)}_{2}}$ , quinoline, the further reduction of an alkene can be stopped for the desired product (Z)-alkene.