Question

Consider the following reactions
(I) $Me-\underset{\left(A\right)}{\equiv }-Me\stackrel{H_2+}{\underset{Pd+BaS{O}_4}{⟶}}\underset{\left(B\right)}{?}\stackrel{D_2/Pd/C}{⟶}\left(C\right)$
(ll) $Me-\underset{\left(A\right)}{\equiv }-Me\stackrel{Na+Liq.N{H}_3}{\underset{+EtOH}{⟶}}\underset{\left(D\right)}{?}\stackrel{D_2/Pd/C}{⟶}\left(E\right)$
Which of the following statements are correct?

• A. (B) is cis-but-2-ene and (D) is trans-but-2-ene
• B. (B) is trans-but-2-ene and (D) is cis-but-2-ene
• C. (C) is meso form and (E) is racentic form
• D. (C) is racentic form and (E) is meso form

Answer 1

Answer: (A), (C)

The correct options are
A (B) is cis-but-2-ene and (D) is trans-but-2-ene
C (C) is meso form and (E) is racentic form

• 2-butyne on reaction with $H_2$/poisoned Pd gives cis-2-butene as syn addition takes place.
• Whereas with Na in liq. ammonia, anti addition takes place and we get a trans product.
• Cis form with syn-addition gives meso form and with anti-addition gives the racemic mixture.
• SImilarly, trans form with anti-addition gives meso form and with syn-addition gives the racemic mixture.
• So (C) is meso form and (E) is the racemic mixture.