Give the products with their stereoisomers, if any.
a) `Me--=-Me overset((Na)//"liq".NH_(3))underset(C_(2)H_(5)OH)to`
b) `Me--=-Meoverset(H_(2)//"Poisoned Pd")to`

RCH=CH_(2) overset(Na//NH_(2)(l))underset(C_(2)H_(5)OH)to RCH_(2)CH_(3) The reaction is called as:

Me - -=-H overset(NaNH_(2))underset(+liq.NH_(3))rarr (C ) overset(H_(3)O^(o+))rarr (D) the final product (D) is:

Me - -=-H overset(NaNH_(2))underset(+liq.NH_(3))rarr (C ) overset(H_(3)O^(o+))rarr (D) the final product (D) is:

In the reaction Me-C-=C-E to overset(Na//liq.NH_(3))(to)Poverset(Br_(2))underset(C Cl_(4))(to)((Q)) , then Q is

R-CH=CH_(2)underset(C_(2)H_(5)OH)overset(Na//NH_(3)(I))toRCH_(2)CH_(3) is called

In the reaction Me –C -=C - Et overset( Na// lig.NH_3) to P underset(C Cl_4) overset(Br_2)to (Q), then Q is :

In the reaction H- C -= CH overset( (1) NaNH_(2) // liq. NH_(3))underset( (2) CH_(3) CH_(2) Br)(to) X overset( (1) Na NH_(2) // liq. NH_(3))underset((2) CH_(3) CH_(2) Br) (to) Y X and Y are -

Arrange the free radicals according to their stability : Me overset(*)(C)H_(2) , overset(*)(C) H_(3) , Me_(2)overset(*)(C)H , Me_(3)overset(*)(C)