Path II is feasible because sodium alkynide (D) `(Me_3C-C-=CNa)` is more nucleophilic than sodium alkynide (B) `(Me-C-=C-Na)` and undergoes `SN^2` reaction with `1^@` `RX(CH_3I)` to give (E), whereas sodium alkynide (B) acts as a base and reacts with `3^@RX(Me_3C-Br)` and undergoes `E2` elimination reaction to give alkyne `(Me-C-=CH)` and alkene
.
Direct method to solve such problems:
Write the product which has to be synthesised and cleave the triple bond from either side and then select the alkynide part having `2^@` or `3^@`-atom chain and the other part having `1^@ C`-atom chain as RX.
Rule:
In ethyne, first introduce `2^@` or `3^@` RX into ethynide ion and then in the second stage introduce `1^@` RX into alkynide ion to get the proper alkyne.
If the reverse order is followed, the elimination product will be obtained.
`HC-=C^(Θ)underset(1^@RX)rarrHC-=C-R`
`underset((ii) 3^@ or 2^@RX)overset((i) NaNH_2)rarrAlken e+Alkyn e`
