Final answer:
The final product after treating 1-heptyne with sodium hydride, iodomethane, and excess Cl2 is 1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,8-heptadecafluoro-8-iodooctane, a perhalogenated compound.
Step-by-step explanation:
When 1-heptyne is treated with sodium hydride, it removes an acidic hydrogen atom from the terminal alkyne to generate an acetylide anion. The addition of iodomethane then alkylates this anion, resulting in a longer carbon chain with a iodomethyl group at the end, transforming 1-heptyne into 8-iodo-1-octyne. The subsequent treatment with excess Cl2 leads to halogenation, where chlorine atoms are added to all available positions. In this case, the triple bond in 8-iodo-1-octyne would be attacked, resulting in a fully substituted alkane, 1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,8-heptadecafluoro-8-iodooctane, as the final product.
This final compound, due to the addition of chlorine atoms at every possible position, is a perhalogenated compound, in which all the hydrogens have been replaced by halogens.