An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br2 to give a compound ‘C’, which on dehydrobromination gives an alkyne ‘D’. On treatment with sodium metal in liquid ammonia one mole of ‘D’ gives one mole of the sodium salt of ‘D’ and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify A,B, C and D. Give the reactions involved.
A react with KOH to give an alkene B
So, A has one Br atom in carbon chain, elimination of HBr by reaction with KOH and give alkene B (C5H10), which is open chain hydrocarbon.
B reacts with Br2 and adds 2 Br atoms to alkene and forming dibromo compound C (C5H10Br2).
C on dehydrobromination gives alkyne D(C5H8).
When D on treatment with sodium metal, liquid NH3 gives sodium salt of D. This shows that alkyne is monsubsituted or side alkyne as shown in the below reaction. D on complete hydrogenation give straight chain alkane C5H12
