An alkyl halide C_(5)H_(11)Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br_(2) to give a compound 'C', which on dehydro-bromination gives an alkyne 'D'. On treatment with sodium metal inliquid 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.

An alkyl halide C_(5)H_(11)Br (A) reacts with ethanolic KOH to give an

1.	An alkyl halide C_(5)H_(11)Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br_(2) to give a compound 'C', which on dehydro-bromination gives an alkyne 'D'. On treatment with sodium metal inliquid 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.
Answer» Solution :There actions involved in identification of A, B, C and D are as follows : `C_(5)H_(11)Br(A) + alc. KOH rarr C_(5)H_(10)(B)` `C_(5)H_(10)(B)+ Br_(2)//CS_(2) rarr C_(5)H_(10)Br_(2)(C)` `C_(5)H_(10)Br_(2)(C) + alc. KOH rarr C_(5)H_(8)(D)` Alkyne. `2C_(5)H_(8) + 2Na rarr 2C_(5)H_(7)Na + H_(2)` HYDROGENATION of alkyne (D) gives straight chain alkane hence all the compounds A, B, C and D must be straight chain compounds. Alkyne gives sodium alkenyde which proves D is terminalalkyne.