An alkyl halide (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.

An alkyl halide (A) reacts with ethanolic KOH to give an alkene ‘B�

1.	An alkyl halide (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.
Answer» e will need to calculate the double bond equivalent for finding the number of bonds in the molecule of Formula to calculate the double bond equivalent isX → Any HalogenHere C = 5, H = 11, X = 1, N = 0So, As the double bond equivalent is 0, this implies that the compound is saturated and has no unsaturation.The below reaction SUGGESTS that (D) is a terminal alkyne. This implies that there should be triple bond at the END of the chain.Alkyne (D) with triple bond could be either formula as shown belowAs it is given that alkyne 'D' would yield straight chain alkane on hydrogenation. Thus (I) is the structure of D.Hence, the STRUCTURES of A, B, C and D is the structure of A and it is 'bromopentane' is the structure of B and it is '1-pentene' is the structure of C and it is '2,2-dibromopentane' is the structure of D and it is '1-pentyne'