Q.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.

Concept: Structural Isomerism and Terminal Alkyne Identification

The key is recognising that alkyne D releases only half a mole of H₂ with sodium in liquid ammonia, diagnostic of a terminal alkyne (one acidic ≡C–H). Complete hydrogenation yields a straight-chain alkane, so D must be 1-pentyne.

Reasoning:

1. Alkyne D = 1-pentyne, HC≡C–CH₂–CH₂–CH₃

   Terminal alkynes react: $\ce{HC≡C-R + Na/NH3(l) -> NaC≡C-R + 1/2 H2}$. Hydrogenation gives n-pentane (straight chain).

2. Alkene B = 1-pentene, CH₂=CH–CH₂–CH₂–CH₃

   Bromination: $\ce{CH2=CH-CH2CH2CH3 + Br2 -> CH2Br-CHBr-CH2CH2CH3}$ (compound C, 1,2-dibromopentane).

3. Compound C loses 2 HBr (dehydrobromination) to form the triple bond in D.

4. Alkyl halide A = 1-bromopentane, CH₃–CH₂–CH₂–CH₂–CH₂Br …

The heart of this problem lies in recognising structural isomerism and the characteristic reactions of alkynes, particularly terminal alkynes. The clue that "one mole of D gives one mole of sodium salt and half a mole of H₂" tells us D is a terminal alkyne—only terminal alkynes have an acidic hydrogen that reacts with strong bases like sodium in liquid ammonia. The requirement that complete hydrogenation yields a straight-chain alkane locks in the carbon skeleton.

Let's trace the reaction sequence backward and forward to identify each compound.

Understanding the Key Clue

The reaction of D with sodium metal in liquid ammonia is diagnostic:

This stoichiometry—one mole of alkyne producing half a mole of hydrogen—means D has exactly one acidic hydrogen. Terminal alkynes ($\ce{RC≡CH}$) have a hydrogen on the sp-hybridised carbon with $\mathrm{p}K_\mathrm{a} \approx 25$, acidic enough to be deprotonated by sodium metal. Internal alkynes ($\ce{RC≡CR'}$) have no such hydrogen and do not react this way.

Therefore, D is a terminal alkyne.

Working Backward from D

Since complete hydrogenation of D gives a straight-chain alkane and D is $\ce{C5H8}$ (an alkyne with five carbons), the only straight-chain terminal alkyne with five carbons is:

Hydrogenation adds two moles of H₂:

Identifying C

C is formed by adding Br₂ to alkene B, and dehydrobromination of C gives D (1-pentyne). Dehydrobromination removes two HBr molecules from a vicinal dibromide to form a triple bond. Working backward, C must be a 1,2-dibromide:

The dehydrobromination proceeds:

Identifying B

B is the alkene formed by adding Br₂ to give C. Since C is 1,2-dibromopentane, B must be:

Bromination of the double bond:

Identifying A …