Preparation of Alkynes
Forge the triple bond. Master double dehydrohalogenation, conquer the Sodamide ($NaNH_2$) reagent trap, and learn to build larger carbon skeletons via acetylide alkylation.
Module Focus: Creating the Triple Bond
Synthesizing an alkyne ($C \equiv C$) requires generating two $\pi$ bonds between adjacent carbon atoms. This usually demands a double elimination reaction, removing four atoms total from a saturated precursor. Because introducing the second $\pi$ bond is significantly harder than the first, typical bases fail, requiring the use of extraordinarily strong reagents.
1. From Calcium Carbide (Industrial Method)
Ethyne (Acetylene) is manufactured on a large scale by the action of water on Calcium Carbide ($CaC_2$). The carbide itself is prepared by heating limestone and coke.
Step 1: Preparation of Quicklime
Step 2: Preparation of Calcium Carbide
Step 3: Hydrolysis to yield Ethyne
2. From Dihalides (Double Dehydrohalogenation)
Treating a vicinal dihalide (halogens on adjacent carbons) or a geminal dihalide (halogens on the same carbon) with a strong base triggers two successive $\beta$-elimination ($E2$) reactions.
Why does alcoholic KOH fail to complete the reaction?
The first elimination with alcoholic KOH easily produces a vinyl halide ($CH_2=CH-Br$). However, alcoholic KOH is not strong enough to remove the second $HBr$ molecule.
Therefore, a much stronger base, Sodamide ($NaNH_2$), is strictly required to force the second elimination.
3. Dehalogenation Methods
Heating a 1,1,2,2-tetrahaloalkane with Zinc dust in alcohol strips all four halogen atoms, forming two new $\pi$ bonds.
Heating two molecules of Iodoform ($CHI_3$) or Chloroform ($CHCl_3$) with Silver powder ($Ag$) yields Ethyne.
4. From Terminal Alkynes (Alkylation)
This is the most important reaction for synthesizing higher alkynes in organic conversions. It exploits the unique acidic nature of terminal alkynes ($RC \equiv C-H$). The $sp$ hybridized carbon is highly electronegative (50% s-character) and can release $H^+$ to a very strong base.
The acetylide ion ($RC \equiv C^-$) is not only a strong nucleophile, but also a very strong base.
- If a Primary ($1^\circ$) Alkyl Halide is used $\rightarrow$ Substitution ($S_N2$) occurs to give higher alkynes.
- If a Secondary ($2^\circ$) or Tertiary ($3^\circ$) Alkyl Halide is used $\rightarrow$ Elimination ($E2$) dominates due to steric hindrance. Instead of an alkyne, you get an Alkene!
NEET Grand Test: Alkyne Preparation
15 High-Yield Questions testing reagents, the $NaNH_2$ trap, and alkylation synthesis flowcharts.
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