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Question

Name the products formed and write an equation when ethyne is added to the following in an inert solvent :
(a) chlorine, (c) iodine,
(b) bromine, (d) hydrogen.
(e) excess of hydrochloric acid.

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Solution

(a) Ethyne in an inert solvent of carbon tetrachloride adds chlorine to change into 1,2- dichloro ethene with carbon-carbon double bond, and then to an 1,1,2,2-tetrachloro ethane with carbon-carbon single bond.


CHCH + Cl2 Inert solvent CCl4−−−−−−−−−− CHCl=CHCl + excess Cl2 1,2Dichloro etheneCHCl2CHCl2 1,1,2,2Tetrachloro ethane


(b) Ethyne in an inert solvent of carbon tetrachloride adds bromine to change into 1,2-dibromo ethene and then to 1,1,2,2 -tetrabromo ethane .
CHCH + Br2 Inert solvent CCl4−−−−−−−−−− CHBr=CHBr + excess Br2 1,2Dibromo etheneCHCl2CHCl2 1,1,2,2Tetrabromo ethane

(c) Iodine reacts slowly in the presence of alcohol to form 1,2-di-iodo ethene.

CHCH + I2 Inert solvent CCl4−−−−−−−−−− CHI=CHI1,2Di iodo ethene

(d) In the presence of nickel, platinum or palladium ethyne change to ethene and then to ethane.

CHCH + H2 Pd/C−− CH2=CH2 + H2 RaneyNi−−−−− CH3CH3

(e) excess of hydrochloric acid

CHCH + HCl Inert solvent CCl4−−−−−−−−−− CH2=CHCl Chloro ethene excess HCl −−−−−−−−−−−−−−−Slow / Markovnikovs ruleCH3CHCl2 1,1Dichloro ethane


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