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Which is the best way of preparing propene from chloroethene?
(A) $C{H_2} = CHCl + C{H_3}Cl\xrightarrow[{Heat}]{{Na/Ether}}$
(B) \[C{H_2} = CHCl + {\left( {C{H_3}} \right)_2}CuLi \to \]
(C) \[C{H_3}Cl + {\left( {C{H_2} = CHCl} \right)_2}CuLi \to \]
(D) \[C{H_2} = CHCl + C{H_3}Li \to \]

Last updated date: 13th Jun 2024
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Hint:We need to know the reactions which involve two molecules (as given in the options) and analyse each given reaction with chloroethane. We have to know that the propene is an alkene with a double bond and chloroethene is a haloalkene with a double bond. Hence it is clear that we need to know the formation of alkene from haloalkene. The haloalkene is allowed to react with another molecule to give propene hence it is known as bimolecular reaction or $Sn_2$ type of reaction. Also it involves chloroethene where Chlorine is a halogen and hence acts as a nucleophile. Also it involves substitution of the chlorine by ethene group hence it is a nucleophilic substitution ($Sn_2$) reaction.

Complete step by step answer:
Let us look at each of the given reactions one by one:
(A) $C{H_2} = CHCl + C{H_3}Cl\xrightarrow[{Heat}]{{Na/Ether}}$
In this reaction, two alkyl halides are allowed to react in Sodium and ether. This is the Wurtz reaction where two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane. Hence propene cannot be formed in this reaction.
(B) \[C{H_2} = CHCl + {\left( {C{H_3}} \right)_2}CuLi \to \]
Clearly, this is the reaction of chloroethene with lithium dimethyl copper. Lithium dimethyl copper is also known as Gilman reagent. They react with organic halides to replace the halide group with an R group to give alkanes. In the above reaction propene is not produced. Therefore, the option B is incorrect.
(C) \[C{H_3}Cl + {\left( {C{H_2} = CHCl} \right)_2}CuLi \to \]
Chloroethene reacts with lithium dichloroethenyl cuprate by $Sn_2$ mechanism to give propene. The $Sn_2$ mechanism is a one-step process in which a nucleophile attacks the substrate, and a leaving group, L departs simultaneously. Cl of chloro acts as a nucleophile.
\[C{H_3}Cl + {\left( {C{H_2} = CHCl} \right)_2}CuLi \to C{H_3} - CH = C{H_2}\]
In the above reaction, propene is formed as a product. Therefore, the option C is correct.
(D) \[C{H_2} = CHCl + C{H_3}Li \to \]
This reaction gives Chloropropene and Li acts as a leaving group.
Hence the correct option is option (C).

We have to note that $Sn_2$ reactions occur most efficiently with $CH_3Cl$ since Cl acts as a strong nucleophile. $Sn_2$ reactions are bimolecular with simultaneous bond-making and bond-breaking steps. Identifying the structure of the molecule formed is also important. As in this case, we needed propene with a double bond. Chloroethene with lithium dimethyl copper gave propane which has a single bond. Therefore, keeping an eye on the bonding is also important and the mechanism involved in it must also be known.