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Reaction mechanism for meso-stilbene dibromide to produce diphenylacetylene?
Wiki User
∙ 14y ago
Its a double elimination.
Since these carbons are secondary they can undergo E1 or E2, depending on the strength of the base used, if the adjacent hydrogen is stericly hindered (blocked by a bulky group).
Using a strong base such as -OH (from K+OH-)
You have the Bromine attack the K+, making a carbocation, then have the H of the adjacent carbon leave (donate) its electron to the carbocation creating a double bond, and have the -OH attach to that leaving Hydrogen; forming water as a result.
This leaves you with a bromine attached to one carbon on the double bond, and a Hydrogen on the other. So you just repeat those steps again. Leaving you with the Diphenlacetlyene.
Which is just PhC(triple bond)CPh + HOH + 2KBr
Wiki User
∙ 14y ago
Meso-stilbene dibromide can undergo a double elimination reaction to form diphenylacetylene. The first step involves deprotonation of one of the bromine atoms by a strong base, producing a carbanion. The carbanion then eliminates the other bromine atom in an E2 elimination reaction, forming the alkyne product.
Wiki User
∙ 13y agoYou have to form a grignard reagent. If you have one bromobenzene react with Mg in ethanol, you will form a grignard reagent that will react with another bromobenzene to form a biphenyl.
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Double dehydrohalogenation on an alkane usually requires a very strong base like NaNH2 why is KOH strong enough to do this reaction with stilbene dibromide to obtain diphenylacetylene?
The reactivity of the hydrogens in stilbene bisbromide is significantly enhanced due to the presence of the highly electronegative bromine atoms. This makes the hydrogens more acidic and thus easier to deprotonate, even in the presence of a weaker base like KOH. Additionally, the aromatic rings in the product, diphenylacetylene, stabilize the intermediate carbanion formed during the dehydrohalogenation reaction, promoting the use of a milder base.
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Double dehydrohalogenation on an alkane usually requires a very strong base like NaNH2 why is KOH strong enough to do this reaction with stilbene dibromide to obtain diphenylacetylene?
The reactivity of the hydrogens in stilbene bisbromide is significantly enhanced due to the presence of the highly electronegative bromine atoms. This makes the hydrogens more acidic and thus easier to deprotonate, even in the presence of a weaker base like KOH. Additionally, the aromatic rings in the product, diphenylacetylene, stabilize the intermediate carbanion formed during the dehydrohalogenation reaction, promoting the use of a milder base.
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Each step in a reaction mechanism is referred to as an elementary step.
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positive ffedback
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You can determine the rate of a reaction mechanism having fast equilibrium by the number of hydrogen ions that are present. If the reaction has a high number of hydrogen ions then the reaction will have fast equilibrium.
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What is the mechanism that your body uses to create polymers from monomers?
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