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Reaction of cyclohexene with hydrochloric acid
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Leolynn Cauthron
Reaction of cyclohexene with hydrochloric acid
Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:
After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:
A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?) B: rearrangement of the carbocation (why did it happen?) C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)
The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).
Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:
After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:
A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?) B: rearrangement of the carbocation (why did it happen?) C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)
The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).
Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:
After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:
A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)
The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).
Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:
After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:
A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)
The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).
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You're nearly there.
I suggest you draw the 3° carbocation out and think a bit more.
Hint: Hyperconjugation
You're nearly there.
I suggest you draw the 3° carbocation out and think a bit more.
Hint: Hyperconjugation
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Then after the said rearrangement, visualize the lose of proton from the second tertiary carbon atom, giving you your desired product.
Then after the said rearrangement, visualize the lose of proton from the second tertiary carbon atom, giving you your desired product.
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