# Reaction of cyclohexene with hydrochloric acid

What happens when cyclohexene reacts with concentrated hydrochloric acid?
I formed the carbocation intermediate by protonating cyclohexene.
But now I am stuck. I know the end product will be a dimer.
I tried to form a bond between the intermediate and cyclohexene like this. Then through rearrangement the plus charge will shift to 3 degree carbon.
In the answer there is a double bond between the two rings. That's where I am stuck . I can not think anything further.

• Were you told that a dimer is the product of the reaction? No chlorocyclohexane formed? – user55119 Mar 10 '18 at 20:36
• @user55119 In my original question the reaction was with Sulphuric acid. I don't know how it changed to HCl. – V J Mar 11 '18 at 2:14
• @VJ: A way to remove traces of alkenes from alkanes (e.g.; cyclohexene from cyclohexane) is to shake the hydrocarbon with conc. sulfuric acid at room temperature, which forms alkyl sulfuric acids. They separate with the sulfuric acid in the oily layer. – user55119 Mar 11 '18 at 4:05
• @user55119 So you are saying that instead of dimer, alkyl sulfuric acid will be formed? Maybe both will be formed and the dimer will be the major one? Though I don't know why. – V J Mar 11 '18 at 4:58
• @VJ: I believe alkylsulfuric acids are stable at lower temperatures. They can ionize to form carbocations. – user55119 Mar 11 '18 at 14:47

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).

Then through rearrangement the plus charge will shift to 3 degree carbon.

You're nearly there.
I suggest you draw the 3° carbocation out and think a bit more.
Hint: Hyperconjugation

Then after the said rearrangement, visualize the lose of proton from the second tertiary carbon atom, giving you your desired product.

• Why will the tertiary carbon lose the proton? Won't the negative charge become unstable due to the +I effect? – V J Mar 10 '18 at 11:35
• Maybe because the product is stable – Jai Arora Mar 10 '18 at 11:42
• Okay I got my answer. The charge will be delocalised due to empty p orbital. Thank you. – V J Mar 10 '18 at 11:42
• Okay, I'll keep that in mind in the future – Jai Arora Mar 11 '18 at 3:29