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.
[![enter image description here](https://i.stack.imgur.com/r8pO1.jpg)](https://i.stack.imgur.com/r8pO1.jpg)
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.




Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:


[![enter image description here](https://i.stack.imgur.com/lrJim.png)](https://i.stack.imgur.com/lrJim.png)


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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> Then through rearrangement the plus charge will shift to 3 degree carbon.
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You're nearly there.  

I suggest you draw the 3° carbocation out and think a bit more.  

Hint: Hyperconjugation 



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>  ![enter image description here](https://i.stack.imgur.com/Na8ml.jpg)
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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.


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