The equation between 2-chloro-2-methylpropane and sodium hydroxide?

Answer 1

C3H7Cl + NaOH -> C3H7OH + NaCl

This is a substitution reaction, OH- ion acts as a nucleophile.

C-Cl bond is polar and Cl leaves as Cl- and carbon has + charge, which the OH- can attack = nucleophilic substitution reaction.

Answer 2

This is a question they seem to ask a lot in organic I labs for preparation of tert-pentyl chloride. You're supposed to wash the product with water and then with sodium bicarbonate to neutralize any unreacted HCL you might have. Then they ask why not just use a nice strong base like sodium hydroxide? Why not indeed. Well, if you look back to your alkene chapter, you'll remember that reacting an alkyl halide (your product) with a strong base will give you an alkene. It's called dehydrohalogenation. You don't want that. All you want to do is mop up any leftover acid. I just did one for the butane variant, but the principle is the same

The formula is: C5H11Cl + NaOH = C5H10 +NaCl +H2O

You go from 2-chloro-2-methylbutane (in my case) to 2-methyl-2-butene. Remember that pulling halogen off this way to form a double bond can lead to more than one product (ie where the new double bond forms). Zaitzev's rule says the predominant product will be the one where the double bond connects to the most substituted carbon.

In my case that means 2-methyl-2-butene rather than 2-methyl-1-butene. In your case, it would be 2-methyl-2-propene. The formula should work out the same too, a 1:1 mole ratio of everything, but I'm too tired to work that out right now. Just repeat what I did for butane-butene and make sure everything adds up.

Answer 3

The chlorine and OH swap

form propyl alcohol and sodium chloride

but the reaction need water and some heat to go on