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gizmo_the_mogwai [7]
3 years ago
7

Propose a mechanism for the following reaction.

Chemistry
1 answer:
raketka [301]3 years ago
5 0
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation. 
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​What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water? 
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The concentration of KBr is  C = 0.07899 \ mol   L^{-1}

Explanation:

From the question we are told that

      The mass of KBr is  m_{KBr}  = 2.35 \ g

       The molar mass of KBr is  M_{KBr} =  119 g/mol

       Volume of water is V = 250 \ mL = 250 *10^{-3} =  0.250 \ L

This implies that the volume of  the solution is  V = 250 mL

The number of moles of KBr is

         n = \frac{m_{KBr}}{M_{KBr}}

Substituting values

         n =  \frac{2.35}{119}

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The concentration of KBr is mathematically represented as

                C = \frac{0.01975}{0.250}

                C = 0.07899 \ mol   L^{-1}

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