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.
Answer:
Explanation:
In CF4 and NF3, the valence electron groups on the central C and N atoms have a tetrahedral arrangement. The shapes of the molecules are determined by the number of bonding and nonbonding of electrons: since CF4 has four bonded atom(s) and zero lone pair(s) of electrons, the shape is tetrahedral.
155,500
I did this to the best of my ability. I have a hard time comprehending things sometimes so I’m so so so sorry if it’s wrong
