Answer:
I think the answer is Toledo
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:
False.
Explanation:
Hexanal is a non-polar compound while water is a polar solvent.
We have the role "Like dissolves like".
So, hexanal is insoluble in water.
Answer:
The reasons why the seemingly floating bubbles disappear was that they tend to loss their latent heat to the water molecules at the surface water.
Explanation:
Heat energy has a considerable effect on the velocity of molecules including water. The water molecules below the container will receive much more heat energy than those above it. This heat energy in the form of specific heat capacity and latent heat that result in the increase in the speed of individual molecules of water and finally to the escape of the molecules to a colder region of the container, in this case the upper region. At the collision of the bottom water to the surface water, they tend to exchange their heat content, the hotter molecules will lose their heat to the cold ones. When the formerly hot molecules encounter this, it will result in lowering the temperature and consequentially to the reduction of their movement, once in the form of bubble, now become ordinary water. This convectional transfer of heat energy will continue until the whole system has a uniform temperature depending on the consistency of the heat source.
