Answer:
South American
Explanation:
When you look at a map of plates, only South American forms a boundary with the African plate out of those specific plates
<u> electrical energy to chemical energy</u>
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Answer:
<h2>0.15 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities.
From the question we have
We have the final answer as
<h3>0.15 moles</h3>
Hope this helps you
This reaction would give rise to two products.
- 2-bromo-3-methylhexane, and
- 3-bromo-3-methylhexane.
However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.
The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton and a bromide ion .
The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.
Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.
Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.
The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.