The collision must have sufficient energy as well. If sufficient energy is not present the activation energy will not be over come and the bonds in the reactants will not be broken.
I hope this helps. Let me know if anything is unclear.
If you look closely at each of the four diagrams you would be able to conclude that
<span>D)
Yes. In B and D. In both cases, there is a net force.
In B, there is a net force to the left; in D there is a net force upward.
In A and C, the forces are in equilibrium both in the horizontal and vertical direction.</span>
Clean? I’m pretty sure not sure what it means but.
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have 
as nucleophile. Also, this compound is also in excess. So, we will have as solvent 
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
