The coefficients of the substances give you the ratio of the number of moles.
The 4 before the H2O and the 5 before the HCl tell you that for every 5 moles of HCl created, 4 moles of H2O had to react.
Therefore:
can be used to find the moles of H2O that react.
-dB/dt = kAB = k(2B)(B) = 2kB^2
-dB/B^2 = 2kdt
Integrating: 1/B - 1/(B_0) = 2kt
At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.
1/45 - 1/50 = 2(k)(10)
k = 1.11 x 10^-4
Then substituting this value of k with t = 40:
1/B - 1/50 = 2(1.11 x 10^-4)(40)
1/B - 1/50 = 0.008889
1/B = 0.028889
B = 34.62 g remaining
Therefore, 50 - 34.62 = 15.38 g of B have been consumed.
Doubling, 30.76 g of A have been consumed.
This means that 15.38 + 30.76 = 46.15 g of C have been formed.
Here we have to draw the major product in the acid catalysed hydration reaction of 4-ethyl-3,3-dimethyl-1-hexene.
The 4-ethyl-3,3-dimethyl-1-hexene converts to 2-hydroxy-4-ethyl-3,3-dimethyl-1-hexane as a major product by acid catalyzed hydration reaction.
The acid catalyzed hydration of an alkene is the Sn¹ reaction. Where in the first step a carbocation is generated. The stability of the carbocation depends upon the position of the neighboring group having +I inductive effect.
In the next step the water molecule attack the carbocation and the corresponding alcohol is produced.
In 4-ethyl-3,3-dimethyl-1-hexene the carbocation formed in the C₂ position which is more stable than the C₁ position due to presence of the dimethyl and ethyl group in the neighboring position which have strong +I inductive effect. This is absence in C₁ position.
In the next step the water molecule attack the C₂ position to form the alcohol.
4-ethyl-3,3-dimethyl-1-hexene converts to 2-hydroxy-4-ethyl-3,3-dimethyl-1-hexane by acid catalyzed hydration reaction which is the major product along with 1-hydroxy-4-ethyl-3,3-dimethyl-1-hexane as a minor product.
The reaction mechanism is shown in the image.
Solution: A Bronsted-Lowry Base is defined as a solution that accept proton/s. hydroxide ion, OH^{-} is a Bronted-Lowry base because it have one negative charge and it can accept 1 proton in the form of H^{+} ion.
Lewis base is defined as an electron rich species that can donate its electron pair to electron deficient species. Here, due to negative charge, hydroxide ion is electron rich and can attack on electron deficient carbon atom or carbocation.
The organic product formed by the reaction of hydroxide ion OH^{-} and carbocation CH_{3}^{+} will be as follows:
OH^{-}+CH_{3}^{+}\rightarrow CH_{3}OH
Thus, the organic product formed will be alcohol.
A displacement reaction Would occur in this situation
