The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
Here we have to draw the four isomers of the compound 3-bromo-4-fluorohexane.
The four isomers of the compound is shown in the figure.
In an organic molecule the chiral -C center is that where four (4) different groups are present. In 3-bromo-4-fluorohexane the 3 and 4 positions are chiral centers. The possible isomers of a molecule can be obtained from the formula 2n. As here 2 chiral centers are present thus number of stereoisomers will be 2×2 = 4.
The four different isomers as shown in the figure are 3R-, 4R-; 3S-, 4S; 3R, 4S and 3S-, 4R- 3-bromo-4-fluorohexane.
In the 3-bromo-4-fluorohexane the functional groups are -Br, C₂H₅, -C₃H₆F and -H for 3-position and -F, -C₂H₅, -C₃H₆ and -H for 4-position respectively.
The priority of the -3 position will be Br > C₃H₆F > C₂H₅ > H and for -4 position F > C₃H₆Br > C₂H₅ > H. If the rotation from the higher priority group to lower is clockwise and anticlockwise then the S- and R- notation are used respectively. However if the -H atom is present at the horizontal position then the notation will be reverse.
Thus the four isomers of the compound is shown.