D) The movement of electrons between atoms
Answer:
Explanation:
In the following reaction we have shown an example of aromatic substitution reaction .
C₆H₆ + RCl = C₆H₅R + HCl
This reaction takes place in the presence of catalyst like AlCl₃ which is a lewis acid .
First of all formation of carbocation is made as follows .
RCl + AlCl₃ = R⁺ + AlCl₄⁻
This R⁺ is carbocation which is also called electrophile . It attacks the ring to get attached with it .
C₆H₆ + R⁺ = C₆H₅R⁺H.
The complex formed is unstable , though it is stabilized by resonance effect . In the last step H⁺ is kicked out of the ring . The driving force that does it is the steric hindrance due to presence of two adjacent group of H and R⁺ at the same place . Second driving force is attack by the base AlCl₄⁻ that had been formed earlier . It acts as base and it extracts proton ( H⁺ ) from the ring .
C₆H₅R⁺H + AlCl₄⁻ = C₆H₆ + AlCl₃ + HCl .
The formation of a stable product C₆H₆ also drives the reaction to form this product .
Answer:
2.8
Explanation:
First, we will calculate the molarity of the acetylsalicylic acid solution.
M = mass of solute (g) / molar mass of solute × volume of solution (L)
M = 0.327 g / 180.158 g/mol × 0.237 L
M = 7.66 × 10⁻³ M
For a weak acid such as acetylsalicylic acid, we can find the concentration of H⁺ using the following expression.
[H⁺] = √(Ca × Ka)
where,
Ca: concentration of the acid
Ka: acid dissociation constant
[H⁺] = √(7.66 × 10⁻³ × 3.3 × 10⁻⁴)
[H⁺] = 1.6 × 10⁻³ M
The pH is:
pH = -log [H⁺]
pH = -log 1.6 × 10⁻³ = 2.8
Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
