Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
Answer:
The answer to your question is: yield = 56.27%
Explanation:
Data
CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br
18.54 ml 1-butanol 15.65 g of 1-bromobutane
% yield = ?
density = 0.81 g/ml
MM = 74 g 1- butanol
MM = 137 g 1-bromobutane
Process
Calculate mass of 1- butanol
density = mass/volume
mass = density x volume
mass = 0.81 x 18.54
mass = 15.02 g of 1-butanol
Theoretical yield
74 g of 1- butanol ----------------- 137 g of 1-bromobutane
15.02 g of 1- butanol ------------- x
x = (15.02 x 137) / 74
x = 27.81 g of 1-bromobutane
% yield = experimental yield / theoretical yield x 100
% yield = 15.65 / 27.81 x 100
% yield = 56.28
Answer:
V₂ = 18.13 L
Explanation:
Given data:
Mole of gas = 1 mol
Initial temperature = 273 K
Initial pressure = 1 atm
Final volume = ?
Final temperature = -41°C (-41+273 = 232 K)
Final pressure = 805 mmHg (805/760 = 1.05 atm)
Solution:
First of all we will calculate the initial volume of gas.
PV = nRT
V = nRT/P
V = 1 mol × 0.0821 mol.L/atm.K × 273 K / 1 atm
V = 22.4 L/atm / 1 atm
V = 22.4 L ( initial volume)
Now we will determine the final volume by using equation,
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Now we will put the values.
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 22.4 L × 232 K / 273 K × 1.05 atm
V₂ = 5196.8 atm .L. K / 286.65 atm.K
V₂ = 18.13 L
Answer:
Mark this as brainliest please
Explanation:
Sulfur trioxide (SO3) gas reacts with water (H2O) to form sulfuric acid (H2SO4).
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices:
A.) an unpaired electron,
<span>B.) a nonbonding electron pair, </span>
<span>C.) a bonding electron pair, </span>
<span>D.) a high value of electronegativity
</span>
The correct answer is B.) a nonbonding electron pair,
