Which hydrocarbon will most likely undergo an addition reaction with br2?
1 answer:
Complete question along with options is attached below,
Answer:
Option-B (Ethene) is the correct answer.
Explanation:
In given options except option-B (an alkene) all options are example of alkanes.
Reason for reactivity of Alkene:
Alkenes are more reactive than alkanes because it consists of a pi bond. Thus, the electron density in this pi bond lies below or above the bond axis, hence making it more exposed and accessible for incoming electrophile.
Secondly, the pi bond is weaker than sigma bond and can easily break down. This weakness originates due to less effective overlap of the orbitals forming pi bond as compared to the overlap resulting in the formation of sigma bond.
Therefore, alkenes can add Br₂ through a electrophillic addition reaction, while alkanes fail to show such reaction.
Answer:
Option-B (Ethene) is the correct answer.
Explanation:
In given options except option-B (an alkene) all options are example of alkanes.
Reason for reactivity of Alkene:
Alkenes are more reactive than alkanes because it consists of a pi bond. Thus, the electron density in this pi bond lies below or above the bond axis, hence making it more exposed and accessible for incoming electrophile.
Secondly, the pi bond is weaker than sigma bond and can easily break down. This weakness originates due to less effective overlap of the orbitals forming pi bond as compared to the overlap resulting in the formation of sigma bond.
Therefore, alkenes can add Br₂ through a electrophillic addition reaction, while alkanes fail to show such reaction.
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Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0