Question

19.(b-2b) qaysi modda eitlen bilan ni katalizatorligida qizdirilganda reaktsiyaga kirishadi?
a) suv bugi b) azot
c) vodorod d) metan

20.(b-2b) pentadiyen-1,4 molekulasida jami nechta pi bog mavjud?
a) 1 b) 2
c) 3 d) 4

21.(q-2,6b) agar 14 g propilen c₂h₆ toliq gidrogenlansa, hosil bolgan propanning massasi qancha boladi?
a) 14,7 g b) 29,6 g
c) 7,4 g d) 14 g

1. (q-2,6b) rux metali tasirida 3-metilbuten-1 hosil qiluvchi c₅h₁₀br₂ tarkibli digalogenli alkan necha xil bolishi mumkin?

a) 3 b) 2
c) 1 d) 4

1. (q-2,6b) quyidagi uglevodorodni sistematik nomenklatura boyicha nomlang:

ch₃ch₂ch(ch₃)c ≡ ch
a) 3-metilpentin-5 b) 3-metilpentin-4
c) 2-etilbutin-1 d) 3-metilpentin-1

1. (q-2,6b) etin c₂h₂ molekulasidagi uglerod atomlari orbitalari qanday gibridlanish holatida boladi?

a) sp² b)sp
c) sp² va sp³ d) sp³

Explanation:

Question 19 (19.(B-2b))

To determine which substance reacts with ethene (C₂H₄) over a Ni catalyst upon heating, we analyze each option:

• A) Water vapor (Suv bug’i): Ethene reacts with water (hydration) typically over acid catalysts (e.g., H₃PO₄), not Ni.
• B) Nitrogen (Azot): N₂ is inert and does not react with ethene under these conditions.
• C) Hydrogen (Vodorod): Ethene undergoes catalytic hydrogenation (addition of H₂) over Ni (or Pd, Pt) at elevated temperatures. This is a standard reaction: \( \text{C}_2\text{H}_4 + \text{H}_2 \xrightarrow[\Delta]{\text{Ni}} \text{C}_2\text{H}_6 \).
• D) Methane (Metan): Methane is a saturated hydrocarbon and does not react with ethene under these conditions.

Thus, hydrogen (Vodorod) reacts with ethene over Ni upon heating.

Pentadiene-1,4 has the structure \( \text{CH}_2=\text{CH}-\text{CH}=\text{CH}-\text{CH}_3 \). A double bond (\( \text{C}=\text{C} \)) contains one \( \sigma \) and one \( \pi \) bond. Pentadiene-1,4 has two double bonds (between C1-C2 and C3-C4). Each double bond contributes one \( \pi \) bond, so total \( \pi \) bonds = 2.

Step 1: Balanced Reaction

Propene (\( \text{C}_3\text{H}_6 \)) undergoes hydrogenation: \( \text{C}_3\text{H}_6 + \text{H}_2
ightarrow \text{C}_3\text{H}_8 \).

Step 2: Molar Masses

• Molar mass of \( \text{C}_3\text{H}_6 \): \( 3(12.01) + 6(1.008) = 42.08 \, \text{g/mol} \).
• Molar mass of \( \text{C}_3\text{H}_8 \): \( 3(12.01) + 8(1.008) = 44.10 \, \text{g/mol} \).

Step 3: Moles of \( \text{C}_3\text{H}_6 \)

Given mass = 14 g. Moles = \( \frac{14 \, \text{g}}{42.08 \, \text{g/mol}} \approx 0.3327 \, \text{mol} \).

Step 4: Moles of \( \text{C}_3\text{H}_8 \)

From the reaction, 1 mol \( \text{C}_3\text{H}_6 \) produces 1 mol \( \text{C}_3\text{H}_8 \). Thus, moles of \( \text{C}_3\text{H}_8 = 0.3327 \, \text{mol} \).

Step 5: Mass of \( \text{C}_3\text{H}_8 \)

Mass = moles × molar mass = \( 0.3327 \, \text{mol} × 44.10 \, \text{g/mol} \approx 14.7 \, \text{g} \).

Answer:

C. Vodorod

Question 20 (20.(B-2b))