Let's solve each question step by step:
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Q1. Which of the following has the longest carbon chain?
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Neopentane: \( \text{C}_5\text{H}_{12} \) (branched structure)
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Isopentane: \( \text{C}_5\text{H}_{12} \) (branched structure)
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2-Methyl pentane: \( \text{C}_6\text{H}_{14} \) (longer carbon chain compared to neopentane and isopentane)
Answer: (c) 2-methyl pentane
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Q2. Which of the following is not an isomer of butyne?
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But-2-yne: \( \text{CH}_3\text{C}\equiv\text{CCH}_3 \) (terminal alkyn)
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But-2-ene: \( \text{CH}_3\text{CH}=\text{CHCH}_3 \) (alkene, not an isomer of butyne)
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But-1,3-diene: \( \text{CH}_2=\text{CHCH}=\text{CH}_2 \) (diene, not an isomer of butyne)
Answer: (d) All of the above
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Q3. Which of the following is not an electrophile?
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Carbocation: Electrophile (positively charged carbon)
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Thioalcohol: Not an electrophile (neutral molecule)
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Carbene: Electrophile (electron-deficient carbon)
Answer: (b) Thioalcohol
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Q4. Which of the following is an electrophile?
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Cyanide: Neutral molecule (not an electrophile)
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Aluminium chloride: Lewis acid (electrophile)
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Ammonia: Base (not an electrophile)
Answer: (b) Aluminium chloride
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Q5. What is the shape of carbocation?
- Carbocations are typically
sp² hybridized, resulting in a
trigonal planar geometry.
Answer: (a) Planar
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Q6. What is the hybridisation of carbon in methyl chloride?
- Methyl chloride (\(\text{CH}_3\text{Cl}\)) has a single bond between carbon and chlorine, and three single bonds between carbon and hydrogen.
- Carbon in this case is
sp³ hybridized because it forms four sigma bonds.
Answer: sp³
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Q7. How many are sigma and pi bonds present in pent-1-ene-4-yne?
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Pent-1-ene-4-yne has the structure: \( \text{CH}_2=\text{CHCH}_2\text{C}\equiv\text{CH} \)
- Sigma bonds:
- Each single bond (\( \text{C}-\text{C} \) or \( \text{C}-\text{H} \)) is a sigma bond.
- Total sigma bonds: \( 8 \) (from single bonds)
- Pi bonds:
- Each double bond (\( \text{C}=\text{C} \)) contributes 1 pi bond.
- Each triple bond (\( \text{C}\equiv\text{C} \)) contributes 2 pi bonds.
- Total pi bonds: \( 1 + 2 = 3 \)
Answer: Sigma bonds = 8, Pi bonds = 3
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Q8. Write the molecular formula of the 2nd and 3rd number of the homologous series whose first member is ethene.
- Ethene (\(\text{C}_2\text{H}_4\)) is the first member of the alkene series.
- The general formula for alkenes is \( \text{C}_n\text{H}_{2n} \).
- 2nd member: \( n = 3 \) → \( \text{C}_3\text{H}_6 \) (propene)
- 3rd member: \( n = 4 \) → \( \text{C}_4\text{H}_8 \) (butene)
Answer:
- 2nd member: \( \text{C}_3\text{H}_6 \)
- 3rd member: \( \text{C}_4\text{H}_8 \)
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Final Answers:
1. (c) 2-methyl pentane
2. (d) All of the above
3. (b) Thioalcohol
4. (b) Aluminium chloride
5. (a) Planar
6. sp³
7. Sigma bonds = 8, Pi bonds = 3
8. 2nd member: \( \text{C}_3\text{H}_6 \), 3rd member: \( \text{C}_4\text{H}_8 \)
Boxed Final Answer:
\[
\boxed{(c), (d), (b), (b), (a), \text{sp}^3, \text{Sigma bonds = 8, Pi bonds = 3}, \text{C}_3\text{H}_6, \text{C}_4\text{H}_8}
\]
Parent Tip: Review the logic above to help your child master the concept of chemistry worksheet pdf.
