Chemistry Worksheet Class 11 on Chapter 12 Organic Chemistry - Some Basic Principles & Techniques - Set 1

Let's solve each question from the Chemistry Worksheet step by step, with clear explanations.

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Q1. Which of the following carbocation is most stable?

(a) (CH₃)₂CH⁺
(b) Ph₃C⁺
(c) CH₃CH₂⁺
(d) None of the above

#### Answer: (b) Ph₃C⁺

#### Explanation:
Carbocation stability depends on:
- Inductive effect
- Resonance
- Hyperconjugation
- Hybridization

Let’s analyze:

- (a) (CH₃)₂CH⁺: This is a secondary carbocation. It has 6 alpha hydrogens for hyperconjugation and inductive stabilization.
- (c) CH₃CH₂⁺: Primary carbocation – least stable due to fewer hyperconjugative effects.
- (b) Ph₃C⁺ (Triphenylmethyl cation): This is highly stabilized by resonance. Each phenyl ring can delocalize the positive charge via resonance, making it one of the most stable carbocations known.

> ✔ Ph₃C⁺ is exceptionally stable due to resonance with three phenyl rings, even more than tertiary alkyl carbocations.

So, (b) is the most stable.

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Q2. The molecule in which the distance between the neighbouring carbon atom is least is an _______.

(a) Alkane
(b) Alkene
(c) Alkyne
(d) All of them have equal distance

#### Answer: (c) Alkyne

#### Explanation:
Bond length decreases as bond order increases.

- Alkane (C–C single bond): Bond order = 1 → Longest bond (~1.54 Å)
- Alkene (C=C double bond): Bond order = 2 → Shorter (~1.34 Å)
- Alkyne (C≡C triple bond): Bond order = 3 → Shortest bond (~1.20 Å)

So, alkynes have the shortest C–C bond length.

✔ Therefore, (c) Alkyne is correct.

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Q3. What is the general formula of cycloalkanes?

(a) CₙH₂ₙ
(b) CₙH₂ₙ₋₁
(c) CₙH₂ₙ₊₁
(d) None of the above

#### Answer: (a) CₙH₂ₙ

#### Explanation:
Cycloalkanes are saturated cyclic hydrocarbons. They have no double bonds, but form a ring.

For example:
- Cyclopropane: C₃H₆ → n=3 → H = 2×3 = 6
- Cyclobutane: C₄H₈ → n=4 → H = 8

So, general formula is CₙH₂ₙ.

Compare with:
- Alkanes: CₙH₂ₙ₊₂ (open chain)
- Cycloalkanes: CₙH₂ₙ (ring structure)

✔ So, (a) is correct.

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Q4. Which of the following is not a cyclic compound?

(a) Neopentane
(b) Phenol
(c) Pyrrole
(d) None of the above

#### Answer: (a) Neopentane

#### Explanation:
- Neopentane (2,2-dimethylpropane): A straight-chain (branched) alkane with no ring structure → acyclic
- Phenol: Contains benzene ring → cyclic
- Pyrrole: Five-membered heterocyclic aromatic compound → cyclic

So, only neopentane is not cyclic.

✔ Answer: (a) Neopentane

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Q5. Which of the following hybridisation has maximum s characteristics?

(a) sp
(b) sp²
(c) sp³
(d) All of them have the same s characteristic

#### Answer: (a) sp

#### Explanation:
Hybridization determines % s-character:

| Hybridization | % s-character | % p-character |
|--------------|----------------|----------------|
| sp³ | 25% | 75% |
| sp² | 33.3% | 66.7% |
| sp | 50% | 50% |

Higher % s-character means more s-orbital contribution → sp has highest s-character.

✔ So, (a) sp is correct.

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Q6. How many sigma and pi bonds are present in 1,2-propadiene?

#### Answer:
First, let's understand 1,2-propadiene.

- IUPAC name: Allene
- Structure: CH₂=C=CH₂

It has:
- 3 carbon atoms
- Two double bonds: C=C=C

Now count bonds:

- Sigma bonds:
- C–C (single): 2 (between C1–C2 and C2–C3)
- C–H: 4 (each terminal CH₂ has 2 H)
- Total σ bonds = 2 (C–C) + 4 (C–H) = 6 sigma bonds

- Pi bonds:
- Each double bond has 1 π bond
- Two double bonds → 2 π bonds

But note: In allene, the central carbon is sp-hybridized, and the two π bonds are perpendicular.

✔ So, 6 sigma bonds and 2 pi bonds

> Answer: 6 sigma and 2 pi bonds

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Q7. What is homologous series?

#### Answer:
A homologous series is a family of organic compounds that have:
- The same functional group
- Similar chemical properties
- Gradually increasing molecular mass
- General formula differing by –CH₂– group

Examples:
- Alkanes: CH₄, C₂H₆, C₃H₈, ...
- Alkenes: C₂H₄, C₃H₆, C₄H₈, ...

Each successive member differs by –CH₂–.

✔ Definition: A series of organic compounds with the same functional group, where each successive member differs by a –CH₂– group.

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Q8. How many isomers are possible for monosubstituted and disubstituted benzene?

#### Answer:

Let’s break this down:

##### Monosubstituted benzene:
- Only one substituent attached to benzene ring.
- All positions are equivalent due to symmetry.
- So, only 1 isomer possible.

##### Disubstituted benzene:
- Two substituents on benzene ring.
- Positions relative to each other matter.
- Possible isomers based on relative positions:
- Ortho (1,2): adjacent
- Meta (1,3): separated by one carbon
- Para (1,4): opposite

So, 3 isomers for disubstituted benzene (assuming both substituents are different).

If both substituents are identical, then:
- Ortho, meta, para → still 3 isomers

But since the question says "disubstituted benzene" without specifying identical or different, we assume two different groups → still 3 positional isomers.

✔ So:
- Monosubstituted: 1 isomer
- Disubstituted: 3 isomers

Total isomers = 1 + 3 = 4 isomers

> Answer: 4 isomers

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✔ Final Answers Summary:

| Q# | Answer |
|----|--------|
| Q1 | (b) Ph₃C⁺ |
| Q2 | (c) Alkyne |
| Q3 | (a) CₙH₂ₙ |
| Q4 | (a) Neopentane |
| Q5 | (a) sp |
| Q6 | 6 sigma, 2 pi bonds |
| Q7 | Homologous series: A series of compounds with same functional group, differing by –CH₂– group |
| Q8 | 4 isomers (1 monosubstituted + 3 disubstituted) |

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