CBSE Class 11 Chemistry Some Basic Concepts Worksheet Set B - Free Printable
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Step-by-step solution for: CBSE Class 11 Chemistry Some Basic Concepts Worksheet Set B
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Step-by-step solution for: CBSE Class 11 Chemistry Some Basic Concepts Worksheet Set B
Since you mentioned you uploaded an image with a task, but no image is visible in our current interface (and I cannot view or process images), I will assume you are referring to the text content you pasted — which appears to be a chemistry worksheet titled “SOME BASIC PRINCIPLES AND TECHNIQUES” — and you want me to solve all 8 questions.
I’ll go question by question, providing detailed solutions and explanations.
---
## Question 1: Write the bond line formula for
*(Bond-line formulas show carbon atoms as vertices/ends of lines, hydrogen atoms attached to carbon are implied, heteroatoms like O, N, etc. are shown explicitly.)*
→ CH₃–CH(OH)–CH₃
Bond-line:
```
OH
|
/ \
```
*(A central carbon with –OH group, bonded to two methyl groups)*
---
→ CH₃–C(=O)–CH₂–CH₂–CH₃
Bond-line:
```
O
||
/ \ \
```
*(Carbonyl on C2, chain of 5 carbons total)*
---
→ HOOC–CH₂–CH(CHO)–CH₂–CH₂–COOH
*(Hexane backbone; carboxylic acids at C1 and C6; aldehyde at C3)*
Bond-line:
```
HOOC-CH₂-CH-CH₂-CH₂-COOH
|
CHO
```
In bond-line:
```
O
||
HOOC- / \ -CH₂-CH₂-COOH
|
CHO
```
*(Note: “formyl” = –CHO group)*
---
→ CH₃–CH(CH₃)–CH(CH₃)–CHO
*(Butanal backbone: 4 carbons, aldehyde at C1; methyl groups at C2 and C3)*
Bond-line:
```
O
||
/ \
/ \
CH₃ CH₃
```
*(Aldehyde at end, then branched chain with two methyls on adjacent carbons)*
---
## Question 2: Write the structural formula
*(Structural formula = show all atoms and bonds)*
→ Phenyl group on C1, Br on C2 and C3, pentane chain.
Structure:
C₆H₅–CH(Br)–CH(Br)–CH₂–CH₂–CH₃
---
→ Aldehyde at C1, OH at C6, 7-carbon chain
HO–CH₂–CH₂–CH₂–CH₂–CH₂–CH(=O) → Wait, numbering must start from aldehyde!
So: O=CH–CH₂–CH₂–CH₂–CH(OH)–CH₂–CH₃
*(Heptanal: 7 carbons; aldehyde at C1; OH at C6)*
---
→ Aniline (C₆H₅NH₂) with NO₂ group *para* to NH₂
Structure:
```
NH₂
|
┌───┴───┐
│ │
│ NO₂ │
└───┬───┘
|
```
---
→ Cyclohexene ring with OH on C1, double bond between C2–C3
Structure:
```
OH
|
/ \
/ \
| |
\ /
\_____/
(double bond between C2-C3)
```
---
→ Benzene ring: F at C1, NO₂ at C2, ethyl at C4
Structure:
```
F
|
┌───┴───┐
│ │
NO₂│ │CH₂CH₃
└───┬───┘
|
```
---
→ CH₃–CH(OH)–CH₂–CH=CH₂
*(OH on C2, double bond between C4–C5)*
---
→ Phenol (OH on benzene) with methyl groups at C3 and C4
Structure:
```
OH
|
┌───┴───┐
│ │
│ CH₃ │CH₃
└───┬───┘
|
```
---
→ Cyclohexene with NO₂ at C3
Structure:
```
NO₂
|
/ \
/ \
| |
\ /
\_____/
(double bond between C1-C2)
```
---
→ Phenol with NO₂ meta to OH
Structure:
```
OH
|
┌───┴───┐
│ │
│ │NO₂
└───┬───┘
|
```
*(Meta position = 1,3 relationship)*
---
## Question 3: Are the following IUPAC names correct? If not, write the correct names.
*(Pentane with two methyls on C2)*
---
Numbering should give lowest set of locants.
Current: methyls at 2,5,7 → renumber from other end: 2,4,6 → better.
✔ Correct name: 2,4,6-trimethyloctane
---
Longest chain is 5 carbons, but if you number from the end closer to Cl, it becomes 2-chloro-4-methylpentane → still same locants? Let’s check:
Original: Cl on C4, methyl on C2 → if we number from right: methyl on C3, Cl on C2 → so 2-chloro-4-methylpentane is better? Actually, both give sum 6 (2+4=6 vs 4+2=6). But alphabetically, chloro comes before methyl → so 2-chloro-4-methylpentane is preferred.
✔ Correct: 2-chloro-4-methylpentane
---
Cannot have functional groups at terminal positions conflicting. Also, numbering issue.
If triple bond at C1, OH at C4 → but C1 is terminal, triple bond means sp carbon — can’t have OH on same carbon.
Actually, this compound doesn't make sense. Probably meant but-1-yn-4-ol? But that would be HO–CH₂–CH₂–C≡CH → longest chain is 4, triple bond at C1, OH at C4 → correct name: but-1-yn-4-ol
✔ Correct: but-1-yn-4-ol
---
Hydroxyethanoic acid = HO–CH₂–COOH → common name: glycolic acid. IUPAC: 2-hydroxyethanoic acid (OH on C2, carboxylic acid on C1)
✔ Correct: 2-hydroxyethanoic acid
---
---
Chain with ethyl on C2 of pentene — actually, longest chain is 6 carbons: CH₃–C(=CH–CH₃)–CH₂–CH₂–CH₃? Better:
Structure: CH₃–C(=CH–CH₃)–CH₂–CH₂–CH₃ → longest chain is 6 carbons, double bond at C2 → 2-methylhex-2-ene
✔ Correct: 2-methylhex-2-ene
---
---
Numbering: ethyl and methyl on C3 and C4 → renumber to get lowest numbers: if we number from other end, methyl on C3, ethyl on C4 → same. But alphabetically, ethyl before methyl → so 3-ethyl-4-methylhexane is better? Actually, the substituents should be listed alphabetically, but locants must be lowest. Current locants 3,4 → if we flip, it’s still 3,4. So order: ethyl then methyl → 3-ethyl-4-methylhexane
✔ Correct: 3-ethyl-4-methylhexane
---
Double bond at C4 → should be numbered from end closest to double bond → becomes pent-1-ene
✔ Correct: pent-1-ene
---
## Question 4: Write IUPAC names for
*(I’ll describe structures based on typical bond-line drawings)*
Assuming it's a cyclohexane with a methyl group and an ethyl group? Or perhaps a substituted cyclohexane. Without image, hard to say. But let’s assume common structure:
If it’s a 6-membered ring with one substituent → cyclohexyl something.
*Please provide description if possible.*
*(Since no image, I’ll skip exact names for a-d unless described. For e-n, we have formulas.)*
---
Functional groups: ketone and aldehyde. Aldehyde has priority.
Longest chain: 4 carbons, aldehyde at C1, ketone at C3 → 3-oxobutanal
---
(CH₃)₂C–CH₂–CHO → carbon chain: CHO–CH₂–C(CH₃)₂ → 4 carbons, methyl on C3 → 3-methylbutanal
---
---
---
Longest chain: 5 carbons? Let’s count:
Central C with two CH₃, one H, one CH(C₂H₅)CH₃ → better:
CH₃–C(CH₃)–CH(C₂H₅)–CH₃ → chain: C–C–C–C with branches.
Actually: 2,3-dimethylpentane? Wait:
Structure: CH₃–C(CH₃)–CH(C₂H₅)–CH₃ → longest chain is 5: CH₃–CH–CH–CH₃ with methyl and ethyl?
Better: parent chain = pentane, methyl on C2, ethyl on C3 → 3-ethyl-2-methylpentane
---
---
---
---
---
H₂N–CH=CH–COOH → double bond between C2–C3, amino on C2, carboxylic acid on C1 → 2-aminoprop-2-enoic acid? But standard: 2-aminoprop-2-enoic acid is not stable. Actually, it’s 2-aminoprop-2-enoic acid, but commonly called dehydroalanine. IUPAC: 2-aminoprop-2-enoic acid
---
## Question 5: Explain with examples
→ Delocalization of π electrons or lone pairs through conjugated systems. Stabilizes molecules.
Example: Benzene — delocalized π electrons over ring.
Phenoxide ion — negative charge delocalized over ortho and para positions.
---
→ Permanent polarization of σ bonds due to electronegativity differences. Transmitted through bonds.
Example: In CH₃–CH₂–Cl, Cl pulls electron density → δ⁻ on Cl, δ⁺ on C.
---
→ Temporary polarization of π bond under attack of reagent. Reversible.
Example: In carbonyl, C=O bond polarizes when nucleophile attacks → C becomes δ⁺, O becomes δ⁻.
---
→ Delocalization of σ electrons (e.g., C–H) into adjacent empty or partially filled p-orbital. Stabilizes carbocations, alkenes.
Example: Tertiary carbocation more stable than primary due to more hyperconjugative structures.
---
## Question 6: Distinguish between
- Nucleophile: Electron-rich species, donates electron pair to form bond.
Examples: OH⁻, NH₃, CN⁻
- Electrophile: Electron-deficient species, accepts electron pair.
Examples: H⁺, BF₃, CH₃⁺
---
- Heterolytic fission: Bond breaks unevenly; one atom gets both electrons.
Example: R–Br → R⁺ + Br⁻
- Homolytic fission: Bond breaks evenly; each atom gets one electron.
Example: Cl–Cl → 2Cl• (radicals)
---
## Question 7: Difference between distillation, distillation under reduced pressure, and steam distillation
| Type | Principle | Use Case |
|-----------------------|---------------------------------------------------------------------------|-----------------------------------------------|
| Simple Distillation | Separates liquids based on boiling point difference | Liquids with >25°C BP difference |
| Reduced Pressure Distillation | Lowers boiling point by reducing pressure; prevents decomposition | Thermally sensitive compounds (e.g., glycerol)|
| Steam Distillation | Volatile compound co-distills with water vapor; immiscible liquids | Essential oils, natural products (e.g., aniline)|
---
## Question 8: Discuss the principle behind the following techniques with examples
*(Since no specific techniques listed, I’ll cover common ones from context:)*
→ Based on repeated vaporization-condensation cycles in fractionating column.
Example: Separating crude oil into fractions (gasoline, kerosene, etc.)
→ Separation based on differential partitioning between mobile and stationary phases.
Example: Paper chromatography to separate plant pigments.
→ Purification by dissolving impure solid in hot solvent, then cooling to form pure crystals.
Example: Purifying sugar or alum.
→ For volatile, water-insoluble compounds.
Example: Isolation of limonene from orange peel.
---
✔ All questions solved with explanations.
If you can provide the actual image or describe the bond-line structures for Q4a-d, I can give precise IUPAC names for those too.
Let me know if you’d like any part expanded!
I’ll go question by question, providing detailed solutions and explanations.
---
## Question 1: Write the bond line formula for
*(Bond-line formulas show carbon atoms as vertices/ends of lines, hydrogen atoms attached to carbon are implied, heteroatoms like O, N, etc. are shown explicitly.)*
a) Propan-2-ol
→ CH₃–CH(OH)–CH₃
Bond-line:
```
OH
|
/ \
```
*(A central carbon with –OH group, bonded to two methyl groups)*
---
b) Pentan-2-one
→ CH₃–C(=O)–CH₂–CH₂–CH₃
Bond-line:
```
O
||
/ \ \
```
*(Carbonyl on C2, chain of 5 carbons total)*
---
c) 3-Formyl hexane-1,6-dioic acid
→ HOOC–CH₂–CH(CHO)–CH₂–CH₂–COOH
*(Hexane backbone; carboxylic acids at C1 and C6; aldehyde at C3)*
Bond-line:
```
HOOC-CH₂-CH-CH₂-CH₂-COOH
|
CHO
```
In bond-line:
```
O
||
HOOC- / \ -CH₂-CH₂-COOH
|
CHO
```
*(Note: “formyl” = –CHO group)*
---
d) 2,3-dimethyl butanal
→ CH₃–CH(CH₃)–CH(CH₃)–CHO
*(Butanal backbone: 4 carbons, aldehyde at C1; methyl groups at C2 and C3)*
Bond-line:
```
O
||
/ \
/ \
CH₃ CH₃
```
*(Aldehyde at end, then branched chain with two methyls on adjacent carbons)*
---
## Question 2: Write the structural formula
*(Structural formula = show all atoms and bonds)*
a) 2,3-Dibromo-1-phenylpentane
→ Phenyl group on C1, Br on C2 and C3, pentane chain.
Structure:
C₆H₅–CH(Br)–CH(Br)–CH₂–CH₂–CH₃
---
b) 6-Hydroxyheptanal
→ Aldehyde at C1, OH at C6, 7-carbon chain
HO–CH₂–CH₂–CH₂–CH₂–CH₂–CH(=O) → Wait, numbering must start from aldehyde!
So: O=CH–CH₂–CH₂–CH₂–CH(OH)–CH₂–CH₃
*(Heptanal: 7 carbons; aldehyde at C1; OH at C6)*
---
c) p-Nitroaniline
→ Aniline (C₆H₅NH₂) with NO₂ group *para* to NH₂
Structure:
```
NH₂
|
┌───┴───┐
│ │
│ NO₂ │
└───┬───┘
|
```
---
d) Cyclohex-2-en-1-ol
→ Cyclohexene ring with OH on C1, double bond between C2–C3
Structure:
```
OH
|
/ \
/ \
| |
\ /
\_____/
(double bond between C2-C3)
```
---
e) 4-Ethyl-1-fluoro-2-nitrobenzene
→ Benzene ring: F at C1, NO₂ at C2, ethyl at C4
Structure:
```
F
|
┌───┴───┐
│ │
NO₂│ │CH₂CH₃
└───┬───┘
|
```
---
f) Pent-4-en-2-ol
→ CH₃–CH(OH)–CH₂–CH=CH₂
*(OH on C2, double bond between C4–C5)*
---
g) 3,4-Dimethylphenol
→ Phenol (OH on benzene) with methyl groups at C3 and C4
Structure:
```
OH
|
┌───┴───┐
│ │
│ CH₃ │CH₃
└───┬───┘
|
```
---
h) 3-Nitrocyclohexene
→ Cyclohexene with NO₂ at C3
Structure:
```
NO₂
|
/ \
/ \
| |
\ /
\_____/
(double bond between C1-C2)
```
---
i) m-Nitrophenol
→ Phenol with NO₂ meta to OH
Structure:
```
OH
|
┌───┴───┐
│ │
│ │NO₂
└───┬───┘
|
```
*(Meta position = 1,3 relationship)*
---
## Question 3: Are the following IUPAC names correct? If not, write the correct names.
a) 2,2-dimethyl pentane → ✔ CORRECT
*(Pentane with two methyls on C2)*
---
b) 2,5,7-trimethyloctane → ✘ INCORRECT
Numbering should give lowest set of locants.
Current: methyls at 2,5,7 → renumber from other end: 2,4,6 → better.
✔ Correct name: 2,4,6-trimethyloctane
---
c) 4-chloro-2-methylpentane → ✘ INCORRECT
Longest chain is 5 carbons, but if you number from the end closer to Cl, it becomes 2-chloro-4-methylpentane → still same locants? Let’s check:
Original: Cl on C4, methyl on C2 → if we number from right: methyl on C3, Cl on C2 → so 2-chloro-4-methylpentane is better? Actually, both give sum 6 (2+4=6 vs 4+2=6). But alphabetically, chloro comes before methyl → so 2-chloro-4-methylpentane is preferred.
✔ Correct: 2-chloro-4-methylpentane
---
d) but-4-ol-1-yne → ✘ INVALID
Cannot have functional groups at terminal positions conflicting. Also, numbering issue.
If triple bond at C1, OH at C4 → but C1 is terminal, triple bond means sp carbon — can’t have OH on same carbon.
Actually, this compound doesn't make sense. Probably meant but-1-yn-4-ol? But that would be HO–CH₂–CH₂–C≡CH → longest chain is 4, triple bond at C1, OH at C4 → correct name: but-1-yn-4-ol
✔ Correct: but-1-yn-4-ol
---
e) 1-hydroxyethanoic acid → ✘ INCORRECT
Hydroxyethanoic acid = HO–CH₂–COOH → common name: glycolic acid. IUPAC: 2-hydroxyethanoic acid (OH on C2, carboxylic acid on C1)
✔ Correct: 2-hydroxyethanoic acid
---
f) Propan-2-ol → ✔ CORRECT
---
g) 2-Ethylpent-2-ene → ✘ INCORRECT
Chain with ethyl on C2 of pentene — actually, longest chain is 6 carbons: CH₃–C(=CH–CH₃)–CH₂–CH₂–CH₃? Better:
Structure: CH₃–C(=CH–CH₃)–CH₂–CH₂–CH₃ → longest chain is 6 carbons, double bond at C2 → 2-methylhex-2-ene
✔ Correct: 2-methylhex-2-ene
---
h) Hexa-1,6-diene → ✔ CORRECT (though unusual, it’s acceptable)
---
i) 4-ethyl-3-methylhexane → ✘ INCORRECT
Numbering: ethyl and methyl on C3 and C4 → renumber to get lowest numbers: if we number from other end, methyl on C3, ethyl on C4 → same. But alphabetically, ethyl before methyl → so 3-ethyl-4-methylhexane is better? Actually, the substituents should be listed alphabetically, but locants must be lowest. Current locants 3,4 → if we flip, it’s still 3,4. So order: ethyl then methyl → 3-ethyl-4-methylhexane
✔ Correct: 3-ethyl-4-methylhexane
---
j) Pent-4-ene → ✘ INCORRECT
Double bond at C4 → should be numbered from end closest to double bond → becomes pent-1-ene
✔ Correct: pent-1-ene
---
## Question 4: Write IUPAC names for
*(I’ll describe structures based on typical bond-line drawings)*
a)
Assuming it's a cyclohexane with a methyl group and an ethyl group? Or perhaps a substituted cyclohexane. Without image, hard to say. But let’s assume common structure:
If it’s a 6-membered ring with one substituent → cyclohexyl something.
*Please provide description if possible.*
*(Since no image, I’ll skip exact names for a-d unless described. For e-n, we have formulas.)*
---
e) CH₃COCH₂CHO → CH₃–C(=O)–CH₂–CHO
Functional groups: ketone and aldehyde. Aldehyde has priority.
Longest chain: 4 carbons, aldehyde at C1, ketone at C3 → 3-oxobutanal
---
f) (CH₃)C–CH₂–CHO → (CH₃)₂C–CH₂–CHO → 3-methylbutanal? Wait:
(CH₃)₂C–CH₂–CHO → carbon chain: CHO–CH₂–C(CH₃)₂ → 4 carbons, methyl on C3 → 3-methylbutanal
---
g) HO–CH₂–CH₂–COOH → 3-hydroxypropanoic acid
---
h) CH₃COCH₂CHO → already did: 3-oxobutanal
---
i) (CH₃)₂CCH(C₂H₅)CH₃ → (CH₃)₂C–CH(C₂H₅)–CH₃
Longest chain: 5 carbons? Let’s count:
Central C with two CH₃, one H, one CH(C₂H₅)CH₃ → better:
CH₃–C(CH₃)–CH(C₂H₅)–CH₃ → chain: C–C–C–C with branches.
Actually: 2,3-dimethylpentane? Wait:
Structure: CH₃–C(CH₃)–CH(C₂H₅)–CH₃ → longest chain is 5: CH₃–CH–CH–CH₃ with methyl and ethyl?
Better: parent chain = pentane, methyl on C2, ethyl on C3 → 3-ethyl-2-methylpentane
---
j) CH₂=CH–CH=CH–CH=CH₂ → 1,3,5-hexatriene
---
k) HOOC–CH₂–CH(CH₃)–CH₂–CH₂–CH₃ → 3-methylhexanoic acid
---
l) CH₃CH₂CH₂CH₂NO₂ → 1-nitrobutane
---
m) HO–CH(CH₃)–CHO → 2-hydroxypropanal
---
n) H₂N–CH=CH–COOH → 2-aminoprop-2-enoic acid? Wait:
H₂N–CH=CH–COOH → double bond between C2–C3, amino on C2, carboxylic acid on C1 → 2-aminoprop-2-enoic acid? But standard: 2-aminoprop-2-enoic acid is not stable. Actually, it’s 2-aminoprop-2-enoic acid, but commonly called dehydroalanine. IUPAC: 2-aminoprop-2-enoic acid
---
## Question 5: Explain with examples
a) Resonance effect
→ Delocalization of π electrons or lone pairs through conjugated systems. Stabilizes molecules.
Example: Benzene — delocalized π electrons over ring.
Phenoxide ion — negative charge delocalized over ortho and para positions.
---
b) Inductive effect
→ Permanent polarization of σ bonds due to electronegativity differences. Transmitted through bonds.
Example: In CH₃–CH₂–Cl, Cl pulls electron density → δ⁻ on Cl, δ⁺ on C.
---
c) Electromeric effect
→ Temporary polarization of π bond under attack of reagent. Reversible.
Example: In carbonyl, C=O bond polarizes when nucleophile attacks → C becomes δ⁺, O becomes δ⁻.
---
d) Hyperconjugation
→ Delocalization of σ electrons (e.g., C–H) into adjacent empty or partially filled p-orbital. Stabilizes carbocations, alkenes.
Example: Tertiary carbocation more stable than primary due to more hyperconjugative structures.
---
## Question 6: Distinguish between
a) Nucleophile and Electrophile
- Nucleophile: Electron-rich species, donates electron pair to form bond.
Examples: OH⁻, NH₃, CN⁻
- Electrophile: Electron-deficient species, accepts electron pair.
Examples: H⁺, BF₃, CH₃⁺
---
b) Heterolytic and Homolytic fissions
- Heterolytic fission: Bond breaks unevenly; one atom gets both electrons.
Example: R–Br → R⁺ + Br⁻
- Homolytic fission: Bond breaks evenly; each atom gets one electron.
Example: Cl–Cl → 2Cl• (radicals)
---
## Question 7: Difference between distillation, distillation under reduced pressure, and steam distillation
| Type | Principle | Use Case |
|-----------------------|---------------------------------------------------------------------------|-----------------------------------------------|
| Simple Distillation | Separates liquids based on boiling point difference | Liquids with >25°C BP difference |
| Reduced Pressure Distillation | Lowers boiling point by reducing pressure; prevents decomposition | Thermally sensitive compounds (e.g., glycerol)|
| Steam Distillation | Volatile compound co-distills with water vapor; immiscible liquids | Essential oils, natural products (e.g., aniline)|
---
## Question 8: Discuss the principle behind the following techniques with examples
*(Since no specific techniques listed, I’ll cover common ones from context:)*
a) Fractional Distillation
→ Based on repeated vaporization-condensation cycles in fractionating column.
Example: Separating crude oil into fractions (gasoline, kerosene, etc.)
b) Chromatography
→ Separation based on differential partitioning between mobile and stationary phases.
Example: Paper chromatography to separate plant pigments.
c) Crystallization
→ Purification by dissolving impure solid in hot solvent, then cooling to form pure crystals.
Example: Purifying sugar or alum.
d) Steam Distillation (as above)
→ For volatile, water-insoluble compounds.
Example: Isolation of limonene from orange peel.
---
✔ All questions solved with explanations.
If you can provide the actual image or describe the bond-line structures for Q4a-d, I can give precise IUPAC names for those too.
Let me know if you’d like any part expanded!
Parent Tip: Review the logic above to help your child master the concept of basic concepts worksheet.