Chapter Exam

Unit 14: Some Basic Principles of Organic Chemistry — Online MCQ Test

Q: A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

Answer: C. Hyperconjugation — Carbocation stability increases with more alkyl groups due to inductive effect and hyperconjugation. Hyperconjugation with adjacent C-H bonds provides additional stability.

Q: The correct order of decreasing acidity of the following aliphatic acids is:

Answer: C. HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH — Acidity decreases as the +I effect of alkyl groups increases, which destabilizes the conjugate base. HCOOH > primary > secondary > tertiary.

Q: Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

Answer: C. 10 — Isomers include oxetane, 2-methyloxetane, 3-methyloxetane, tetrahydrofuran, and oxirane derivatives including stereoisomers (cis/trans). Total count is 10.

Q: Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

Answer: C. 10 — Cyclic ethers with formula C4H8O include oxetane (1 isomer), 2-methyloxetane (2 isomers), 3-methyloxetane (1 isomer), tetrahydrofuran (1 isomer), and oxirane derivatives (e.g., ethyl oxirane, 2,2-dimethyl oxirane, 2,3-dimethyl oxirane). Accounting for stereoisomers, there are 10 total.

Q: The reaction R–OH + NaH -> R–ONa + H2; R–ONa + Me–I -> R–O–Me + NaI can be classified as:

Answer: D. Williamson ether synthesis reaction — It is the Williamson ether synthesis reaction, which involves the reaction of an alkoxide (R-ONa) with an alkyl halide (R-X) to form an ether (R-O-R) and a salt (NaX).

Q: The pair of electron in the given carbanion, CH3C ≡ C–, is present in which of the following orbitals?

Answer: C. sp — In the acetylide carbanion (CH3-C≡C⁻), the negatively charged carbon is sp-hybridized because it forms a triple bond.

Q: With respect to the conformers of ethane, which of the following statements is true?

Answer: A. Both bond angles and bond length remains same — Conformations of ethane are generated by rotation around the C-C single bond. This rotation does not affect the bond lengths or bond angles between atoms.

Q: Which of the following molecules represents the order of hybridisation sp², sp², sp, sp from left to right atoms?

Answer: B. CH₂=CH–C≡CH — In CH₂=CH–C≡CH: C1 (=CH₂) is sp², C2 (=CH–) is sp², C3 (≡C–) is sp, C4 (≡CH) is sp. This gives the hybridisation sequence sp², sp², sp, sp from left to right.

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Practice Unit 14: Some Basic Principles of Organic Chemistry with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: Tetravalency of carbon: Shapes of simple molecules - hybridization (s and p): Classification of organic compounds based on functional groups: and those containing halogens oxygen n.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 14: Some Basic Principles of Organic Chemistry — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 14: Some Basic Principles of Organic Chemistry, with answers and explanations. These are sample questions; the exam has its own separate question set.

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

A. +R effect of -CH3 groups
B. -R effect of -CH3 groups
C. Hyperconjugation ✓
D. -I effect of -CH3 groups

Answer: C. Hyperconjugation
Carbocation stability increases with more alkyl groups due to inductive effect and hyperconjugation. Hyperconjugation with adjacent C-H bonds provides additional stability.

The correct order of decreasing acidity of the following aliphatic acids is:

A. (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH > HCOOH
B. CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH > HCOOH
C. HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH ✓
D. HCOOH > (CH₃)₃CCOOH > (CH₃)₂CHCOOH > CH₃COOH

Answer: C. HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
Acidity decreases as the +I effect of alkyl groups increases, which destabilizes the conjugate base. HCOOH > primary > secondary > tertiary.

Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

A. 6
B. 8
C. 10 ✓
D. 11

Answer: C. 10
Isomers include oxetane, 2-methyloxetane, 3-methyloxetane, tetrahydrofuran, and oxirane derivatives including stereoisomers (cis/trans). Total count is 10.

Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:

A. 6
B. 8
C. 10 ✓
D. 11

Answer: C. 10
Cyclic ethers with formula C4H8O include oxetane (1 isomer), 2-methyloxetane (2 isomers), 3-methyloxetane (1 isomer), tetrahydrofuran (1 isomer), and oxirane derivatives (e.g., ethyl oxirane, 2,2-dimethyl oxirane, 2,3-dimethyl oxirane). Accounting for stereoisomers, there are 10 total.

For the following reactions: (a) CH3CH2CH2Br + KOH → CH3CH=CH2 + KBr + H2O, (b) (CH3)2CHBr + KOH → (CH3)2CHOH + KBr, (c) CH2=CH2 + Br2 → CH2BrCH2Br. Which of the following statements is correct?

A. (a) is elimination, (b) is substitution and (c) addition reaction. ✓
B. (a) is elimination, (b) and (c) are substitution reactions.
C. (a) is substitution, (b) and (c) are addition reactions.
D. (a) and (b) are elimination reactions and (c) is addition reaction.

Answer: A. (a) is elimination, (b) is substitution and (c) addition reaction.
Reaction (a) is a dehydrohalogenation (elimination), (b) is a nucleophilic substitution of Br by OH, and (c) is an electrophilic addition of bromine across the double bond.

The reaction R–OH + NaH -> R–ONa + H2; R–ONa + Me–I -> R–O–Me + NaI can be classified as:

A. Alcohol formation reaction
B. Dehydration reaction
C. Williamson alcohol synthesis reaction
D. Williamson ether synthesis reaction ✓

Answer: D. Williamson ether synthesis reaction
It is the Williamson ether synthesis reaction, which involves the reaction of an alkoxide (R-ONa) with an alkyl halide (R-X) to form an ether (R-O-R) and a salt (NaX).

The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon, is:

A. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration.
B. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation.
C. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism. ✓
D. a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol.

Answer: C. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.
Carbonyl compounds with alpha-hydrogen atoms undergo rapid equilibration with their corresponding enol forms, a process specifically referred to as keto-enol tautomerism.

The pair of electron in the given carbanion, CH3C ≡ C–, is present in which of the following orbitals?

A. sp3
B. sp2
C. sp ✓
D. 2p

Answer: C. sp
In the acetylide carbanion (CH3-C≡C⁻), the negatively charged carbon is sp-hybridized because it forms a triple bond.

With respect to the conformers of ethane, which of the following statements is true?

A. Both bond angles and bond length remains same ✓
B. Bond angle remains same but bond length changes
C. Bond angle changes but bond length remains same
D. Both bond angle and bond length change

Answer: A. Both bond angles and bond length remains same
Conformations of ethane are generated by rotation around the C-C single bond. This rotation does not affect the bond lengths or bond angles between atoms.

Which of the following molecules represents the order of hybridisation sp², sp², sp, sp from left to right atoms?

A. CH₂=CH–CH=CH₂
B. CH₂=CH–C≡CH ✓
C. HC≡C–C≡CH
D. CH₃–CH=CH–CH₃

Answer: B. CH₂=CH–C≡CH
In CH₂=CH–C≡CH: C1 (=CH₂) is sp², C2 (=CH–) is sp², C3 (≡C–) is sp, C4 (≡CH) is sp. This gives the hybridisation sequence sp², sp², sp, sp from left to right.

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