Chapter Exam

Unit 15: Hydrocarbons — Online MCQ Test

Q: Which one of the following compounds can exist as cis-trans isomers?

Answer: D. 1,2-Dimethylcyclohexane — 1,2-Dimethylcyclohexane exists as cis and trans isomers due to the rigidity of the ring system with substituents on different carbons.

Q: In the reaction sequence: HC≡CH --(NaNH2/liq. NH3, then CH3CH2Br)--> X --(NaNH2/liq. NH3, then CH3CH2Br)--> Y. X and Y are:

Answer: D. X = 1-Butyne ; Y = 3-Hexyne — Acetylene reacts with NaNH2 and CH3CH2Br to form 1-butyne (X). 1-butyne further reacts to form 3-hexyne (Y).

Q: In the reaction sequence: HC≡CH --(NaNH2/liq. NH3)--> X --(CH3CH2Br)--> Y --(NaNH2/liq. NH3)--> Z --(CH3CH2Br)--> 3-Hexyne. X and Y are :

Answer: D. X = 1-Butyne ; Y = 3-Hexyne — The reaction uses NaNH2 to deprotonate the terminal alkyne, followed by an SN2 reaction with ethyl bromide. First step produces 1-butyne (X), and subsequent reaction produces 3-hexyne.

Q: Match List I with List II. List I (Reaction) A. B. C. D. List II (Reagents/Condition) I. Anhyd. AlCl3 II. CrO3 III. KMnO4/KOH, Δ IV. (i) O3 (ii) Zn-H2O Choose the correct answer from the options given below:

Answer: A. A-IV, B-I, C-II, D-III — The reaction schemes are not fully visible in the excerpt, but the correct reagent matching indicated by the options is A-IV, B-I, C-II, D-III.

Q: Match List I with List II. List I (Molecule) A. ethane B. ethene C. carbon molecule, C2 D. ethyne List II (Number and types of bond/s between two carbon atoms) I. one σ-bond and two π-bonds II. two π-bonds III. one σ-bond IV. one σ-bond and one π-bond Choose the correct answer from the options given below:

Answer: A. A-III, B-IV, C-II, D-I — Ethane has a single C-C sigma bond; ethene has one sigma and one pi bond; C2 is taken here as having two pi bonds; ethyne has one sigma and two pi bonds. Therefore A-III, B-IV, C-II, D-I.

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

Answer: C. Both bond angles and bond length remains same — In ethane conformations, only the dihedral angle changes; bond angles and bond lengths remain constant.

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

Answer: C. Both bond angles and bond length remains same — Conformations of ethane involve rotation about the C-C bond; this changes the dihedral angle, but the internal bond angles and bond lengths of the molecule remain unchanged.

Q: Predict the correct intermediate and product in the reaction of propyne with H2O/H2SO4/HgSO4.

Answer: C. A: CH3-C(OH)=CH2, B: CH3-C(=O)CH3 — Hydration of propyne follows Markovnikov's rule to form the enol intermediate (prop-1-en-2-ol), which tautomerizes to form acetone (propanone).

Q: Predict the correct intermediate and product in the following reaction: CH3-C ≡ CH + H2O, H2SO4, HgSO4 -> intermediate -> product

Answer: A. A : CH3-C(OH)=CH2, B : CH3-CO-CH3 — Hydration of propyne in the presence of HgSO4/H2SO4 forms an enol intermediate (propen-2-ol) which tautomerizes to acetone (propanone).

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Practice Unit 15: Hydrocarbons with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: Classification isomerism IUPAC nomenclature general methods of preparation properties and reactions. Alkanes - Conformations: Sawhorse and Newman projections (of ethane): Mechanism.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 15: Hydrocarbons — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 15: Hydrocarbons, with answers and explanations. These are sample questions; the exam has its own separate question set.

Which one of the following compounds can exist as cis-trans isomers?

A. Pent-1-ene
B. 2-Methylhex-2-ene
C. 1,1-Dimethylcyclopropane
D. 1,2-Dimethylcyclohexane ✓

Answer: D. 1,2-Dimethylcyclohexane
1,2-Dimethylcyclohexane exists as cis and trans isomers due to the rigidity of the ring system with substituents on different carbons.

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is:

A. The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain.
B. The eclipsed conformation of ethane is more stable than staggered conformation, even though the eclipsed conformation has torsional strain.
C. The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain. ✓
D. The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.

Answer: C. The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
The staggered conformation is more stable than the eclipsed conformation due to the absence of torsional strain in the staggered form.

In the reaction sequence: HC≡CH --(NaNH2/liq. NH3, then CH3CH2Br)--> X --(NaNH2/liq. NH3, then CH3CH2Br)--> Y. X and Y are:

A. X = 2-Butyne ; Y = 3-Hexyne
B. X = 2-Butyne ; Y = 2-Hexyne
C. X = 1-Butyne ; Y = 2-Hexyne
D. X = 1-Butyne ; Y = 3-Hexyne ✓

Answer: D. X = 1-Butyne ; Y = 3-Hexyne
Acetylene reacts with NaNH2 and CH3CH2Br to form 1-butyne (X). 1-butyne further reacts to form 3-hexyne (Y).

In the reaction sequence: HC≡CH --(NaNH2/liq. NH3)--> X --(CH3CH2Br)--> Y --(NaNH2/liq. NH3)--> Z --(CH3CH2Br)--> 3-Hexyne. X and Y are :

A. X = 2-Butyne ; Y=3-Hexyne
B. X = 2-Butyne ; y=2-Hexyne
C. X = 1-Butyne; y = 2-Hexyne
D. X = 1-Butyne ; Y = 3-Hexyne ✓

Answer: D. X = 1-Butyne ; Y = 3-Hexyne
The reaction uses NaNH2 to deprotonate the terminal alkyne, followed by an SN2 reaction with ethyl bromide. First step produces 1-butyne (X), and subsequent reaction produces 3-hexyne.

Match List I with List II. List I (Reaction) A. B. C. D. List II (Reagents/Condition) I. Anhyd. AlCl3 II. CrO3 III. KMnO4/KOH, Δ IV. (i) O3 (ii) Zn-H2O Choose the correct answer from the options given below:

A. A-IV, B-I, C-II, D-III ✓
B. A-I, B-IV, C-II, D-III
C. A-IV, B-I, C-III, D-II
D. A-III, B-I, C-II, D-IV

Answer: A. A-IV, B-I, C-II, D-III
The reaction schemes are not fully visible in the excerpt, but the correct reagent matching indicated by the options is A-IV, B-I, C-II, D-III.

Match List I with List II. List I (Molecule) A. ethane B. ethene C. carbon molecule, C2 D. ethyne List II (Number and types of bond/s between two carbon atoms) I. one σ-bond and two π-bonds II. two π-bonds III. one σ-bond IV. one σ-bond and one π-bond Choose the correct answer from the options given below:

A. A-III, B-IV, C-II, D-I ✓
B. A-III, B-IV, C-I, D-II
C. A-I, B-IV, C-II, D-III
D. A-IV, B-III, C-II, D-I

Answer: A. A-III, B-IV, C-II, D-I
Ethane has a single C-C sigma bond; ethene has one sigma and one pi bond; C2 is taken here as having two pi bonds; ethyne has one sigma and two pi bonds. Therefore A-III, B-IV, C-II, D-I.

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

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

Answer: C. Both bond angles and bond length remains same
In ethane conformations, only the dihedral angle changes; bond angles and bond lengths remain constant.

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

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

Answer: C. Both bond angles and bond length remains same
Conformations of ethane involve rotation about the C-C bond; this changes the dihedral angle, but the internal bond angles and bond lengths of the molecule remain unchanged.

Predict the correct intermediate and product in the reaction of propyne with H2O/H2SO4/HgSO4.

A. A: CH3-C(OH)=CH2, B: CH3-CH2-SO4
B. A: CH3-C(=O)CH3, B: CH3-C(OH)=CH2
C. A: CH3-C(OH)=CH2, B: CH3-C(=O)CH3 ✓
D. A: CH3-CH2-SO4, B: CH3-C(=O)CH3

Answer: C. A: CH3-C(OH)=CH2, B: CH3-C(=O)CH3
Hydration of propyne follows Markovnikov's rule to form the enol intermediate (prop-1-en-2-ol), which tautomerizes to form acetone (propanone).

Predict the correct intermediate and product in the following reaction: CH3-C ≡ CH + H2O, H2SO4, HgSO4 -> intermediate -> product

A. A : CH3-C(OH)=CH2, B : CH3-CO-CH3 ✓
B. A : CH3-C(SO4)=CH2, B : CH3-CO-CH3
C. A : CH3-C(OH)=CH2, B : CH3-CH2-SO4
D. A : CH3-CO-CH3, B : CH3-C ≡ CH

Answer: A. A : CH3-C(OH)=CH2, B : CH3-CO-CH3
Hydration of propyne in the presence of HgSO4/H2SO4 forms an enol intermediate (propen-2-ol) which tautomerizes to acetone (propanone).

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