Unit 15: Hydrocarbons — Online MCQ Test
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Practice Unit 15: Hydrocarbons with a free chapter-wise online MCQ test.
This chapter covers: Classification isomerism IUPAC nomenclature general methods of preparation properties and reactions. Alkanes - Conformations: Sawhorse and Newman projections (of ethane): Mechanism....
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Unit 15: Hydrocarbons — Important Questions & Answers
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.
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.
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).
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.
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.
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.