Unit 15: Hydrocarbons — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
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) mechanismo....
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Unit 15: Hydrocarbons — Important Questions & Answers
The incorrect statements regarding geometrical isomerism are : (A) Propene shows geometrical isomerism. (B) Trans isomer has identical atoms/groups on the opposite sides of the double bond. (C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene. (D) 2-methylbut-2-ene shows two geometrical isomers. (E) Trans-isomer has lower melting point than cis isomer. Choose the correct answer from the options given below:
- A. (A) and (E) Only
- B. (A), (D) and (E) Only
- C. (B) and (C) Only
- D. (C), (D) and (E) Only
Answer: B. (A), (D) and (E) Only
Propene cannot show geometrical isomerism (A is incorrect). 2-methylbut-2-ene cannot show geometrical isomerism due to two identical methyl groups on one carbon (D is incorrect). Trans-isomers are usually more symmetrical and pack better in solids, leading to higher melting points than cis-isomers (E is incorrect).
Propene cannot show geometrical isomerism (A is incorrect). 2-methylbut-2-ene cannot show geometrical isomerism due to two identical methyl groups on one carbon (D is incorrect). Trans-isomers are usually more symmetrical and pack better in solids, leading to higher melting points than cis-isomers (E is incorrect).
Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of hydrogen gas. Statement II : Four g of propyne reacts with sodium to liberate hydrogen gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct
- C. Statement I is correct but Statement II is incorrect
- D. Both Statement I and Statement II are incorrect
Answer: C. Statement I is correct but Statement II is incorrect
Propyne (CH3-C≡CH) has one acidic hydrogen. Reaction: CH3-C≡CH + Na -> CH3-C≡CNa + 0.5 H2. Statement I is correct. For Statement II: 4g propyne = 4/40 = 0.1 mol. This should liberate 0.05 mol H2. At STP, 0.05 mol H2 occupies 0.05 * 22400 = 1120 mL, not 224 mL. Statement II is incorrect.
Propyne (CH3-C≡CH) has one acidic hydrogen. Reaction: CH3-C≡CH + Na -> CH3-C≡CNa + 0.5 H2. Statement I is correct. For Statement II: 4g propyne = 4/40 = 0.1 mol. This should liberate 0.05 mol H2. At STP, 0.05 mol H2 occupies 0.05 * 22400 = 1120 mL, not 224 mL. Statement II is incorrect.
Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H2 gas. Statement II: Four g of propyne reacts with NaNH2 to liberate gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct
- C. Statement I is correct but Statement II is incorrect
- D. Both Statement I and Statement II are incorrect
Answer: C. Statement I is correct but Statement II is incorrect
Statement I is correct (terminal alkyne reacts with Na). Statement II is incorrect: 4g propyne is 0.1 mol, which would liberate 0.05 mol H2, occupying 1120 mL at STP, not 224 mL.
Statement I is correct (terminal alkyne reacts with Na). Statement II is incorrect: 4g propyne is 0.1 mol, which would liberate 0.05 mol H2, occupying 1120 mL at STP, not 224 mL.
Buna-S can be represented as:
- A. $\left[-\mathrm{CH_2-CH=C(C_6H_5)-CH=CH-CH_2-}\right]_n$
- B. $\left[-\mathrm{CH_2-CH=CH-CH=C(C_6H_5)-CH_2-}\right]_n$
- C. $\left[-\mathrm{CH=CH-CH=CH-CH(C_6H_5)-CH_2-}\right]_n$
- D. $\left[-\mathrm{CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2-}\right]_n$
Answer: D. $\left[-\mathrm{CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2-}\right]_n$
Buna-S is a copolymer of buta-1,3-diene and styrene. Its commonly written repeating unit is formed by addition polymerization of the diene fragment with styrene incorporation, matching the structure in option D.
Buna-S is a copolymer of buta-1,3-diene and styrene. Its commonly written repeating unit is formed by addition polymerization of the diene fragment with styrene incorporation, matching the structure in option D.
Arrange the following compounds in increasing order of rate of aromatic electrophilic substitution reaction
- A. c, a, b, d
- B. b, c, a, d
- C. d, b, c, a
- D. d, b, a, c
Answer: A. c, a, b, d
The rate of electrophilic aromatic substitution increases with activation of the benzene ring. Compound $c$ has a carbonyl group conjugated with the ring, strongly deactivating it, so it is slowest. Compound $a$ has an alcohol substituent on the saturated ring carbon and is less activating than $b$. Compound $b$ has a phenolic $\mathrm{-OH}$ directly on the aromatic ring, which activates strongly. Compound $d$ has both a phenolic $\mathrm{-OH}$ and an electron-donating oxygen-containing ring substituent, making it the most activated. Thus the increasing order is $c < a < b < d$.
The rate of electrophilic aromatic substitution increases with activation of the benzene ring. Compound $c$ has a carbonyl group conjugated with the ring, strongly deactivating it, so it is slowest. Compound $a$ has an alcohol substituent on the saturated ring carbon and is less activating than $b$. Compound $b$ has a phenolic $\mathrm{-OH}$ directly on the aromatic ring, which activates strongly. Compound $d$ has both a phenolic $\mathrm{-OH}$ and an electron-donating oxygen-containing ring substituent, making it the most activated. Thus the increasing order is $c < a < b < d$.