aromatic-hydrocarbons MCQs for UPSC Prelims

34 practice questions on aromatic-hydrocarbons from the Hydrocarbons section of the UPSC Prelims syllabus. 34 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

10 Easy 17 Medium 7 Hard

Sample questions

Q1
easy

Benzene is known to have equal bond lengths due to which of the following phenomena?

  1. A Resonance
  2. B Inductive effect
  3. C Ionic bonding
  4. D Hyperconjugation
Show answer and explanation

Correct answer: A - Resonance

Resonance in benzene leads to the delocalization of electrons across all the carbon atoms, resulting in equal bond lengths. Ionic bonding, inductive effect, and hyperconjugation do not result in equal bond lengths in benzene.

Q2
medium

In Friedel-Crafts alkylation, why is a Lewis acid catalyst like AlCl3 necessary?

  1. A To reduce the alkyl halide
  2. B To generate the electrophile from the alkyl halide
  3. C To stabilize the benzene ring
  4. D To increase the reaction temperature
Show answer and explanation

Correct answer: B - To generate the electrophile from the alkyl halide

AlCl3 is necessary to generate the electrophile (carbocation) from the alkyl halide by accepting a chlorine atom. It does not increase temperature or stabilize the benzene ring, nor does it reduce the alkyl halide.

Q3
medium

Assertion (A): Polynuclear aromatic hydrocarbons tend to be carcinogenic. Reason (R): Polynuclear aromatic hydrocarbons are produced when tobacco or coal is burned incompletely at high temperature.

  1. A Both A and R are true, and R is the correct explanation of A.
  2. B Both A and R are true, but R is not the correct explanation of A.
  3. C A is true, but R is false.
  4. D A is false, but R is true.
Show answer and explanation

Correct answer: B - Both A and R are true, but R is not the correct explanation of A.

Both statements are true. Compounds such as benzpyrene are known carcinogens, and they are indeed formed by the incomplete combustion of tobacco, coal and other organic matter. But R only says where these compounds come from, and the source of a substance does not make it carcinogenic. Their cancer-causing action comes from what they do once inside a cell, so R does not explain A.

Q4
medium

A student mistakenly believes that Friedel-Crafts alkylation can be performed on a nitrobenzene. Why is this incorrect?

  1. A Nitrobenzene is deactivated and unreactive in Friedel-Crafts reactions.
  2. B Friedel-Crafts alkylation requires a basic catalyst.
  3. C Nitrobenzene directly forms a carbocation without a catalyst.
  4. D AlCl3 used in Friedel-Crafts will rapidly react with the nitro group.
Show answer and explanation

Correct answer: A - Nitrobenzene is deactivated and unreactive in Friedel-Crafts reactions.

Nitrobenzene is highly deactivated towards Friedel-Crafts reactions due to the electron-withdrawing nature of the -NO2 group, making it unreactive. AlCl3 doesn't react specifically with the nitro group, and the reaction does not require a basic catalyst.

Q5
medium

Calculate the theoretical heat of hydrogenation for benzene if the hydrogenation of cyclohexene releases 120 kJ/mol and compare it with benzene’s actual heat of hydrogenation, which is 208 kJ/mol.

  1. A 240 kJ/mol; benzene is more stable by 32 kJ/mol.
  2. B 360 kJ/mol; benzene is more stable by 32 kJ/mol.
  3. C 240 kJ/mol; benzene is more stable by 152 kJ/mol.
  4. D 360 kJ/mol; benzene is more stable by 152 kJ/mol.
Show answer and explanation

Correct answer: D - 360 kJ/mol; benzene is more stable by 152 kJ/mol.

The theoretical heat of hydrogenation for benzene (3 double bonds) is 3 x 120 = 360 kJ/mol. Benzene’s actual heat of hydrogenation is 208 kJ/mol, indicating a stability gain of 360 - 208 = 152 kJ/mol. Other calculations misinterpret the number of double bonds or the actual heat value.

Q6
medium

Cyclobutadiene is a flat four-membered carbon ring containing two carbon-carbon double bonds. Applying Huckel's rule, is this molecule aromatic?

  1. A No, because a four-membered ring is too small ever to be planar
  2. B No, its ring holds $4$ pi electrons, and $4$ does not fit $4n+2$ for any whole number $n$
  3. C Yes, because the ring is cyclic, flat and made only of $sp^2$ carbons
  4. D Yes, its ring holds $4$ pi electrons, which fits $4n+2$ with $n=1$
Show answer and explanation

Correct answer: B - No, its ring holds $4$ pi electrons, and $4$ does not fit $4n+2$ for any whole number $n$

Two double bonds contribute two pi electrons each, so the ring holds four delocalised pi electrons. Huckel's rule needs $4n+2$ electrons with $n$ a whole number, and $4n+2=4$ would require $n=0.5$, so the count fails and cyclobutadiene is not aromatic. Putting $n=1$ into $4n+2$ gives $6$, not $4$, and being cyclic, planar and conjugated is necessary but not sufficient without the right electron count.

Q7
hard

In an electrophilic aromatic substitution mechanism, how is the arenium ion stabilized?

  1. A By resonance, delocalizing positive charge over the ring.
  2. B By hyperconjugation in the conjugated system.
  3. C By the formation of a transient carbocation.
  4. D By inductive effects from electron-withdrawing groups.
Show answer and explanation

Correct answer: A - By resonance, delocalizing positive charge over the ring.

The arenium ion is stabilized by resonance, which allows the positive charge to be delocalized across the aromatic ring, maintaining stability. Inductive effects and hyperconjugation are not the primary means of stabilization for arenium ions, and it is not merely a transient carbocation.

Q8
medium

Benzene is nitrated using a mixture of concentrated nitric acid and concentrated sulphuric acid. What is the role of the sulphuric acid in this mixture?

  1. A It acts as a base, taking a proton from benzene before the nitric acid attacks
  2. B It supplies a sulphonium ion that is the species actually attacking the benzene ring
  3. C It only mops up the water formed later, and no new species is produced from the nitric acid
  4. D It protonates nitric acid so that water leaves, generating the nitronium ion $\text{NO}_2^+$
Show answer and explanation

Correct answer: D - It protonates nitric acid so that water leaves, generating the nitronium ion $\text{NO}_2^+$

Sulphuric acid is the stronger acid, so it protonates nitric acid; the protonated acid then loses water to give the nitronium ion, and it is $\text{NO}_2^+$ that attacks the ring. Removing water afterwards helps the equilibrium but is not what creates the electrophile. No sulphonium ion is involved in nitration, and benzene is the nucleophile here, so nothing removes a proton from it before the attack.

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