alkenes MCQs for UPSC Prelims

35 practice questions on alkenes from the Hydrocarbons section of the UPSC Prelims syllabus. 35 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

10 Easy 19 Medium 6 Hard

Sample questions

Q1
easy

Which condition is necessary for an alkene to exhibit geometrical (cis-trans) isomerism?

  1. A The alkene must be cyclic and saturated.
  2. B The alkene must have identical groups attached to each carbon of the double bond.
  3. C The alkene must have different groups attached to each carbon of the double bond.
  4. D The alkene must have only hydrogen atoms attached to the carbons of the double bond.
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Correct answer: C - The alkene must have different groups attached to each carbon of the double bond.

Geometrical isomerism in alkenes requires different groups attached to each carbon atom of the double bond. Incorrect options reflect common misconceptions about hydrogen presence or the necessity of a cyclic or saturated structure.

Q2
medium

Assertion (A): Alkenes like ethene can rotate freely around the carbon-carbon double bond. Reason (R): The pi bond in the double bond restricts rotation due to its sideways overlap.

  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
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Correct answer: D - A is false, but R is true

The assertion is false because alkenes cannot rotate freely due to the pi bond. The pi bond restricts rotation, which is correctly pointed out in the reason, making only R true.

Q3
medium

Assertion (A): In the absence of peroxides, propene reacts with $\text{HBr}$ to give $1$-bromopropane as the major product. Reason (R): By Markovnikov's rule, when $\text{HX}$ adds to an unsymmetrical alkene the halogen goes to the carbon carrying the smaller number of hydrogen atoms.

  1. A Both A and R are true, R explains A
  2. B Both A and R are true, R does not explain A
  3. C A is true, R is false
  4. D A is false, R is true
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Correct answer: D - A is false, R is true

A is false. Without peroxides the addition follows Markovnikov's rule and the major product is $2$-bromopropane. R is true, and applying it to propene shows why: the middle carbon of the double bond carries one hydrogen and the terminal carbon two, so bromine attaches to the middle carbon. $1$-Bromopropane becomes the major product only when peroxides are present.

Q4
hard

In a reaction, 2-butanol is dehydrated to form butenes. Which product is favored according to Saytzeff's rule?

  1. A 2-butene
  2. B 1-butene
  3. C 1,3-butadiene
  4. D butane
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Correct answer: A - 2-butene

Saytzeff's rule predicts that the more substituted alkene, 2-butene, will be favored during the dehydration of 2-butanol. 1-butene is less substituted, and butane and 1,3-butadiene are not dehydration products.

Q5
medium

Which of the following alkenes will exhibit geometrical (cis-trans) isomerism?

  1. A Ethene
  2. B 1-butene
  3. C 2-butene
  4. D Propene
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Correct answer: C - 2-butene

2-butene exhibits geometrical isomerism because it has two different groups on each carbon of the double bond. Propene and ethene lack this characteristic as one of the carbons does not have two different substituents. 1-butene does not have the necessary conditions for cis-trans isomerism.

Q6
easy

Which type of bond characterizes the double bond in ethene?

  1. A Two pi bonds
  2. B One sigma bond only
  3. C One sigma and one pi bond
  4. D Two sigma bonds
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Correct answer: C - One sigma and one pi bond

In ethene, the double bond consists of one sigma bond, which is formed by the head-on overlap of orbitals, and one pi bond, resulting from the side-on overlap. Options with two sigma or two pi bonds are incorrect as these do not occur in a double bond.

Q7
easy

The structure of ethene involves a double bond. Which of the following correctly describes the components of this double bond?

  1. A Two sigma bonds
  2. B One sigma bond and one pi bond
  3. C Two pi bonds
  4. D One pi bond only
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Correct answer: B - One sigma bond and one pi bond

The double bond in ethene consists of one sigma bond formed by end-to-end overlap and one pi bond formed by side-to-side overlap. Incorrect options arise from misunderstanding the bond structure.

Q8
medium

When 1 mole of an unknown alkene is ozonized and then reductively cleaved, 1 mole of formaldehyde and 1 mole of acetaldehyde are formed. Identify the alkene.

  1. A Ethene
  2. B Propene
  3. C But-1-ene
  4. D But-2-ene
Show answer and explanation

Correct answer: B - Propene

The cleavage products indicate that the alkene was Propene (CH3-CH=CH2), which upon ozonolysis gives formaldehyde (from the terminal CH2) and acetaldehyde (from the internal carbon-carbon double bond). Butenes would not produce formaldehyde.

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