alkynes MCQs for UPSC Prelims

29 practice questions on alkynes from the Hydrocarbons section of the UPSC Prelims syllabus. 29 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

11 Easy 13 Medium 5 Hard

Sample questions

Q1
hard

But-1-ene is to be converted into but-1-yne. Which sequence of reagents achieves this?

  1. A $\text{HBr}$, then alcoholic KOH
  2. B $\text{Br}_2$ in $\text{CCl}_4$, then zinc dust
  3. C Hot concentrated $\text{KMnO}_4$, then sodamide
  4. D $\text{Br}_2$ in $\text{CCl}_4$, then excess sodamide
Show answer and explanation

Correct answer: D - $\text{Br}_2$ in $\text{CCl}_4$, then excess sodamide

Bromine adds across the double bond to give 1,2-dibromobutane, and excess sodamide then removes two molecules of HBr to leave a triple bond. Adding HBr and eliminating with alcoholic KOH simply returns an alkene. Bromine followed by zinc dust also regenerates but-1-ene, and hot $\text{KMnO}_4$ cleaves the double bond into carboxylate fragments instead.

Q2
hard

But-2-yne is partially reduced by two methods: method X uses hydrogen over Lindlar's catalyst, and method Y uses sodium in liquid ammonia. Which pair of alkenes is obtained?

  1. A X gives cis-but-2-ene and Y gives trans-but-2-ene
  2. B X gives trans-but-2-ene and Y gives cis-but-2-ene
  3. C Both methods give butane
  4. D X gives butane and Y gives cis-but-2-ene
Show answer and explanation

Correct answer: A - X gives cis-but-2-ene and Y gives trans-but-2-ene

Lindlar's catalyst is a poisoned palladium surface, so both hydrogen atoms are delivered to the same face of the triple bond and the cis alkene results. Sodium in liquid ammonia adds electrons and protons one at a time through an intermediate that settles into the less crowded trans arrangement. Neither method reduces past the alkene stage, so any option giving butane is wrong.

Q3
easy

Ethyne can be prepared by reacting calcium carbide (CaCâ‚‚) with water. Apart from ethyne, what is the other product of this reaction?

  1. A Carbon dioxide gas
  2. B Oxygen gas
  3. C Nitrogen gas
  4. D Calcium hydroxide
Show answer and explanation

Correct answer: D - Calcium hydroxide

The reaction is $\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$, so the only other product is calcium hydroxide, which stays behind as a slurry rather than escaping as a gas. No oxygen, nitrogen or carbon dioxide is formed; the carbon leaves entirely as ethyne.

Q4
easy

How many sigma bonds and how many pi bonds are present in one molecule of ethyne, $\text{C}_2\text{H}_2$?

  1. A $2$ sigma and $3$ pi
  2. B $3$ sigma and $2$ pi
  3. C $5$ sigma and $2$ pi
  4. D $4$ sigma and $1$ pi
Show answer and explanation

Correct answer: B - $3$ sigma and $2$ pi

Ethyne is $\text{H}-\text{C}\equiv\text{C}-\text{H}$, so there are two C-H sigma bonds and one C-C sigma bond, making three sigma bonds in all. A triple bond is one sigma plus two pi, so the two pi bonds come from the sideways overlap of the unhybridised p orbitals on the two $sp$ carbons. Counting every line of the triple bond as a pi bond gives three pi, and the larger counts add bonds the molecule does not have.

Q5
medium

Which treatment converts 1,2-dibromoethane into ethyne?

  1. A Aqueous KOH at room temperature
  2. B Dilute $\text{H}_2\text{SO}_4$ with heating
  3. C Alcoholic KOH, then sodamide ($\text{NaNH}_2$)
  4. D Zinc dust in ethanol
Show answer and explanation

Correct answer: C - Alcoholic KOH, then sodamide ($\text{NaNH}_2$)

Two molecules of HBr have to be removed, and this needs two bases of different strength. Alcoholic KOH removes the first to give bromoethene, and the much stronger base sodamide removes the second to give ethyne. Aqueous KOH substitutes the bromines to give a glycol, zinc dust strips both halogens to give ethene, and dilute acid causes no elimination at all.

Q6
medium

1,2-Dibromopropane is warmed with one equivalent of alcoholic KOH. What is the organic product of this first step, before any sodamide is added?

  1. A Propan-1-ol
  2. B Propene
  3. C Propyne
  4. D A bromopropene
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Correct answer: D - A bromopropene

One equivalent of alcoholic KOH removes one molecule of HBr, so only one of the two bromines is lost. The product is a vinylic bromide, a bromopropene, which still carries bromine on a doubly bonded carbon. Propyne requires a second elimination by the stronger base sodamide, propene would need both bromines replaced by hydrogen, and an alcohol would need substitution by aqueous alkali.

Q7
easy

Ethyne is passed into dilute sulphuric acid containing mercuric sulphate at about $333\ \text{K}$. Which compound is obtained?

  1. A Propanone
  2. B Ethanal
  3. C Methanal
  4. D Ethanol
Show answer and explanation

Correct answer: B - Ethanal

Water adds across the triple bond to give an unstable enol, $\text{CH}_2\text{=CHOH}$, which rearranges at once to the carbonyl compound ethanal. Ethanol would require the addition of hydrogen rather than water. Propanone has three carbons, which ethyne cannot supply, and methanal has only one.

Q8
medium

Assertion (A): Ethyne undergoes cyclic polymerization to form benzene. Reason (R): Ethyne molecules react together under high temperature and pressure conditions to undergo polymerization.

  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: A - Both A and R are true, and R is the correct explanation of A

The assertion and reason are both true and interconnected. Ethyne does polymerize to benzene under specific conditions, such as high temperature and pressure, demonstrating that the reason explains the assertion.

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