Alkynes- Acidic character – Addition of hydrogen: NEET MCQs with Answers

15 free questions on Alkynes- Acidic character – Addition of hydrogen for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Addition of hydrogen, halogens, water, and hydrogenhalides: Polymerization

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Question 1

Previous-year question
When water is added to an alkyne containing more than two carbon atoms in the presence of HgSO₄ and dilute H₂SO₄ at 333 K, the product is [NEET 2022 (Overseas)]
  1. an aldehyde
  2. an alcohol
  3. a ketone
  4. a vicinal diol
Show answer
Answer: (C) a ketone
Explanation: Hydration of alkynes (Hg²⁺/H₂SO₄, 333 K) follows Markovnikov's rule: the enol formed first tautomerises to a carbonyl compound. Only ethyne gives an aldehyde (ethanal); higher alkynes such as propyne give ketones (propyne → propanone). Alcohols and vicinal diols are not the final products.

Question 2

Previous-year question
Which of the following hydrocarbons forms a white precipitate with ammoniacal silver nitrate solution? [NEET 2021 (Paper-2)]
  1. C₂H₆
  2. C₃H₄
  3. C₃H₈
  4. C₄H₁₀
Show answer
Answer: (B) C₃H₄
Explanation: Terminal alkynes have an acidic ≡C–H hydrogen which is replaced by silver in ammoniacal AgNO₃ (Tollens' reagent), giving a white precipitate of silver alkynide. C₃H₄ (propyne, CH₃C≡CH) is a terminal alkyne: CH₃C≡CH → CH₃C≡CAg↓. C₂H₆, C₃H₈ and C₄H₁₀ are alkanes with no acidic hydrogen.

Question 3

Previous-year question
The most suitable reagent for the conversion H₃C–C≡C–CH₃ → cis-2-butene is [NEET 2019]
  1. Na/liquid NH₃
  2. H₂, Pd/C with quinoline
  3. Zn/HCl
  4. Hg²⁺/H⁺, H₂O
Show answer
Answer: (B) H₂, Pd/C with quinoline
Explanation: Lindlar's catalyst — palladium on calcium carbonate or barium sulphate, poisoned with quinoline — adds one mole of hydrogen from a single face of the alkyne. This syn addition gives the cis alkene and stops there.

Question 4

Previous-year question
Predict the correct intermediate (A) and product (B) in the reaction H₃C–C≡CH —H₂O, H₂SO₄, HgSO₄→ (A) → (B). [NEET 2017]
  1. A: H₃C–C(OH)=CH₂ ; B: H₃C–C(OSO₃H)=CH₂
  2. A: H₃C–CO–CH₃ ; B: H₃C–C≡CH
  3. A: H₃C–C(OH)=CH₂ ; B: H₃C–CO–CH₃
  4. A: H₃C–C(OSO₃H)=CH₂ ; B: H₃C–CO–CH₃
Show answer
Answer: (C) A: H₃C–C(OH)=CH₂ ; B: H₃C–CO–CH₃
Explanation: Markownikoff hydration puts the –OH on the middle carbon, giving the enol prop-1-en-2-ol as the unstable intermediate. This immediately tautomerises to the more stable keto form, propan-2-one.

Question 5

Previous-year question
Which one is the correct order of acidity? [NEET 2017]
  1. CH≡CH > CH₃–C≡CH > CH₂=CH₂ > CH₃–CH₃
  2. CH≡CH > CH₂=CH₂ > CH₃–C≡CH > CH₃–CH₃
  3. CH₃–CH₃ > CH₂=CH₂ > CH₃–C≡CH > CH≡CH
  4. CH₂=CH₂ > CH₃–CH=CH₂ > CH₃–C≡CH > CH≡CH
Show answer
Answer: (A) CH≡CH > CH₃–C≡CH > CH₂=CH₂ > CH₃–CH₃
Explanation: Acidity of a C–H bond increases with the s-character of the orbital holding the electron pair, so sp > sp² > sp³. Among the two sp cases, the electron-releasing methyl group of propyne destabilises the anion, so acetylene is the most acidic.

Question 6

Previous-year question
In the reaction HC≡CH —(i) NaNH₂/liq. NH₃, (ii) CH₃CH₂Br→ X —(i) NaNH₂/liq. NH₃, (ii) CH₃CH₂Br→ Y, X and Y are [NEET 2016]
  1. X = 1-butyne; Y = 3-hexyne
  2. X = 2-butyne; Y = 3-hexyne
  3. X = 2-butyne; Y = 2-hexyne
  4. X = 1-butyne; Y = 2-hexyne
Show answer
Answer: (A) X = 1-butyne; Y = 3-hexyne
Explanation: Sodamide converts the terminal alkyne into its acetylide, which then reacts with ethyl bromide in an SN2 alkylation. The first alkylation gives but-1-yne and the second, on the remaining terminal hydrogen, gives hex-3-yne.

Question 7

Previous-year question
Which of the following organic compounds has the same hybridisation as its combustion product CO₂? [AIPMT 2014]
  1. Ethane
  2. Ethyne
  3. Ethene
  4. Ethanol
Show answer
Answer: (B) Ethyne
Explanation: Carbon in CO₂ is sp hybridised because it is bonded to two atoms with no lone pair. Among the compounds listed, only ethyne has sp hybridised carbons.
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Question 8

Which reaction route correctly converts sodium ethanoate into benzene?
  1. Kolbe electrolysis to ethane; Cl₂/hν to chloroethane; aqueous KOH to ethene; Br₂ to 1,2-dibromoethane; NaNH₂ to ethyne; red-hot iron tube to benzene
  2. Soda lime to ethane; Cl₂/hν to chloroethane; alcoholic KOH to ethene; Br₂ to 1,2-dibromoethane; NaNH₂ to ethyne; red-hot iron tube to benzene
  3. Kolbe electrolysis to ethane; Cl₂/hν to chloroethane; alcoholic KOH to ethene; Br₂ to 1,2-dibromoethane; NaNH₂ to ethyne; red-hot iron tube to benzene
  4. Kolbe electrolysis to ethane; Cl₂/hν to chloroethane; alcoholic KOH to ethene; Br₂ to 1,2-dibromoethane; Zn dust to ethyne; red-hot iron tube to benzene
Show answer
Answer: (C) Kolbe electrolysis to ethane; Cl₂/hν to chloroethane; alcoholic KOH to ethene; Br₂ to 1,2-dibromoethane; NaNH₂ to ethyne; red-hot iron tube to benzene
Explanation: Kolbe electrolysis of sodium ethanoate gives ethane, whereas soda lime would give methane. Chloroethane loses HCl only with alcoholic KOH, because aqueous KOH gives ethanol. A vicinal dibromide needs a strong base such as NaNH₂ to lose two HBr and form ethyne, because zinc dust merely removes Br₂ and regenerates ethene. Ethyne then trimerises to benzene in a red-hot iron tube.

Question 9

But-2-ene shows geometrical isomerism, but but-2-yne does not, even though its triple bond also restricts rotation. What is the reason?
  1. The triple bond in but-2-yne contains no π bonds
  2. The two methyl groups in but-2-yne are not identical
  3. The C–C≡C–C unit is linear, so the methyl groups cannot lie on the same or opposite sides of the bond
  4. But-2-yne has sp³ hybridised carbons at the triple bond
Show answer
Answer: (C) The C–C≡C–C unit is linear, so the methyl groups cannot lie on the same or opposite sides of the bond
Explanation: Each triply bonded carbon is sp hybridised with a bond angle of 180°, so the four-carbon chain is linear. Each triply bonded carbon carries only one substituent, placed in line with the triple bond. No cis or trans arrangement is therefore possible.

Question 10

2.0 g of propyne is treated with excess bromine in CCl₄ until no further reaction occurs. What is the final organic product and its mass, assuming complete reaction? (C = 12, H = 1, Br = 80)
  1. 1,1,2,2-Tetrabromopropane, 18.0 g
  2. 1,2,2,3-Tetrabromopropane, 18.0 g
  3. 1,1,2,2-Tetrabromopropane, 9.0 g
  4. 1,2-Dibromopropene, 10.0 g
Show answer
Answer: (A) 1,1,2,2-Tetrabromopropane, 18.0 g
Explanation: The two π bonds of the triple bond each add one Br₂, placing two bromine atoms on each of the original alkyne carbons, which gives CH₃CBr₂CHBr₂. Propyne has M = 40 g mol⁻¹, so 2.0 g is 0.05 mol. The product has M = 360 g mol⁻¹, so its mass is 0.05 × 360 = 18.0 g.

Question 11

Sodium ethynide reacts with water to give ethyne and NaOH, and ethyne reacts with sodamide to give sodium ethynide and ammonia. What order of acid strength do these two reactions show?
  1. HC≡CH > H₂O > NH₃
  2. H₂O > NH₃ > HC≡CH
  3. H₂O > HC≡CH > NH₃
  4. NH₃ > HC≡CH > H₂O
Show answer
Answer: (C) H₂O > HC≡CH > NH₃
Explanation: An acid–base reaction goes towards the weaker acid. Water protonates the ethynide ion, so water is a stronger acid than ethyne. Ethyne protonates the amide ion, so ethyne is a stronger acid than ammonia.

Question 12

Assertion: Although a triple bond has more π electrons than a double bond, alkynes are generally less reactive than alkenes towards electrophilic addition. Reason: The π electrons of an alkyne are held more tightly by sp carbons, and the vinylic cation formed is less stable.
  1. Assertion is false, Reason is true
  2. Assertion is true, Reason is true, and Reason correctly explains Assertion
  3. Assertion is true, Reason is true, but Reason does not explain Assertion
  4. Assertion is true, Reason is false
Show answer
Answer: (B) Assertion is true, Reason is true, and Reason correctly explains Assertion
Explanation: sp carbons have greater s character and hold the π electrons more firmly, so these electrons are less available to an electrophile. Attack on an alkyne also gives a vinylic cation, which is less stable than the alkyl carbocation formed from an alkene. Both factors raise the barrier for addition, so the Reason explains the Assertion.

Question 13

3.4 g of pent-1-yne is treated with excess sodium metal until the reaction is complete. What volume of hydrogen is evolved at STP? (Take the molar volume as 22.4 L mol⁻¹; C = 12, H = 1)
  1. 560 mL
  2. 280 mL
  3. 1120 mL
  4. 2240 mL
Show answer
Answer: (A) 560 mL
Explanation: The molar mass of C₅H₈ is 68 g mol⁻¹, so 3.4 g is 0.05 mol. The terminal alkyne has one acidic hydrogen: 2 RC≡CH + 2 Na → 2 RC≡CNa + H₂. So 0.05 mol alkyne gives 0.025 mol H₂, which is 0.025 × 22400 = 560 mL at STP.

Question 14

Cyclic polymerisation in a red-hot iron tube at 873 K joins three alkyne molecules into a benzene ring. Which alkyne gives hexamethylbenzene by this route?
  1. But-1-yne
  2. Ethyne
  3. Propyne
  4. But-2-yne
Show answer
Answer: (D) But-2-yne
Explanation: Each alkyne molecule supplies two adjacent ring carbons, so the substituents on the triple-bond carbons end up on the ring. But-2-yne carries one methyl group on each triple-bond carbon, so three molecules place six methyl groups on the ring. Ethyne gives benzene, and propyne gives mainly 1,3,5-trimethylbenzene.

Question 15

The reactivity of hydrogen halides in addition to propyne follows the order HI > HBr > HCl > HF. The main reason for this order is that, from HF to HI,
  1. the electronegativity of the halogen increases
  2. the H–X bond becomes longer and weaker, so a proton is donated to the π bond more easily
  3. the size of the halide ion decreases
  4. the H–X bond becomes stronger, so the molecule collides more effectively
Show answer
Answer: (B) the H–X bond becomes longer and weaker, so a proton is donated to the π bond more easily
Explanation: The first step of the addition is transfer of H⁺ from H–X to the π electrons. The H–X bond weakens down the group as the halogen gets larger, so HI gives up its proton most easily. HF has the strongest bond and reacts the least readily.

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