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MHT-CET Chemistry · Alkynes

Making Alkynes, Partial Hydrogenation, and Adding a Carbon

An alkyne is made by removing two HX from a dihalide, lengthened by alkylating its acetylide, and reduced to a cis-alkene over Lindlar's catalyst; a cyanide adds one carbon to an alkyl halide in the same spirit.

All Alkynes notesMHT-CET Chemistry strategy

Why this matters

7 PYQs, all but one MODERATE. Three are elimination and acetylide sequences, two are Lindlar's catalyst, and two are the KCN-then-reduce route to an amine. Three cards.

Concept 1 of 3: Double Dehydrohalogenation and Acetylide Alkylation

Removing HX twice turns a C–C single bond into a triple bond: alcoholic KOH takes the first HX, and the stronger base sodamide takes the second. A terminal alkyne's H is weakly acidic, so a strong base removes it and the acetylide attacks an alkyl halide, adding carbons to the chain.

Definition

  • Vicinal or geminal dihalide + alc. KOH → vinyl halide; + NaNH₂ → alkyne. 1,2-dibromoethane → ethyne.
  • Addition runs backwards too: HC≡CH + 2HBr → CH₃CHBr₂; alc. KOH then NaNH₂ give ethyne back.
  • Acetylide: HC≡CH + LiNH₂ → HC≡C⁻Li⁺; + CH₃CH₂Br → but-1-yne.
SequenceProduct
1,2-Dibromoethane → alc. KOH → A → NaNH₂ → BB = ethyneQ
A → LiNH₂ → ethynyl lithium → C₂H₅Br → but-1-yneA = ethyneQ
HC≡CH → HBr → HBr → alc. KOH → NaNH₂ → DD = ethyneQ
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q52Moderate

Example 1 · Alkynes · Reactions of Alkynes

Identify product 'B' in the following sequence of reactions. 1,2-Dibromoethane →alcoholKOHA→sodamideB\xrightarrow[\text{alcohol}]{\text{KOH}} A \xrightarrow{\text{sodamide}} B

Reading hydration into a sequence that has no water

Ethyne → ethanal needs H₂O with Hg²⁺/H₂SO₄. The 2021 sequence is HBr, HBr, alcoholic KOH, NaNH₂ — add two HBr and take them off again, and you are back at ethyne (the official key).
Drill 2 more on double dehydrohalogenation and acetylide alkylation

Concept 2 of 3: Partial Hydrogenation to a cis-Alkene

An ordinary catalyst would hydrogenate all the way to the alkane. Lindlar's catalyst is poisoned so it stops at the alkene, and both H atoms are delivered from the metal surface on the same side — cis.

Definition

  • Lindlar's catalyst: Pd–C (or Pd/CaCO₃) poisoned with quinoline → cis-alkene.
  • Na / liquid NH₃ → trans-alkene.
  • Pt, Pd or Ni with excess H₂ → alkane.
ReagentProduct from R–C≡C–R
Pd–C / quinoline (Lindlar)cis-alkeneQ
Na / liquid NH₃trans-alkeneQ
Ni or Pt, excess H₂Alkane
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q52Easy

Example 2 · Alkynes · Reactions of Alkynes

Which of the following reagents is used to convert C≡CC \equiv C triple bond to C=CC = C double bond to give Cis isomer of alkene?

Na in liquid ammonia

It stops at the alkene too, but it gives the TRANS isomer. Cis needs Lindlar's catalyst.
Drill 1 more on partial hydrogenation to a cis-alkene

Concept 3 of 3: Adding One Carbon with Cyanide

Cyanide substitutes a halide and brings its carbon with it, so CH₃X becomes CH₃CN. Reducing the C≡N with sodium in ethanol gives a primary amine one carbon longer than the halide — ethylamine from a methyl halide.

Definition

  • CH₃X + KCN → CH₃CN (ethanenitrile).
  • Na / C₂H₅OH (Mendius reduction) → CH₃CH₂NH₂.
SequenceB
CH₃I + KCN → A; A + Na/C₂H₅OH → BCH₃CH₂NH₂Q
CH₃Br → KCN → A → Na/C₂H₅OH → BCH₃CH₂NH₂Q
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q93Moderate

Example 3 · Alkynes · Reactions of Alkynes

Identify product 'B' in following reaction. CH3-I+KCN→A→Na, C2H5OHB\text{CH}_3\text{-I} + \text{KCN} \to A \xrightarrow{\text{Na, C}_2\text{H}_5\text{OH}} B
Drill 1 more on adding one carbon with cyanide

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (3)

Double Dehydrohalogenation and Acetylide Alkylation3 rows
SequenceProduct
1,2-Dibromoethane → alc. KOH → A → NaNH₂ → BB = ethyneQ
A → LiNH₂ → ethynyl lithium → C₂H₅Br → but-1-yneA = ethyneQ
HC≡CH → HBr → HBr → alc. KOH → NaNH₂ → DD = ethyneQ
Partial Hydrogenation to a cis-Alkene3 rows
ReagentProduct from R–C≡C–R
Pd–C / quinoline (Lindlar)cis-alkeneQ
Na / liquid NH₃trans-alkeneQ
Ni or Pt, excess H₂Alkane
Adding One Carbon with Cyanide2 rows
SequenceB
CH₃I + KCN → A; A + Na/C₂H₅OH → BCH₃CH₂NH₂Q
CH₃Br → KCN → A → Na/C₂H₅OH → BCH₃CH₂NH₂Q

Watch out for (2)

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