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Organic Chemistry · Alkynes: An Introduction to Organic Synthesis

Reactions of Alkynes: Addition of HX and X2

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All product predictions follow standard Markovnikov/anti-addition rules; no experimental procedures or safety data are included.
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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Alkynes, like alkenes, undergo electrophilic addition across their π bonds — but a triple bond is two π bonds, so the addition can happen once or twice. This topic covers the two simplest addition families: hydrogen halides (HX: HCl, HBr, HI) and halogens (X2: Br2, Cl2).

The master pattern is "alkene addition, then vinyl addition":

  1. The first addition converts the alkyne into an alkene (a — halogen on a double-bond carbon).
  2. A second addition then converts that alkene into a saturated compound.

The regiochemistry in every step follows , and the stereochemistry of halogen addition follows the anti (opposite-face) pattern you know from alkene bromination.

Two outcomes deserve special attention because they are classic exam traps:

  • *HX adds twice to give a geminal* dihalide** (R-CX2-R') — both halogens end up on the same carbon, not on adjacent carbons. Students routinely predict the vicinal product and lose the point.
  • X2 adds once to give an (E)-dihaloalkene — the two halogens add anti to each other, so they sit on opposite faces of the double bond, forcing the E geometry. A second addition of X2 gives the .

Controlling how much reagent you add (1 equivalent vs. excess) controls how far the reaction goes — the same molecule can stop at the vinyl halide or run all the way to the saturated dihalide.

Why this matters

  • You can't build alkynes' chemistry without knowing their additions. These reactions appear in every synthesis problem in this chapter and later ones. If you don't know that "HBr (2 equiv)" means geminal dibromide, you will misdraw the product of half the reactions you meet.
  • The Markovnikov + anti pattern is a thinking skill. Once you see that alkyne additions are just alkene additions run twice, you can predict new reactions you've never seen. This topic is the practice ground for that reasoning.
  • Vinyl halides are synthetic building blocks. The mono-addition products (R-CX=CH2 and (E)-R-CX=CHX) are useful intermediates in cross-coupling and polymer chemistry — the very reactions that put halogens on sp2 carbons where substitution is impossible.
  • Exam frequency is extremely high. "What product forms when 1 equiv/2 equiv of HBr reacts with propyne?" is a canonical question. This topic gives you the framework to answer it cold.
  • Industrial relevance: vinyl chloride (CH2=CHCl, from HCl + acetylene) is the monomer for PVC; halogenated alkynes and vinyl halides appear throughout materials and pharmaceutical chemistry.

The college version

Core Concepts

Addition 1: alkyne → vinyl halide

The triple bond acts like a concentrated version of an alkene: it donates π-electron density to an electrophile. For HX, the proton adds first, forming a vinylic carbocation, which is then trapped by the halide:

R-C#CH + HX → R-CX=CH2

Markovnikov's rule applies: the proton goes to the alkyne carbon that already has more hydrogens, and the halogen goes to the more substituted carbon. For a terminal alkyne R-C#CH, the product is the more substituted vinyl halide, R-CX=CH2.

For X2 (e.g., Br2), addition goes through a halonium-ion-like intermediate and delivers the two halogens to opposite faces (). For a symmetric alkyne like 2-butyne:

CH3-C#C-CH3 + Br2 → (E)-CH3-CBr=CHBr-CH3

The anti addition means the two Br atoms are on opposite sides of the double bond → E geometry. For terminal alkynes, the halogen ends up on the more substituted carbon first, again following Markovnikov.

Addition 2: vinyl halide → saturated product (or tetrahalide)

The vinyl halide from step 1 is still an alkene, so it reacts again:

  • HX (excess): Markovnikov addition to the vinyl halide puts the second halogen on the carbon that already carries the first halogen — because that carbon is the more substituted one. Both halogens end up on the same carbon:

R-CX=CH2 + HX → R-CX2-CH3

This is the — the trap to remember. (It's also exactly how you make geminal dihalides from alkynes, which Topic 2's preparation section uses in reverse.)

  • X2 (excess): anti addition to the (E)-dihaloalkene gives the tetrahaloalkane:

CH3-CBr=CHBr-CH3 + Br2 → CH3-CBr2-CHBr2-CH3

Why HX gives geminal, not vicinal, dihalides

Both additions obey Markovnikov. The first halogen lands on the more substituted carbon; the second addition to the vinyl halide also places the new halogen on the more substituted carbon — which is now the same carbon bearing the first halogen. The result is both halogens on one carbon (R-CX2-R'), the geminal arrangement. Vicinal dihalides (halogens on adjacent carbons) are the product of alkene halogenation, not of alkyne HX addition — a distinction examiners love.

Stoichiometry is the control knob

  • 1 equiv HX (or X2) → mono-addition product (vinyl halide / dihaloalkene).
  • Excess HX (2 equiv) → geminal dihalide.
  • Excess X2 (2 equiv) → tetrahaloalkane.

In practice, excess reagent drives the second addition; using exactly 1 equivalent requires careful control (and the mono-addition product is often contaminated), but the conceptual point is exact: the equivalents determine the product.

Common Confusions

Do Not ConfuseWithDifference
Alkyne + 2 HXAlkene + X2Alkyne + 2 HX → geminal dihalide; alkene + X2 → vicinal dihalide. Different starting materials, different products.
Markovnikov product of first additionMarkovnikov product of second additionBoth add H to the H-richer carbon; in the second addition that means the new X lands on the already-halogenated carbon.
1 equiv vs. excess"Same product either way"1 equiv stops at the vinyl halide/dihaloalkene; excess runs to geminal dihalide/tetrahalide. Equivalents change the answer.
(E)-dihaloalkene(Z)-dihaloalkeneAnti addition forces the two X's to opposite faces → E. Predict E, never Z, for X2 addition to alkynes.
Vinyl halide reactivityAlkyl halide reactivityVinyl halides resist substitution but undergo addition/elimination; that's why a second addition (not substitution) follows.
Regiochemistry (Markovnikov)Stereochemistry (anti)Markovnikov: which carbon gets X. Anti: which face. Both apply simultaneously to X2 reactions.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

A triple bond is like a rope made of three strands. A pair of scissors (one molecule of HBr) snips one strand, and the rope becomes a double bond with a little flag (the halogen) on it. Snip again with another pair of scissors, and the rope becomes a single bond with two flags. Here's the sneaky part: the scissors always cut so both flags land on the same side — same carbon for HBr, but for Br2 the two flags land on opposite sides, so the double bond is twisted into the "E" shape. Count the scissors (equivalents) and you know how many strands get cut.

Worked example

Example 1: Propyne + HBr, 1 equiv then excess (the canonical problem)

Setup: Predict the product when propyne, CH3-C#CH, reacts with (a) 1 equiv HBr and (b) excess HBr.

(a) 1 equiv — mono-addition: The proton adds to the terminal carbon (more hydrogens — Markovnikov), giving a vinylic carbocation at the internal carbon; Br− attacks there:

CH3-C#CH + HBr → CH3-CBr=CH2

Product: 2-bromopropene (the more substituted vinyl bromide).

(b) excess — second addition: The vinyl halide CH3-CBr=CH2 is still an alkene. HBr adds Markovnikov again: H to the CH2 carbon, Br to the carbon already bearing Br (more substituted):

CH3-CBr=CH2 + HBr → CH3-CBr2-CH3

Product: 2,2-dibromopropane — a geminal dibromide, both bromines on C2.

Answer: (a) 2-bromopropene; (b) 2,2-dibromopropane. The second halogen lands on the same carbon as the first — geminal, not vicinal.

Example 2: 2-butyne + Br2 (stereochemistry in words)

Setup: Predict the product of 2-butyne, CH3-C#C-CH3, with 1 equiv Br2, then with excess Br2, describing the geometry.

(1 equiv): Br2 adds anti across one π bond. The two Br atoms end up on opposite faces of the resulting double bond, so the two CH3 groups are on opposite sides too — the (E)-isomer:

CH3-C#C-CH3 + Br2 → (E)-CH3-CBr=CHBr-CH3

Product: (E)-2,3-dibromo-2-butene (the E geometry is the signature of anti addition).

(excess): The remaining double bond adds Br2 (again anti) to give the saturated tetrahalide:

(E)-CH3-CBr=CHBr-CH3 + Br2 → CH3-CBr2-CHBr2-CH3

Product: 2,2,3,3-tetrabromobutane.

Answer: 1 equiv → (E)-2,3-dibromo-2-butene; excess → 2,2,3,3-tetrabromobutane. The E geometry comes straight from anti addition.

Example 3: Stoichiometry planning (dimensional analysis)

Setup: A synthesis calls for converting 0.25 mol of 1-hexyne, CH3(CH2)3-C#CH, to the geminal dibromide with excess HBr. How many moles of HBr must be delivered to satisfy the 1:2 alkyne:HBr stoichiometry?

Formula first (each alkyne consumes 2 mol HBr — one per addition):

n(HBr) = n(alkyne) × 2 mol HBr1 mol alkyne

Substitute:

n(HBr) = 0.25 mol × 21 = 0.50 mol

Answer: 0.50 mol HBr minimum (units check: mol alkyne × mol HBr/mol alkyne → mol HBr). "Excess" in the lab means delivering more than this, but 2 equivalents is the stoichiometric requirement.

Key takeaways

  • Alkyne = two π bonds = up to two additions. 1 equiv → alkene-type product; excess → saturated product.
  • HX additions are Markovnikov at every step; 2 equiv HX gives a geminal dihalide R-CX2-R' — never predict vicinal.
  • X2 additions are anti; 1 equiv Br2 gives the (E)-dihaloalkene, excess gives the tetrahaloalkane.
  • Terminal alkynes: first HX lands the halogen on the more substituted (internal) carbon — R-C#CH + HBr → R-CBr=CH2.
  • The vinyl halide intermediate is still an alkene — it reacts again; that's the whole game.
  • Markovnikov + anti + stoichiometry predict every product in this topic; no memorization of individual examples needed.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

  1. What is the product of propyne + 1 equiv HBr? Of propyne + excess HBr?

    Show answer

    2-bromopropene (CH3-CBr=CH2); then 2,2-dibromopropane (CH3-CBr2-CH3, geminal).

  2. Why does HBr addition give a geminal dihalide rather than a vicinal one?

    Show answer

    Both additions follow Markovnikov. The first halogen goes to the more substituted carbon; in the second addition the more substituted carbon is the one already bearing the first halogen — so both halogens collect on the same carbon.

  3. 1 equiv Br2 on 2-butyne gives which product, and what geometry?

    Show answer

    (E)-2,3-dibromo-2-butene — anti addition puts the two Br atoms on opposite faces, giving E geometry.

  4. What is the product of excess Br2 on 2-butyne?

    Show answer

    2,2,3,3-tetrabromobutane, CH3-CBr2-CHBr2-CH3.

  5. How many moles of Cl2 are needed to fully convert 0.10 mol of acetylene, HC#CH, to the tetrachloride?

    Show answer

    Stoichiometry is 2 mol Cl2 per mol alkyne (one per π bond): 0.10 × 2 = 0.20 mol Cl2.

  6. A student predicts 1,2-dibromobutane from 1-butyne + excess HBr. What did they get wrong?

    Show answer

    They predicted the alkene-type (vicinal) product. Excess HBr on an alkyne gives the geminal dibromide: 2,2-dibromobutane, CH3CH2-CBr2-CH3. Vicinal dibromides come from Br2 + alkene, not HBr + alkyne.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Vinyl halide
A halide on an sp2 (alkene) carbon, R-CX=CH2.
Geminal dihalide
Two halogens on the same carbon, R-CX2-R'.
Vicinal dihalide
Two halogens on adjacent carbons.
Markovnikov's rule
The electrophile's H adds to the carbon with more hydrogens; the other group goes to the more substituted carbon.
Anti addition
The two added groups come from opposite faces of the π bond.
Tetrahaloalkane
A saturated alkane bearing four halogens.
(E) isomer
The higher-priority groups on opposite sides of a double bond.

Sources & references

  1. openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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