Source: CHM 255 PSO Exam 2 Practice Problems
Alkynes are hydrocarbons with a carbon-carbon triple bond. For this exam, you need to name them using IUPAC rules, predict products of their reactions with HBr, Br2, and H2/Pd, and understand how alkynes relate to alkene chemistry (they undergo similar additions, sometimes twice).
Difficulty: Intermediate
Prerequisites: Alkene addition reactions (see companion notes), IUPAC nomenclature basics, functional group priority.
Alkynes sit at the end of the unsaturation spectrum: a double bond reacts once with electrophilic addition reagents, a triple bond can react once or twice depending on equivalents. The naming rules follow the same pattern as alkenes but use the suffix "-yne." This section also covers predicting starting materials from products (retrosynthesis with Br2) and complete reduction with H2/Pd. If you are comfortable with alkene reactions, alkyne reactions are a natural extension.
Alkyne
A hydrocarbon containing a carbon-carbon triple bond (one sigma bond and two pi bonds). The triple-bond carbons are sp-hybridised with 180° bond angles.
Terminal alkyne
An alkyne where the triple bond is at the end of the carbon chain (one of the triple-bond carbons bears a hydrogen). Think of it as having an exposed C-H on the triple bond. This C-H is relatively acidic (pKa ~25) because of the high s-character of the sp orbital.
Internal alkyne
An alkyne where the triple bond is between two carbon groups (no triple-bond C-H). These are generally more stable than terminal alkynes.
Lindlar catalyst
A "poisoned" palladium catalyst (Pd on CaCO3 with lead acetate and quinoline) that reduces alkynes to cis-alkenes only, stopping at the double bond stage. Not specifically tested on this worksheet, but often paired with alkyne material.
Vinyl halide
A halogen (Br, Cl) attached directly to a doubly bonded carbon. Formed when 1 equivalent of HBr adds to an alkyne.
Geminal dihalide
Two halogens on the same carbon. Formed when 2 equivalents of HX add to a terminal alkyne (both halogens go to the same carbon via Markovnikov addition).
Find the longest carbon chain that includes the triple bond.
Number from the end that gives the triple bond the lowest possible locant.
The suffix is "-yne" and the position of the triple bond is indicated by the number of the first carbon in the triple bond.
Substituents (methyl, fluoro, hydroxyl, etc.) are named and numbered as usual.
When both a double bond and a triple bond are present, use "-en-...-yne" and give the triple bond the lower number if there is a tie.
Hydroxyl (-OH) groups as substituents: if the OH is the highest-priority functional group, the parent chain uses the suffix "-ol" and the compound is named as an alkynol (e.g. 3-butyn-1-ol).
Examples from the worksheet:
A terminal alkyne with a methyl substituent: identify the longest chain containing the triple bond, number to give the triple bond the lowest locant, and name accordingly (e.g. 1-butyne, 3-methyl-1-butyne).
An alkyne with a fluorine and an OH group: the OH takes naming priority as "-ol," and fluorine is named as a prefix "fluoro-."
An alkynol (alkyne + alcohol): the chain is named to include both the triple bond and the OH, with the OH getting the lowest possible number.
1 equivalent of HBr adds across the triple bond following Markovnikov's rule, giving a vinyl bromide.
The bromine goes to the more substituted carbon of the triple bond.
2 equivalents of HBr would give a geminal dibromide (both bromines on the same carbon), but the worksheet specifies 1 equiv.
For a conjugated enyne (both a double bond and a triple bond), 1 equiv. of HBr adds preferentially to the more reactive site. The product depends on the specific substrate.
Excess H2 with Pd reduces both double bonds and triple bonds completely to single bonds.
A substrate with both a double bond and a triple bond is fully saturated.
Stereochemistry of hydrogenation: syn addition at each stage. On a ring, this can create cis-fused stereocentres.
1 equivalent of Br2 adds across the triple bond to give a dibromo alkene (anti addition, giving a trans-dibromoalkene).
2 equivalents of Br2 add to give a tetrabromoalkane (all four bromines across what was the triple bond).
Working backwards (Q15c): If you are given the tetrabromide product and told Br2 (2 equiv.) was used, remove the four bromines in pairs and reconstruct the triple bond to find the starting alkyne.
Students forget that alkynes can react with 1 or 2 equivalents of reagent. Always check how many equivalents the question specifies.
Students apply Markovnikov's rule to Br2 addition. When both atoms being added are the same, regioselectivity labels do not apply.
Students confuse terminal and internal alkynes when naming. A terminal alkyne has the triple bond at position 1; an internal alkyne has it elsewhere in the chain.
Students assume H2/Pd stops at the alkene stage. Without a Lindlar catalyst (or similar), Pd with excess H2 reduces all the way to the alkane.
⚠️ IUPAC naming of alkynes (Q14) is straightforward marks if you know the rules. Practise naming compounds with multiple functional groups (OH, F, triple bond) and get the priority right.
⚠️ Q15 tests three different alkyne reactions in one question. Be systematic: identify the reagent, determine how many equivalents, apply the correct regiochemistry and stereochemistry.
⚠️ Working backwards from a tetrabromide to the starting alkyne (Q15c) is a retrosynthesis problem. Remove bromines in pairs and restore the triple bond.
⚠️ If the substrate has both a double bond and a triple bond, H2/Pd reduces both. Draw the fully saturated product.
True or false: The suffix for an alkyne in IUPAC naming is "-ene." (False. It is "-yne.")
Fill in the blank: 1 equivalent of HBr adds to a terminal alkyne to give a ______ bromide. (vinyl)
True or false: 2 equivalents of Br2 added to an alkyne give a geminal dibromide. (False. They give a tetrabromoalkane with bromines on both carbons.)
Fill in the blank: H2/Pd reduces a triple bond all the way to a ______. (single bond / alkane)
True or false: Terminal alkyne C-H bonds are more acidic than typical C-H bonds. (True. pKa ~25 vs ~50 for sp3 C-H.)
Q: Name the following compound: HC≡C-CH(CH3)-CH3 (a terminal alkyne with a methyl branch).
A: 3-Methyl-1-butyne. The longest chain containing the triple bond is four carbons, numbered from the triple-bond end.
Q: What is the product of treating 1-hexyne with 1 equivalent of HBr?
A: 2-Bromo-1-hexene (a vinyl bromide). Markovnikov addition places the Br on the internal carbon of the triple bond.
Q: An unknown alkyne is treated with 2 equivalents of Br2 to give 1,1,2,2-tetrabromoethane. What was the starting alkyne?
A: Ethyne (acetylene, HC≡CH). Remove the four bromines in pairs and restore the triple bond.
Q: What is the product of treating a compound containing both a C=C double bond and a C≡C triple bond with excess H2/Pd?
A: Both unsaturations are fully reduced to single bonds. The product is the fully saturated alkane.
Q: Why is the C-H bond of a terminal alkyne more acidic than a typical alkyl C-H?
A: The carbon is sp-hybridised, with 50% s-character. The s orbital holds electrons closer to the nucleus, stabilising the conjugate base (the acetylide anion) more effectively than an sp3 carbon would.
Alkyne reactions extend the addition chemistry learned with alkenes. The key new idea is that a triple bond can undergo one or two rounds of addition depending on the equivalents of reagent provided.
Terminal alkyne acidity connects back to the acid-base material: the sp C-H (pKa ~25) sits between typical C-H bonds (~50) and alcohols (~16) on the pKa scale.
Alkynes also serve as synthetic intermediates in later chapters, where partial reduction (Lindlar catalyst for cis-alkenes, Na/NH3 for trans-alkenes) is used to set double-bond geometry.
Tags: alkyne, terminal alkyne, internal alkyne, IUPAC nomenclature, triple bond, HBr addition to alkynes, Br2 addition to alkynes, catalytic hydrogenation, vinyl bromide, geminal dihalide, tetrabromoalkane, Lindlar catalyst, sp hybridisation, acetylide, CHM 255, Purdue organic chemistry