301. The carbon atoms directly involved in the \( \mathrm{C\equiv C} \) bond of an alkyne are generally
ⓐ. \(sp^3\)-hybridised and tetrahedral
ⓑ. \(sp\)-hybridised and linear
ⓒ. \(sp^2\)-hybridised and trigonal planar
ⓓ. unhybridised and square planar
Correct Answer: \(sp\)-hybridised and linear
Explanation: Each carbon atom of an alkyne triple bond is generally \(sp\)-hybridised. The two \(sp\) hybrid orbitals are arranged linearly, giving a bond angle of \(180^\circ\) around the triple-bonded carbon. The remaining two unhybridised \(p\)-orbitals on each carbon form two \( \pi \)-bonds. This bonding arrangement makes the \( \mathrm{C\equiv C} \) unit linear. The geometry is different from \(sp^2\) alkene carbon and \(sp^3\) alkane carbon.
302. A hydrocarbon has formula \( \mathrm{C_5H_8} \) and is known to be an open-chain compound with one triple bond. Its family is
ⓐ. alkyne
ⓑ. alkane
ⓒ. alkene
ⓓ. arene
Correct Answer: alkyne
Explanation: The formula \( \mathrm{C_5H_8} \) matches \( \mathrm{C_nH_{2n-2}} \) when \(n=5\). The question also states that the compound is open-chain and contains one triple bond. These conditions identify it as an open-chain monoalkyne. An alkane with \(5\) carbon atoms would be \( \mathrm{C_5H_{12}} \), while an open-chain monoalkene would be \( \mathrm{C_5H_{10}} \). The structural condition is important because formula comparison alone can sometimes be incomplete.
303. A table compares local bonding in three hydrocarbon families.
| Row | Hydrocarbon site | Hybridisation | Geometry |
|---|
| P | alkane carbon with four single bonds | \(sp^3\) | tetrahedral |
| Q | alkene carbon of \( \mathrm{C=C} \) | \(sp^2\) | trigonal planar |
| R | alkyne carbon of \( \mathrm{C\equiv C} \) | \(sp\) | linear |
| S | alkyne carbon of \( \mathrm{C\equiv C} \) | \(sp^3\) | tetrahedral |
The row that needs correction is
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Alkane carbon with four single bonds is \(sp^3\)-hybridised and tetrahedral. Alkene carbon in a double bond is \(sp^2\)-hybridised and trigonal planar. Alkyne carbon in a triple bond is \(sp\)-hybridised and linear. Row S incorrectly assigns alkane-type hybridisation and geometry to an alkyne carbon. The number and type of electron regions around carbon decide the hybridisation and shape.
304. The suitable IUPAC name for \( \mathrm{CH_3C\equiv CCH_3} \) is
ⓐ. but-\(1\)-yne
ⓑ. but-\(2\)-yne
ⓒ. but-\(2\)-ene
ⓓ. butane
Correct Answer: but-\(2\)-yne
Explanation: The structure \( \mathrm{CH_3C\equiv CCH_3} \) has a four-carbon chain, so the root is \( \mathrm{but} \). The carbon-carbon triple bond requires the suffix \( \mathrm{-yne} \). The triple bond lies between carbon \(2\) and carbon \(3\), so the locant is \(2\). The name is therefore but-\(2\)-yne. But-\(1\)-yne would have the triple bond at the end of the carbon chain.
305. A chain compound written as \( \mathrm{HC\equiv CCH_2CH_3} \) is named
ⓐ. but-\(2\)-yne
ⓑ. but-\(1\)-ene
ⓒ. but-\(1\)-yne
ⓓ. pent-\(1\)-yne
Correct Answer: but-\(1\)-yne
Explanation: The structure \( \mathrm{HC\equiv CCH_2CH_3} \) contains four carbon atoms in a continuous chain. It has a carbon-carbon triple bond, so the suffix is \( \mathrm{-yne} \). Numbering from the terminal \( \mathrm{HC\equiv} \) end gives the triple bond the locant \(1\). Numbering from the other end would give a higher locant, which is not preferred. The correct name is but-\(1\)-yne because the triple bond begins at carbon \(1\).
306. In naming an alkyne, the parent chain should be numbered so that the triple bond receives
ⓐ. the highest possible locant
ⓑ. no locant under any condition
ⓒ. a locant equal to the total number of hydrogens
ⓓ. the lowest possible locant
Correct Answer: the lowest possible locant
Explanation: Alkynes are named by selecting a parent chain containing the \( \mathrm{C\equiv C} \) bond. The chain is then numbered so that the triple bond gets the lowest possible locant. The locant tells where the triple bond begins in the selected chain. This is the same general idea used for simple alkene naming with double bonds. A higher locant is not preferred when the same structure can be numbered from the other end.
307. The pair \( \mathrm{HC\equiv CCH_2CH_3} \) and \( \mathrm{CH_3C\equiv CCH_3} \) is best described as
ⓐ. position isomers
ⓑ. chain isomers
ⓒ. geometrical isomers
ⓓ. successive homologues
Correct Answer: position isomers
Explanation: Both compounds have the same molecular formula \( \mathrm{C_4H_6} \). In \( \mathrm{HC\equiv CCH_2CH_3} \), the triple bond starts at carbon \(1\), giving but-\(1\)-yne. In \( \mathrm{CH_3C\equiv CCH_3} \), the triple bond starts at carbon \(2\), giving but-\(2\)-yne. The carbon skeleton remains a four-carbon chain in both cases. The difference is the position of the triple bond, so the pair shows position isomerism.
308. Alkynes generally do not show cis-trans isomerism around the triple bond because the \( \mathrm{C\equiv C} \) unit is
ⓐ. tetrahedral like an alkane carbon
ⓑ. trigonal planar with two groups on each carbon
ⓒ. linear with one substituent direction on each carbon
ⓓ. a benzene-like ring system
Correct Answer: linear with one substituent direction on each carbon
Explanation: A triple-bonded carbon in an alkyne is \(sp\)-hybridised and linear. Each carbon of the \( \mathrm{C\equiv C} \) bond has only one substituent outside the triple bond. Therefore, there is no pair of groups on each double-bonded-like carbon that can be arranged as cis or trans. Cis-trans isomerism is typical for suitable alkenes because a \( \mathrm{C=C} \) bond can hold two groups on each carbon in a fixed planar arrangement. The linear geometry of alkynes removes that possibility.
309. A table of alkyne names and structures is shown below.
| Row | Name | Structure |
|---|
| P | ethyne | \( \mathrm{HC\equiv CH} \) |
| Q | propyne | \( \mathrm{CH_3C\equiv CH} \) |
| R | but-\(1\)-yne | \( \mathrm{HC\equiv CCH_2CH_3} \) |
| S | but-\(2\)-yne | \( \mathrm{HC\equiv CCH_2CH_3} \) |
The row that needs correction is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Ethyne is correctly represented as \( \mathrm{HC\equiv CH} \). Propyne is correctly represented as \( \mathrm{CH_3C\equiv CH} \), which is equivalent to \( \mathrm{HC\equiv CCH_3} \). But-\(1\)-yne is correctly represented as \( \mathrm{HC\equiv CCH_2CH_3} \). Row S is wrong because that same structure has the triple bond at carbon \(1\), not carbon \(2\). But-\(2\)-yne should be represented as \( \mathrm{CH_3C\equiv CCH_3} \).
310. Ethyne is prepared in the laboratory by adding water to calcium carbide. The reaction is
ⓐ. \( \mathrm{CH_4+Cl_2\rightarrow CH_3Cl+HCl} \)
ⓑ. \( \mathrm{CH_2=CH_2+H_2\rightarrow CH_3CH_3} \)
ⓒ. \( \mathrm{CH_3COONa+NaOH\rightarrow CH_4+Na_2CO_3} \)
ⓓ. \( \mathrm{CaC_2+2H_2O\rightarrow C_2H_2+Ca(OH)_2} \)
Correct Answer: \( \mathrm{CaC_2+2H_2O\rightarrow C_2H_2+Ca(OH)_2} \)
Explanation: Calcium carbide reacts with water to produce ethyne, also called acetylene. The inorganic product is calcium hydroxide, \( \mathrm{Ca(OH)_2} \). The balanced equation is \( \mathrm{CaC_2+2H_2O\rightarrow C_2H_2+Ca(OH)_2} \). The other equations describe chlorination, hydrogenation, and soda-lime decarboxylation. The appearance of \( \mathrm{CaC_2} \) and water is the identifying clue for ethyne preparation.
311. If \(1.0\,\text{mol}\) of \( \mathrm{CaC_2} \) reacts completely with water, the amount of ethyne produced is
ⓐ. \(0.5\,\text{mol}\)
ⓑ. \(1.0\,\text{mol}\)
ⓒ. \(2.0\,\text{mol}\)
ⓓ. \(4.0\,\text{mol}\)
Correct Answer: \(1.0\,\text{mol}\)
Explanation: \( \textbf{Balanced equation:} \)
\[
\mathrm{CaC_2+2H_2O\rightarrow C_2H_2+Ca(OH)_2}
\]
\( \textbf{Mole ratio:} \) \(1\,\text{mol}\) of \( \mathrm{CaC_2} \) gives \(1\,\text{mol}\) of \( \mathrm{C_2H_2} \).
\( \textbf{Given amount of calcium carbide:} \) \(1.0\,\text{mol}\).
\( \textbf{Water condition:} \) Complete reaction means enough water is available.
\( \textbf{Ethyne formed:} \)
\[
1.0\,\text{mol}\ \mathrm{CaC_2}\times\frac{1\,\text{mol}\ \mathrm{C_2H_2}}{1\,\text{mol}\ \mathrm{CaC_2}}=1.0\,\text{mol}
\]
\( \textbf{Final answer:} \) \(1.0\,\text{mol}\) of ethyne is produced.
The coefficient \(2\) belongs to water, not to ethyne.
312. A laboratory gas preparation gives \( \mathrm{C_2H_2} \) when a solid reacts with water and leaves \( \mathrm{Ca(OH)_2} \). The solid is most likely
ⓐ. \( \mathrm{CaCO_3} \)
ⓑ. \( \mathrm{NaCl} \)
ⓒ. \( \mathrm{CH_3COONa} \)
ⓓ. \( \mathrm{CaC_2} \)
Correct Answer: \( \mathrm{CaC_2} \)
Explanation: Ethyne, \( \mathrm{C_2H_2} \), is prepared by the reaction of calcium carbide with water. The reaction produces calcium hydroxide as the inorganic product. \( \mathrm{CaCO_3} \) reacts with acids to give \( \mathrm{CO_2} \), not ethyne under this condition. \( \mathrm{NaCl} \) and \( \mathrm{CH_3COONa} \) do not match the described gas preparation. The combination of \( \mathrm{C_2H_2} \) gas and \( \mathrm{Ca(OH)_2} \) points directly to \( \mathrm{CaC_2} \).
313. Dehydrohalogenation of a vicinal dihalide can prepare an alkyne when
ⓐ. two molecules of hydrogen halide are eliminated
ⓑ. one molecule of water is added across a triple bond
ⓒ. carbon dioxide is removed from a carboxylate
ⓓ. two alkyl halide molecules couple with sodium
Correct Answer: two molecules of hydrogen halide are eliminated
Explanation: A vicinal dihalide has halogen atoms on adjacent carbon atoms. Strong base can remove two molecules of hydrogen halide from such a compound in successive elimination steps. The first elimination may form a haloalkene, and the second can form an alkyne. This is a preparation route based on increasing unsaturation. It is different from Wurtz coupling, which builds a larger alkane, and from decarboxylation, which removes a carboxyl carbon.
314. The conversion of \( \mathrm{CH_2BrCH_2Br} \) into \( \mathrm{HC\equiv CH} \) requires removal of
ⓐ. \( \mathrm{H_2O} \)
ⓑ. \( \mathrm{CO_2} \)
ⓒ. \( \mathrm{2HBr} \)
ⓓ. \( \mathrm{Br_2} \) only, with no hydrogen loss
Correct Answer: \( \mathrm{2HBr} \)
Explanation: \( \textbf{Starting compound:} \) \( \mathrm{CH_2BrCH_2Br} \) is a vicinal dibromide.
\( \textbf{Target compound:} \) \( \mathrm{HC\equiv CH} \) is ethyne.
\( \textbf{Change in bonding:} \) The carbon-carbon bond changes from single-bonded in the dibromide framework to a triple bond in ethyne.
\( \textbf{Eliminated groups:} \) Each elimination removes one \( \mathrm{HBr} \).
\( \textbf{Total elimination:} \)
\[
\mathrm{CH_2BrCH_2Br\rightarrow HC\equiv CH+2HBr}
\]
\( \textbf{Final answer:} \) Two molecules of \( \mathrm{HBr} \) are removed.
Removing only \( \mathrm{Br_2} \) would ignore the hydrogen loss needed to create the extra unsaturation.
315. A table compares preparation routes for unsaturated hydrocarbons.
| Row | Starting material | Main route | Product type |
|---|
| P | alcohol | dehydration | alkene |
| Q | alkyl halide | dehydrohalogenation | alkene |
| R | vicinal dihalide | double dehydrohalogenation | alkyne |
| S | calcium carbide | reaction with water | ethyne |
The fully suitable set of rows is
ⓐ. P and Q only
ⓑ. R and S only
ⓒ. P only
ⓓ. P, Q, R, and S
Correct Answer: P, Q, R, and S
Explanation: Alcohol dehydration is a common route to alkenes. Alkyl halide dehydrohalogenation can also prepare alkenes by loss of \( \mathrm{HX} \). A vicinal dihalide can form an alkyne when two molecules of \( \mathrm{HX} \) are eliminated. Calcium carbide reacts with water to give ethyne. The table correctly separates alkene and alkyne preparation routes instead of treating all eliminations as identical.
316. A claim says, “The formula \( \mathrm{C_4H_6} \) proves a compound is but-\(1\)-yne.” The better conclusion is that
ⓐ. formula alone does not fix the exact alkyne name
ⓑ. \( \mathrm{C_4H_6} \) must be an alkane
ⓒ. \( \mathrm{C_4H_6} \) cannot contain only carbon and hydrogen
ⓓ. every compound with \( \mathrm{C_4H_6} \) must be ethyne
Correct Answer: formula alone does not fix the exact alkyne name
Explanation: \( \mathrm{C_4H_6} \) matches the open-chain monoalkyne formula \( \mathrm{C_nH_{2n-2}} \) for \(n=4\). However, it does not by itself reveal whether the triple bond is at carbon \(1\) or carbon \(2\). But-\(1\)-yne and but-\(2\)-yne have the same molecular formula but different triple-bond positions. The exact name requires structural information. Formula recognition is useful, but it should not be stretched beyond what it actually shows.
317. Alkynes, like alkanes and alkenes, are generally poorly soluble in water because they are
ⓐ. highly ionic salts
ⓑ. strong mineral acids
ⓒ. non-polar hydrocarbons
ⓓ. compounds containing many \( \mathrm{-OH} \) groups
Correct Answer: non-polar hydrocarbons
Explanation: Alkynes contain only carbon and hydrogen, so they are hydrocarbons. Their molecules are generally non-polar and do not interact strongly with polar water molecules. Water dissolves substances well when strong polar interactions or hydrogen bonding are possible, which simple alkynes do not provide. Alkynes are more soluble in non-polar organic solvents than in water. The presence of a \( \mathrm{C\equiv C} \) bond changes reactivity, but it does not make the molecule strongly water-soluble.
318. For straight-chain alkynes in a homologous series, the boiling point generally increases as the carbon number increases mainly because
ⓐ. the compounds become ionic
ⓑ. every higher member forms hydrogen bonds with water
ⓒ. the triple bond disappears in higher members
ⓓ. molecular size and dispersion forces increase
Correct Answer: molecular size and dispersion forces increase
Explanation: Successive members of an alkyne homologous series differ by \( \mathrm{-CH_2-} \). As the carbon chain becomes longer, molecular size, surface area, and number of electrons increase. These changes strengthen London dispersion forces between molecules. Stronger intermolecular attractions require more energy to overcome during boiling. The trend is a physical-property effect, not a change from covalent bonding to ionic bonding.
319. The relatively higher acidity of terminal alkynes compared with alkenes is mainly due to the
ⓐ. presence of oxygen in all alkynes
ⓑ. free rotation around the \( \mathrm{C\equiv C} \) bond
ⓒ. greater \(s\)-character of the \(sp\)-carbon bonded to hydrogen
ⓓ. formation of \( \mathrm{CO_2} \) during ordinary ionisation
Correct Answer: greater \(s\)-character of the \(sp\)-carbon bonded to hydrogen
Explanation: In a terminal alkyne, the acidic hydrogen is attached to an \(sp\)-hybridised carbon. An \(sp\)-hybrid orbital has \(50\%\) \(s\)-character, which holds electrons closer to the nucleus than \(sp^2\) or \(sp^3\) orbitals. This helps stabilise the acetylide ion formed after loss of \( \mathrm{H^+} \). Alkenes and alkanes have lower \(s\)-character at the corresponding carbon and are much less acidic. The acidity is still weak compared with mineral acids, but it is significant enough for terminal alkynes to react with strong bases or certain metal ions.
320. The compound that contains an acidic terminal alkyne hydrogen is
ⓐ. \( \mathrm{CH_3C\equiv CCH_3} \)
ⓑ. \( \mathrm{HC\equiv CCH_3} \)
ⓒ. \( \mathrm{CH_2=CHCH_3} \)
ⓓ. \( \mathrm{CH_3CH_2CH_3} \)
Correct Answer: \( \mathrm{HC\equiv CCH_3} \)
Explanation: A terminal alkyne has a hydrogen atom directly attached to a triple-bonded carbon. In \( \mathrm{HC\equiv CCH_3} \), the \( \mathrm{H-C\equiv C} \) end contains such a hydrogen. \( \mathrm{CH_3C\equiv CCH_3} \) is an internal alkyne and has no hydrogen attached directly to the triple-bonded carbon. Alkenes and alkanes do not contain the same terminal \(sp\)-carbon hydrogen. The position of the triple bond and the presence of the terminal hydrogen must both be checked.
