Can you get this USNCO question?

Organic and biochemistryNational 2018

How many distinct acyclic compounds have the formula $\ce{C5H10}$ ?

18% of national qualifiers got this right

Explanation

draft

A saturated acyclic $\ce{C5H12}$ (pentane) has formula $\ce{C5H12}$ ; $\ce{C5H10}$ has two fewer H's, giving one degree of unsaturation. Since the compound is acyclic, that unsaturation must be a C=C double bond — these are the pentenes.

Enumerate all skeletal + positional + stereo isomers of pentene:

Straight-chain (pentyl) skeleton:
1-pentene: $\ce{CH2=CH-CH2-CH2-CH3}$
2-pentene: $\ce{CH3-CH=CH-CH2-CH3}$ — this has cis and trans (E/Z) forms, giving 2 distinct compounds

Methylbutane (branched) skeleton:
3-methyl-1-butene: $\ce{CH2=CH-CH(CH3)-CH3}$
2-methyl-1-butene: $\ce{CH2=C(CH3)-CH2-CH3}$
2-methyl-2-butene: $\ce{CH3-C(CH3)=CH-CH3}$

Counting: 1-pentene (1) + cis/trans-2-pentene (2) + 3-methyl-1-butene (1) + 2-methyl-1-butene (1) + 2-methyl-2-butene (1) = 6 distinct acyclic $\ce{C5H10}$ isomers.

Counts only the three distinct carbon skeletons (n-pentene, methylbutene variants treated as one each) while ignoring that the double bond can occupy different positions and give different structures within each skeleton.

Finds 4 by listing 1-pentene, 2-pentene (as a single entry, not splitting cis/trans), and two of the three branched methylbutenes, missing one branched isomer and the E/Z distinction.

Correctly identifies all five constitutional (positional/skeletal) isomers — 1-pentene, 2-pentene, 3-methyl-1-butene, 2-methyl-1-butene, 2-methyl-2-butene — but forgets that 2-pentene exists as separate cis and trans stereoisomers.

Correctly counts all five constitutional isomers and recognizes that 2-pentene has cis (Z) and trans (E) forms, giving 5 + 1 = 6 total distinct acyclic $\ce{C5H10}$ compounds.

Concept · Structural and E/Z isomerism in alkenes (degree of unsaturation)