Structural Isomerism Of Haloalkanes
Counting every distinct constitutional structure, how many isomeric monobromides share the molecular formula C4H9Br for a saturated open-chain compound?
Select the correct option:
Solution
Four
The molecular formula C4H9Br corresponds to a saturated four-carbon chain bearing one bromine, and constitutional isomers differ in how the carbon skeleton is arranged and where the bromine sits. Starting from the straight chain n-butane, bromine can occupy a terminal carbon to give 1-bromobutane or an internal carbon to give 2-bromobutane, yielding two isomers. From the branched skeleton isobutane, bromine on a primary carbon gives 1-bromo-2-methylpropane, while bromine on the central carbon gives 2-bromo-2-methylpropane, the tertiary bromide. These four arrangements, 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane, are all the distinct constitutional isomers. The answer three is wrong because it overlooks one of the branched possibilities. The answer five is incorrect because it would double-count equivalent positions related by symmetry. The answer two is wrong because it considers only the straight chain and ignores branching. A useful way to be systematic is to first enumerate the distinct carbon skeletons for four carbons, the straight chain and the single branched isobutane frame, and then mark every chemically non-equivalent carbon on each skeleton as a possible site for bromine. Symmetry must be respected so that the two terminal carbons of n-butane, for instance, are treated as one position rather than two. This follows the NCERT approach to skeletal and positional isomerism. Sanity check: two straight-chain plus two branched structures total exactly four isomers.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing halogens
- Topic
- structural isomerism of haloalkanes
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Four
The molecular formula C4H9Br corresponds to a saturated four-carbon chain bearing one bromine, and constitutional isomers differ in how the carbon skeleton is arranged and where the bromine sits. Starting from the straight chain n-butane, bromine can occupy a terminal carbon to give 1-bromobutane or an internal carbon to give 2-bromobutane, yielding two isomers. From the branched skeleton isobutane, bromine on a primary carbon gives 1-bromo-2-methylpropane, while bromine on the central carbon gives 2-bromo-2-methylpropane, the tertiary bromide. These four arrangements, 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane, are all the distinct constitutional isomers. The answer three is wrong because it overlooks one of the branched possibilities. The answer five is incorrect because it would double-count equivalent positions related by symmetry. The answer two is wrong because it considers only the straight chain and ignores branching. A useful way to be systematic is to first enumerate the distinct carbon skeletons for four carbons, the straight chain and the single branched isobutane frame, and then mark every chemically non-equivalent carbon on each skeleton as a possible site for bromine. Symmetry must be respected so that the two terminal carbons of n-butane, for instance, are treated as one position rather than two. This follows the NCERT approach to skeletal and positional isomerism. Sanity check: two straight-chain plus two branched structures total exactly four isomers.
This hard difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of structural isomerism of haloalkanes. It appeared in the 2025 exam.
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