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NCERT Exemplar · Q25

Q.Consider the compounds I to VII:
I. CH3—CH2—CH2—CH2—OH
II. CH3—CH2—CH(OH)—CH3
III. (CH3)3C—OH
IV. CH3—CH(CH3)—CH2—OH
V. CH3—CH2—O—CH2—CH3
VI. CH3—O—CH2—CH2—CH3
VII. CH3—O—CH(CH3)2
Identify the pairs of compounds that represents position isomerism.

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Position isomerism occurs when compounds have the same functional group and carbon skeleton but differ in the location of the functional group on the chain. For the given set, the pairs showing position isomerism are I & II (alcohols with –OH at C1 vs C2) and V & VI (ethers with the oxygen atom at different positions along the chain).

The key to identifying position isomers is to first ignore everything except the functional group and the carbon skeleton. If two compounds share the same molecular formula, the same functional group, and the same carbon backbone, but the functional group is attached at a different carbon, they are position isomers.

Let’s examine each compound systematically.

  1. Identify the functional group and molecular formula for each compound.

    • I: Primary alcohol, C4H10OC_4H_{10}O (butan-1-ol).
    • II: Secondary alcohol, C4H10OC_4H_{10}O (butan-2-ol).
    • III: Tertiary alcohol, C4H10OC_4H_{10}O (2-methylpropan-2-ol).
    • IV: Primary alcohol, C4H10OC_4H_{10}O (2-methylpropan-1-ol).
    • V: Ether, C4H10OC_4H_{10}O (diethyl ether).
    • VI: Ether, C4H10OC_4H_{10}O (methyl propyl ether).
    • VII: Ether, C4H10OC_4H_{10}O (methyl isopropyl ether).

    All seven compounds have the same molecular formula C4H10OC_4H_{10}O, so they are all functional group isomers of one another (alcohols vs ethers). But within each functional group class, we look for position isomerism.

  2. Focus on the alcohols (I, II, III, IV).

    • I and II share a straight chain of four carbons. The –OH group is on carbon-1 in I and on carbon-2 in II. Same skeleton, same functional group, different position → position isomers.
    • III has a branched skeleton (a tertiary carbon with three methyl groups). Its carbon skeleton is different from I and II (it’s a 3-carbon chain with a branch, not a straight 4-carbon chain). So III is a chain isomer of I and II, not a position isomer.
    • IV also has a branched skeleton (a 3-carbon chain with a methyl branch at carbon-2). Again, different carbon backbone from I and II, so it’s a chain isomer.
    • Between III and IV: both have branched skeletons but different branching patterns — they are chain isomers of each other, not position isomers.
    Watch out

    A common mistake is to think that moving the –OH group anywhere on a C4C_4 chain gives position isomers. But if the carbon skeleton itself changes (e.g., from straight to branched), the relationship becomes chain isomerism, not position isomerism. Position isomerism requires the same carbon skeleton.

  3. Now examine the ethers (V, VI, VII).

    • V: CH3CH2–O–CH2CH3CH_3CH_2–O–CH_2CH_3 — the oxygen is between two ethyl groups. The carbon chain is effectively C–C–O–C–CC–C–O–C–C.
    • VI: CH3–O–CH2CH2CH3CH_3–O–CH_2CH_2CH_3 — the oxygen is between a methyl and a propyl group. The chain is C–O–C–C–CC–O–C–C–C.
    • VII: CH3–O–CH(CH3)2CH_3–O–CH(CH_3)_2 — the oxygen is between a methyl and an isopropyl group. The chain is C–O–C(C)CC–O–C(C)C.

    For ethers, the “functional group” is the C–O–CC–O–C linkage. Position isomerism in ethers means the oxygen atom is placed at a different location along the same carbon skeleton. …

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