Question:

Write the structures of all possible isomers of alcohol having molecular formula C$_4$H$_{10$O.}

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Count alcohol isomers by arranging the \(\mathrm{C_4}\) skeleton (straight vs branched) and then placing the OH group at unique positions (primary, secondary, tertiary).
Updated On: Sep 3, 2025
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Solution and Explanation


All alcohol isomers (no ethers) with formula C$_4$H$_{10}$O are four in number.
1. Butan-1-ol (n-butanol): \(\mathrm{CH_3CH_2CH_2CH_2OH}\).
2. Butan-2-ol (sec-butanol): \(\mathrm{CH_3CH_2CH(OH)CH_3}\).
3. 2-Methylpropan-1-ol (isobutanol): \((\mathrm{CH_3})_2\mathrm{CHCH_2OH}\).
4. 2-Methylpropan-2-ol (tert-butanol): \((\mathrm{CH_3})_3\mathrm{COH}\).
\[ \boxed{\text{Total alcohol isomers of C}_4\text{H}_{10}\text{O} = 4} \]
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