NCERT Exemplar · Q18
Q.Write the IUPAC name of each of the following two compounds. Compound A is the open-chain diol CH3-CH(CH3)-CH(OH)-CH(C2H5)-CH(OH)-CH3, i.e. a six-carbon main chain bearing a methyl branch on the second carbon, an ethyl branch on the fourth carbon, and hydroxyl groups on the third and fifth carbons. Compound B is a cyclohexane ring carrying a -NO2 group on one carbon and an -OCH3 (methoxy) group on the carbon that is meta to it (a 1,3 arrangement).
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Start your 14-day free trial to unlock the full solution →For A, the six-carbon chain is the parent and the two -OH groups are the principal characteristic group (suffix -diol), so numbering gives OH the lowest locants (2 and 4); the branches are an ethyl at C-3 and a methyl at C-5, giving 3-ethyl-5-methylhexane-2,4-diol. For B, both -NO2 (nitro) and -OCH3 (methoxy) are prefix substituents on cyclohexane placed 1,3; alphabetical priority gives methoxy the lower number, so it is 1-methoxy-3-nitrocyclohexane.
Compound A
- Longest chain through both hydroxyl-bearing carbons: six carbons -> hexane; two -OH -> hexane-diol.
- Number to give the -OH groups the lowest locants. From the right-hand end the OH groups fall on C-2 and C-4 (locant set {2,4}), versus {3,5} from the left; choose {2,4}.
- With that numbering the ethyl branch is on C-3 and the methyl branch on C-5.
- Assemble alphabetically (ethyl before methyl): 3-ethyl-5-methylhexane-2,4-diol.
Compound B
- Parent ring: cyclohexane (saturated six-membered ring). …
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