Exercises — IUPAC nomenclature — alkanes, alkenes, alkynes, aromatics, alcohols, ethers, aldehydes, ketones, acids, esters, amines,

Count carbons in the single continuous chain: $C_1-C_2-C_3-C_4-C_5$ = 5 carbons → root pent (see the chain table in the parent). All bonds single → saturated → suffix -ane. Answer: pentane.

The middle carbon carries a $C=O$ with carbons on both sides → a ketone. From the seniority table its suffix is -one. (Root: 3 carbons → prop. Position of $C=O$ is C2. Full name for the curious: propan-2-one, common name acetone.)

Bromine's prefix is bromo. Halides are on the "servants only" row of the seniority table — they are never a suffix, always a prefix.

1. Senior group: only $-OH$ present → suffix -ol.
2. Longest chain containing OH: 4 carbons → but.
3. Number for lowest OH locant: from the right the OH sits on C2; from the left it sits on C3. Choose the direction giving 2.
4. No other substituents.
5. Assemble → butan-2-ol. ✅

1. No senior suffix group other than the double bond → suffix -ene.
2. Longest chain = 4 C → but.
3. With only a C=C present, number to give the double bond the lowest locant. From the left the $C=C$ starts at C1; from the right it would start at C2. Pick 1.
4. Answer: but-1-ene.

Ethers are never a suffix. Pick the larger carbon group as the parent chain — here both sides are ethyl, so parent = ethane. The $-O-C{H}_{2}C{H}_3$ hanging off it is the ethoxy prefix. Answer: ethoxyethane (common name diethyl ether).

1. Two features: a $C=C$ and an $-OH$. The alcohol is senior to the double bond, so -ol is the suffix and -en- stays inside the chain.
2. Longest chain containing OH = 4 C → but.
3. Numbering: the senior group ($-OH$) must get the lowest locant. Number from the OH end: $OH$ on C1, the $C=C$ runs from C3 to C4 (cite lower → 3).
4. Assemble root + unsaturation locant + suffix locant → but-3-en-1-ol. ✅ The red numbering in the figure shows why the other direction ("but-1-en-4-ol") is wrong: it hands the low number to the double bond, not the boss group.

1. Groups present: aldehyde $-CHO$ and carboxylic acid $-COOH$. From the seniority table acid outranks aldehyde, so acid = suffix -oic acid; the aldehyde is demoted to the prefix oxo-.
2. Chain = 4 C → but. The acid carbon is necessarily C1.
3. Counting from that C1: $C1$ = COOH, $C4$ = CHO → the oxo group is at position 4.
4. Answer: 4-oxobutanoic acid. ✅

1. No functional suffix group → suffix -ane.
2. Longest chain: trace all paths. The straight run gives 5 carbons; branching off the $C{H}_3$ still tops out at 5. Root = pent.
3. Number for the lowest substituent locant: from the left the methyl branch is on C2; from the right it is on C4. Choose 2.
4. Answer: 2-methylpentane.

1. No suffix group → suffix -ane.
2. Longest chain = 5 C → pent.
3. Numbering (lowest locant set): from the left the substituents sit at {2 (Cl), 3 (methyl)}; from the right at {3, 4}. The set $\{2,3\}$ is lower → number from the left.
4. Substituents alphabetically: chloro (c) before methyl (m).
5. Assemble: 2-chloro-3-methylpentane. ✅

Root but = 4-carbon chain: $C1-C2-C3-C4$. Suffix -ol at 2 → $-OH$ on C2. Prefix 3-methyl → a $C{H}_3$ branch on C3. $C{H}_3-\underset{OH}{\mathrm{CH}}-\underset{C{H}_3}{\mathrm{CH}}-C{H}_3$ i.e. $C{H}_3-CH(OH)-CH(C{H}_3)-C{H}_3$. Consistency check: senior group $-OH$ must have the lowest locant. From this end it is 2; from the other end it would be 3. So 2 is correct and the name reconstructs a single molecule. ✅

1. Suffix -ane.
2. Longest chain — trace carefully. The horizontal run is $C{H}_{3}C{H}_2-CH-C{H}_{2}C{H}_{2}C{H}_3$ = 6 carbons (hexane). But the branch is an ethyl ($-C{H}_{2}C{H}_3$): going into that branch, the longest path is $3+1+...$ — checking all routes, the genuine longest continuous chain is 6 carbons with an ethyl branch, or equivalently we could pick a chain that turns the corner. Comparing candidate chains, the maximal one is 6 C. Root = hex.
3. The ethyl substituent sits on C3 either way (the molecule is symmetric about that carbon: propyl on one side, ethyl+carbon on the other — number for lowest locant → 3).
4. Answer: 3-ethylhexane.

1. Groups: two $-COOH$ and one $-OH$. Acid is senior → suffix. With two acid groups the suffix is -dioic acid, and both acid carbons are chain ends.
2. Chain = 5 C → pent. Acid carbons are C1 and C5.
3. The $-OH$ is on the central carbon C3 → prefix 3-hydroxy.
4. Assemble: 3-hydroxypentanedioic acid. ✅ Every carbon is accounted for; the two acids symmetrically pin C1 and C5, so numbering direction doesn't matter — the OH is at 3 either way.

1. Benzene + $-COOH$ is the retained name benzoic acid; the acid carbon defines position 1 of the ring.
2. "Directly opposite / para" = position 4.
3. Chlorine is a prefix → 4-chloro.
4. Answer: 4-chlorobenzoic acid. ✅ (See Aromatic Compounds — Benzene Derivatives for why benzoic acid, phenol and aniline keep their historic names.)

1. Groups: a ketone ($C=O$ with carbons both sides) and an alcohol. From seniority, ketone > alcohol, so the ketone is the suffix -one and the alcohol becomes the prefix hydroxy-.
2. Chain = 5 C → pent.
3. Numbering: senior group (the ketone) gets the lowest locant. From the ketone end: $C=O$ on C2, $-OH$ on C5. From the OH end: $C=O$ would be C4. Choose the direction giving the ketone 2.
4. Assemble with the hydroxy prefix at C5 → 5-hydroxypentan-2-one. ✅

Root but = 4 carbons. -ol at 1 → $-OH$ on C1. -yn- at 3 → a $C\equiv C$ triple bond starting at C3 (i.e. between C3 and C4). $HO-C{H}_2-C{H}_2-C\equiv CH$ A triple bond = two degrees of unsaturation. Using $DoU=\frac{2C+2+N-H}{2}$ for $C_4{H}_{6}O$: $\frac{2(4)+2-6}{2}=2$. ✅ (Oxygen doesn't affect the count — revisit Degree of Unsaturation.)

The –OH (it is the senior principal group); the C=C only sets direction when nothing senior is present.

$HOOC-C{H}_2-CH(OH)-C{H}_2-COOH$? 3-hydroxypentanedioic acid

$C{H}_3-CO-C{H}_2-C{H}_2-C{H}_2-OH$? 5-hydroxypentan-2-one

chloro before methyl (c before m) → e.g. 2-chloro-3-methylpentane.