Q.Name the following compounds according to IUPAC system of nomenclature:

Concept: IUPAC Nomenclature of organic compounds — identifying the principal functional group, the longest carbon chain containing it, and numbering to give the lowest locants.

(i) $CH_3CH(CH_3)CH_2CH_2CHO$

The longest chain with the aldehyde group has 5 carbons (pentanal). A methyl substituent is at C-4.

4-methylpentanal

(ii) $CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl$

The principal group is a ketone. The longest chain containing the carbonyl is 6 carbons (the ethyl group is a substituent, not part of the main chain): numbering from the methyl end gives the ketone the lower locant, C-3. Substituents: an ethyl group at C-4 and a chloro group at C-6.

6-chloro-4-ethylhexan-3-one

(iii) $CH_3CH=CHCHO$

The aldehyde takes priority; the 4-carbon chain has a double bond at C-2.

But-2-enal

(iv) $CH_3COCH_2COCH_3$

A diketone with the principal group as the ketone; the longest chain has 5 carbons with carbonyls at C-2 and C-4.

Pentane-2,4-dione

(v) $CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3$

The ketone is the principal group. Numbering from the methyl-ketone end gives the carbonyl the lower locant, C-2. The quaternary carbon two positions away (C-3) carries two methyl substituents, and the branched carbon further along the chain (C-5) carries one more.

3,3,5-trimethylhexan-2-one

(vi) $(CH_3)_3CCH_2COOH$

The carboxylic acid gives a 4-carbon chain (butanoic acid) by counting one of the three equivalent methyl groups on the quaternary carbon as part of the main chain. The remaining two methyls are substituents at C-3.

3,3-dimethylbutanoic acid

(vii) $OHCC_6H_4CHO$-p

Two aldehyde groups on a benzene ring at positions 1 and 4.

Benzene-1,4-dicarbaldehyde

IUPAC nomenclature is a systematic method for naming organic compounds: identify the principal functional group, find the longest chain that includes it, number the chain so the group gets the lowest locant, then name substituents alphabetically as prefixes.

(i) $CH_3CH(CH_3)CH_2CH_2CHO$

The aldehyde carbon is always C1. Numbering from the $-CHO$ end: C1 ($-CHO$), C2 ($-CH_2-$), C3 ($-CH_2-$), C4 ($-CH(CH_3)-$), C5 ($-CH_3$). The methyl is on C4.

Answer: 4-methylpentanal

(ii) $CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl$

Write out the atoms in order: $CH_3-CH_2-CO-CH(C_2H_5)-CH_2-CH_2-Cl$. Excluding the ethyl branch, the main chain has 6 carbons, with the ketone as the 3rd carbon counting from the $CH_3$ end (that numbering gives the ketone locant 3, lower than numbering from the $Cl$ end, which would give it locant 4). So the parent is hexan-3-one.

Numbering from the $CH_3$ end: C1 ($CH_3$), C2 ($CH_2$), C3 ($CO$), C4 ($CH$, bearing the ethyl branch), C5 ($CH_2$), C6 ($CH_2Cl$). Substituents: ethyl at C4, chloro at C6.

Answer: 6-chloro-4-ethylhexan-3-one

(iii) $CH_3CH=CHCHO$

Numbering from the aldehyde: C1 (CHO), C2 (CH), C3 (CH), C4 (CH3), double bond between C2-C3.

Answer: but-2-enal

(iv) $CH_3COCH_2COCH_3$

A symmetrical 5-carbon chain with ketones at C2 and C4.

Answer: pentane-2,4-dione

(v) $CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3$

Write out the main-chain atoms in order: $CH_3-CH(CH_3)-CH_2-C(CH_3)_2-CO-CH_3$ — 6 carbons. Numbering from the $CO-CH_3$ end gives the ketone the lower locant (C2, versus C5 from the other end), so this is hexan-2-one.

Numbering from that end: C1 ($CH_3$), C2 ($CO$), C3 ($C(CH_3)_2$, two methyl substituents), C4 ($CH_2$), C5 ($CH(CH_3)$, one methyl substituent), C6 ($CH_3$).

Answer: 3,3,5-trimethylhexan-2-one

(vi) $(CH_3)_3CCH_2COOH$

The carboxyl carbon is always C1. One of the three equivalent methyl groups on the quaternary carbon is counted as continuing the main chain (to maximise chain length), making this a 4-carbon (butanoic acid) parent: C1 (COOH), C2 ($CH_2$), C3 (the quaternary carbon, now bearing the two remaining methyls), C4 ($CH_3$, the methyl chosen to extend the chain).

Answer: 3,3-dimethylbutanoic acid

(vii) $OHCC_6H_4CHO$-p

Two $-CHO$ groups at the para (1,4) positions of a benzene ring.

Answer: benzene-1,4-dicarbaldehyde

IUPAC Nomenclature — Step-by-Step Method

Method: Longest Carbon Chain with Principal Functional Group Priority

This is the standard IUPAC approach for naming organic compounds. The steps are:

Steps

1. Identify the principal functional group (highest priority group) — this determines the suffix.
2. Select the longest carbon chain that includes the principal functional group.
3. Number the chain so that the principal functional group gets the lowest possible locant.
4. Name substituents (alkyl groups, halogens, etc.) with their positions.
5. Arrange alphabetically (ignoring prefixes like di-, tri-).
6. Write the name as: (locant)-substituent(s) + parent chain + suffix

Solutions

(i) $CH_3CH(CH_3)CH_2CH_2CHO$

• Principal group: aldehyde ($-CHO$) → suffix -al
• Longest chain: 5 carbons including the aldehyde carbon → pentan-
• Numbering: start from aldehyde carbon (C1)
• Substituent: methyl at C4

Name: 4-methylpentanal

(ii) $CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl$

• Principal group: ketone ($-CO-$) → suffix -one
• Longest chain: 6 carbons including the carbonyl → hexan-
• Numbering: carbonyl gets lowest locant → C3
• Substituents: ethyl at C4, chloro at C6

Name: 6-chloro-4-ethylhexan-3-one

(iii) $CH_3CH=CHCHO$

• Principal group: aldehyde → suffix -al
• Longest chain: 4 carbons including aldehyde and double bond → but-
• Numbering: start from aldehyde carbon (C1)
• Double bond: between C2 and C3 → -2-en-

Name: but-2-enal

(iv) $CH_3COCH_2COCH_3$

• Principal group: ketone (two carbonyls) → suffix -dione
• Longest chain: 5 carbons → pentan-
• Numbering: carbonyls at C2 and C4

Name: pentane-2,4-dione

(v) $CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3$

• Principal group: ketone → suffix -one
• Longest chain: 7 carbons including carbonyl → heptan-
• Numbering: carbonyl gets lowest locant → C2
• Substituents: methyl at C3, two methyls at C5

Name: 3,5,5-trimethylheptan-2-one

(vi) $(CH_3)_3CCH_2COOH$

• Principal group: carboxylic acid ($-COOH$) → suffix -oic acid
• Longest chain: 4 carbons including carboxyl → butan-
• Numbering: start from carboxyl carbon (C1)
• Substituent: three methyls at C3 → tert-butyl group

Name: 3,3-dimethylbutanoic acid

(vii) $OHCC_6H_4CHO$-p

• Principal group: two aldehyde groups → suffix -dial
• Parent: benzene ring → benzene-
• Position: para (1,4-)

Name: benzene-1,4-dicarbaldehyde

(Common name: terephthalaldehyde)

Quick Priority Order (for reference)

Carboxylic acid > Aldehyde > Ketone > Alcohol > Alkene > Alkyne > Alkane

(i) $CH_3CH(CH_3)CH_2CH_2CHO$

Correct IUPAC name:

4-methylpentanal

Common mistakes:

• Mistake 1: Numbering from the wrong end. Students often start from the methyl branch instead of the functional group ($-CHO$).

  Avoid: The aldehyde carbon is always carbon 1. Number the chain so that $-CHO$ gets the lowest number.

• Mistake 2: Writing “pentan-1-al” or “pentanal-1”.

  Avoid: For aldehydes, the “1” position is implied — just write pentanal. Only specify position if the aldehyde is not at the end (rare).

• Mistake 3: Forgetting the methyl substituent’s position.

  Avoid: After numbering from the aldehyde carbon, locate the methyl group — here it’s on carbon 4.

(ii) $CH_3CH_2COCH(C_2H_5)CH_2CH_2Cl$

Correct IUPAC name:

6-chloro-4-ethylhexan-3-one

Common mistakes:

• Mistake 1: Not identifying the longest chain correctly. Students often pick a chain that doesn’t include the carbonyl ($C=O$) or the chloro group.

  Avoid: The parent chain must include the carbonyl carbon. Count carbons in the longest continuous chain that includes $C=O$.

• Mistake 2: Numbering from the wrong end.

  Avoid: The carbonyl carbon should get the lowest possible number. Compare both ends — here, numbering from the chloro side gives the carbonyl carbon number 3 (not 4).

• Mistake 3: Writing ethyl as “$C_2H_5$” in the name.

  Avoid: Always use the substituent name ethyl, not the formula.

• Mistake 4: Forgetting to name the halogen as a prefix (chloro).

  Avoid: Halogens are treated as substituents — use fluoro, chloro, bromo, iodo.

(iii) $CH_3CH=CHCHO$

Correct IUPAC name:

But-2-enal (or 2-butenal)

Common mistakes:

• Mistake 1: Naming it as an alkene first (e.g., “but-2-en-1-al”).

  Avoid: The aldehyde functional group takes priority over the double bond. The suffix is -al, and the double bond is indicated by the infix -en-. No need to write “-1-al” — aldehyde carbon is always 1.

• Mistake 2: Wrong numbering — starting from the double bond side.

  Avoid: Number from the aldehyde carbon (C1). The double bond then gets the lower locant (here, C2–C3).

• Mistake 3: Writing “but-2-en-1-al” or “but-2-enal-1”.

  Avoid: The “1” for aldehyde is implied. Just but-2-enal is correct.

(iv) $CH_3COCH_2COCH_3$

Correct IUPAC name:

Pentane-2,4-dione

Common mistakes:

• Mistake 1: Calling it a “diketone” without proper numbering.

  Avoid: Use the suffix -dione for two carbonyl groups. Number the chain so that the carbonyls get the lowest possible locants.

• Mistake 2: Writing “pentan-2,4-dione” vs “pentane-2,4-dione”.

  Avoid: The parent alkane is pentane; drop the ‘e’ before a vowel (dione) → pentane-2,4-dione.

• Mistake 3: Forgetting that both carbonyls are ketones — no aldehyde present.

  Avoid: If both are $-CO-$, it’s a dione, not an aldehyde-ketone mix.

(v) $CH_3CH(CH_3)CH_2C(CH_3)_2COCH_3$

Correct IUPAC name:

5-methyl-3,3-dimethylhexan-2-one

(Or 3,3,5-trimethylhexan-2-one)

Common mistakes:

• Mistake 1: Not identifying the longest chain correctly.

  Avoid: The chain must include the carbonyl carbon. Count carefully — here the longest chain is 6 carbons (hexane), not 5.

• Mistake 2: Numbering from the wrong end.

  Avoid: The carbonyl carbon should get the lowest number. Compare: numbering from the ketone side gives C=O at position 2; from the other side gives position 5. So choose the ketone side.

• Mistake 3: Writing substituents in wrong alphabetical order.

  Avoid: ethyl comes before methyl alphabetically — but here both are methyl. Write as 3,3,5-trimethyl (combine identical substituents).

• Mistake 4: Forgetting to use di, tri for identical substituents.

  Avoid: Two methyls on C3 → 3,3-dimethyl; three total methyls → trimethyl.

(vi) $(CH_3)_3CCH_2COOH$

Correct IUPAC name:

3,3-dimethylbutanoic acid

Common mistakes:

• Mistake 1: Naming it as “tert-butylacetic acid” (common name).

  Avoid: IUPAC requires systematic naming. The parent chain is butanoic acid (4 carbons including carboxyl carbon).

• Mistake 2: Numbering from the wrong end.

  Avoid: The carboxyl carbon ($-COOH$) is always carbon 1. Number from there.

• Mistake 3: Forgetting that the carboxyl carbon is part of the parent chain.

  Avoid: Count the $-COOH$ carbon as part of the chain. Here, the chain is 4 carbons long (butanoic), not 3.

(vii) $OHCC_6H_4CHO-p$

Correct IUPAC name:

Benzene-1,4-dicarbaldehyde

(or terephthalaldehyde — common name, but IUPAC prefers the systematic name)

Common mistakes:

• Mistake 1: Writing “p-benzenedialdehyde” or “1,4-diformylbenzene”.

  Avoid: The correct IUPAC suffix for two aldehyde groups on a benzene ring is -dicarbaldehyde. Use benzene-1,4-dicarbaldehyde.

• Mistake 2: Forgetting to number the positions.

  Avoid: For disubstituted benzene, use locants 1,2- (ortho), 1,3- (meta), or 1,4- (para). Here it’s 1,4-.

• Mistake 3: Writing “benzene-1,4-dial” — incorrect.

  Avoid: The suffix -dial is not standard for two aldehydes on a ring. Use -dicarbaldehyde.

Quick Summary: Top 5 Mistakes to Avoid in IUPAC