Q.Give the IUPAC names of the following compounds:

Concept: Structural Isomerism & IUPAC Nomenclature – The key is to identify the longest carbon chain, number it to give the lowest locants to substituents (using the alphabetical tiebreak when two numbering directions give the same locant set), and name substituents alphabetically (ignoring multiplying prefixes like di, tri).

(i) $CH_3CH(Cl)CH(Br)CH_3$

Longest chain is 4 carbons (butane). The locant set {2,3} is the same from either end, so the alphabetical tiebreak decides: bromo outranks chloro, so bromo gets the lower locant.

2-Bromo-3-chlorobutane

(ii) $CHF_2CBrClF$

Two-carbon chain (ethane). Substituents: F, Cl, Br on C1; two F on C2. Alphabetical order: bromo, chloro, difluoro.

1-Bromo-1-chloro-1,2,2-trifluoroethane

(iii) $ClCH_2C \equiv CCH_2Br$

Four-carbon chain with a triple bond (but-2-yne). The locant set {1,4} is tied either way, so the alphabetical tiebreak applies again: bromo outranks chloro.

1-Bromo-4-chlorobut-2-yne

(iv) $(CCl_3)_3CCl$

The longest chain runs THROUGH the central carbon and into TWO of its three trichloromethyl branches at once (a genuine 3-carbon chain), leaving the third CCl₃ as a trichloromethyl substituent and the remaining Cl as a plain chloro substituent on the middle carbon. All 7 chain chlorines (3 + 1 + 3) are cited together under one prefix.

1,1,1,2,3,3,3-Heptachloro-2-(trichloromethyl)propane

(v) $CH_3C(p\text{-}ClC_6H_4)_2CH(Br)CH_3$

Longest chain: 4 carbons (butane). Numbering from the end nearer the two aryl groups gives locant set {2,2,3}, which beats the flipped {2,3,3} at the first point of difference.

3-Bromo-2,2-bis(4-chlorophenyl)butane

(vi) $(CH_3)_3CCH=CClC_6H_4I\text{-}p$

The longest chain absorbs ONE of the tert-butyl group's three methyls (methyl–C–CH=CCl, 4 carbons), leaving the other two methyls as substituents on C-3. Numbered from the C=CCl end so the double bond gets the lowest locant (1, not 3).

1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

(i) $CH_3CH(Cl)CH(Br)CH_3$ - parent butane; Br and Cl fall on C2/C3 either way (locants tie), so the alphabetically-first substituent (bromo) takes the lower number -> 2-bromo-3-chlorobutane.

(ii) $CHF_2CBrClF$ - parent ethane; numbering from the $CBrClF$ end gives the lower locant set $\{1,1,1,2,2\}$ (vs $\{1,1,2,2,2\}$) -> 1-bromo-1-chloro-1,2,2-trifluoroethane.

(iii) $ClCH_2C\equiv CCH_2Br$ - parent but-2-yne; substituent locants $\{1,4\}$ tie, so bromo takes position 1 -> 1-bromo-4-chlorobut-2-yne.

(iv) $(CCl_3)_3CCl$ - the longest chain runs through the central carbon between two $CCl_3$ groups, giving a propane parent; C1 and C3 each carry three Cl, the central C2 carries one Cl plus the third $CCl_3$ as a (trichloromethyl) substituent -> 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane.

(v) $CH_3C(p\text{-}ClC_6H_4)_2CH(Br)CH_3$ - parent butane; numbering with the two aryl groups at C2 gives the locant set $\{2,2,3\}$ (aryl, aryl, bromo), which beats numbering the other way with Br at C2 ($\{2,3,3\}$) at the first point of difference (2 < 3) -> 3-bromo-2,2-bis(4-chlorophenyl)butane. (The alphabetical rule only breaks a TRUE tie in the locant set -- here the sets themselves already differ, so the lower-locant-set rule decides first.)

(vi) $(CH_3)_3CCH=CClC_6H_4I\text{-}p$ - the longest chain through the C=C includes one methyl of the tert-butyl group, giving a four-carbon but-1-ene chain; numbering from the doubly-bonded end that bears Cl and the aryl group -> 1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene.

Structural Isomerism — IUPAC Naming Method

Method: Systematic IUPAC Nomenclature (Substitutive Naming)

This method follows the IUPAC Blue Book rules: identify the longest parent chain, number it to give lowest locants to substituents (breaking ties alphabetically when two numbering directions give the same locant set), name substituents alphabetically, and specify multiple bonds where present.

Steps for each compound:

1. Identify the longest carbon chain (parent alkane/alkene/alkyne) — check every direction a chain can run, including through a branch point in more than one direction.
2. Number the chain to give the lowest possible locants to substituents; if tied, the substituent cited first alphabetically gets the lower locant.
3. Name substituents (halogens, alkyl groups, etc.) in alphabetical order.
4. Specify multiple bonds with appropriate suffixes (-ene, -yne), giving them priority for low locants.
5. Combine identical substituents under one multiplying prefix, citing all their locants together.

Solutions

(i) $CH_3CH(Cl)CH(Br)CH_3$

• Parent chain: 4 carbons → butane
• Locant set for the two substituents is {2,3} from either end (symmetric chain) — the alphabetical tiebreak applies: bromo (b) outranks chloro (c), so bromo gets the lower locant.

Answer: 2-bromo-3-chlorobutane

(ii) $CHF_2CBrClF$

• Parent chain: 2 carbons → ethane
• Substituents: F (×2) at C-1, Br, Cl, F at C-2
• Alphabetical: bromo, chloro, difluoro

Answer: 1-bromo-1-chloro-1,2,2-trifluoroethane

(iii) $ClCH_2C \equiv CCH_2Br$

• Parent chain: 4 carbons with triple bond → but-2-yne
• Locant set {1,4} tied either way — alphabetical tiebreak: bromo before chloro, so bromo gets locant 1.

Answer: 1-bromo-4-chlorobut-2-yne

(iv) $(CCl_3)_3CCl$

• The central carbon is bonded to Cl and three CCl₃ groups. The longest chain does not stop at the centre — it runs through the centre and into two of the three CCl₃ branches at once, giving a genuine 3-carbon chain (propane), with the third CCl₃ left over as a substituent.
• C-1 and C-3: three chlorines each (from their own CCl₃ origin). C-2 (the original centre): one chlorine + one trichloromethyl branch.
• All 7 chain chlorines (3 + 1 + 3, at locants 1,1,1,2,3,3,3) are cited together as heptachloro.

Answer: 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane

(v) $CH_3C(p\text{-}ClC_6H_4)_2CH(Br)CH_3$

• Parent chain: 4 carbons → butane
• Numbering from the end nearer the two aryl groups gives locant set {2,2,3} for (aryl, aryl, bromo) — lower than the flipped {2,3,3} at the first point of difference.
• Substituents: two 4-chlorophenyl groups at C-2, bromo at C-3
• Alphabetical: bromo, chlorophenyl

Answer: 3-bromo-2,2-bis(4-chlorophenyl)butane

(vi) $(CH_3)_3CCH=CClC_6H_4I\text{-}p$

• The longest chain is NOT the tert-butyl group treated as an intact substituent on a short chain — it absorbs one of the tert-butyl's three methyls into the chain itself: methyl–C–CH=CCl, 4 carbons (but-1-ene), leaving the other two methyls as substituents on C-3.
• Numbering must start from the C=CCl end so the double bond gets locant 1 (not 3) — the double bond outranks ordinary substituents for low-locant priority.
• Substituents: chloro and 4-iodophenyl at C-1, two methyl at C-3.
• Alphabetical: chloro, iodophenyl, methyl

Answer: 1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

Key exam tip: Always check for the lowest locant set first — if two numbering directions give the same set, only then compare alphabetically. And always check whether the "obvious" chain (e.g. treating a tert-butyl or CCl₃ group as a fixed substituent) is really the LONGEST one — a chain can run through a branch point in more than one direction.

This is a classic area where even good students lose marks. Let's go compound by compound, listing the common mistakes and how to avoid each.

(i) $CH_3CH(Cl)CH(Br)CH_3$

✗ Common Mistakes

• Wrong parent chain selection — students sometimes number from the wrong end, giving a higher locant sum.
• Alphabetical tiebreak applied backwards — writing "chlorobromo" or defaulting to whichever halogen is drawn first, instead of checking which one is alphabetically first.
• Missing commas and hyphens — e.g., 2-chloro 3-bromobutane instead of the correctly punctuated name.

✓ How to Avoid

• Find the longest carbon chain — here it's 4 carbons (butane).
• Check the locant set first — both Cl and Br sit at positions 2 and 3 regardless of which end you number from, so the locant SET is tied at {2,3}.
• Break the tie alphabetically — the substituent cited first alphabetically (bromo, "b") gets the lower locant, not chloro.
• Write substituents in alphabetical order (ignoring di-, tri- prefixes for alphabetizing).

Correct IUPAC name:

2-bromo-3-chlorobutane

(ii) $CHF_2CBrClF$

✗ Common Mistakes

• Not recognizing this is a haloalkane with multiple different halogens — students often misorder the substituents.
• Wrong parent chain — some pick a 1-carbon chain instead of 2-carbon (ethane).
• Alphabetizing "difluoro" by its "d" rather than remembering that a simple multiplying prefix (di-, tri-) is normally ignored for alphabetization — except that here it doesn't matter, since "difluoro" and "fluoro" both alphabetize as "fluoro" and are tied with each other, not with bromo/chloro.

✓ How to Avoid

• Identify the longest carbon chain — 2 carbons → ethane.
• Number the chain so that substituents get the lowest locants. Comparing the two numbering directions, the one that puts three chlorine-family substituents (Br, Cl, F) on C-1 and both remaining F's on C-2 gives the lower locant set — so C1 = CBrClF, C2 = CHF₂.
• List substituents alphabetically — bromo, chloro, difluoro.

Correct IUPAC name:

1-bromo-1-chloro-1,2,2-trifluoroethane

(iii) $ClCH_2C \equiv CCH_2Br$

✗ Common Mistakes

• Not recognizing the triple bond — students sometimes treat it as an alkane.
• Miscounting the parent chain length — this is a common trap: there are only 4 carbons here (but-2-yne), not 5 — don't count a substituent atom (Cl or Br) as if it were a chain carbon.
• Alphabetical tiebreak applied backwards — same trap as compound (i): the locant set {1,4} is tied either way, so bromo (alphabetically first) must get the lower locant.

✓ How to Avoid

• Identify the functional group — triple bond (alkyne). Parent chain must include it.
• Count only carbon atoms in the chain — Cl–CH₂–C≡C–CH₂–Br has exactly 4 carbons → but-2-yne.
• Break the tied locant set alphabetically — bromo before chloro, so bromo gets locant 1.

Correct IUPAC name:

1-bromo-4-chlorobut-2-yne

(iv) $(CCl_3)_3CCl$

✗ Common Mistakes

• Stopping the chain at the central carbon — students often name this as a substituted methane (with three "trichloromethyl" branches), missing that the chain can run straight THROUGH the centre and into two of the branches at once.
• Splitting identical substituents into separate citations — e.g. writing "2-chloro...hexachloro..." as two separate chlorine citations instead of combining every chain chlorine under one multiplying prefix.

✓ How to Avoid

• Draw the structure — central C bonded to Cl and three CCl₃ groups.
• Look for a chain that passes through the branch point — going from one CCl₃'s carbon, through the centre, into a second CCl₃'s carbon gives a genuine 3-carbon chain (propane). This beats stopping at 1 carbon (methane).
• The third CCl₃ becomes a substituent — named "trichloromethyl" — and the Cl on the (now middle) chain carbon is a plain chloro substituent.
• Combine every identical substituent — all 7 chain chlorines (three from each end carbon, one from the middle) are cited together as heptachloro with all their locants listed.

Correct IUPAC name:

1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane

(v) $CH_3C(p\text{-}ClC_6H_4)_2CH(Br)CH_3$

✗ Common Mistakes

• Not recognizing the aromatic rings — students forget to name the p-chlorophenyl groups.
• Numbering from the wrong end — giving bromo locant 2 and the aryl groups locant 3, instead of checking which direction actually gives the lower locant SET.

✓ How to Avoid

• Identify the longest carbon chain — 4 carbons (butane). The two aromatic rings are substituents, not chain extensions (each aryl ring attaches by a single bond, it isn't a continuation of the carbon chain).
• Compare both numbering directions — locants {aryl, aryl, bromo} = {2,2,3} from one end vs {2,3,3} from the other. {2,2,3} is lower at the first point of difference.
• Alphabetical order — bromo, then chlorophenyl (since "b" < "c"). Use "bis" for two identical aryl groups.

Correct IUPAC name:

3-bromo-2,2-bis(4-chlorophenyl)butane

(vi) $(CH_3)_3CCH=CClC_6H_4I\text{-}p$

✗ Common Mistakes

• Treating "tert-butyl" as a fixed, intact substituent — this hides a longer chain: one of the tert-butyl's three methyls can be absorbed into the main chain itself, since it's directly bonded to the chain carbon.
• Numbering the double bond from the wrong end — a double bond gets priority for the lowest possible locant, so the chain must be numbered starting from the C=CCl end, not from the tert-butyl end.

✓ How to Avoid

• Look for a longer chain through the "obvious" substituent — methyl–C(CH₃)₂–CH=CCl– is a genuine 4-carbon chain (but-1-ene) once one methyl of the tert-butyl group is counted as part of the chain; the other two methyls remain as substituents.
• Give the double bond the lowest locant — numbering from the C=CCl end puts it at C-1, not C-3.
• Substituents — chloro and 4-iodophenyl both on C-1, two methyl groups on C-3.
• Alphabetical order — chloro, iodophenyl, methyl.

Correct IUPAC name:

1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

Summary Table of Key Mistakes

Final tip: Always draw the structure first, then apply IUPAC rules step-by-step — longest chain (checking every direction, including through branch points) → numbering (lowest locant set, alphabetical tiebreak) → substituents combined and alphabetized. This eliminates most errors.