PRELIMINARY STAGE 2026
Contest 40
Konongo Odumase SHS: 45 points
Our Lady of Mt. Carmel Girls’ SHS: 31 points
Juaben SHS: 28 points
Contest 41
St. Monica’s SHS: 40 points
Amenfiman SHS: 37 points
Tema Methodist Day SHS: 15 points
Contest 42
Accra High School: 33 points
Ghana SHS, Tamale: 32 points
Sunyani SHS: 31 points
ROUND 1
PREAMBLE
Give the systematic name of one constitutional isomer of the compound with molecular formula $C_5H_{11}Cl$, given that it possesses an alkyl substituent (i.e. it is a branched, not straight-chain, isomer).
FORMULA (SHORTCUT METHOD)
$C_5H_{11}Cl$ is a monochlorinated pentane.
Pentane ($C_5H_{12}$) has three carbon skeletons: n-pentane (straight chain — excluded here since it has no alkyl substituent), 2-methylbutane / "isopentane" (one branch), and 2,2-dimethylpropane / "neopentane" (two branches).
Any monochloro derivative of the isopentane or neopentane skeleton qualifies as a valid answer.
---
1. 1-chloro-2-methylbutane.
WHY THIS WORKS: Structure $ClCH_2-CH(CH_3)-CH_2-CH_3$ — a 2-methylbutane (isopentane) skeleton with Cl on C1, the carbon directly adjacent to the branch point.
Carbon count: 4 in the main chain + 1 in the methyl branch = 5 C.
Hydrogen count: $2+1+3+2+3=11\text{ H}$.
Formula $C_5H_{11}Cl$ ✓, and the methyl branch on C2 is the required alkyl substituent.
---
2. 2-chloro-2-methylbutane.
WHY THIS WORKS: Structure $CH_3-CCl(CH_3)-CH_2-CH_3$ — the same isopentane skeleton, but with Cl on the branch (tertiary) carbon itself, C2.
Carbon count: 5 (same skeleton).
Hydrogen count: $3+0+3+2+3=11\text{ H}$.
Formula $C_5H_{11}Cl$ ✓, with the methyl branch on C2 satisfying the alkyl-substituent requirement.
(This is the tertiary chloride commonly called tert-amyl chloride.)
---
3. 1-chloro-3-methylbutane.
WHY THIS WORKS: Structure $ClCH_2-CH_2-CH(CH_3)-CH_3$ — the isopentane skeleton with Cl on the terminal carbon farthest from the branch point.
Numbering from the Cl end gives substituent locants {1,3}, which is lower than numbering from the other end ({2,4} for "4-chloro-2-methylbutane"), so {1,3} is the correct IUPAC choice.
Carbon count: 5, Hydrogen count: $2+2+1+3+3=11\text{ H}$.
Formula $C_5H_{11}Cl$ ✓, with the methyl branch on C3 as the alkyl substituent.
---
PRACTICE QUESTIONS
1. QUESTION: Give the systematic name of one branched isomer of $C_4H_9Cl$.
ANSWER: 2-chloro-2-methylpropane (or 1-chloro-2-methylpropane)
SOLUTION:
Both are built on the 2-methylpropane skeleton.
2. QUESTION: Give the systematic name of the isomer of $C_5H_{11}Cl$ built on the 2,2-dimethylpropane skeleton.
ANSWER: 1-chloro-2,2-dimethylpropane
SOLUTION:
All twelve hydrogens of 2,2-dimethylpropane are equivalent, so there is only one such isomer.
3. QUESTION: Give the systematic name of the isomer of $C_5H_{11}Cl$ in which Cl is on a tertiary carbon.
ANSWER: 2-chloro-2-methylbutane
SOLUTION:
C2 of 2-methylbutane is bonded to three other carbons.
4. QUESTION: Give the systematic name of the branched isomer of $C_4H_9Br$ with Br on a primary carbon.
ANSWER: 1-bromo-2-methylpropane
SOLUTION:
It is built on the 2-methylpropane skeleton.
5. QUESTION: Give the systematic name of one isomer of $C_5H_{11}Cl$ built on the 2-methylbutane skeleton with Cl on an end carbon.
ANSWER: 1-chloro-2-methylbutane (or 1-chloro-3-methylbutane)
SOLUTION:
Cl is on a primary carbon at either end of the chain.
6. QUESTION: Give the systematic name of the isomer of $C_3H_7Cl$ with Cl on the middle carbon.
ANSWER: 2-chloropropane
SOLUTION:
C2 is the secondary carbon of propane.