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"How many monochlorinated products are possible?" appears in JEE papers in many forms: structural isomers only, all stereoisomers, or the number of fractions after distillation. Students lose marks by miscounting equivalent hydrogens or forgetting a stereocentre. This post gives a two-step method, applies it to six alkanes and shows how to estimate the percentage of each product.
What does the syllabus say about alkane halogenation?
The JEE (Main) 2026 syllabus lists the "mechanism of halogenation of alkanes", and the JEE (Advanced) 2026 brochure lists halogenation among the reactions of alkanes. The NEET (UG)-2026 syllabus has the same line as JEE Main. The NCERT Class 11 chapter Hydrocarbons describes the free radical chain mechanism (initiation, propagation, termination) and states that the rate of replacing hydrogens is 3° > 2° > 1°. Check your exam year's syllabus.
How do you count the structural isomers?
Step 1: find sets of equivalent hydrogens. Two hydrogens are equivalent if a symmetry of the molecule swaps them. Replacing any hydrogen in a set with Cl gives the same compound, so each set gives one structural isomer.
Step 2: check each product for a new stereocentre. If the carbon carrying Cl now has four different groups, that product exists as a pair of enantiomers.
Worked examples on open-chain alkanes
Propane, CH3CH2CH3: two sets (six CH3 hydrogens, two CH2 hydrogens). Products: 1-chloropropane and 2-chloropropane. Neither is chiral. 2 products.
Butane: two sets give 1-chlorobutane and 2-chlorobutane. C2 in 2-chlorobutane carries H, Cl, CH3 and C2H5, so it is chiral. 2 structural, 3 including stereoisomers.
2-Methylpropane: nine equivalent CH3 hydrogens and one tertiary hydrogen. Products: 1-chloro-2-methylpropane and 2-chloro-2-methylpropane, neither chiral. 2 products.
Pentane: three sets (C1/C5, C2/C4, C3). Products: 1-, 2- and 3-chloropentane. Only 2-chloropentane is chiral. 3 structural, 4 including stereoisomers.
2-Methylbutane, (CH3)2CHCH2CH3: four sets.
Six hydrogens on the two equivalent CH3 groups at C2 give 1-chloro-2-methylbutane (chiral: C2 has H, CH3, CH2Cl, C2H5).
One tertiary hydrogen gives 2-chloro-2-methylbutane (not chiral).
Two hydrogens at C3 give 2-chloro-3-methylbutane (chiral).
Three hydrogens at C4 give 1-chloro-3-methylbutane (not chiral).
4 structural, 6 including stereoisomers.
2,2-Dimethylpropane (neopentane): all twelve hydrogens are equivalent. 1 product.
A harder one: methylcyclohexane
There are five sets of hydrogens: CH3, C1, C2 (same as C6), C3 (same as C5) and C4. So there are 5 structural isomers. Now the stereoisomers:
(Chloromethyl)cyclohexane: no stereocentre, 1.
1-Chloro-1-methylcyclohexane: no stereocentre, 1.
1-Chloro-2-methylcyclohexane: two stereocentres, no symmetry, so cis and trans each exist as a pair of enantiomers, 4.
1-Chloro-3-methylcyclohexane: same reasoning, 4.
1-Chloro-4-methylcyclohexane: cis and trans, each with a mirror plane through C1 and C4, 2.
Total: 1 + 1 + 4 + 4 + 2 = 12 including stereoisomers.
How much of each product forms?
Chlorination is not very selective, so every set reacts, but at different rates. The open textbook OpenStax Organic Chemistry shows how to get the relative rates from experiment. Butane gives 1-chlorobutane and 2-chlorobutane in a 30 : 70 ratio. Per hydrogen, that is 30/6 = 5% for each primary hydrogen and 70/4 = 17.5% for each secondary one, so a secondary hydrogen is 3.5 times as reactive. The same method with 2-methylpropane (35 : 65) gives a tertiary hydrogen about 5 times as reactive as a primary one.
So the relative reactivity per hydrogen is about 1 : 3.5 : 5 for primary, secondary and tertiary.
Worked example: 2-methylbutane. Multiply the number of hydrogens by the reactivity:
1-chloro-2-methylbutane: 6 × 1 = 6
2-chloro-2-methylbutane: 1 × 5 = 5
2-chloro-3-methylbutane: 2 × 3.5 = 7
1-chloro-3-methylbutane: 3 × 1 = 3
Total = 21. Estimated percentages: 6/21 = 28.6%, 5/21 = 23.8%, 7/21 = 33.3%, 3/21 = 14.3%. They add to 100%. The single tertiary hydrogen gives almost a quarter of the product, but the major product comes from the secondary hydrogens, because there are two of them.
If a question gives different reactivity values, for a different temperature or for bromination, use the values in the question.
Why does the order run 3° > 2° > 1°?
OpenStax gives the bond dissociation energies: about 400 kJ/mol for a tertiary C–H bond, 410 kJ/mol for secondary and 421 kJ/mol for primary. The weaker the bond, the easier it is for a chlorine radical to remove that hydrogen, and the more stable the radical formed.
Which mistakes cost marks?
Treating mirror-image carbons as different. In pentane, C1 and C5 give the same product.
Missing a stereocentre. Check every product, not just the obvious one.
Counting enantiomers in distillation questions. They do not separate; diastereomers can.
Using the reactivity ratio without the hydrogen count. The number of hydrogens matters as much as their type.
Our posts on types of structural isomerism and stereoisomerism cover the background.
A practice routine
For hexane, 2-methylpentane, 3-methylpentane and 2,3-dimethylbutane, mark the hydrogen sets in different colours.
Count structural products, then stereoisomers.
Estimate the percentages with 1 : 3.5 : 5 and check that they add to 100.
Write one line for each answer saying what was counted.
When your count does not match the key, compare the logic, not just the number. The Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try it now: how many monochlorination products does 3-methylpentane give, counting stereoisomers? Check each new carbon for four different groups.
