Free Radical Halogenation of Alkanes: How to Count Monochloro Products
Chemistry · Hydrocarbons · NEET
When an alkane meets Cl2 in sunlight, a chlorine atom swaps places with a hydrogen atom. This happens through free radicals, and you get one product for each DIFFERENT kind of hydrogen in the molecule. Memory hook: "Count the different H's, that is your number of products." For NEET, also add extra products if any product has a chiral carbon (mirror-image forms).
Free radical chlorination runs in three steps (initiation, propagation, termination). The number of monochloro products equals the number of different kinds of hydrogen, plus extra mirror-image forms for any chiral product.
Your doubts, answered
How do I actually count the number of monochloro products?
Count how many DIFFERENT kinds of hydrogen the alkane has. Two hydrogens are the same kind only if replacing each one gives the exact same molecule. Each different kind of hydrogen gives one structural product. Trick: draw the molecule, then group hydrogens that are related by symmetry. The number of groups = number of structural (constitutional) products.
Which alkane gives only ONE monochloro product? This confuses me.
An alkane gives only one product when ALL its hydrogens are the same kind (equivalent by symmetry). The classic example is neopentane (2,2-dimethylpropane), C(CH3)4. All 12 hydrogens sit on identical CH3 groups, so replacing any one gives the same molecule: neopentyl chloride. Other one-product alkanes: ethane, and cyclohexane. If any hydrogen is different, you get more than one product.
Why does NEET sometimes ask 'including stereoisomers'? What changes?
Normally you count structural products only. But when the question says 'including stereoisomers', you must also check each product for a chiral carbon. A chiral carbon has four different groups attached. Every chiral product exists as two mirror-image forms (R and S), so it counts as 2, not 1. Add these extra forms to your structural count. This is exactly why 2-methylbutane gives 6, not 4.
What is the difference between structural products and total products with stereoisomers?
Structural (constitutional) products differ in WHERE the chlorine sits, giving different connectivity. Stereoisomers have the same connectivity but different 3D arrangement (mirror images from a chiral carbon). For 2-methylbutane: 4 structural products, but 2 of them are chiral, so total = 4 + 2 = 6.
Do 1 degree, 2 degree, 3 degree hydrogens matter for COUNTING or only for the major product?
For COUNTING the number of products, you only care about how many different kinds of hydrogen there are, not their type. The type (1 degree, 2 degree, 3 degree) matters only when you are asked which product forms MOST. Because 3 degree H is easiest to remove, its product usually dominates. But for the count, ignore reactivity.
Why is this a free RADICAL reaction and not ionic?
Sunlight (or heat) breaks the Cl-Cl bond evenly, giving two chlorine ATOMS, each with one unpaired electron. These are free radicals. The reaction runs in three steps: initiation (Cl2 splits), propagation (radicals grab H, make alkyl radical, then grab Cl), and termination (radicals join). No ions form, so it is called free radical substitution.
⚠️ The NEET trap ✗ For monochlorination of 2-methylbutane, students count only where chlorine can go and answer 4. ✓ The NEET 2025 question said 'including stereoisomers', so you must add the two chiral products' mirror forms: 4 structural + 2 = 6 (option B). 🧠 Read the question: if it says 'including stereoisomers', check every product for a chiral carbon and double it.
Real NEET questions
NEET 2025
How many products (including stereoisomers) are expected from monochlorination of 2-methylbutane, (CH3)2CH-CH2-CH3?
A · 5
B · 6 ✓
C · 2
D · 3
Solution: 2-methylbutane has 4 kinds of hydrogen, so 4 structural products: (1) 1-chloro-2-methylbutane, (2) 2-chloro-2-methylbutane, (3) 2-chloro-3-methylbutane, (4) 1-chloro-3-methylbutane. Now check for chiral carbons. Product (1) 1-chloro-2-methylbutane has a carbon bonded to 4 different groups (chiral) so it counts as 2 (R and S). Product (3) 2-chloro-3-methylbutane is also chiral, so it counts as 2. Products (2) and (4) are not chiral. Total = 2 + 1 + 2 + 1 = 6. Answer: B.
NEET 2019 Odisha
The alkane that gives only one mono-chloro product on chlorination with Cl2 in presence of diffused sunlight is:
A · 2,2-dimethylbutane
B · neopentane ✓
C · n-pentane
D · isopentane
Solution: Only one product means all hydrogens must be identical. Neopentane, C(CH3)4, has four identical CH3 groups, so all 12 hydrogens are the same kind. Chlorinating any of them gives the same molecule, neopentyl chloride, only ONE product. The others have several different kinds of hydrogen: 2,2-dimethylbutane gives 4, n-pentane gives 3, isopentane (2-methylbutane) gives 4. Answer: B.
NEET 2018
The compound C7H8 (toluene) undergoes: C7H8 ->[3Cl2, heat] A ->[Br2/Fe] B ->[Zn/HCl] C. The product C is:
A · 3-bromo-2,4,6-trichlorotoluene
B · o-bromotoluene
C · m-bromotoluene ✓
D · p-bromotoluene
Solution: With Cl2 and heat (light, no catalyst), the reaction is free radical substitution at the side chain (the CH3 of toluene), not the ring. So A = C6H5-CCl3 (trichloromethylbenzene). The CCl3 group is meta-directing, so Br2/Fe puts bromine at the meta position: B is the meta-bromo compound. Finally Zn/HCl reduces CCl3 back to CH3, giving C = m-bromotoluene. Answer: C. Key point: light + heat means side-chain free radical halogenation, while a halogen carrier like Fe means ring substitution.
Solved Hydrocarbons NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the replacement of a hydrogen atom of an alkane by a halogen atom (Cl or Br), driven by sunlight or heat through free radical intermediates.
How many monochloro products does n-pentane give?
Three. n-Pentane has three kinds of hydrogen (on C1, C2, and C3), giving 1-chloropentane, 2-chloropentane, and 3-chloropentane.
Why does chlorination give a mixture but bromination is more selective?
Chlorine radical is very reactive and grabs almost any hydrogen, so it makes many products. Bromine radical is less reactive and picky, so it mostly attacks the weakest bond (3 degree H), giving one main product. Remember: bromine is selective.
Does the order of reactivity of hydrogens change the COUNT of products?
No. The count depends only on how many different kinds of hydrogen exist. Reactivity (3 degree greater than 2 degree greater than 1 degree) only decides which product is formed the most.
Why does this topic matter for NEET?
NEET regularly asks you to count monochloro products or spot the alkane giving one product. It tests symmetry and chirality together, so one careful count earns a sure mark.
