Determining Dibromination Products of Propane

The reaction involves the free-radical substitution of propane ($CH_3CH_2CH_3$) with bromine ($Br_2$) in the presence of sunlight. Dibromination means two hydrogen atoms are replaced by bromine atoms.

Propane Structure and Symmetry

Propane has three carbon atoms. Due to symmetry, the two terminal methyl groups (C1 and C3) are equivalent, and their hydrogens are primary. The central methylene group (C2) hydrogens are secondary.

Identifying Possible Constitutional Isomers

Dibromination can occur by replacing two hydrogen atoms. The possible distinct constitutional isomers are:

• 1,1-dibromopropane: Both bromine atoms replace hydrogens on the same terminal carbon (C1 or C3). Structure: $CH_3CH_2CHBr_2$.
• 1,2-dibromopropane: One bromine replaces a hydrogen on a terminal carbon (C1 or C3) and the other replaces a hydrogen on the central carbon (C2). Structure: $CH_3CHBrCH_2Br$.
• 1,3-dibromopropane: Bromine atoms replace hydrogens on the two different terminal carbons (C1 and C3). Structure: $BrCH_2CH_2CH_2Br$.
• 2,2-dibromopropane: Both bromine atoms replace the hydrogens on the central carbon (C2). Structure: $CH_3CBr_2CH_3$.

These are the 4 unique structural arrangements (constitutional isomers).

Considering Stereoisomers

We must check if any of these products have chiral centers.

• 1,1-dibromopropane: No chiral center.
• 1,2-dibromopropane ($CH_3CHBrCH_2Br$): Carbon-2 is attached to four different groups ($CH_3$, $H$, $Br$, and $CH_2Br$), making it a chiral center.
• 1,3-dibromopropane: No chiral center.
• 2,2-dibromopropane: No chiral center.

Since 1,2-dibromopropane possesses a chiral center, it exists as a pair of enantiomers (non-superimposable mirror images, denoted as R and S forms). Free radical reactions typically produce racemic mixtures (both enantiomers in equal amounts) unless specific chiral influences are present.

Calculating the Total Number of Different Products

The term "different products" typically includes stereoisomers.

• 1,1-dibromopropane contributes 1 product.
• 1,2-dibromopropane contributes 2 products (R and S enantiomers).
• 1,3-dibromopropane contributes 1 product.
• 2,2-dibromopropane contributes 1 product.

Total number of different products = 1 + 2 + 1 + 1 = 5.

Conclusion

There are 5 different dibromination products formed when propane reacts with bromine in sunlight, considering both constitutional isomers and stereoisomers.