Question

How many monobrominated product(s) (including stereoisomers) would form in the free radical bromination of n-butane?

• A. 2
• B. 1
• C. 3
• D. 4

Answer 1

The Correct Option is C

We need to consider the possible positions where bromine can substitute a hydrogen atom in n-butane, and also account for any stereoisomers formed.

Step 1: Draw the Structure of n-Butane

n-Butane has the following structure: \[CH_3-CH_2-CH_2-CH_3\]

Step 2: Consider Possible Monobromination Products

Bromine can substitute a hydrogen atom on either a primary carbon (\(CH_3\)) or a secondary carbon (\(CH_2\)). Let's consider each case:

1. Bromination at a primary carbon (C1 or C4):

Replacing a hydrogen on either of the terminal carbons (C1 or C4) will yield the same product, 1-bromobutane: \[CH_2Br-CH_2-CH_2-CH_3\] Since C1 and C4 are chemically equivalent, we only get one distinct product.

1. Bromination at the C2 carbon:

Replacing a hydrogen on C2 will yield 2-bromobutane: \[CH_3-CHBr-CH_2-CH_3\] This carbon is a chiral center (it is bonded to four different groups: H, CH₃, CH₂CH₃, and Br). Therefore, 2-bromobutane exists as two stereoisomers (enantiomers).

1. Bromination at the C3 carbon:

Replacing a hydrogen on C3 will yield 3-bromobutane: \[CH_3-CH_2-CHBr-CH_3\] this is exactly the same as 2-bromobutane in stereoisomers

Step 3: Count the Monobrominated Products

1. 1-bromobutane: one isomer
2. 2-bromobutane: two isomers due to chiral carbon

1 distinct product at C1 and two stereoisomers at C2, so a total of 3 products

Step 4: Count the Total Number of Stereoisomers

We have:

• 1-bromobutane: 1 isomer (not chiral)
• 2-bromobutane: 2 stereoisomers (enantiomers)

So, in total, there are \(1 + 2 = 3\) monobrominated products (including stereoisomers).

Conclusion

There are 3 monobrominated products formed in the free radical bromination of n-butane.