Question

Compound of molecular formula C₅H₁₀ can not decolorize Baeyer's reagent, how many monohalo products (structural) are obtained by all isomers of C₅H₁₀.

Hint:

In organic chemistry, when a compound does not react with Baeyer's reagent, it typically indicates the absence of double bonds. The monohalo products formed depend on the number of distinct hydrogen atoms available in the molecule for substitution.

Answer 1

Correct Answer: 17

The molecular formula C₅H₁₀ represents an alkane, meaning that it contains only single bonds between carbon atoms. Since Baeyer's reagent (potassium permanganate, KMnO₄) is used to test for the presence of double bonds, a compound that does not decolorize Baeyer's reagent must be an alkane, which does not have any C=C double bonds.

Now, let's consider the structural isomers of C₅H₁₀:

- The straight-chain alkane, pentane. - The branched-chain isomers: 2-methylbutane, 3-methylbutane, and isopentane.

For each isomer of C₅H₁₀, we consider the number of possible monohalo products (i.e., products obtained by replacing one hydrogen atom with a halogen atom, such as chlorine or bromine).

- For pentane, we can substitute a halogen at any of the 5 positions, giving us 5 monohalo products. - For 2-methylbutane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products. - For 3-methylbutane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products. - For isopentane, we can substitute a halogen at 4 different positions, giving us 4 monohalo products.

Thus, the total number of monohalo products for all isomers of C₅H₁₀ is:

\[5 + 4 + 4 + 4 = 17\]

Hence, the number of monohalo products is 17.