Compound of molecular formula C5H10 can not decolorize Baeyer's reagent, how many monohalo products (structural) are obtained by all isomers of C5H10.
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Correct Answer: 17
Solution and Explanation
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.
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