Organic compound C₅ H₁₀ does not give Baeyer’s reagent test. Calculate total number of structural monobromo isomers when react with Br₂ /hν

The correct answer is 14.

Let's break down the step-by-step derivation to find the total number of structural monobromo isomers:

Step 1: Determine the nature of the organic compound C₅H₁₀
The degree of unsaturation (DU) for the molecular formula $C_5{H}_{10}$ is calculated as follows:
$\mathrm{DU}=C+1-\frac{H}{2}=5+1-\frac{10}{2}=1$
A degree of unsaturation of 1 indicates the presence of either one double bond (alkene) or one ring (cycloalkane). Since the compound does not give the Baeyer's reagent test (alkaline potassium permanganate test), it does not contain a carbon-carbon double bond. Therefore, the compound must be a cycloalkane.

Step 2: Identify all possible structural isomers (cycloalkanes) of C₅H₁₀
There are five possible structural isomers of cycloalkanes with the formula $C_5{H}_{10}$:
1. Cyclopentane
2. Methylcyclobutane
3. Ethylcyclopropane
4. 1,1-Dimethylcyclopropane
5. 1,2-Dimethylcyclopropane

Step 3: Calculate the structural monobromo isomers formed by each cycloalkane upon reaction with Br₂/hν

1. Cyclopentane:
All 5 carbons are equivalent. Monobromination gives only 1 structural isomer:
• Bromocyclopentane
Subtotal = 1

2. Methylcyclobutane:
There are 4 chemically non-equivalent hydrogen-bearing carbon positions where bromination can occur:
• C1 (tertiary ring carbon): 1-bromo-1-methylcyclobutane
• C2 (secondary ring carbon adjacent to the methyl group): 2-bromo-1-methylcyclobutane
• C3 (secondary ring carbon opposite to the methyl group): 3-bromo-1-methylcyclobutane
• Methyl carbon: (bromomethyl)cyclobutane
Subtotal = 4

3. Ethylcyclopropane:
There are 4 chemically non-equivalent carbon positions containing hydrogen:
• C1 of the cyclopropane ring (tertiary): 1-bromo-1-ethylcyclopropane
• C2/C3 of the cyclopropane ring (secondary): 2-bromo-1-ethylcyclopropane
• C1 of the ethyl group: (1-bromoethyl)cyclopropane
• C2 of the ethyl group: (2-bromoethyl)cyclopropane
Subtotal = 4

4. 1,1-Dimethylcyclopropane:
There are 2 chemically non-equivalent carbon positions with hydrogen (the quaternary ring carbon C1 has no hydrogen atoms):
• C2/C3 of the ring (secondary): 2-bromo-1,1-dimethylcyclopropane
• Methyl carbons: 1-(bromomethyl)-1-methylcyclopropane
Subtotal = 2

5. 1,2-Dimethylcyclopropane:
There are 3 chemically non-equivalent carbon positions with hydrogen:
• C1/C2 of the ring (tertiary): 1-bromo-1,2-dimethylcyclopropane
• C3 of the ring (secondary): 3-bromo-1,2-dimethylcyclopropane
• Methyl carbons: 1-(bromomethyl)-2-methylcyclopropane
Subtotal = 3

Step 4: Sum the structural monobromo isomers
Adding the structural monobromo isomers from all possible cycloalkanes gives:
$\mathrm{Total}=1+4+4+2+3=14$