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Q6.6:
Write the isomers of the compound having formula ( C_{4}H_{9}Br ).

Solution :

Step 1: Determining the Degree of Unsaturation

To identify the structural isomers, we first calculate the Degree of Unsaturation (DoU), also known as the Double Bond Equivalent (DBE), for the molecular formula $C_{4}H_{9}Br$.

$DoU = C + 1 - \frac{H + X - N}{2}$

$DoU = 4 + 1 - \frac{9 + 1 - 0}{2} = 5 - 5 = 0$

[Per the Degree of Unsaturation Principle, a value of 0 indicates a saturated acyclic compound, meaning all bonds are single bonds.]

Step 2: Identifying Chain and Positional Isomers

Based on the carbon skeleton, we categorize the isomers into two groups: those derived from the $n$-butane chain and those derived from the isobutane (2-methylpropane) chain.

Isomer Structure IUPAC Name Classification
$CH_{3}CH_{2}CH_{2}CH_{2}Br$ 1-Bromobutane Primary ($1^\circ$)
$CH_{3}CH_{2}CH(Br)CH_{3}$ 2-Bromobutane Secondary ($2^\circ$)
$(CH_{3})_{2}CHCH_{2}Br$ 1-Bromo-2-methylpropane Primary ($1^\circ$)
$(CH_{3})_{3}CBr$ 2-Bromo-2-methylpropane Tertiary ($3^\circ$)

Step 3: Visual Representation of Isomeric Skeletons

CH3-CH2-CH2-CH2Br 1-Bromobutane CH3-CH2-CH(Br)-CH3 2-Bromobutane (CH3)2CH-CH2Br 1-Bromo-2-methylpropane (CH3)3C-Br 2-Bromo-2-methylpropane

Step 4: Analytical Summary of Isomerism

  • Chain Isomerism: 1-Bromobutane and 1-bromo-2-methylpropane exhibit chain isomerism as they possess different carbon skeletons [per the definition of constitutional isomers].
  • Positional Isomerism: 1-Bromobutane and 2-bromobutane exhibit positional isomerism due to the varying location of the functional group ($-Br$) on the butane chain [per IUPAC nomenclature rules].
  • Total Count: There are four distinct structural isomers for the molecular formula $C_{4}H_{9}Br$.

Final Solution: The four structural isomers of $C_{4}H_{9}Br$ are 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, and 2-bromo-2-methylpropane.


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