Question

The number of sigma (\( \sigma \)) and pi (\( \pi \)) bonds present in 3-Methylbut-1-ene are respectively

• A. \(1 \text{ and } 14 \)
• B. \(18 \text{ and } 2 \)
• C. \(16 \text{ and } 2 \)
• D. \(17 \text{ and } 1 \)
• E. \(14 \text{ and } 1 \)

Hint:

Double bond always contributes one $\pi$ bond.

Answer 1

Answer: (E)

Step 1: Drawing the Structure of the Molecule:
The name "3-Methylbut-1-ene" tells us the following:
but-: The main chain has 4 carbon atoms.
-1-ene: There is a double bond starting at carbon-1.
3-Methyl: There is a methyl group (CH\(_3\)) attached to carbon-3.
Let's number the carbon chain: C1-C2-C3-C4. The structure is: CH\(_2\)=CH-CH(CH\(_3\))-CH\(_3\).
For clarity, let's draw the expanded structural formula:
\[ \begin{array}{ccccc} & \text{H} & \text{H} & \text{H} &
& | & | & | &
\text{H}_2\text{C} & = & \text{C} & - & \text{C}
& & | & & |
& & \text{H} & & \text{C} - \text{H}
& & & & |
& & & & \text{H}
\end{array} \quad\rightarrow\quad \begin{array}{ccccc} \text{H} & & \text{H} & \text{H} & \text{H}
| & & | & | & |
\text{C}_1 & = & \text{C}_2 & - & \text{C}_3 & - & \text{H}
| & & & & |
\text{H} & & & & \text{C}_4 & - & \text{H}
& & & & | & & |
& & & & \text{H} & & \text{H}
\end{array} \] The correct expanded structure is:
\( \text{CH}_2 = \text{CH} - \text{CH(CH}_3\text{)} - \text{CH}_3 \)
Which expands to:
H\(_2\)C=CH-CH(CH\(_3\))-CH\(_3\)
H\(_2\)C\textsubscript{1}=C\textsubscript{2}H-C\textsubscript{3}H(CH\(_3\))-C\textsubscript{4}H\(_3\) (Incorrect representation)
Let's draw it correctly: The main chain is C1=C2-C3-C4. A methyl group is on C3. \[ \text{CH}_2 = \text{CH} - \underset{\text{CH}_3}{\underset{|}{ \text{CH}}} - \text{CH}_3 \] Step 2: Counting the Pi (\(\pi\)) Bonds:
Pi bonds are present in multiple bonds.
A single bond consists of 1 \(\sigma\) bond.
A double bond consists of 1 \(\sigma\) bond and 1 \(\pi\) bond.
A triple bond consists of 1 \(\sigma\) bond and 2 \(\pi\) bonds.
The structure has one double bond (C=C). Therefore, there is 1 \(\pi\) bond.
Step 3: Counting the Sigma (\(\sigma\)) Bonds:
We can count the sigma bonds in two ways: Method 1: Direct Counting from the expanded structure
C-H bonds: 2 (on C1) + 1 (on C2) + 1 (on C3) + 3 (in methyl group) + 3 (in terminal CH3) = 10 C-H bonds.
C-C bonds: 1 (in C=C) + 1 (C2-C3) + 1 (C3-C4) + 1 (C3-methyl C) = 4 C-C bonds.
Total \(\sigma\) bonds = 10 (C-H) + 4 (C-C) = 14 \(\sigma\) bonds.
Method 2: Using the Formula for Acyclic Hydrocarbons For an acyclic (non-cyclic) molecule with N atoms, the number of sigma bonds is \(N-1\). Number of Carbon atoms = 5 Number of Hydrogen atoms = 10 Total atoms \(N = 5 + 10 = 15\). Number of \(\sigma\) bonds = \(15 - 1\) = 14.
This formula works for acyclic molecules. Since 3-methylbut-1-ene is acyclic, this is a valid quick check. Step 4: Final Answer:
The molecule has 14 sigma (\(\sigma\)) bonds and 1 pi (\(\pi\)) bond.