The structures of cis-2-butene and trans-2-butene are as follows:
cis-2-butene: \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \) (Polar)
trans-2-butene: \( \text{H} - \text{CH} = \text{CH} - \text{CH}_3 \) (Non-Polar)
Cis-but-2-ene has a higher dipole moment than trans-but-2-ene due to the arrangement of the substituent groups. The cis configuration results in a net dipole moment, while the trans configuration cancels out the dipole due to symmetry.
The question asks us to identify the incorrect statement regarding the geometrical isomers of 2-butene. Let's go through the options one by one to find the incorrect statement:
Option 1: "cis-2-butene and trans-2-butene are not interconvertible at room temperature."
This statement is true. Geometrical isomers such as cis and trans forms are typically not interconvertible at room temperature due to the energy barrier required to break and reform the double bond.
Option 2: "cis-2-butene has less dipole moment than trans-2-butene."
This statement is false. In fact, cis-2-butene has a higher dipole moment because the polarities of the two methyl groups add up, whereas in the trans isomer, the dipoles cancel each other out due to symmetry.
Option 3: "trans-2-butene is more stable than cis-2-butene."
This statement is true. Generally, trans isomers are more stable than cis isomers due to less steric hindrance and lower energy configuration.
Option 4: "cis-2-butene and trans-2-butene are stereoisomers."
This statement is true. cis-2-butene and trans-2-butene are stereoisomers, specifically a type of stereoisomer known as geometrical isomers.
In conclusion, the incorrect statement is “cis-2-butene has less dipole moment than trans-2-butene.”
Consider the following reaction sequence.
Which of the following hydrocarbons reacts easily with MeMgBr to give methane? 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}