Q61 / 180 · NEET 2026 (Re-NEET) · Chemistry
Given below are two statements:
Statement I:
trans-But-2-ene upon treatment with Br$_2$ in CCl$_4$ gives the following product
Statement II:
cis-But-2-ene upon treatment with alkaline KMnO$_4$ gives the following product
In the light of the above statements, choose the most appropriate answer from the options given below.
Statement I:
trans-But-2-ene upon treatment with Br$_2$ in CCl$_4$ gives the following product
Statement II:
cis-But-2-ene upon treatment with alkaline KMnO$_4$ gives the following product
In the light of the above statements, choose the most appropriate answer from the options given below.
- Correct answerAStatement I is incorrect but Statement II is correct
- BBoth Statement I and Statement II are correct
- CBoth Statement I and Statement II are incorrect
- DStatement I is correct but Statement II is incorrect
Solution
Reaction of trans-but-2-ene with Br$_2$ in CCl$_4$ proceeds via anti-addition to give meso-2,3-dibromobutane.
In the Newman projection given in Statement I, the front carbon has Br at $0^{\circ}$, Me at $120^{\circ}$, and H at $240^{\circ}$. The back carbon has Br at $180^{\circ}$, Me at $60^{\circ}$, and H at $300^{\circ}$. For this staggered conformation to be meso, it must have a center of inversion, meaning a group at angle $\theta$ on the front carbon must have an identical group at $\theta + 180^{\circ}$ on the back carbon. While the Br atoms are opposite ($0^{\circ}$ and $180^{\circ}$), the front Me at $120^{\circ}$ is not opposite to the back Me at $60^{\circ}$. Thus, it lacks a center of inversion and represents a chiral enantiomer. Hence, Statement I is incorrect.
Reaction of cis-but-2-ene with alkaline KMnO$_4$ proceeds via syn-addition to give meso-butane-2,3-diol.
In the Newman projection given in Statement II, the front carbon has OH at $0^{\circ}$, Me at $120^{\circ}$, and H at $240^{\circ}$. The back carbon has OH at $60^{\circ}$, Me at $180^{\circ}$, and H at $300^{\circ}$. Rotating the back carbon by $60^{\circ}$ counter-clockwise perfectly eclipses all identical groups (OH with OH at $0^{\circ}$, Me with Me at $120^{\circ}$, H with H at $240^{\circ}$). This indicates the presence of a plane of symmetry, confirming it is the meso isomer. Hence, Statement II is correct.
Therefore, Statement I is incorrect but Statement II is correct.
Answer: Statement I is incorrect but Statement II is correct
In the Newman projection given in Statement I, the front carbon has Br at $0^{\circ}$, Me at $120^{\circ}$, and H at $240^{\circ}$. The back carbon has Br at $180^{\circ}$, Me at $60^{\circ}$, and H at $300^{\circ}$. For this staggered conformation to be meso, it must have a center of inversion, meaning a group at angle $\theta$ on the front carbon must have an identical group at $\theta + 180^{\circ}$ on the back carbon. While the Br atoms are opposite ($0^{\circ}$ and $180^{\circ}$), the front Me at $120^{\circ}$ is not opposite to the back Me at $60^{\circ}$. Thus, it lacks a center of inversion and represents a chiral enantiomer. Hence, Statement I is incorrect.
Reaction of cis-but-2-ene with alkaline KMnO$_4$ proceeds via syn-addition to give meso-butane-2,3-diol.
In the Newman projection given in Statement II, the front carbon has OH at $0^{\circ}$, Me at $120^{\circ}$, and H at $240^{\circ}$. The back carbon has OH at $60^{\circ}$, Me at $180^{\circ}$, and H at $300^{\circ}$. Rotating the back carbon by $60^{\circ}$ counter-clockwise perfectly eclipses all identical groups (OH with OH at $0^{\circ}$, Me with Me at $120^{\circ}$, H with H at $240^{\circ}$). This indicates the presence of a plane of symmetry, confirming it is the meso isomer. Hence, Statement II is correct.
Therefore, Statement I is incorrect but Statement II is correct.
Answer: Statement I is incorrect but Statement II is correct
Hence, the correct answer is option: 'A'.