Question

Given below are two statements:
Statement I: trans-But-2-ene upon treatment with $\text{Br}_2$ in $\text{CCl}_4$ gives the following product:

Statement II: cis-But-2-ene upon treatment with alkaline $\text{KMnO}_4$ gives the following product:

In the light of the above statements, choose the most appropriate answer from the options given below.

• A. Statement I is incorrect but Statement II is correct
• B. Both Statement I and Statement II are correct
• C. Both Statement I and Statement II are incorrect
• D. Statement I is correct but Statement II is incorrect

Hint:

Keep this classic mnemonic matrix handy for alkene additions: - CAR: Cis + Anti addition $\longrightarrow$ Racemic mixture - TAM: Trans + Anti addition $\longrightarrow$ Meso compound - CIS: Cis + Syn addition $\longrightarrow$ Meso compound Comparing the given structures against these rules makes it straightforward to spot incorrect stereoisomers!

Answer 1

The Correct Option is A

Concept: The stereochemical outcome of electrophilic addition reactions across carbon-carbon double bonds depends inherently on both the configuration of the starting alkene (cis or trans) and the stereospecific mechanism of the addition pathway (syn or anti):

• Bromination (\(\text{Br}_2 / \text{CCl}_4\)): Proceeds via an

anti-addition mechanism due to the intermediate formation of a cyclic bromonium ion, which blocks the front-side attack and forces the second bromide ion to attack from the opposite side.

• Hydroxylation with cold alkaline \(\text{KMnO}_4\) (Baeyer's Reagent): Proceeds via a coordinated

syn-addition pathway because a cyclic manganese ester complex forms simultaneously across the same face of the double bond.
We can use standard stereochemical memory rules to predict outcomes: \[\begin{aligned} \text{\textit{trans} alkene} + \text{\textit{anti} addition} &\longrightarrow \text{meso compound} \\ \text{\textit{cis} alkene} + \text{\textit{syn} addition} &\longrightarrow \text{meso compound} \end{aligned}\]

Step 1: Evaluating Statement I.
Let us analyze the reaction of trans-but-2-ene with $\text{Br}_2$ in $\text{CCl}_4$. Because a trans alkene undergoing an anti-addition yields a symmetrically substituted

meso stereoisomer, the resulting product, meso-2,3-dibromobutane, must possess an internal plane of symmetry ($\sigma$) or a center of inversion ($i$). Let us inspect the Newman projection provided in Statement I of the image: - The front carbon has substituents arranged as: Br (top), Me (right), H (left). - The back carbon has substituents arranged as: Br (bottom), Me (right), H (left). Let us check if this projection represents the meso form. In this staggered conformation, if we find a center of inversion, it is meso. Moving from the front Br (top) through the center leads directly to the back Br (bottom) $\rightarrow$ inversion matches. However, moving from the front Me (bottom-right) through the center points towards the top-left, but the back Me is located at the bottom-right! Therefore, this structure does not have a center of inversion. If we rotate the back carbon by $180^\circ$ to check the eclipsed conformation: the top Br would eclipse the bottom Br, meaning the bromine atoms would be on opposite sides rather than matching. This shows that the provided structure is a chiral enantiomer (d or l pair), not the meso isomer. Since trans-but-2-ene must produce the optically inactive meso form upon anti-bromination, Statement I gives the wrong stereoisomer and is incorrect.

Step 2: Evaluating Statement II.
Let us examine the reaction of cis-but-2-ene with cold dilute alkaline $\text{KMnO}_4$. A cis alkene undergoing a stereospecific syn-addition yields a symmetrically substituted

meso compound, which is meso-butane-2,3-diol. Let us analyze the Newman projection displayed under Statement II: - The front carbon has OH pointing straight up ($12\text{ o'clock}$), Me pointing to the bottom-right ($4\text{ o'clock}$), and H pointing to the bottom-left ($8\text{ o'clock}$). - The back carbon has OH pointing to the top-right ($2\text{ o'clock}$), Me pointing straight down ($6\text{ o'clock}$), and H pointing to the top-left ($10\text{ o'clock}$). Let us rotate the back carbon by $60^\circ$ counter-clockwise to see the eclipsed form: - The back OH (from $2\text{ o'clock}$) moves to $12\text{ o'clock}$, perfectly eclipsing the front OH. - The back Me (from $6\text{ o'clock}$) moves to $4\text{ o'clock}$, perfectly eclipsing the front Me. - The back H (from $10\text{ o'clock}$) moves to $8\text{ o'clock}$, perfectly eclipsing the front H. Because every single substituent on the front carbon perfectly matches and eclipses its identical twin group on the back carbon when rotated into an eclipsed conformation, this molecule contains a clear internal plane of symmetry. This proves that the structure is exactly the expected meso-butane-2,3-diol. Therefore, Statement II is completely correct.

Step 3: Conclusion.
Since Statement I is incorrect and Statement II is correct, the matching choice is Option (1).