Question

Select the set of correct statements: Statement-I : The dipole moment of trans-but-2-ene is zero. Statement-II : The boiling points of 2-methylpropane and butane are equal. Statement-III : Chlorobenzene on Friedel-Crafts acylation produces 2-chloroacetophenone as the main product. Statement-IV : 1-iodobutane undergoes \(S_N2\) reaction faster than 1-chlorobutane.

• A. I, II
• B. II, III
• C. III, IV
• D. I, IV

Hint:

• Trans alkenes often have lower or zero dipole moments due to symmetry.

• Branching decreases boiling point.

• Chlorine is ortho-para directing but para product is generally major.

• \(S_N2\) reactivity follows: \[ RI \gt RBr \gt RCl \gt RF \]

Answer 1

The Correct Option is D

Concept: The correctness of the statements can be checked using molecular symmetry, boiling point trends, electrophilic substitution orientation and nucleophilic substitution reactivity.

Step 1: Analyze Statement-I. Trans-but-2-ene is symmetrical. The bond dipoles cancel each other. \[\begin{aligned} \mu = 0 \end{aligned}\] Hence, \[\begin{aligned} \text{Statement-I is Correct} \end{aligned}\]

Step 2: Analyze Statement-II. Branching decreases boiling point. \[ \begin{aligned} \text{n-Butane} &:\quad \text{Higher boiling point} \\ \text{2-Methylpropane} &:\quad \text{Lower boiling point} \end{aligned} \] Therefore, \[\begin{aligned} \text{Statement-II is Incorrect} \end{aligned}\]

Step 3: Analyze Statement-III. Chlorine is ortho-para directing. However, due to steric effects, the para product predominates. Major product: \[\begin{aligned} p\text{-chloroacetophenone} \end{aligned}\] Hence, \[\begin{aligned} \text{Statement-III is Incorrect} \end{aligned}\]

Step 4: Analyze Statement-IV. For \(S_N2\) reactions, \[\begin{aligned} RI \gt RBr \gt RCl \gt RF \end{aligned}\] because iodide is the best leaving group. Therefore, \[\begin{aligned} \text{1-Iodobutane reacts faster than 1-chlorobutane} \end{aligned}\] Thus, \[\begin{aligned} \text{Statement-IV is Correct} \end{aligned}\]

Step 5: Identify the correct set. \[ \begin{aligned} \text{Statement I} &:\quad \text{Correct} \\ \text{Statement II} &:\quad \text{Incorrect} \\ \text{Statement III} &:\quad \text{Incorrect} \\ \text{Statement IV} &:\quad \text{Correct} \end{aligned} \] Therefore, \[\begin{aligned} \boxed{\text{I, IV}} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.