See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The dipole moment of a para-substituted donor-acceptor benzene depends on the conjugation between the electron-donating group (NMe2) and the electron-withdrawing group (NO2) through the aromatic ring. Greater conjugation leads to a larger dipole moment due to increased charge separation. Step 1 - Compound (A): 4-(dimethylamino)nitrobenzene has NMe2 (strong donor) para to NO2 (strong acceptor). There are no steric substituents on the ring. The NMe2 nitrogen is free to adopt a planar geometry, allowing full conjugation of the lone pair with the aromatic ring and hence with the NO2 group. This through-conjugation creates significant charge separation (quinonoid resonance structure), resulting in a large dipole moment of 6.87 D. Step 2 - Compound (B): In the sterically crowded analog (2,3,5,6-tetramethyl derivative), four methyl groups flank both the NMe2 and NO2 substituents. The ortho methyl groups adjacent to NMe2 force the NMe2 group out of the plane of the ring (steric inhibition of resonance). This disrupts the conjugation of the nitrogen lone pair with the pi system. Without effective conjugation, there is no quinonoid charge separation, and the dipole moment is significantly reduced to 4.11 D (closer to a simple vector sum of localized group moments without through-conjugation enhancement). Step 3 - Why other options fail: - Option (b) reverses the values, but compound A (unhindered) must have the higher dipole moment. - Option (c) makes both equal at 4.11 D, ignoring that compound A has full conjugation. - Option (d) makes both equal at 6.87 D, ignoring the steric inhibition in compound B. Therefore, the correct answer is A.