Read the passage given below and answer the questions based on it.
Amines constitute an important class of organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s). Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore sp3 hybridised and the geometry of amines is pyramidal. Amines are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule. Due to the electron releasing nature of alkyl group, it (R) pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid. Moreover, the substituted ammonium ion formed from the amine gets stabilised due to dispersal of the positive charge by the +I effect of the alkyl group. Hence, alkyl amines are stronger bases than ammonia. Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups. The order of basicity of amines in the gaseous phase follows the expected order: 3° > 2° > 1° > NH3. Adrenaline and ephedrine, both containing a secondary (2°) amino group, are used to increase blood pressure. The well-known antihistaminic drug Benadryl contains a tertiary (3°) amino group.
Find out the hybridization of nitrogen in N, N-dimethyl methanamine.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(sp^3\)
In amines, as in ammonia, the nitrogen atom forms three σ bonds and carries one lone pair. These four electron pairs occupy four sp3 hybrid orbitals arranged tetrahedrally, and the molecule is pyramidal, with the C–N–C angle (108° in trimethylamine) slightly less than 109.5° because of the lone pair.
N,N-Dimethylmethanamine, \( \ce{(CH3)3N} \), has three N–C bonds and one lone pair on N, so its nitrogen is sp3 hybridised.
Why the other options are incorrect
Option (1) \(sp^2\) — Incorrect. Would need N to form a double bond (three regions of electron density), as in an imine.
Option (2) \(sp\) — Incorrect. Would need a triple bond or two regions, as in a nitrile (–C≡N).
Option (3) \(d^2sp^3\) — Incorrect. Uses d orbitals (six pairs); nitrogen has no d orbitals in its valence shell.