Read the passage given and answer the questions.

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 alkyl groups. The order of basicity of amines of amines in the gaseous phase follows the expected order: 3° > 2° > 1° > NH3. Adrenaline and ephedrine both containing 2° amino group used to increase blood pressure. Well known antihistamine drug Benadryl also contains 3.

Find out the hybridization of nitrogen in N, N-dimethyl methanamine.

\(sp^3\)

The correct answer is option 4. \(sp^3\).

The nitrogen in N, N-dimethyl methanamine (also known as trimethylamine) is \(sp^3\) hybridized. Here's why:

\(N\) has 5 valence electrons in its ground state = \(1s^2 2s^2 2p^3\)

In N, N-dimethyl methanamine, the nitrogen atom forms bonds with three carbon atoms (through the methyl groups) and has a lone pair of electrons.

To accommodate all these bonding orbitals (3 sigma bonds and 1 lone pair) and maintain tetrahedral geometry around the nitrogen, it needs 4 hybridized orbitals.

Combining 1s, 2s, and 2p orbitals, we can generate 4 \(sp^3\) hybrid orbitals, each containing one unpaired electron.

These hybrid orbitals form sigma bonds with the carbon atoms and accommodate the lone pair, resulting in the \(sp^3\) hybridization of the nitrogen.

Therefore, the correct answer is (4) \(sp^3\).