Question

Which isomer among the following has the highest boiling point?

• A. n-Butylamine
• B. tert-Butylamine
• C. Ethyldimethylamine
• D. Diethylamine

Hint:

Chemistry Tip: The general boiling point order for isomeric amines is: Primary ($1^\circ$) $>$ Secondary ($2^\circ$) $>$ Tertiary ($3^\circ$). Within the same degree, Straight Chain $>$ Branched Chain.

Answer 1

The Correct Option is A

Concept: Chemistry (Organic Chemistry) - Physical Properties of Amines (Boiling Points).

Step 1: Classify each of the given isomeric amines. First, determine the degree of each amine ($C_4H_{11}N$ isomers): - n-Butylamine: $CH_3-CH_2-CH_2-CH_2-NH_2$ (Primary, $1^\circ$) - tert-Butylamine: $(CH_3)_3C-NH_2$ (Primary, $1^\circ$, highly branched) - Ethyldimethylamine: $CH_3-CH_2-N(CH_3)_2$ (Tertiary, $3^\circ$) - Diethylamine: $CH_3-CH_2-NH-CH_2-CH_3$ (Secondary, $2^\circ$)

Step 2: Analyze the effect of hydrogen bonding classes. The boiling point heavily depends on intermolecular hydrogen bonding. Primary ($1^\circ$) and secondary ($2^\circ$) amines have N-H bonds, allowing them to form hydrogen bonds. Tertiary ($3^\circ$) amines lack N-H bonds and cannot form intermolecular hydrogen bonds, giving them the lowest boiling points among isomers. Thus, Ethyldimethylamine has the lowest boiling point.

Step 3: Compare primary versus secondary amines. Primary amines possess two hydrogen atoms attached to the highly electronegative nitrogen atom ($-NH_2$), enabling them to form more extensive hydrogen bonding networks than secondary amines, which only have one such hydrogen atom ($-NH-$). Therefore, the $1^\circ$ amines (n-Butylamine and tert-Butylamine) will have higher boiling points than the $2^\circ$ amine (Diethylamine).

Step 4: Analyze the effect of steric hindrance and branching. We are left with two primary amines: n-Butylamine (straight chain) and tert-Butylamine (heavily branched). Branching forces molecules into a more compact, spherical shape, which significantly reduces the effective surface area for intermolecular Van der Waals (dispersion) forces.

Step 5: Conclude which isomer maximizes all intermolecular forces. Because n-Butylamine has a long, straight carbon chain, it maximizes its surface area for strong Van der Waals interactions while simultaneously possessing the optimal $-NH_2$ group for extensive hydrogen bonding. Combining these two dominating factors results in n-Butylamine having the highest boiling point of the group. $$ \therefore \text{n-Butylamine has the highest boiling point.} $$