Isomerism

Compound with molecular formula has the following metamers :

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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 ( isomers): - n-Butylamine: (Primary, ) - tert-Butylamine: (Primary, , highly branched) - Ethyldimethylamine: (Tertiary, ) - Diethylamine: (Secondary, )

Step 2: Analyze the effect of hydrogen bonding classes. The boiling point heavily depends on intermolecular hydrogen bonding. Primary ( ) and secondary ( ) amines have N-H bonds, allowing them to form hydrogen bonds. Tertiary ( ) 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 ( ), enabling them to form more extensive hydrogen bonding networks than secondary amines, which only have one such hydrogen atom ( ). Therefore, the amines (n-Butylamine and tert-Butylamine) will have higher boiling points than the 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 group for extensive hydrogen bonding. Combining these two dominating factors results in n-Butylamine having the highest boiling point of the group. $ $

Understanding Geometrical Isomerism:
Geometrical (cis-trans) isomerism is a type of stereoisomerism that arises due to the restricted rotation around a double bond (C=C).
For geometrical isomerism to occur:
• The molecule must contain a C=C double bond.
• Each carbon involved in the double bond must be attached to two different groups.

Let’s evaluate each option:

1) 2-butene (CH₃CH=CHCH₃)
- Each double-bonded carbon has different groups attached (CH₃ and H).
Can exhibit geometrical isomerism (cis and trans forms).

2) 1-butene (CH₂=CHCH₂CH₃)
- The terminal carbon of the double bond has two hydrogen atoms attached.
Cannot exhibit geometrical isomerism.

3) Pent-2-ene (CH₃CH=CHCH₂CH₃)
- Both double-bonded carbons are bonded to different groups.
Can exhibit geometrical isomerism.

4) But-2-yne (CH₃C≡CCH₃)
- Triple bonds do not allow geometrical isomerism as they are linear and have no restricted rotation about C≡C.
Cannot exhibit geometrical isomerism.

Therefore:
Only compounds 1 (2-butene) and 3 (pent-2-ene) can show geometrical isomerism.

Final Answer: Option 2: Only 1 and 3