Ch. 2 - Molecular RepresentationsWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

Solution: The compounds n-butane, CH3(CH2)2CH3 and trimethylamine, N(CH3)3 have very similar molecular weights. However, their melting points are appreciably different. Select the compound with the lowest melti

Problem

The compounds n-butane, CH3(CH2)2CH3 and trimethylamine, N(CH3)3 have very similar molecular weights. However, their melting points are appreciably different. Select the compound with the lowest melting point, and the correct reason why.

a. trimethylamine, because it will experience dipolar forces between molecules 

b. n-butane, because it is unbranched 

c. trimethylamine, because it is more highly O branched than n-butane 

d. n-butane, because it will only experience dispersion forces between molecules