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(a)
Interpretation:
From the given compounds, compounds with higher value of solubility has to be identified and the reason for the higher value of solubility has to be explained.
The given compounds are chloroethane and methylethylamine.
Concept Introduction:
The physical properties are determined by the intermolecular forces, polarity of the compound and hydrogen bonding. If a compound is polar and have hydrogen bonding, the solubility will be more for that compound than in non-polar compounds.
Intermolecular forces:
The forces which acts in between the atoms to hold a molecule are said to be intermolecular forces. They are weak forces. Intermolecular forces are categorized as hydrogen bonding, ionic bonding, ion-induced dipole forces, ion-dipole forces and Vanderwaal forces.
Hydrogen bonding:
The attraction between a lone pair of hydrogen atom with that of an electronegative atom forms hydrogen bonding. Two water molecules are held together by hydrogen bond.
(b)
Interpretation:
From the given compounds, compounds with higher value of melting point has to be identified and the reason for the higher value of melting point has to be explained.
The given compounds are diethyl ether and 1-butanol.
Concept Introduction:
The physical properties are determined by the intermolecular forces, polarity of the compound and hydrogen bonding. If a compound is polar and have hydrogen bonding, the melting point will be more for that compound than in non-polar compounds.
Intermolecular forces:
The forces which acts in between the atoms to hold a molecule are said to be intermolecular forces. They are weak forces. Intermolecular forces are categorized as hydrogen bonding, ionic bonding, ion-induced dipole forces, ion-dipole forces and Vanderwaal forces.
Hydrogen bonding:
The attraction between a lone pair of hydrogen atom with that of an electronegative atom forms hydrogen bonding. Two water molecules are held together by hydrogen bond.
(c)
Interpretation:
From the given compounds, compounds with higher value of boiling point has to be identified and the reason for the higher value of boiling point has to be explained.
The given compounds are trimethylamine and propylamine.
Concept Introduction:
The physical properties are determined by the intermolecular forces, polarity of the compound and hydrogen bonding. If a compound is polar and have hydrogen bonding, the boiling point will be more for that compound than in non-polar compounds.
Intermolecular forces:
The forces which acts in between the atoms to hold a molecule are said to be intermolecular forces. They are weak forces. Intermolecular forces are categorized as hydrogen bonding, ionic bonding, ion-induced dipole forces, ion-dipole forces and Vanderwaal forces.
Hydrogen bonding:
The attraction between a lone pair of hydrogen atom with that of an electronegative atom forms hydrogen bonding. Two water molecules are held together by hydrogen bond.
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Chapter 15 Solutions
CHEMISTRY: THE MOLECULAR NATURE OF MATTE
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- Please correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forwardThe following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forward
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