Textbook: Wade 7th Ed. (2009)

Chapter 2: Structure and Properties of Organic Molecules

Practice Problems (No matching mendel sets were found.)

Short link: mendelset.com/chapters/1020

Printer-friendly version

Individual Problems

Individual Problems

Problem # 1287

Rank each set of compounds in order of decreasing boiling point (1 = highest boiling point):

a) ethane, n-octane, n-pentane

b) n-butane, 1-butanol, 1-chlorobutane.

c) n-octane, 2-methylheptane, 2,5-dimethylhexane

(Note that the n- prefix before an alkane just means that it's one chain, without any branching.)

Solution

Remember the rules of what gives compounds a higher boiling point (and melting point):

Van der Waals Forces

More electrons = more Van der Waals interactions = higher boiling point

So in general, the more "stuff" in a molecule (the higher the molecular weight), the higher its b.p. Example: C₁₀H₂₂ will have a higher b.p. than C₅H₁₂.

How to remember this trend? What do you grill with? propane gas. (C₃H₈)

What's in a lighter? butane liquid (C₄H₁₀)

Hydrogen Bonding

Compounds that can hydrogen bond have higher b.p./m.p/ than those that don't.

Compounds can hydrogen bond is they have a N, O, or F bonded directly to an H.

In organic chemistry, the best examples are alcohols (ROH) and amines (RNH₂).

Branching

Branching makes it harder for molecules to pack together, which makes it harder to form Van der Waals interactions, and so tends to lower b.p./m.p.

a) (highest b.p.) n-octane > n-pentane > ethane (lowest b.p.)

Reason: octane has the most "stuff" (higher molecular weight).

b) 1-butanol > 1-chloroethane > n-butane

Reason: 1-butanol can hydrogen bond. 1-chloroethane has a higher molecular weight than n-butane.

c) n-octane > 2-methylheptane > 2,5-dimethylhexane

Reason: Branching. n-octane has no branching. 2,5-dimethylhexane has the most branching. Notice that each compound has the same molecular formua- C₈H₁₈. (and therefore the same weight).

Problem Details

MendelSet practice problem # 1287 submitted by Matt on October 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Intermolecular Forces and Functional Groups

Question Type: Concept (explain why this is so..)

Keyword(s): boiling point, intermolecular forces

Problem # 1286

For the molecule shown below, indicate the hybridization (sp³, sp², sp, etc.) of atoms A through H, and the bond angles of X, Y, and Z.

Solution

The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.

Also remember these rules:

If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp³ hybridized (s¹ + p³ = 4).
If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp² hybridized (s¹ + p² = 3).
If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s¹ + p¹= 2).

A sp² This oxygen has one double bond and two lone pairs. Three things total = sp².

B sp² This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp².

C sp³ This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp³.

D sp³ This carbon has one implicit hydrogen and three single bonds. Four things total = sp³.

E sp³ This oxygen has two lone pairs and two bonding pairs. Four things total = sp³.

F sp² This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp².

G sp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.

H sp This nitrogen has a lone pair and a triple bond. Two things total = sp.

For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:

Best way to separate two things is a straight line (180°).
Best way to separate three things is a "trifecta" (120°).
Best way to separate four things in three dimensional space is a tetrahedron (109.5°).

X 120° The carbon between the two oxygens is sp²hybridized, so the shape is trigonal planar and the bond angle is 120°.

Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.

Z 109.5° The carbon is sp³ hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.

Problem Details

MendelSet practice problem # 1286 submitted by Matt on October 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Bonding, Geometries, and Hybridization

Question Type: Concept (explain why this is so..)

Keyword(s): hybridization, bond angles

Problem # 697

Rank the following compounds in order of decreasing boiling point.

Also, make a guess about their relative solubilities in water. Explain your reasoning.

Solution

Water is very polar, so the more polar a molecule is, the more soluble it will be in water.

Alcohols can hydrogen bond, and so will be the most polar, have the highest boiling point, and be the most soluble in water.

Aldehydes (carbonyls) can't hydrogen bond but the C=O double bond is very polar, so aldehydes are somewhat soluble in water.

The ether is the least polar of the three functional groups, and so will have the lowest boiling point, and be the least soluble in water.

So the overal trend for polarity, water solubility, and boiling point is:

alcohol > aldehyde > ether

Problem Details

MendelSet practice problem # 697 submitted by Matt on July 20, 2011.

Subject: Organic Chemistry

Subject Topic(s): Ethers and Epoxides, Intermolecular Forces and Functional Groups

Question Type: Concept (explain why this is so..)

Keyword(s): boiling point, polarity, solubility

Textbook: Wade 7th Ed. (2009)

Chapter 2: Structure and Properties of Organic Molecules

Practice Problems (No matching mendel sets were found.)

Individual Problems

Login