Course Grades (December 14, 2017)

  Lab Grades    (December 13, 2017)
MCTC Principles of Chemistry 2 Hybrid
Fall 2017

Syllabus
Lab Manual

Mastering Chemistry
 Online Office Hours:    TBA    click HERE

On-Campus Office Hours (Room S-3580): 
Monday: 11-12PM and 1 - 2PM
Wednesday: 4-5 PM

Instructor: Kirk Boraas
Email Kirk

Course Lectures

CHEM1151 Website

D2L
This is NOT an "online" course.    
Campus visits are required (weekly labs and periodic exams).

Withdrawing? Click HERE first.
Sample Exams

Exam #2 Fall 2017
Exam #4 Fall 2017
Course Announcements

December 7 ...Exam #2 ONLINE Review Session:   .pdf   video

Click here for entrance.

9/18 OR 9/19 7:00 PM Online Exam 1              10/17  Exam 2  6-9 PM Rm S-3500    
11/13
OR 11/14 7:00 PM Online Exam 3              12/8 Exam 4   6-9 PM  Rm L-3000        
        
 
12/13 Optional Cumulative Final Exam 6-9 PM Rm S-1400

Previous Weeks 

Weeks 1 & 2            Weeks  3 & 4            Weeks 5 & 6         Weeks 7 & 8      |    Weeks 9 & 10     Weeks 11 & 12       Weeks 13 & 14     Weeks 15 & 16

December 6, 2017    Wednesday   Week 16
Chapter 20: Organic Reactions and Synthesis of Soap


Answer these questions.  Click and drag over the space below for answers.

1. What best describes the term saponification? (Source)
    a. Cleaving of an ester molecule into carboxylic acid and alcohol.
    b. Dehydration synthesis by removing water
    c. Hydrolysis of a salt by adding a weak acid
    d. Synthesis of two alkyl groups to make an ether

2. Which of the following is considered a useful alkali in
    saponification reactions?
(Source)
     a. CCL      b. Cl      c. NaOH        d. Ca+2

3. Where does the hydroxide ion attach in the saponification reaction?
    a. the carbonyl carbon
    b. the hydroxyl carbon
    c. the alkyl carbon
    d. the methoxyl carbon

Organic Reactions: SaponificationKB
Organic Reaction: SaponificationKB
4.  How does one identify the carbonyl carbon?
      a. The carbonyl carbon is attached to a singly bonded oxygen and two other alkyl groups
      b. The carbonyl carbon is attached to an oxygen atom via a double  bond and two other "R" groups.
      c. The carbonyl carbon is attached to an oxygen that is attached to several other alkyl groups.


The following questions refer to the ethyl tetradecanoate molecule show at right.
ethyl tetradecanoate

5. If a strong base interacts with the molecule, identify the letter corresponding
     to the attack of the hydroxide ion.    a.   b.   c.   d.   e. 

6. Which bond is broken as the soap molecule forms?
      a-b bon        b-c bond        c-d bond      d-e bond       

7. Which letter identifies a carbon that is part of the ethyl group in this  molecule?

8.  How many carbons are there in the soap molecule that forms?

9. What is the name of the alcohol that is released once the reaction is complete?

Answers:  Click and drag in the space below
1. a     2. c     3.      4. b      5. c     6. c-d bond           7.e    8.    15    9. ethanol


December 5, 2017    Tuesday   Week 17
Chapter 20: Polarimetry


Answer these questions.

1. What is plane polarized light?  What is non-polarized light?

2. What does it mean to say that an optically active material rotates
    the polarization plane of light?

3. What type of molecule is optically active?

4. How is polarized light affected if it passes through a 50:50 mix
    of two enantiomers?

5. If vertically polarized light strikes a horizontally oriented
    polarizing filter, how much light passes through the filter?

6. How is non-polarized light transformed into polarized light?
Polarimetry (up to time mark 4 min 50 sec)
Polarimetry (up to time 4 min 50 sec)
The following questions refer to the diagram at right.

7.  What is type of light is initially produced by the light source?
8.   How does the polarizing filter "b" affect the light from the
       light source?
9.   If light passes through the polarizing filter "e", what must the
      polarization of the light be?

10. Light from the first polarizing filter can't pass through the second
      polarizing filter because it is vertically polarized.  How does an
      optically active sample change this?

Polarimeter
Answers:  Click and drag in the space below

1. Light that has an electric field that vibrates in only one plane.  Non-polarized light is a mixture of all polarizations from horizontal to vertical.
2. When passing through an optically active material, the polarized light plane is rotated.  For example, if vertically polarized light passes through
     an optically active material, it's plane will be rotated from vertical to something in between vertical and horizontal.  The longer the sample tube,
     the more the light plane is rotated.
3. An optically active molecule is one that is chiral.  Being chiral means to have a chiral carbon with 4 DIFFERENT entities attached to it.
4. A 50:50 mixture contains half of one enantiomer and half of the other.  That is, an even mixture of both variation of the optically active molecule.
    Each of the two enantiomers rotates the light polarization in opposite directions.  Thus, in a 50:50 sample the effects cancel out and no change
    to the light polarization is observed.
5. Ideally, no light should pass through although a little always leaks through.
6. When passed through a polarizing filter, non-polarized light is transformed into polarized light corresponding to the orientation of the
     polarizing filter.
7. Unpolarized ("a")
8. Filter "b" is vertically oriented.  It passes only the vertically polarized light coming from the light source.
9. "e" is oriented horizontally and only passes (ideally) horizontally polarized light
10. Optically active materials rotate the polarization of the incident light.  In this case, the sample located at "d" if optically active, would rotate the
      vertically polarized light coming from the first filter "b".  Having done this, the light now has horizontal character and some passes through the
      second filter.


December 4,  2017    Monday   Week 16
Chapter 20: Optical Isomers


Answer these questions.  Click and drag over the space below for answers.

1. For each of the two molecules shown below, decide whether the
    the molecule is optically active (Chiral vs. Achiral) and if it can
    form enantiomers (Source)
Two molecules.  optical isomers?

Stereochemistry: Enantiomers
Stereochemistry: Enantiomers
2. How many optically active (chiral) carbons are there in 2-butanol (figure below)
2 butanol

3. Examine the molecule below and identify the carbon atoms (a - j) as either chiral (optically active) or achiral (not optically active).

optically active carbons
Answers:  Click and drag in the space below
1. a. Not optically active (i.e. achiral) since the carbon is attached to two methyl groups
    b. Optically active (chiral)  since there is a carbon attached to 4 different substituents.

2. Only one optically active carbon.  (the one attached to the OH group)
3. a.
3. a. achiral    b. achiral    c. chiral    d. achiral    e. achiral    f. achiral     g. chiral    h. achiral   i. achiral    j. achiral


December 1, 2017    Friday   Week 15
Chapter 20: Functional Groups
Lecture 25.2 Functional Groups Part 1KB
Functional Groups Part 1KB
Lecture 25.3 Functional Groups Part 2KB
Functional Groups Part 2KB
For each of the molecules at right provide the type of functional group represented and the name of the molecule.   (Source)


Click below for answers:


a. Amine                     methylamine
b. alcohol                    methanol
c. ether                       diethylether
d. alcohol                    2-propanol
e. carboxylic acid       propanoic acid
f.  ester                      ethyl propanoate
g. aldehyde                butanal
h. ketone                   3-hexanone
structural group nomenclature
Answer these questions.  Click and drag over the space below for answers.


Answers:  Click and drag in the space below


November 30, 2017    Thursday   Week 15
Chapter 20 Aromatic Hydrocarbons


Delocalized Electrons: Benzene
Delocalized Electrons: Benzene
Basic Nomenclature: m, o and p
Basic Nomenclature: m, o and p
Answer these questions:benzene delocalized electrons
1. In the picture at right, what is the significance of
    the two rings shown on top and bottom of the
     benzene molecule?

2. Which of the Lewis Structures below is correct
      for the benzene molecule?

benzene resonance

3. What are the possible names for the compounds shown at right?
   
Benzene structures
Answers:  Click and drag in the space below
1. The rings represent the overlap of pi bonds on the top and bottom of the
     benzene molecule.  The electrons in the six pi bonds are delocalized or
     spread out over the top and bottom of the molecule.

2. All are correct.  The first two demonstrate the resonance that exists as the double bonds shift to produce two equivalent Lewis Structures.  The third
    structure indicates the delocalization of pi electrons on the top and bottom of the molecule.

3. a. methyl benzene    
    b. 1,4-dimethylbenzene or p-dimethylbenzene     (p is for para)    or       p-xylene
    c. 1,3-dimethyl benzene or m-dimethylbenzene   (m is for meta)    or      m-xylene
    d. 1,2-dimethyl benzene or o-dimethyl benzene   (o is for ortho)    or      p-xylene

November 29 2017    Wednesday   Week 15
Chapter 20: Alkene Nomenclature and Geometric Isomers


Alkene and Alkyne Nomenclature
Alkene and Alkyne Nomenclature
Geometric Isomers: Cis/Trans
Geometric Isomers: Cis/Trans
Answer these questions.  Click and drag over the space below for answers.

1. Name the compounds at right. (Source)

 Click and drag below for correct answers.

a. 1-pentene
b. 1-heptene
c.  cis-2-butene
d.  trans-4-nonene
e. cis-3-heptene
Name these alkenes


November 28, 2017    Tuesday   Week 15
Chapter 20 Structural Isomers


Answer These Questions:

1. How many structural isomers are there for C4H10?  Name them.

2. The following structures are isomers.  True or False?
    more possible structural isomers

3. Draw and name the isomers of C7H16.  (
Source)
  


Click and drag here for answers.

1.   TWO     butane and isobutane
2.  a. False    b. True    c. True
3. Click here for answers.
Structural Isomers
Structural Isomers

November 27, 2017    Monday   Week 16.
Chapter 20 Alkane Nomenclature


Alkane Nomenclature Part 1
Alkane Nomenclature Part 1
Alkane Nomenclature Part 2
Alkane Nomenclature Part 2
Answer these questions.  Click and drag over the space below for answers.

1. Provide the correct IUPAC name for the structures at right.  (Source)

2. Provide structures for the following organic compounds:

     a. 4-ethyloctane
     b. 2-methylnonane
     c. 3,3-dimethylpentane
     d. 3-ethylpentane
     e. 3-ethyl-2-methylheptane
     f. 3-ethyl-4-isopropyl-2,2-dimethyloctane

Click and drag below for answers...

1.      a.  2-methylbutane
         b.   3-ethyl-4-methylheptane
         c.  4-ethylheptane
         d.  2,2,5-trimethyl-3-ethylhexane
         e. 5-chloro-4-ethyloctane
         f.  4-ethyl-3-methyl-5-tert butyloctane
2. Answers Click here
alkane structures