Matrix Algebra

 On using a combination of tricks to solve matrix algbra.  One solving this matrix involves the use
of multiplying by a fraction and two for one of the new rows you have
to multiply row 1 by something and row 2 by something THEN add them
together.  For your math pleasure included below in the row operations
I did to solve the matrix.  Follow along and verify that I did my math
right.

The original equations:

X + Y -Z = 6
3X -2Y +Z = -5
X + 3Y -2Z =14

Which produces the augemented matrix

[1  1 -1 :  6]
[3 -2  1 : -5]
[1  3 -2 :14]

I want my matrix to look like

[1 0 0 : a]
[0 1 0 : b]
[0 0 1 : c]

So the first thing I need a 1 in the upper left corner.  Check.  That was easy.
I now need 0s.  How do I get zeros?  If I used the operations:

R2 = -3r1 +r2    &
R3 = r1 + (-1)r3

I would end up with the matrix
[1  1 -1 :    6]
[0 -5  4 : -23]
[0 -2  1 :   -8]

Remember to double check what I am doing.  It does you NO good if I
understand how to do this.

This next part I can see where you would be confused where to go next.
 I need a zero on the bottom.  How do I get it there?  What is a
common denominator between 5 and 2?  Would 10 work?  How do I make
each fisrt number 10?

What if I times row 2 by -2 and row 3 by +5  Then I could add -10 to
+10 and would that = 0?

So try the row operation:

R3 = -2r2 + 5r3

If you do, you should get the matrix
[1  1 -1 :    6]
[0 -5  4 : -23]
[0  0  3 :    6]

Email me if you have any questions about that last step and how I got
it. If you get it we'll continue on.

At this point I can either get ones along the diagonal or go for
zeros.  Also at this point if the teacher didn't ask for reduced row
echelon (r.r.e) form I am actually done.  Because I know that

3Z = 6 solve for Z

-5Y + 4Z = -23 (I know Z from above so plug in and solve for Y).

& X + Y -Z (Again from above I now know Y and Z so plug in and get X).

But assuming your teacher was a jerk and actually wanted r.r.e form
we'll continue our dreary task.

Because it is super easy I am going to make the diagonal 1s.

let
R2 = (-1/5)r2  &
R3 = (-1/3)r3

If we do this our new matrix is

[1 1   -1    :       6]
[0 1 (-4/5) : (23/5)]
[0 0    1    :      -2]

Now let R3 = r2+ (4/5)r3.  I leave it to you to figure out why.  It is
also left up to you to see if there was another way of proceeding if I
hadn't changed the diagonals to ones.  I'll give you a hint.  It
involves +/- 12 and its common factors.

[ 1 1 -1 : 6]
[ 0 1  0 : 3]
[ 0 0  1 : 2]

I purposely dropped a negative sign in the above matrix, see if you
can find it.  Don't go on until you find it.  I want to make sure you
are paying attention and not just mindlessly reading my email.

Found it yet?  Good!

For my next trick I will let R1 = r1 + r3.  Why?  To get a zero where I want it.

[1 1 0 :  4]
[0 1 0 :  3]
[0 0 1 : -2]

I am almost there.  I only have one more set of row operations.  I am
going to let you figure out what they are.  But if done right you
should get the matrix

[1 0 0 :  1]
[0 1 0 :  3]
[0 0 1 : -2]

Did you get the same thing I did?  Do you understand what I did and
more importantly, why I did it?  Is there anything that didn't make
sense that I need to go over one more time.  Speak now or forever hold
your peace, well at least until next time.  I hope that helped.  If
you need anything else you know how to reach me.

Questions regarding O-Chem

1. Hydrogen is the antithesis to double and triple bonds, i.e. hydrogen is kryptonite to the double and triple bonds superman.  Since there is only 1 mole of H2 not all of the triple bond can break only one of them.  Therefore the major product should be a.  A chain with seven carbons with a double bond in the middle i.e. choice a.  C. lost a carbon somewhere.  i would go with a.  If there were lots and lots of hydrogen d is possible but unlikely given only 1 mole.

 

2. Aromatic means that there is an even number alternating single and double bonds equal to 4n+2 where n=0,1,2,3,...

http://en.wikipedia.org/wiki/Toluene  All of these compounds are real but they go by aliases (you'll probably have to memorize these.) Counting carbons toluene has only 7 so it won't work.

http://en.wikipedia.org/wiki/Xylene  the "o" is NOT oxygen it actually refers to the placement of the methyl groups.  If two groups are next to each other it is called ortho-whatever or o-whatever.  If there is a carbon between the two side groups then its called para or p-whatever.  And if there are two carbons between side groups then its called meta or m-whatever.  Looking at the picture this clearly has 8 carbons.  C. Will NOT work because dichloro added two chlorine atoms chlorine is NOT carbon therefore dichlorobenzene has only 6 carbons.

D. phenyl means it has 6 carbons butan means means it has 4 carbons for a grand total of 10 carbons i.e. too many.

E. Ethyl means two carbons.  Benzene means 6 with three double bonds i.e. 8 carbons so this one is good.

F. Has 12 carbons.

 

Thus for number 2 I would pick B and E. (Using Wiki is ok as long as its not during a test.)

 

3. How long is the longest chain.  6.  Does it have any double bonds?  Yes.  How many?  Three.  Does this structure have a common name?  Yes.  Benzene.  Thus whatever it is it ends in something benzene.  Which is more important, boring hydro carbon chains or exotic halides.  Correct Halides.  Is either halide more important? No.  So which one comes first alphabetically, chloro or bromo?  Bromo good.  So number starting at bromine.  Now I can either number clock wise or anti clockwise.  Which choice will give us smaller numbers on more pieces.  Good so let's number them clockwise.  How long is that carbon chain?  Good 3.  What do they call a three carbon chain?  Prop.  Good now is it the main chain or a side chain? Is it bigger than six or smaller?  Its smaller.  Good so it is a side chain.  What do side chains end with?  "yl"  Good.  Thus we start numbering and naming.  Put the halides first then the alkane group. And voila! 1-bromo-3-chloro-2-propylbenzene.

 

4.  I think they all will react, but I would check the chapter to make certain.

 

5.  http://en.wikipedia.org/wiki/File:Benzene_Representations.svg

Hybridization is a lumping together of two electron orbitals.  Think of them as the race tracks electrons follow.  S orbitals look like spheres.  P orbitals look like figure 8s or dumbells.  an sp orbital is bulbous on one end and skinky on the other as though a p orbital swallowed the spherical s orbital.  Have you ever had room mates.  There is what is distinctly your stuff.  Then there is the stuff that is distinctly their stuff, but then there's some of the stuff that you have NO idea whose it is.  When dealing with electrons its the same way.  When the two orbitals merge they call them hybridized orbitals because its a combination of s stuff and p stuff and has kind of a unique shape much like a common room in a shared apartment.  Anyway you have only a few choices for hybridized orbitals.  sp3 has four lobes so something like methane

 

   H

    |

H-C-H

    |

   H

 

One central atom and four thingamajigs coming off of it be they hydrogens, lone pair electrons, bromines, big long hair carbon chain or what not there are four things, its doesn't matter what they are all coming off of the central atom.  we call this sp3 hyridization and it is really easy to remember because you can think of it as s1p3 1+3=4 so an sp3 hybrid should have four lobes.  Now its your turn.  How many lobes will a sp2 hybrid have?  If you are paying attention it should have one less than an sp3 so sp2 have a total of not four but 3 lobes.  This is easily double checked by thinking of sp2 as s1p2 where 1+2= you guessed it--3.  So an sp2 hybrid should have three lobes not four.  Carbons in double bonds often have this arrangement.  Consider the following:

 

    H        H

      \      /

#1   C=C  # 2 

      /       \

   Br        CH2CH2CH2CH3

 

 

Carbon #1 has a C, a H, and a Br bonded to it forming three lobes.  Yes one of the carbons is double bonded so there are really two attachments but if you were to circle all of the stuff attached to carbon #1 you would draw three circles.  Thus the hybridization must add up to three lobes sp2.  Similarly Carbon#2 also has three things attached to it.  One double bonded carbon forming one lobe.  One hydrogen atom forming the second lobe.  And one LONG chain of crap I don't care about forming a third lobe.  (If the question had asked about some of the other carbons I might care and carefully draw out the carbons and hydrogens and I would see they all have four things bonded to them making them sp3 like we learned about earlier.)  Notice that when dealing with hybridizations its not the total number of bonds because in carbon that is always 4, but rather the important part is how many different directions are the bonds facing.

 

Last example.

 

Consider the molecule aceylene aka ethyne

 

H-C*C-H

 

Where * is a triple bond between the carbons.  In this case the bonds go only two different directions I would thus say it has two lobes.  Therefore whatever hybridization is going on must add up to 2.  Well, if sp3 adds up to 4.  And sp2 adds up to 3.  Then sp must add up to two.  Sure enough  we think of it as s1p1 1+1=2 Thus if a molecule has only two lobes, i.e. a triple bond it must be sp hybridized.

 

Brief recap.  sp3=4 lobes i.e. all single bonds.  sp2=3 lobes i.e. a double bond.  sp=2 lobes and is usually a triple bond.  NOTE:  This only applies to the carbons.  Hydrogens only have 1 thing attached to them so they will ALWAYS be unhybridized s orbitals.  Hydrogen is like your underwear.  As much stuff as you share and hybridize with your room mates your underwear is probably not among those things.  Hydrogen stays unhybridized carbon's a bit of a swinger with its electrons.

 

Now let's go back to the question. 

 

Benzene is often depicted as something like this:

 

 /  \

| O |

 \  /

 

but you should think of it as something like

                                                                                      

                                                                   

       H 

        |

       C

      /  \\     

H-C     C-H

   ||       |

   C     C-H

  /   \   //

H      C

         |

        H

or    

       H

        |    

       C

      //  \                                                                               

H-C     C-H                                                                                                             

   |       ||                                                              

   C     C-H                                                                           

  /   \\   /                                                         

H      C                                                             

         |

        H    

 

With the double bond bouncing everywhere faster than is possible to measure.  Since each of the carbons has three lobes we could call it sp2 hybridized, but there is a better description.  Delocalized electrons.  I would call it e.  There are no triples, not sp orbitals.  The hydrogens are on the outside not on the ring, there are no chlorines or other halogens thus e is probably the best choice.

Significant Figures

Dear Tutees,

For your viewing and studying pleasure, today's post is an excerpt from my Master Tutor Project.  Enjoy.

This is an idea I got from my high school chemistry teacher for remembering placement of the decimal in significant figures.    It is called the Atlantic and Pacific rule.

Most people will recognize that the “P” acific Ocean is on the left, while the “A” tlantic Ocean is on the right.  If a decimal is “P” resent count all non-zeros starting at the left.  If a decimal is “A” bsent count all non-zeros starting at the right.  Any bodies of water between the Atlantic and Pacific are ignored, just as any zeros in the middle of a number are ignored.  (Image Source: http://www.theodora.com/maps/new3/north_america_color.gif).   Examples:

503,000  Note that a decimal point is absent.  Therefore, start on the right count all non-zeros e.g. 503 thus 3 significant figures.  But, 503,000.00 has a decimal point present, so start at the Pacific or the left side or 8 significant figures. But 0.001 has only 1 because the first three zeros are non-significant.