1 radian = 180 π degrees 1.75 radians = 1.75× 180 π = 100.268 Note the following commonly met angles: 30 = π 6 radians 45 = π 4 radians 60 = 3 radians 90 = π 2 radians 135 = 3π 4 radians 180 = π radians 30 = 6 radians 45 = 4 radians60 = 3 90 = 2 radians o o o o Your calculator should be able to work with angles measured in both radians. A 2:1 degree is sometimes nicknamed an Attila the Hun (two-one) in the UK. The term Bren gun is also sometimes used as rhyming slang. citation needed Lower Second Class Honours. This is the lower division of Second Class degrees and is abbreviated as ’2:2’ or ‘II.ii’ (pronounced two-two).
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of (pi) bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
- The polynomial of degree 4, P(x) has a root of multiplicity 2 at z = 3 and roots of multiplicity 1 at x = 0 and a = Find a formula for P(x). It goes through the point (5, 36).
- Pi radians are equal to 180 degrees: π rad = 180° One radian is equal 13 degrees: 1 rad = 180°/π = 13° The angle α in degrees is equal to the angle α in radians times 180 degrees divided by pi constant: α (degrees) = α (radians) × 180° / π. Degrees = radians × 180° / π. Convert 2 radians angle to.
- This is the UK system of degree classification. This is not the same as a GPA or other international grading systems. UK degrees are classified as follows. Highest to Lowest degree classification (marks out of 100): 1st / Class I - 70 or above; 2:1 / Class II Division I - 60 to 69; 2:2 / Class II Division II - 50 to 59; 3rd / Class III - 40 to 49.
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. Microsoft office 2016 16 12 0. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 | 1-methyoxypentane |
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 | 3-chloro-5-octyne |
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
[ DoU= dfrac{2C+2+N-X-H}{2} ]
- (C) is the number of carbons
- (N) is the number of nitrogens
- (X) is the number of halogens (F, Cl, Br, I)
- (H) is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
- One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 ( pi ) bond).
- Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 ( pi ) bonds).
Possible combinations of rings/ bonds | |||
---|---|---|---|
# of rings | # of triple bonds | ||
1 | 0 | ||
0 | 0 | ||
2 | 0 | ||
0 | 0 | ||
0 | 1 | ||
1 | 0 | ||
3 | 3 | 0 | 0 |
2 | 1 | 0 | |
1 | 2 | 0 | |
0 | 1 | 1 | |
0 | 3 | 0 | |
1 | 0 | 1 |
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
SOLUTION
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
References
- Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5thEd.). New York: W. H. Freeman. (473-474)
- Shore, N. (2007). Study Guide and Solutions Manual for Organic Chemistry (5th Ed.). New York: W.H. Freeman. (201)
Problems
- Are the following molecules saturated or unsaturated:
- (b.) (c.) (d.) C10H6N4
- Using the molecules from 1., give the degrees of unsaturation for each.
- Calculate the degrees of unsaturation for the following molecular formulas:
- (a.) C9H20(b.) C7H8(c.) C5H7Cl (d.) C9H9NO4
- Using the molecular formulas from 3, are the molecules unsaturated or saturated.
- Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
Degrees Pro 4 2 14
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
Degrees Pro 4 2 13
(b.) The molecule can contain any of these combinations(i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv).. (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Degrees are the unit of measure used for angles, just as feet are used to measure distance. The symbol for degrees is '°'. There are 360° in a full circle, and 90° in a right angle. A protractor is used for measuring angles. As shown below, a protractor is a half or full circle measuring device, marked in degrees along the outer edge, with a straight line running from 0° to 180°. There is a small hole in the center of the protractor. To use a protractor:
- Place the hole of the protractor over the point of the angle where the lines meet.
- Make sure one side of the angle is on the zero line.
- Read the degrees off the protractor where the line of the other side is or mark a point along the edge at the appropriate angle measurement.
Degrees Pro 4 2 125
Example 1 - Part A. Measure angles 1, 2, and 3 using a protractor to follow the steps above.
Degrees Pro 4 2 17
You should find that the angles measure as follows:
Part B. What is the sum of angles 1 and 2? How does that number compare to angle 3?
The sum of angles 1 and 2 is 180 degrees. Angle 3 also equals 180 degrees. Angle 3 is a straight line, so it follows that the sum of angles 1 and 2, 180°, is also a straight line.
Part C. Add angles 1, 2, and 3. The sum of angles 1, 2, and 3 equals 360 degrees.
There are 360 degrees in a circle and angles 1, 2, and 3 together complete the circle.
Part B. What is the sum of angles 1 and 2? How does that number compare to angle 3?
The sum of angles 1 and 2 is 180 degrees. Angle 3 also equals 180 degrees. Angle 3 is a straight line, so it follows that the sum of angles 1 and 2, 180°, is also a straight line.
Part C. Add angles 1, 2, and 3. The sum of angles 1, 2, and 3 equals 360 degrees.
There are 360 degrees in a circle and angles 1, 2, and 3 together complete the circle.