Nomenclature of organic compounds: Naming Alkanes

These are saturated hydrocarbons i.e., contain all single C - C bonds and only C and H atoms.

1. Naming straight chain hydrocarbons:

Prefix: indicates the number of carbons

Suffix: ‘-ane’

2. Branched chain alkanes:

• Identify the longest carbon chain
• Number the carbons in the parent carbon chain such that the substituents get minimum numbers
• If two substituents are found in equivalent positions, the lower number is given to the one coming first in the alphabetical listing
• If the same substituent occurs more than once, the location of each point on which the substituent occurs is given. In addition, the number of times the substituent group occurs is indicated by a prefix
• Write the name of the substituent along with its position before the name of the parent chain in alphabetical order

Nomenclature of Alkenes: These have the general formula C_nH_2n. These unsaturated hydrocarbons are isomeric with the saturated cycloalkanes. The IUPAC rules for naming alkenes are similar in many respects to those for naming alkanes.

1. Determine the root word by selecting the longest chain that contains the double bond and change the ending of the name of the alkane of identical length from ane to ene.

Number the chain so to include both carbon atoms of the double bond, and begin numbering at the end of the chain nearer the double bond. Designate the location of the double bond by using the number of the first atom of the double bond as prefix: Indicate the locations of this substituent groups by the numbers of the carbon atoms to which they are attached.

2. Two frequently encountered alkenyl groups are the vinyl group and the allyl group.

CH₂ = CH —                   CH₂ = CH CH₂ —

vinyl group                      allyl group (are not included in IUPAC system)

The following examples illustrate how these names are employed

CH₂ = CH — Br                  CH₂ = CH — CH₂ Cl

vinyl bromide                      allyl chloride (are not IUPAC name)

3. The geometry of the double bond of a disubstituted alkene is designated with the prefixes, cis and trans. If two identical group are on the same side of the double bond, it is cis, if they are on opposite sides; it is trans.

Nomenclature of Alkyne: In the IUPAC system the compounds are named as alkynes in which the final – ane of the parent alkane is replaced by the suffix – yne. The position of the triple bond is indicated by a number when necessary.

When both a double and triple bond are present, the hydrocarbon is named an alkenyne with numbers as low as possible given to the multiple bonds. In case of a choice, the double bond gets the lower number.
In complex structures the alkynyl group is used as a modifying prefix.

Clock Puzzles 1

Puzzle - 1: What time should the last watch show?

Puzzle - 2: Where should the hour hand point to on the bottom clock?

Puzzle - 3: Where should the missing hour hand point to?

Puzzle - 4: Where should the minute hand point to on the bottom clock?

Puzzle - 5: Where should the minute hand point to on the bottom clock?

Riddles 3 [With Answers]

Riddle 1: What invention lets you look right through a wall?

Answer: A window

Riddle 2: What can you catch but not throw?

Answer:  A cold

Riddle 3: What is at the end of a rainbow?

Answer:  ‘w’

Riddle 4: What is always coming but never arrives?

Answer:  Tomorrow

Riddle 5: Which weighs more, a pound of feathers or a pound of bricks?

Answer:  Neither, they both weigh one pound

COMBINATION OR PROPAGATION OF ERRORS

Generally, the result of an experiment is obtained by doing mathematical operations on several measurements. Obviously, the final error depends not only upon the errors in individual measurements but also on the nature of mathematical operations.

Following are the rules for combination of errors.

(a) Errors in a Sum or Difference: Let there be two quantities A and B. If ΔA and ΔB are the corresponding absolute errors in their measurements, then

Measured value of A = A ± ΔA

Measured value of B = B ± ΔB

(i) Let Z denote the sum of A and B and ΔZ be the corresponding absolute error.

Clearly, Z = A + B and

Z ± ΔZ (A + ΔA) + (B + ΔB)

Z ± ΔZ (A + B) + (ΔA ± ΔB)

ΔZ = ± (ΔA ± ΔB) … (1)

Thus, the maximum error in Z, i.e., ΔZ = (ΔA ± ΔB), i.e., the maximum error in the sum is the sum of individual errors.

(ii) In case Z denotes the difference of A and B, and ΔZ is the corresponding absolute error,

Z = A – B

Clearly, Z ± ΔZ (A ± ΔA) - (B ± ΔB)

= A ± ΔA - B ± ΔB

Or ± ΔZ = ± (ΔA + ΔB) … (2)

Thus, the maximum error in Z, i.e., ΔZ = (ΔA ± ΔB) the maximum error in the difference is again the sum of the individual errors.

Hence, when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors associated with the measurements of the quantities to be added or subtracted.

(b) Errors in a Product or Quotient: Let there be two quantities A and B and let AA and AB be the corresponding absolute errors in their measurements. Clearly,

Measured value of A = A ± ΔA

Measured value of B = B ± ΔB

(i) Let Z denote the product of A and B and let ΔZ be the corresponding absolute error.

Clearly, Z = AB and

Z + ΔZ = (A ± ΔA) (B ± ΔB)

Or Z + ΔZ = AB ± BΔA ± AΔB ± ΔAΔB)

Dividing both sides by Z (= AB), we get

Or 1 ± ΔZ/Z = 1 ΔA/A ± ΔB/B

(We have ignored ΔAΔB/AB as it contains the product of two small quantities ΔA and ΔB)

Thus, ±ΔZ/Z = ± (ΔA/A + ΔB/B)

Or ΔZ/Z = ΔA/A + ΔB/B … (3)

ΔZ/Z, ΔA/A, ΔB/B are the relative errors in the measurement of Z, A and B respectively.

(ii) Let Z denote the quotient of A and B and let ΔZ be the corresponding absolute error.

Clearly, Z = A/B

[Neglecting the last term s it is the product of two small quantities ΔA and ΔB]

Or ±ΔZ = ±Z ΔA/A ± ΔB/B

Or ±ΔZ/Z = ±ΔA/A ± ΔB/B

Or ΔZ/Z = ΔA/A + ΔB/B ... (4)

We find that the results (3) and (4) are the same.

Thus, when two quantities are multiplied (or divided), the relative error in the product (or quotient, i.e., division) is the sum of the relative errors in the quantities to be multiplied or divided.

(c) Error Due to the Power of a Measured Quantity: Let there be a quantity A and let ΔA be the absolute errors in its measurement.

Measured values of A = A ± ΔA

Let Z = A²

Clearly, Z ± ΔZ = (A ± ΔA)²

= A² + ΔA² ± 2A ΔA

Neglecting (ΔA)² as it is exceedingly small,

Z ± ΔZ = A² + ΔA² ± 2A ΔA

Or ± ΔZ = ± 2A ΔA

Or ΔZ = 2A ΔA

Dividing both sides by Z, we get

ΔZ/ Z = 2A ΔA/ Z

= 2A ΔA/ A² = 2 ΔA/ A

Thus, the relative error in Z (i.e. A2) is twice the relative error in A.

(d) General Relation for Error involving Powers of Measured Quantities:
If,

Riddles 3

Riddle 1: What invention lets you look right through a wall?

Riddle 2: What can you catch but not throw?

Riddle 3: What is at the end of a rainbow?

Riddle 4: What is always coming but never arrives?

Riddle 5: Which weighs more, a pound of feathers or a pound of bricks?

Centroid of triangle & Centroid of tetrahedron

Centroid of triangle: The coordinates of the centroid of the triangle with vertices A (x₁, y₁, z₁) B (x₂, y₂, z₂) and C (x₃, y₃, z₃)

[(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3]

Proof: Let D be the mid-point of AC then coordinates of D are

[(x₂ + x₃)/ 2, (y₂ + y₃)/ 2, (z₂ + z₃)/ 2]

Let G be the centroid of ΔABC

Then, G divides AD in the ratio 2:1

So, coordinate of D are

[{1.x₁ + 2(x₂ + x₃/ 2)}/ 1 + 2, {1.y₁ + 2(y₂ + y₃/ 2)}/ 1 + 2, {1.z₁ + 2(z₂ + z₃/ 2)}/ 1 + 2]

i.e, [(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3]

Centroid of tetrahedron: Let ABCD be a tetrahedron such that the coordinates of its vertices are A (x₁, y₁, z₁) B (x₂, y₂, z₂) C (x₃, y₃, z₃) and D (x₄, y₄, z₄) the coordinates of the centroid of faces ABC, DAB, DBC and DCA are respectively

G₁ = [(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3, (z₁ + z₂ + z₃)/ 3],

G₂ = [(x₁ + x₂ + x₄)/ 3, (y₁ + y₂ + y₄)/ 3, (z₁ + z₂ + z₄)/ 3],

G₃ = [(x₂ + x₃ + x₄)/ 3, (y₂ + y₃ + y₄)/ 3, (z₂ + z₃ + z₄)/ 3],

G₄ = [(x₄ + x₃ + x₁)/ 3, (y₄ + y₃ + y₁)/ 3, (z₄ + z₃ + z₁)/ 3].

Now, coordinates of point G dividing DG, in the ratio 3:1 are

[{1.x₄ + 3(x₁ + x₂ + x₃/ 3)}/ 1 + 3, {1.y₄ + 3(y₁ + y₂ + y₃/ 3)}/ 1 + 3, {1.z₄ + 3(z₁ + z₂ + z₃/ 3)}/ 1 + 3]

= [(x₁ + x₂ + x₃ + x₄)/ 4, (y₁ + y₂ + y₃ + y₄)/ 4, (z₁ + z₂ + z₃ + z₄)/ 4]

Similarly the point dividing CG₁, CG₂, AG₃ and BG₄ in the ratio 3:1 has the same coordinates.

Hence, the point G [(x₁ + x₂ + x₃ + x₄)/ 4, (y₁ + y₂ + y₃ + y₄)/ 4, (z₁ + z₂ + z₃ + z₄)/ 4]

Math Puzzle 4 [With Answers]

Puzzle 1: Which number replaces the question mark?

Answer: 45

Consider diagonally

15 + 15 = 30    35 + 5 = 40

30 + 15 = 45    40 + 5 = 45

60 + 15 = 75    50 + 5 = 55

Puzzle 2: Which number replaces the question mark?

3, 5, 13, 43, 177, ?, 5353

Answer: 891

3 x 1+ 2 = 5

5 x 2 + 3 = 13

13 x 3 + 4 = 43

43 x 4 + 5 = 177

177 x 5 + 6 = 891

891 x 6 + 7 = 5353

Puzzle 3: Which number replaces the question mark?

Answer: 7

Sum of the numbers in each part is 25.

8 + 5 + 5 + ? = 25

? = 25 – 18

? = 7

Puzzle 4: Which number replaces the question mark?

If 2 + 3 = 8

3 + 7 = 27

4 + 5 = 32

5 + 8 = 60

6 + 7 = 72

Then 7 + 8 =?

Answer: 98

This equations follow the following logic;

2 + 3 = (2 x 3) + (2 x 1) = 6 + 2 = 8;

3 + 7 = (3 x 7) + (3 x 2) = 21 + 6 = 27;

4 + 5 = (4 x 5) + (4 x 3) = 20 + 12 = 32;

5 + 8 = (5 x 8) + (5 x 4) = 40 + 20 = 60;

6 + 7 = (6 x 7) + (6 x 5) = 42 + 30 = 72;

So to find the missing number;

7 + 8 = (7 x 8) + (7 x 6) = 56 + 42 = 98.