Organic Chemistry Notes

Organic Chemistry Notes

Calculation of Formal Charges

First we should understand that the Lewis-dot structure dots are replaced by line (i.e. bond) in Kekule line structure.
Now the task is to assign the formal charge for the nitrogen atom.
It is very simple.

We should know the number of valence electrons of the atom’s formal charge to be assigned. In case of nitrogen, it has 5 valence electrons.

Count only the number of electrons of nitrogen atom’s share (i.e. shown in blue color). Here 4 electrons. (Don’t count other atom’s electron share)

5 – 4 = +1
Then we should draw the structure like this

Practice problems

Assign the formal charges
for given molecule

Nomenclature of Alkanes

To begin the nomenclature of alkane, we should learn the parent name of the compound which is based on the number of carbons present in the compound:

No. of carbonsNameStructure

To name a branched alkane system, we should know about the word “alkyl”. Alkyl group is formed by removal of a
hydrogen from a corresponding alkane. It is named by replacing the –ane in the corresponding alkane by-yl (i.e. alkyl).

Step 1: Identify the longest chain and name it.

Points to remember: The majority of the time the beginners assume that the horizontal one would be a longest chain.
We should check it with all terminals.

Sometimes, a compound may have two or more chains of equal length. In those cases, the chain with the maximum
a number of substituents are considered as the longest chain.

Step 2:Name all the substituents attached to the longest chain as an alkyl substituent. For example:

Step 3:Number the longest chain in such a way the substituent should get a least position.

Preference will be given to the alphabet in case of two substituents share the same position. For example:

Step 4:Arrange the substituents in alphabetical order. Place locants in front of each substituent. For identical substituents, use di, tri, or tetra, and their alphabetical order is ignored. For example:

Practice problems

Isomeric alkanes

The heat of combustion of straight-chain alkane and its branched isomer

The enthalpies of the heat of combustion of two isomeric alkanes are different, even though the products of the reactions are identical. The results of the heat of combustion provide information about the stability of the isomeric alkanes.

The boiling point of straight-chain alkane and its branched isomer

The shape of branched alkane becomes more compact and its surface gets decreased. As a result of decreased surface area, the strength of van-der Waals force also get decreases due to minimal or no collision with adjacent molecules, the boiling points decrease. For example:


Molecules change their shape by rotation of C ̶ C single bonds and adopt various three-dimensional shapes. This is called as conformations. Some of the conformations are higher in energy, while others are lower in energy. Newman projection is a special type of drawing to draw and understand the various types of confirmations.

Eclipsed, skew, or gauche and staggered conformation of ethane is represented in Newman projection. Any dihedral angle other than 0° and 180° is called as skew or gauche conformation. The staggered is the most stable conformation and lowest in energy, while the eclipsed is the highly unstable conformation and highest in energy.

Next, we learn how the line structure of the molecule should is converted to Newman projection. The Newman projection could be drawn according to the view of the observer.
Step 1: According to the view of the observer, the groups connected to the front carbon are identified and placed in respective positions of “Y”.

Step 2: According to the view of the observer, the groups connected to the back carbon are identified and placed in respective positions of the circle.

Step 3: Merge the front carbon and back carbon drawings

Note: In the line structure, both the chlorine and fluorine atoms are cis to each other (both are wedge bonds). So, in the Newman projection also both atoms should be represented on the same side.

Practice Problems

1. Draw a Newman projection for the following compounds

2. Newman projections P, Q, R, and S are shown below: (IIT-JEE Advance-2020)

Which one of the following options represents identical molecules?

(A) P and Q (B) Q and S
(C) Q and R (D) R and S

3. Convert the following Newman projections into corresponding line structure compounds

4. Determine whether the following compounds are constitutional isomers: