2. For each of the following molecular formulas, calculate the number of elements of unsaturation (show all work). a) CzoH1s b) CH,Bro c) CH&NI
Reactive Intermediates
In chemistry, reactive intermediates are termed as short-lived, highly reactive atoms with high energy. They rapidly transform into stable particles during a chemical reaction. In specific cases, by means of matrix isolation and at low-temperature reactive intermediates can be isolated.
Hydride Shift
A hydride shift is a rearrangement of a hydrogen atom in a carbocation that occurs to make the molecule more stable. In organic chemistry, rearrangement of the carbocation is very easily seen. This rearrangement can be because of the movement of a carbocation to attain stability in the compound. Such structural reorganization movement is called a shift within molecules. After the shifting of carbocation over the different carbon then they form structural isomers of the previous existing molecule.
Vinylic Carbocation
A carbocation where the positive charge is on the alkene carbon is known as the vinyl carbocation or vinyl cation. The empirical formula for vinyl cation is C2H3+. In the vinyl carbocation, the positive charge is on the carbon atom with the double bond therefore it is sp hybridized. It is known to be a part of various reactions, for example, electrophilic addition of alkynes and solvolysis as well. It plays the role of a reactive intermediate in these reactions.
Cycloheptatrienyl Cation
It is an aromatic carbocation having a general formula, [C7 H7]+. It is also known as the aromatic tropylium ion. Its name is derived from the molecule tropine, which is a seven membered carbon atom ring. Cycloheptatriene or tropylidene was first synthesized from tropine.
Stability of Vinyl Carbocation
Carbocations are positively charged carbon atoms. It is also known as a carbonium ion.
![### Calculating the Number of Elements of Unsaturation
#### Question 2: For each of the following molecular formulas, calculate the number of elements of unsaturation (show all work).
**a) \( C_{10}H_{18} \)**
First, recall the formula for calculating elements of unsaturation (also known as degrees of unsaturation):
\[ \mathrm{U} = \frac{2C + 2 - (H + X - N)}{2} \]
Where:
- \( \mathrm{U} \) is the number of elements of unsaturation.
- \( C \) is the number of carbons.
- \( H \) is the number of hydrogens.
- \( X \) is the number of halides (F, Cl, Br, I).
- \( N \) is the number of nitrogens.
Substituting \( C = 10 \) and \( H = 18 \) into the equation:
\[ \mathrm{U} = \frac{2(10) + 2 - 18}{2} \]
\[ \mathrm{U} = \frac{20 + 2 - 18}{2} \]
\[ \mathrm{U} = \frac{4}{2} = 2 \]
So, the number of elements of unsaturation for \( C_{10}H_{18} \) is **2**.
**b) \( C_5H_7BrO \)**
Substituting \( C = 5 \), \( H = 7 \), and \( X = 1 \) (Br is a halide) into the equation:
\[ \mathrm{U} = \frac{2(5) + 2 - (7 + 1)}{2} \]
\[ \mathrm{U} = \frac{10 + 2 - 8}{2} \]
\[ \mathrm{U} = \frac{4}{2} = 2 \]
So, the number of elements of unsaturation for \( C_5H_7BrO \) is **2**.
**c) \( C_5H_6NI \)**
Substituting \( C = 5 \), \( H = 6 \), \( N = 1 \), and \( X = 1 \) (I is a halide) into the equation:
\[ \mathrm{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F804acf52-6fb5-4da7-a67d-ae9379337409%2F30fd3e27-35d0-4e85-9493-3db57957219e%2F07lmaq8_processed.png&w=3840&q=75)
Step by step
Solved in 2 steps with 2 images
