Hydrocarbons

2-n-Propylpent-1-ene is a hydrocarbon with the following structure:
- There are 5 carbon atoms in the main chain and 3 more in the propyl group, totaling 8 carbon atoms.
- The bonds are single bonds (sigma) between all atoms except for the double bond between C1 and C2, which contains one bond and one bond.

The number of bonds is calculated as follows:
- 8 C-C single bonds (each contributing one bond)
- 9 C-H bonds (each contributing one bond)
- 1 C=C double bond (which contributes 1 bond)
Thus, the total number of bonds = 23.

The number of bonds is calculated as:
- 1 bond in the C=C double bond.

Therefore, the number of bonds is 23 and the number of bonds is 1.

The correct option is (B) : 23 σ bonds, 1π bonds

In 2, 3-Dimethylbutane, the carbon atoms in the main chain are primary and tertiary. The carbon atoms attached to the methyl groups (at positions 2 and 3) are primary carbons, while the carbon at position 3 (in the backbone chain) is a tertiary carbon. This molecule contains only primary and tertiary carbon atoms, making it the correct choice. Other options include secondary carbon atoms, which do not meet the condition.

The correct option is (C) : 2,3-Dimethylbutane

a) Aliphatic compound is .
b) Alicyclic compound is .
c) Benzenoid aromatic compound is
d) Non-benzenoid aromatic compound is .
e) Heterocyclic compound is .

The correct option is (B) : a)-(iii); b)-(v); c)-(i); d)-(ii); e)-(iv)

Aromatic compounds follow these criteria (Hückel's Rule):

• Cyclic: The molecule must be cyclic (a ring).
• Planar: The molecule must be planar (all atoms in the same plane).
• Conjugated: The molecule must have a continuous ring of overlapping p-orbitals (alternating single and double bonds).

Hückel's Rule (4n + 2 π electrons): The molecule must have (4n + 2) π electrons in the ring, where n is an integer (0, 1, 2, 3...).

Non-Benzenoid Aromatic Compounds

These are aromatic compounds that do not contain a benzene ring.

Analyzing the Given Compounds:

i (Aniline): This is benzene with an -NH2 substituent. It is a benzenoid aromatic compound.

ii (Phenol): This is benzene with an -OH substituent. It is a benzenoid aromatic compound.

iii (Cyclopentadienyl Anion): This is a cyclic, planar, and conjugated system. It has 6 π electrons (2 from each double bond and 2 from the negative charge), which satisfies Hückel's rule (4n + 2, where n = 1). It is aromatic, and it is non-benzenoid.

iv (Pyridine): This is benzene with one CH replaced by N. It is an aromatic, and it is non-benzenoid.

Therefore, the non-benzenoid aromatic compounds are iii and iv.

The correct answer is (A) iii and iv

Reaction Mechanism: Electrophilic Addition

• Protonation: The hydrogen from HBr (the electrophile) adds to one of the carbons of the double bond, forming a carbocation intermediate.
• Bromide Attack: The bromide ion (Br-) acts as a nucleophile and attacks the carbocation, forming the alkyl halide product.

Markovnikov's Rule

In the addition of HX (where X is a halogen) to an alkene, the hydrogen adds to the carbon with more hydrogens already attached (the less substituted carbon), and the halogen adds to the carbon with fewer hydrogens attached (the more substituted carbon). This rule is also stated as hydrogen will add to the carbon which will produce the most stable (most substituted) carbocation.

Applying Markovnikov's Rule to the Given Reaction

In the given reaction:

CH3-CH2-CH2-CH=CH2 + HBr

The double bond is between carbons 4 and 5 (numbering from the left).

Carbon 4 (CH=) has one hydrogen attached.

Carbon 5 (=CH2) has two hydrogens attached.

According to Markovnikov's rule, the hydrogen from HBr will add to carbon 5 (the carbon with more hydrogens), and the bromine will add to carbon 4 (the carbon with fewer hydrogens). This gives the following major product:

CH3-CH2-CH2-CHBr-CH3

Therefore, the correct answer is (B) CH3-CH2-CH2-CHBr-CH3

Friedel-Crafts Acylation

Friedel-Crafts acylation is an electrophilic aromatic substitution reaction where an acyl group (R-C=O) is added to an aromatic ring. The reaction requires a Lewis acid catalyst, such as AlCl3.

Factors Affecting Friedel-Crafts Reactions:

Activating and Deactivating Groups: Electron-donating groups (activating groups) on the aromatic ring increase the reactivity of the ring towards electrophilic attack. Electron-withdrawing groups (deactivating groups) decrease the reactivity.

Strongly Deactivating Groups: Aromatic rings with strongly deactivating groups do not undergo Friedel-Crafts reactions. The deactivating group reduces the electron density of the ring, making it less susceptible to electrophilic attack. Also, the Lewis acid catalyst can complex with the deactivating group, further deactivating the ring.

Analyzing the Compounds:

Compound i (Benzene): Benzene readily undergoes Friedel-Crafts acylation.

Compound ii (Nitrobenzene): The nitro group (-NO2) is a strongly deactivating group. Nitrobenzene does not undergo Friedel-Crafts reactions.

Compound iii (Toluene): The methyl group (-CH3) is an activating and ortho/para directing group. Toluene undergoes Friedel-Crafts acylation.

Therefore, the correct answer is (D) only ii