General Organic Chemistry (GOC) - 1 - JEE MCQ

If a molecule contains both carbon-carbon double or triple bonds, the two are treated as per in seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chains, then the lowest number is given to the C=C double bond.

Thus, the correct name is pent-3-en-1-yne.

There are four double bonds. Hence, the number of π-electrons is 2 × 4 = 8.

- The correct order of priority for functional groups in the IUPAC system is determined by their reactivity and presence in naming conventions.
- Carboxylic acids -COOH have the highest priority due to their strong acidic nature.
- Sulfonic acids -SO₃H  come next because of similar acidity.
- Amides -CONH₂ follow due to their derivation from carboxylic acids.
- Aldehydes -CHO have lower priority than the above groups.
- Therefore, the correct order is: -COOH, -SO₃H, -CONH₂, -CHO, which corresponds to option D.

1º, 2º & 3º H’s are that which is attach at 1º , 2º & 3º carbon respectively.

2-Chloro-4, 4-dimethyl hexane

a, c, f & i are incorrect
 

2-Bromobenzene-1,4-dicarboxylic acid

2, 2, 5-Trimethyl-4-(1-methylpropyl) nonane

Sigma bond is stronger than π-bond because of better overlap. All single bonds are σ bonds and all multiple bonds contain one σ and other π bonds.

Structure 1 (Benzene):

• Benzene has alternating double bonds in a cyclic structure, which can delocalize electrons.
• Shows resonance.

Structure 2 (Cyclohexadienyl cation):

• A positively charged cyclic structure with alternating double bonds can delocalize electrons.
• Shows resonance.

Structure 3 (Cyclohexadienyl anion):

• A negatively charged cyclic structure with alternating double bonds can delocalize electrons.
• Shows resonance.

Structure 4 (Cyclohexadienyl radical):

• A radical in a cyclic structure with alternating double bonds can delocalize electrons.
• Shows resonance.

Structure 5 (Cyclohexane with oxygen and positive charge):

• A positively charged oxygen attached to a saturated cyclic structure cannot delocalize electrons.
• Does not show resonance.

Structure 6 (CH₂=CH–Cl):

• The double bond adjacent to a halogen (chlorine) allows delocalization of lone pair electrons.
• Shows resonance.

Structure 7 (Thiophene):

• Sulfur in a cyclic structure with alternating double bonds can delocalize electrons.
• Shows resonance.

Structure 8 (Phenoxide ion, C₆H₅O⁻):

• The negative charge on oxygen is delocalized into the benzene ring.
• Shows resonance.

Structure 9 (CH₂=CH–CH₂–Cl):

• A chlorine atom attached to a non-conjugated structure cannot delocalize electrons.
• Does not show resonance.

Structure 10 (Pyridine):

• A nitrogen atom in a cyclic structure with alternating double bonds allows delocalization of electrons.
• Shows resonance.

Understanding Hyperconjugation: Hyperconjugation involves the interaction of sigma (σ) bonds (typically C-H bonds) of alkyl groups directly attached to a positively charged carbon (C⁺) or an unsaturated system (like a double bond).

Structure Analysis:

• In the given structure, the positively charged carbon (C⁺) is directly attached to two methyl groups (−CH₃) and a hydrogen atom.
• Each −CH₃ group contributes 3 hydrogens for hyperconjugation.
• The single hydrogen atom attached to the positively charged carbon also participates in hyperconjugation.

Total Hydrogens Involved:

• Contribution from the two methyl groups: 3 + 3 = 6
• Contribution from the single hydrogen atom on the positively charged carbon: 1
• Total: 6 + 1 = 7

Lack of vacant orbital

In phenols, the presence of electron releasing groups decreases the acidity, whereas the presence of electron withdrawing groups increases the acidity, compared to phenol. Among the meta and para-nitrophenols, the latter is more acidic as the presence of -NO₂ group at the para position stabilizes the phenoxide ion to a greater extent than when it is present at the meta position. Thus, the correct order of acidity is:

Para-nitrophenol > meta-nitrophenol > phenol > methyl phenol

(B, C & G).