Given below are two statements: Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with CH₃COCl / AlCl₃$CH_{3}COCl / AlCl_{3}$ will give

Statement II: NO₂$NO_{2}$ group is a m-directing, and deactivating group. In the light of the above statements, choose the most appropriate answer from the options given below.

Related Formula

Orienting effects in electrophilic aromatic substitution:

• Bromine (-Br$-Br$) is ortho/para-directing (para-dominated due to steric hindrance).
• Sulfonic acid group (-SO₃H$-SO_3H$) is a strong deactivating, meta-directing group.

Core Logic

Analyzing the first step:

• Sulfonation of bromobenzene with SO₃/H₂SO₄$\mathrm{SO}_3/\mathrm{H}_2\mathrm{SO}_4$ yields 4-bromobenzenesulfonic acid as the major product (A) due to steric hindrance at the ortho-position.

Step 1: Determine orientation for the second step

In 4-bromobenzenesulfonic acid, we have two substituents:

• -Br$-Br$ (ortho/para director)
• -SO₃H$-SO_3H$ (meta director)

The positions meta to the deactivating -SO₃H$-SO_3H$ group correspond to the positions ortho to the -Br$-Br$ group. Both directing effects align on the same position (carbon-3/carbon-5). Since -Br$-Br$ is activating relative to -SO₃H$-SO_3H$, it controls the orientation.

Step 2: Identify Product B

Halogenation with Br₂/Fe$\mathrm{Br}_2/\mathrm{Fe}$ introduces a bromine atom ortho to the existing bromine atom (meta to -SO₃H$-SO_3H$):

This matches Option (2).

Pattern Recognition

When an activating group (-Br$-Br$) and a deactivating group (-SO₃H$-SO_3H$) compete on a benzene ring, the orienting influence of the activating group wins. Position ortho to the bromine atom is favored over meta positions of the sulfonic acid group.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

Statement I: Cis-2-butene (CH₃-CH=CH-CH₃$CH_3-CH=CH-CH_3$) undergoes reductive ozonolysis to cleave the double bond and yield two molecules of acetaldehyde (ethanal, CH₃CHO$CH_3CHO$). This statement is true.

Statement II: 3,6-dimethyloct-4-ene (CH₃-CH₂-CH(CH₃)-CH=CH-CH(CH₃)-CH₂-CH₃$CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3$) undergoes reductive ozonolysis to yield 2-methylbutanal (CH₃-CH₂-CH(CH₃)-CHO$CH_3-CH_2-CH(CH_3)-CHO$). This product contains a chiral carbon atom (the C2$C2$ carbon bonded to -H$-H$, -CH₃$-CH_3$, -CHO$-CHO$, and -C₂H₅$-C_2H_5$). Thus, the statement that it has no chiral carbon atom is false.

Pattern Recognition

To check chirality after ozonolysis, draw the cleaved fragment structure first. Any carbon with four different groups is chiral. 2-methylbutanal has four unique groups attached to C2$C2$, making Statement II clearly false.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Core Logic

Statement I is true: In m-xylene, both methyl groups direct incoming electrophiles to position 4 (synergistic ortho/para effect). Nitration gives 4-nitro-m-xylene. Subsequent strong oxidation of both -CH₃$-CH_3$ groups yields 4-nitrobenzene-1,3-dicarboxylic acid.

Statement II is true: Alkyl groups (-CH₃$-CH_3$) act as ortho/para directors via hyperconjugation, whereas nitro groups (-NO₂$-NO_2$) are strongly deactivating meta directors.

Pattern Recognition

When evaluating multi-substituted benzenes, identify whether directors reinforce the same positions. In m-xylene, position 4 is ortho to one methyl and para to the other.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Core Logic

A negative Baeyer's test confirms that the hydrocarbon structural isomers contain no aliphatic alkene or alkyne unsaturations, establishing that they are purely aromatic benzene derivatives with side alkyl chains.

To find structures possessing four distinct ring positions available for electrophilic substitution, we examine the symmetries of specific tri-substituted configurations. There are exactly 2$2$ such structural isomers satisfying these spatial conditions.

Pattern Recognition

Negative test = aromatic ring constraint. Calculate positional substitution patterns meticulously to ensure ring symmetry matches the required counts.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Related Formula

Core Logic

Reductive ozonolysis (O₃, Zn/H₂O$O_3, Zn/H_2O$) cleaves the double bond of 1,4-dimethylcyclohexene and replaces each carbon of the alkene with a carbonyl group (C=O$C=O$).

3-methyl-6-oxoheptanal has a 7-carbon principal chain containing an aldehyde group at C-1, a methyl substituent at C-3, and a ketone group at C-6.

Step 1: Ozonolysis Verification

Cleaving the endocyclic double bond in structure (2) produces:

Numbering from the aldehyde functional group (C-1):

• C-1: -CHO$-\mathrm{CHO}$
• C-3: -CH₃$-\mathrm{CH_3}$ methyl branch
• C-6: -C=O$-\mathrm{C=O}$ keto group
• C-7: terminal methyl group

IUPAC Name: 3-methyl-6-oxoheptanal.

Pattern Recognition

Reverse ozonolysis: Connect aldehyde C-1 and keto C-6 back into a ring double bond.

Chapter Mix

Class 11 Chemistry: Hydrocarbons