The problem asks to identify the reaction sequences that produce o-xylene (1,2-dimethylbenzene) as the major product. We will analyze each option step-by-step.

o-Xylene Structure

o-Xylene has the following structure:

It is a benzene ring with two methyl groups at adjacent positions (1 and 2).

Analysis of Option A

Reactant: o-Bromotoluene (1-bromo-2-methylbenzene) Reagents: (i) Mg, ether (ii) $CH_3Br$

1. Step (i): Formation of Grignard Reagent o-Bromotoluene, an aryl halide, reacts with magnesium in dry ether to form a Grignard reagent, o-tolylmagnesium bromide. $o-CH_3-C_6H_4-Br + Mg \xrightarrow{ether} o-CH_3-C_6H_4-MgBr$

2. Step (ii): Reaction with Methyl Bromide The Grignard reagent acts as a nucleophile and attacks the electrophilic methyl carbon of methyl bromide. This results in the formation of a new carbon-carbon bond, replacing the -MgBr group with a $-CH_3$ group. $o-CH_3-C_6H_4-MgBr + CH_3Br \rightarrow o-CH_3-C_6H_4-CH_3 + MgBr_2$ The product is 1,2-dimethylbenzene, which is o-xylene. This reaction sequence is a standard method for alkylation and gives o-xylene as the major product. Conclusion: Option A is correct.

Analysis of Option B

Reactant: 3,4-Dimethylaniline (4-amino-1,2-dimethylbenzene) Reagents: (i) $NaNO_2, HCl, 273 K$ (ii) $H_3PO_2$

1. Step (i): Diazotization The reactant is a primary aromatic amine. It undergoes diazotization upon reaction with nitrous acid ($HNO_2$, generated in situ from $NaNO_2$ and HCl) at low temperature (273 K) to form a diazonium salt. $3,4-(CH_3)_2-C_6H_3-NH_2 + NaNO_2 + 2HCl \xrightarrow{273 K} 3,4-(CH_3)_2-C_6H_3-N_2^+Cl^- + NaCl + 2H_2O$ The product is 3,4-dimethylbenzenediazonium chloride.

2. Step (ii): Deamination The diazonium salt is treated with hypophosphorous acid ($H_3PO_2$), which is a reducing agent that replaces the diazonium group $(-N_2^+)$ with a hydrogen atom (-H). This reaction is a deamination process. $3,4-(CH_3)_2-C_6H_3-N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_4(CH_3)_2 + N_2 + H_3PO_3 + HCl$ The starting material has methyl groups at positions 3 and 4 relative to the amino group. Removing the amino group (at position 1) and replacing it with hydrogen results in a benzene ring with two methyl groups at adjacent positions. This product is 1,2-dimethylbenzene, or o-xylene. Conclusion: Option B is correct.

Analysis of Option C

Reactant: o-Methylacetophenone (1-(2-methylphenyl)ethan-1-one) Reagents: (i) $CH_3MgBr$ (ii) $H_3O^+$ (iii) $H_2SO_4$, heat

1. Step (i) & (ii): Grignard Reaction and Workup o-Methylacetophenone is a ketone. It reacts with the Grignard reagent $CH_3MgBr$. The nucleophilic methyl group of the Grignard reagent attacks the carbonyl carbon. Subsequent acidic workup ($H_3O^+$) protonates the resulting alkoxide to form a tertiary alcohol. $o-CH_3-C_6H_4-C(=O)CH_3 \xrightarrow{(i) CH_3MgBr, (ii) H_3O^+} o-CH_3-C_6H_4-C(OH)(CH_3)_2$ The product is 2-(o-tolyl)propan-2-ol.

2. Step (iii): Dehydration The tertiary alcohol is heated with a strong acid like $H_2SO_4$. This causes dehydration (elimination of water) to form an alkene. A proton is removed from an adjacent carbon (from one of the methyl groups). $o-CH_3-C_6H_4-C(OH)(CH_3)_2 \xrightarrow{H_2SO_4, \Delta} o-CH_3-C_6H_4-C(=CH_2)CH_3 + H_2O$ The major product is 1-isopropenyl-2-methylbenzene (or 2-(o-tolyl)propene), not o-xylene. Conclusion: Option C is incorrect.

Analysis of Option D

Reactant: o-Chlorotoluene (1-chloro-2-methylbenzene) Reagents: (i) NaOH, 623 K, 300 atm (ii) $CH_3Cl, AlCl_3$

1. Step (i): Nucleophilic Aromatic Substitution (Dow's Process) o-Chlorotoluene is reacted with NaOH under harsh conditions (high temperature and pressure). This is a nucleophilic aromatic substitution (Dow's process) where the chloro group is replaced by a hydroxyl group, forming a phenoxide which upon workup (assumed) gives a phenol. $o-CH_3-C_6H_4-Cl \xrightarrow{NaOH, 623K, 300atm} o-CH_3-C_6H_4-OH \,(o-cresol)$

2. Step (ii): Friedel-Crafts Alkylation The product, o-cresol (2-methylphenol), is then subjected to Friedel-Crafts alkylation with $CH_3Cl$ and $AlCl_3$. Both -OH and $-CH_3$ are ortho, para-directing groups. The -OH group is a much stronger activator than $-CH_3$, so it controls the position of substitution. It directs the incoming electrophile ($CH_3^+$) to its ortho and para positions. In o-cresol, the position para to the -OH group is open (position 4). The ortho position (position 6) is sterically hindered by the adjacent $-CH_3$ group. Therefore, alkylation will occur predominantly at the para position. The major product will be 4-methyl-o-cresol, which is 2-hydroxy-1,4-dimethylbenzene, not o-xylene. Conclusion: Option D is incorrect.

Final Summary

• Reaction A produces o-xylene.
• Reaction B produces o-xylene.
• Reaction C produces an alkene.
• Reaction D produces a substituted phenol.

Therefore, the reaction sequences that lead to o-xylene as the major product are A and B.