In the following series of reactions identify the major products A & B respectively:

### Core Logic The reaction given is the reductive ozonolysis of an alkene, 2-methylbut-2-ene (CH_3-CH=C(CH_3)_2$CH_3-CH=C(CH_3)_2$). In reductive ozonolysis (O_3$O_3$ followed by Zn/H_2O$Zn/H_2O$), the carbon-carbon double bond is completely cleaved. An oxygen atom is placed on each carbon atom of the broken double bond to form carbonyl compounds (aldehydes or ketones). ### Step 1: Identifying the Products CH_3-CH=C(CH_3)_2 xrightarrowO_3 / Zn, H_2O CH_3-CHO + O=C(CH_3)_2$$CH_3-CH=C(CH_3)_2 \xrightarrow{O_3 / Zn, H_2O} CH_3-CHO + O=C(CH_3)_2$$ The reaction yields two distinct product molecules: 1. Acetaldehyde (CH_3CHO$CH_3CHO$) - contains 1 oxygen atom. 2. Acetone (CH_3COCH_3$CH_3COCH_3$) - contains 1 oxygen atom. The question asks for the number of oxygen atoms present **per molecule** of the product (P). Since any resulting product molecule (either acetaldehyde or acetone) contains exactly 1 oxygen atom, the answer is 1. ### Pattern Recognition Reductive ozonolysis of a simple alkene without other oxygenated functional groups always creates simple aldehydes or ketones. Each resulting discrete molecule formed from the cleaved double bond will have exactly 1 carbonyl group (1 oxygen atom) unless it's a cyclic alkene opening up (which would have 2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

### Core Logic Terminal alkynes possess acidic hydrogen. When treated with a strong base or active metal like Sodium (Na$Na$), they form sodium acetylide salts. CH_3-Cequiv C-H + Na rightarrow CH_3-Cequiv C^-Na^+ + frac12H_2$CH_3-C\equiv C-H + Na \rightarrow CH_3-C\equiv C^-Na^+ + \frac{1}{2}H_2$ So, A is Sodium propynide (CH_3-Cequiv C^-Na^+$CH_3-C\equiv C^-Na^+$). ### Step 1: Analyzing the second step The product is CH_3-Cequiv C-CH_2-CH_2-CH_3$CH_3-C\equiv C-CH_2-CH_2-CH_3$. This indicates an S_N2$S_N2$ substitution reaction between the acetylide ion (nucleophile) and an alkyl halide (electrophile). Since NaBr$NaBr$ is a byproduct, B must be a propyl bromide. CH_3-Cequiv C^-Na^+ + CH_3-CH_2-CH_2-Br rightarrow CH_3-Cequiv C-CH_2-CH_2-CH_3 + NaBr$CH_3-C\equiv C^-Na^+ + CH_3-CH_2-CH_2-Br \rightarrow CH_3-C\equiv C-CH_2-CH_2-CH_3 + NaBr$ ### Step 2: Conclusion A is CH_3-Cequiv C^-Na^+$CH_3-C\equiv C^-Na^+$ and B is CH_3-CH_2-CH_2-Br$CH_3-CH_2-CH_2-Br$ (1-Bromopropane). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic The given reaction is an intramolecular Friedel-Crafts alkylation. 1) The alkyl chloride reacts with anhydrous AlCl_3$AlCl_3$ to form a carbocation intermediate. 2) The carbocation generated will act as an electrophile and attack the adjacent phenyl ring. 3) The intermediate carbocation will undergo electrophilic aromatic substitution to form a new six-membered ring, as a 6-membered ring is highly stable and preferred over other ring sizes.

### Pattern Recognition Intramolecular Friedel-Crafts usually prefers forming 5 or 6 membered rings due to lesser angle strain. Here, the tether length is perfect for closing into a 6-membered tetralin-like system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Core Logic Addition of D-Cl$D-Cl$ across the double bond takes place via an electrophilic addition mechanism following Markovnikov's rule. 1) The electrophile D^+$D^+$ attacks the double bond to generate the most stable carbocation. The tertiary carbocation formed at the methyl-substituted carbon is more stable than the secondary carbocation. 2) The nucleophile Cl^-$Cl^-$ then attacks the planar tertiary carbocation from either face (top or bottom), resulting in a racemic mixture if a new chiral center is fully free, but here the stereochemistry depends on the relative anti/syn addition logic or thermodynamic stability. A mixture of diastereomers can be formed, but standard electrophilic additions often yield predominantly the trans-product due to steric reasons or via a bridged intermediate depending on conditions, though pure HCl/DCl$HCl/DCl$ addition is non-stereospecific. However, based on the official answer key (Option 3), we identify the correct stereochemical representation provided by the examining body.

### Note on Discrepancy According to the official NTA key, Option 3 is correct. According to our experts, both options 3 and 4 can be formed as a mixture since the carbocation is planar and Cl^-$Cl^-$ can attack from both sides. We proceed with the officially accepted answer (3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Core Logic Electrophilic substitution reactivity depends on the electron density of the aromatic ring, which is influenced by the inductive (I$I$) and mesomeric (M$M$) effects of the substituents. - Compound A (Benzene): Standard reference. - Compound B (Toluene): The -CH_3$-CH_3$ group shows +I$+I$ and hyperconjugation effects, activating the ring. Most reactive. - Compound C (Chlorobenzene): The -Cl$-Cl$ group shows +M$+M$ and -I$-I$ effects, but the -I$-I$ effect dominates, mildly deactivating the ring. - Compound D (Nitrobenzene): The -NO_2$-NO_2$ group shows strong -M$-M$ and -I$-I$ effects, heavily deactivating the ring. Least reactive. Order of reactivity: Toluene (B) > Benzene (A) > Chlorobenzene (C) > Nitrobenzene (D). ### Step 1: Final Order Reactivity: B > A > C > D$B > A > C > D$. This matches option (4). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes