### 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 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

### Core Logic The structure of 2-methylbutane is CH_3-CH(CH_3)-CH_2-CH_3$CH_3-CH(CH_3)-CH_2-CH_3$. It has four different types of hydrogen atoms, which can be substituted to form structural isomers: 1) 1-chloro-2-methylbutane: Chlorination at terminal CH_3$CH_3$ near branch. Yields a chiral center at C2 (2 enantiomers). 2) 2-chloro-2-methylbutane: Chlorination at the tertiary carbon (1 achiral product). 3) 2-chloro-3-methylbutane: Chlorination at the CH_2$CH_2$ group. Yields a chiral center at C2 (2 enantiomers). 4) 1-chloro-3-methylbutane: Chlorination at the far terminal CH_3$CH_3$. No chiral center (1 achiral product).

### Step 1: Counting Stereoisomers Total isomeric products = 2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$. ### Pattern Recognition When asked for "isomeric products" in halogenation without specifying "structural isomers", you must count stereoisomers (enantiomers) as well. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes