### Related Formula textRate of Electrophilic Addition propto textStability of Intermediate Carbocation$\text{Rate of Electrophilic Addition} \propto \text{Stability of Intermediate Carbocation}$ ### Core Logic Addition of H-Br(aq)$H-Br(aq)$ follows an electrophilic addition pathway where a carbocation intermediate is formed in the rate-determining step. Among the given structures, compound Q forms a resonance-stabilized allylic/benzylic carbocation, rendering it highly stable compared to the others.

### Pattern Recognition Look for conjugated or allylic systems that stabilize the positive charge dynamically via resonance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Related Formula Partial hydrogenation of alkynes using Pd/C$Pd/C$ yields alkenes: R-Cequiv CH + H_2 xrightarrowPd/C R-CH=CH_2$R-C\equiv CH + H_2 \xrightarrow{Pd/C} R-CH=CH_2$ Ozonolysis cleaves the double bond to form carbonyls: R-CH=CH_2 xrightarrow(i)O_3, (ii)Zn/H_2O R-CHO + HCHO$R-CH=CH_2 \xrightarrow{(i)O_3, (ii)Zn/H_2O} R-CHO + HCHO$ ### Core Logic Step 1: Controlled reduction of propyne gives propene: CH_3-Cequiv CH xrightarrowPd/C, H_2 CH_3-CH=CH_2 quad [A]$CH_3-C\equiv CH \xrightarrow{Pd/C, H_2} CH_3-CH=CH_2 \quad [A]$ Step 2: Reductive ozonolysis of propene ([A]$[A]$) splits the alkene at the C=C$C=C$ bond, creating ethanal ([B]$[B]$) and methanal ([C]$[C]$): CH_3-CH=CH_2 xrightarrowO_3, text then Zn/H_2O CH_3CHO\ [B] + HCHO\ [C]$CH_3-CH=CH_2 \xrightarrow{O_3, \text{ then } Zn/H_2O} CH_3CHO\ [B] + HCHO\ [C]$ ### Step 1: Final Identification Hence, [A] = CH_3-CH=CH_2$[A] = CH_3-CH=CH_2$ [B] = CH_3CHO$[B] = CH_3CHO$ [C] = HCHO$[C] = HCHO$ ### Pattern Recognition Whenever an alkyne undergoes partial hydrogenation with regular catalysts, count the carbons to trace the matching alkene framework. Cleaving a terminal alkene like propene always results in formaldehyde (HCHO$HCHO$) as one of the fragment products. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Related Formula textAlkene + textHX rightarrow textAlkyl Halide quad text(Markovnikov's Rule)$\text{Alkene} + \text{HX} \rightarrow \text{Alkyl Halide} \quad \text{(Markovnikov's Rule)}$ ### Core Logic The reaction involving the addition of textHI$\text{HI}$ to the double bond follows electrophilic addition mechanism governed by Markovnikov's rule. The proton adds to the less substituted carbon to generate the more stable tertiary carbocation intermediate, which is then attacked by textI^-$\text{I}^-$ to yield the corresponding major tertiary alkyl iodide. ### Step 1: Verification of Choices Option (2) perfectly captures the sound execution of Markovnikov addition mechanics whereas the remaining choices present invalid product distributions or faulty reaction stoichiometry templates. ### Pattern Recognition Electrophilic addition cleanly forms the most substituted, most stable carbocation before halide attack takes place. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Core Logic The substrate is cyclohexene, which can undergo two distinct types of reactions with chlorine (Cl_2$Cl_2$) depending on the reaction conditions. **Reaction 1: Condition hv$hv$ (Product A)** In the presence of light (hnu$h\nu$) or high temperature, halogens undergo homolytic cleavage to generate free radicals. This triggers allylic substitution (a free radical substitution mechanism). The allylic position is targeted because the resulting allylic free radical is resonance stabilized. Thus, substitution occurs at the carbon adjacent to the double bond, yielding 3-chlorocyclohexene as Product A. **Reaction 2: Condition CCl_4$CCl_4$ (Product B)** In the presence of a non-polar solvent like CCl_4$CCl_4$ and without light/heat, Cl_2$Cl_2$ undergoes an electrophilic addition reaction across the carbon-carbon double bond. A cyclic chloronium ion intermediate is formed, leading to anti-addition of two chlorine atoms. This yields 1,2-dichlorocyclohexane as Product B. ### Step 1: Structures of A and B

Product A preserves the double bond and substitutes a Cl at the allylic position. Product B loses the double bond and adds two Cl atoms adjacently. ### Pattern Recognition X_2$X_2$ + light (hnu$h\nu$) = Free radical Allylic Substitution. X_2$X_2$ + dark/solvent (CCl_4$CCl_4$) = Electrophilic Addition across double bond. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### 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