Following are the four molecules "P", "Q", "R" and "S":

Which one among the four molecules will react with H-Br(aq)$H-Br(aq)$ at the fastest rate?

Related Formula

Partial hydrogenation of alkynes using Pd/C$Pd/C$ yields alkenes:

Ozonolysis cleaves the double bond to form carbonyls:

Core Logic

Step 1: Controlled reduction of propyne gives propene:

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]$[C]$):

Step 1: Final Identification

Hence, [A] = CH₃-CH=CH₂$[A] = CH_3-CH=CH_2$ [B] = CH₃CHO$[B] = CH_3CHO$ [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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Related Formula

Core Logic

The reaction involving the addition of HI$\text{HI}$ 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 I^-$\text{I}^-$ 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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Core Logic

The substrate is cyclohexene, which can undergo two distinct types of reactions with chlorine (Cl₂$Cl_2$) depending on the reaction conditions.

Reaction 1: Condition hv$hv$ (Product A) In the presence of light (hν$h\nu$) 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 CCl₄$CCl_4$ (Product B) In the presence of a non-polar solvent like CCl₄$CCl_4$ and without light/heat, Cl₂$Cl_2$ 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

X₂$X_2$ + light (hν$h\nu$) = Free radical Allylic Substitution. X₂$X_2$ + dark/solvent (CCl₄$CCl_4$) = Electrophilic Addition across double bond.

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 (CH₃-CH=C(CH₃)₂$CH_3-CH=C(CH_3)_2$).

In reductive ozonolysis (O₃$O_3$ followed by Zn/H₂O$Zn/H_2O$), 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

The reaction yields two distinct product molecules:

• Acetaldehyde (CH₃CHO$CH_3CHO$) - contains 1 oxygen atom.
• Acetone (CH₃COCH₃$CH_3COCH_3$) - 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).

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. CH₃-C≡ C-H + Na arrow CH₃-C≡ C^-Na^+ + (1)/(2)H₂$CH_3-C\equiv C-H + Na \rightarrow CH_3-C\equiv C^-Na^+ + \frac{1}{2}H_2$ So, A is Sodium propynide (CH₃-C≡ C^-Na^+$CH_3-C\equiv C^-Na^+$).

Step 1: Analyzing the second step

The product is CH₃-C≡ C-CH₂-CH₂-CH₃$CH_3-C\equiv C-CH_2-CH_2-CH_3$. This indicates an SN2$S_N2$ substitution reaction between the acetylide ion (nucleophile) and an alkyl halide (electrophile). Since NaBr$NaBr$ is a byproduct, B must be a propyl bromide. CH₃-C≡ C^-Na^+ + CH₃-CH₂-CH₂-Br arrow CH₃-C≡ C-CH₂-CH₂-CH₃ + 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 CH₃-C≡ C^-Na^+$CH_3-C\equiv C^-Na^+$ and B is CH₃-CH₂-CH₂-Br$CH_3-CH_2-CH_2-Br$ (1-Bromopropane).

Chapter Mix

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