Identify product [A], [B] and [C] in the following reaction sequence : CH_3-Cequiv CHfracPd/CH_2rightarrow[A]frac(i)O_3(ii)Zn,H_2Orightarrow[B]+[C]

Solution & Explanation

### Related Formula Partial hydrogenation of alkynes using Pd/C yields alkenes: R-Cequiv CH + H_2 xrightarrowPd/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 ### 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] Step 2: Reductive ozonolysis of propene ([A]) splits the alkene at the C=C bond, creating ethanal ([B]) and methanal ([C]): CH_3-CH=CH_2 xrightarrowO_3, text then Zn/H_2O CH_3CHO\ [B] + HCHO\ [C] ### Step 1: Final Identification Hence, [A] = CH_3-CH=CH_2 [B] = CH_3CHO [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) as one of the fragment products. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

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Q84 jee_main_2024_31_jan_evening Halogenation of Alkanes (Isomers)
Number of isomeric products formed by mono-chlorination of 2-methylbutane in presence of sunlight is ________
Numerical Answer. Answer: 6 to 6

Solution

### Core Logic The structure of 2-methylbutane is 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 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 group. Yields a chiral center at C2 (2 enantiomers). 4) 1-chloro-3-methylbutane: Chlorination at the far terminal CH_3. No chiral center (1 achiral product).
Halogenation of Alkanes (Isomers) diagram for Q84 - JEE Main 2024 Evening
Halogenation of Alkanes (Isomers) diagram for Q84 - JEE Main 2024 Evening
### Step 1: Counting Stereoisomers Total isomeric products = 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
Q87 jee_main_2024_31_jan_morning Kolbe's Electrolysis
Number of alkanes obtained on electrolysis of a mixture of CH_3COONa and C_2H_5COONa is
Numerical Answer. Answer: 3 to 3

Solution

### Core Logic Kolbe's electrolytic method generates free radicals at the anode, which then combine to form alkanes. The given mixture yields two types of carboxylate radicals which decarboxylate to form alkyl radicals: CH_3COONa rightarrow dotCH_3 C_2H_5COONa rightarrow dotC_2H_5 These radicals can couple in three different ways: 1. Cross coupling: dotCH_3 + dotC_2H_5 rightarrow CH_3-CH_2-CH_3 (Propane) 2. Self-coupling 1: dotCH_3 + dotCH_3 rightarrow CH_3-CH_3 (Ethane) 3. Self-coupling 2: dotC_2H_5 + dotC_2H_5 rightarrow CH_3-CH_2-CH_2-CH_3 (Butane) Thus, a total of 3 different alkanes are formed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

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