Consider the following compound (X)
beginarrayc mathrm I \\ mathrm H - mathrm C equiv mathrm C - mathrm C H _ 2 - mathrm C H - mathrm C H _ 3 \\ mathrm I \\ mathrm C H _ 3 endarray$$\begin{array}{c} \mathrm {I} \\ \mathrm {H} - \mathrm {C} \equiv \mathrm {C} - \mathrm {C H} _ {2} - \mathrm {C H} - \mathrm {C H} _ {3} \\ \mathrm {I} \\ \mathrm {C H} _ {3} \end{array}$$
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding mathrmC - H$\mathrm{C - H}$ bond are :
A.(1)\ textII, IV$(1)\ \text{II, IV}$
B.(2)\ textIII, II$(2)\ \text{III, II}$
C.(3)\ textI, IV$(3)\ \text{I, IV}$
D.(4)\ textII, I$(4)\ \text{II, I}$
Solution & Explanation
### Related Formula
Free radical stability structural hierarchy sequence:
textResonance Stabilized (Propargyl/Allyl) > 3^circ > 2^circ > 1^circ > textVinylic/Alkyne Center$$\text{Resonance Stabilized (Propargyl/Allyl)} > 3^\circ > 2^\circ > 1^\circ > \text{Vinylic/Alkyne Center}$$
### Core Logic
Let's analyze individual cleavage points across the carbon backbone skeleton:
* **Position II** yields a propargyl intermediate radical directly adjacent to the alkyne bond. This allows strong resonance stabilization across the pi$\pi$ system, making it the most stable radical position.
* **Position I** places the radical directly on an mathrmsp$\mathrm{sp}$-hybridized carbon center. The high electronegativity of mathrmsp$\mathrm{sp}$ orbitals tightly holds the unpaired electron, making homolytic cleavage extremely difficult and rendering this intermediate the least stable radical position.
Free Radical Stability
### Step 1: Verdict
Therefore, the most stable and least stable positions are II and I, respectively.
### Pattern Recognition
Radicals located on mathrmsp$\mathrm{sp}$ carbons (vinylic/alkynic) are highly unstable due to poor orbital overlap, while positions next to triple bonds (propargylic) are exceptionally stable due to active resonance delocalization.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 13
Q85jee_main_2024_27_jan_morningStereoisomerism
3-Methylhex-2-ene on reaction with textHBr$\text{HBr}$ in presence of peroxide forms an addition product (A). The number of possible stereoisomers for 'A' is textquadquad$\text{\quad\quad}$.
Numerical Answer.Answer: 4 to 4
Solution
### Core Logic
In the presence of peroxide, textHBr$\text{HBr}$ undergoes an anti-Markovnikov free-radical addition mechanism across the alkene bond of 3-methylhex-2-ene. Bromine appends to carbon-2 while hydrogen attaches at carbon-3.
The generated dynamic addition product (A) corresponds structurally to 2-bromo-3-methylhexane. Addition product chiral identification map for Q85 - JEE Main 2024 Morning
### Step 1: Check chiral centers count
Inspecting the structure of 2-bromo-3-methylhexane reveals two asymmetric chiral centers:
- Carbon-2 carrying text-H, -CH_3text, -Br, and -CH(CH_3text)CH_2textCH_2textCH_3$\text{-H, -CH}_3\text{, -Br, and -CH(CH}_3\text{)CH}_2\text{CH}_2\text{CH}_3$
- Carbon-3 carrying text-H, -CH_3text, -CH(Br)CH_3text, and -CH_2textCH_2textCH_3$\text{-H, -CH}_3\text{, -CH(Br)CH}_3\text{, and -CH}_2\text{CH}_2\text{CH}_3$
Since both stereocenters are unsymmetrical (n=2$n=2$):
textTotal Stereoisomers = 2^n = 2^2 = 4$$\text{Total Stereoisomers} = 2^n = 2^2 = 4$$
### Pattern Recognition
Anti-Markovnikov hydrobromination often generates multiple stereocenters. Count unsymmetrical C^*$C^*$ elements systematically.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Hydrocarbons
Q86jee_main_2024_27_jan_morningAromaticity
Among the given organic compounds, the total number of aromatic compounds is textquadquad$\text{\quad\quad}$.
Four distinctive hydrocarbon ring structures depicted as options A, B, C, D.Four distinctive hydrocarbon ring structures depicted as options A, B, C, D.
Numerical Answer.Answer: 3 to 3
Solution
### Core Logic
To satisfy Huckel's rules for aromaticity, a system must be planar, cyclic, completely conjugated, and possess (4n+2)pi$(4n+2)\pi$ delocalized electrons. Evaluating structures B, C, and D against these criteria shows they follow the (4n+2)pi$(4n+2)\pi$ electron counting rules successfully. Thus, three structures are fully aromatic.
### Pattern Recognition
Verify ring conjugation continuity and apply Huckel's electron count constraints (2, 6, 10, 14dots$2, 6, 10, 14\dots$).
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Hydrocarbons
Among the following, total number of meta directing functional groups is (Integer based):
-textOCH_3$-\text{OCH}_3$, -textNO_2$-\text{NO}_2$, -textCN$-\text{CN}$, -textCH_3$-\text{CH}_3$, -textNHCOCH_3$-\text{NHCOCH}_3$, -textCOR$-\text{COR}$, -textOH$-\text{OH}$, -textCOOH$-\text{COOH}$, -textCl$-\text{Cl}$
Numerical Answer.Answer: 4 to 4
Solution
### Core Logic
Meta-directing functional groups are electron-withdrawing groups via resonance or induction effects (-M$-M$ or -I$-I$). Let's audit the list:
- -textOCH_3 rightarrow textortho/para$-\text{OCH}_3 \rightarrow \text{ortho/para}$ (+M$+M$)
- -textNO_2 rightarrow textmeta$-\text{NO}_2 \rightarrow \text{meta}$ (-M, -I$-M, -I$)
- -textCN rightarrow textmeta$-\text{CN} \rightarrow \text{meta}$ (-M, -I$-M, -I$)
- -textCH_3 rightarrow textortho/para$-\text{CH}_3 \rightarrow \text{ortho/para}$ (+I$+I$, hyperconjugation)
- -textNHCOCH_3 rightarrow textortho/para$-\text{NHCOCH}_3 \rightarrow \text{ortho/para}$ (+M$+M$)
- -textCOR rightarrow textmeta$-\text{COR} \rightarrow \text{meta}$ (-M, -I$-M, -I$)
- -textOH rightarrow textortho/para$-\text{OH} \rightarrow \text{ortho/para}$ (+M$+M$)
- -textCOOH rightarrow textmeta$-\text{COOH} \rightarrow \text{meta}$ (-M, -I$-M, -I$)
- -textCl rightarrow textortho/para$-\text{Cl} \rightarrow \text{ortho/para}$ (+M$+M$ counteracted by strong -I$-I$)
The meta-directing groups are -textNO_2$-\text{NO}_2$, -textCN$-\text{CN}$, -textCOR$-\text{COR}$, and -textCOOH$-\text{COOH}$.
### Pattern Recognition
Groups whose attachment atom carries a double or triple bond to an electronegative element typically function as meta-directors.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Haloalkanes and Haloarenes
Q64jee_main_2024_29_jan_morningResonance and Resonance Energy
### Core Logic
By definition, a resonance hybrid (the actual structure) is always more stable than any of its contributing canonical structures (resonance structures). The difference in potential energy between the most stable contributing resonance structure and the actual resonance hybrid is called the resonance energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q69jee_main_2024_29_jan_morningQualitative Analysis of Organic Compounds
Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with FeSO_4$FeSO_4$ in presence of concentrated H_2SO_4$H_2SO_4$ indicates the presence of element/s
A.textBr$\text{Br}$
B.textN$\text{N}$
C.textN and S$\text{N and S}$
D.textS$\text{S}$
Solution
### Core Logic
In the Lassaigne's test for elemental analysis, if an organic compound contains both Nitrogen and Sulphur together, the sodium fusion extract contains sodium thiocyanate (NaSCN$NaSCN$) instead of sodium cyanide (NaCN$NaCN$) and sodium sulphide (Na_2S$Na_2S$).
Na + C + N + S rightarrow NaSCN$$Na + C + N + S \rightarrow NaSCN$$
When this extract is treated with Iron (II) sulphate (FeSO_4$FeSO_4$) and acidified with concentrated H_2SO_4$H_2SO_4$ (which oxidizes some Fe^2+$Fe^{2+}$ to Fe^3+$Fe^{3+}$), the Fe^3+$Fe^{3+}$ ions react with the thiocyanate ions to form a blood-red colored complex.
### Step 1: The Reaction
Fe^2+ xrightarrow[textConc. H_2SO_4,
H^+ Fe^3+$$Fe^{2+} \xrightarrow[\text{Conc. } H_2SO_4,
{H^+} Fe^{3+}$$Fe^3+ + 3SCN^- rightarrow Fe(SCN)_3 quad text(blood red colour)$$Fe^{3+} + 3SCN^- \rightarrow Fe(SCN)_3 \quad \text{(blood red colour)}$$
The appearance of the blood red colour specifically confirms the simultaneous presence of both Nitrogen and Sulphur.
### Pattern Recognition
N alone rightarrow$\rightarrow$ Prussian Blue (Fe_4[Fe(CN)_6]_3$Fe_4[Fe(CN)_6]_3$).
S alone rightarrow$\rightarrow$ Purple colour with sodium nitroprusside or black ppt of PbS$PbS$.
N + S together rightarrow$\rightarrow$ Blood red colour (Fe(SCN)_3$Fe(SCN)_3$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
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