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 :

### Core Logic Let's evaluate each statement based on electrophilic aromatic substitution guidelines: * Statement (A): Both -mathrmOCH_3$-\mathrm{OCH}_{3}$ and -mathrmNHCOCH_3$-\mathrm{NHCOCH}_{3}$ have lone pairs on the atom directly attached to the benzene ring. These lone pairs undergo resonance delocalization into the ring, increasing electron density and activating it toward substitution. This statement is true. * Statement (B): While -mathrmCN$-\mathrm{CN}$ is a deactivating meta-directing group, -mathrmOH$-\mathrm{OH}$ is a strongly activating ortho/para-directing group due to resonance. This statement is false. * Statement (C): Both -mathrmCN$-\mathrm{CN}$ and -mathrmSO_3mathrmH$-\mathrm{SO}_{3}\mathrm{H}$ withdraw electron density via inductive and resonance effects (-I, -M$-I, -M$). This deactivates the ring and directs substitution to the meta position. This statement is true. * Statement (D): Activating groups increase electron density primarily at the ortho and para positions via resonance, directing incoming electrophiles to those sites. This statement is true. * Statement (E): Halogens are an exception: they are deactivating due to their strong inductive effect (-I$-I$), but ortho/para-directing due to resonance (+M$+M$). This statement is false. Therefore, statements (A), (C), and (D) are correct, which matches Option (1). ### Pattern Recognition Groups that activate the aromatic ring by donating electrons via resonance always direct incoming electrophiles to the ortho and para positions. Halogens are a unique exception: they deactivate the ring but still direct to the ortho and para positions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula For an unsymmetrical molecule with n$n$ distinct stereogenic units (chiral centers or stereogenic double bonds): textTotal Stereoisomers = 2^n$$\text{Total Stereoisomers} = 2^n$$ ### Core Logic Let's examine the structure of 5-phenylpent-4-en-2-ol: mathrmPh-CH=CH-CH_2-CH(OH)-CH_3 $$\mathrm{Ph-CH=CH-CH_2-CH(OH)-CH_3} $$ Identify the stereogenic units: 1. **Double Bond (mathrm-CH=CH-$\mathrm{-CH=CH-}$):** Positioned between carbons 4 and 5, this alkene group can exist in 2 distinct geometric configurations: *cis* (Z$Z$) or *trans* (E$E$). 2. **Chiral Carbon Center (*mathrmC$*\mathrm{C}$):** Carbon-2 is attached to four distinct groups: -mathrmH$-\mathrm{H}$, -mathrmOH$-\mathrm{OH}$, -mathrmCH_3$-\mathrm{CH}_3$, and -mathrmCH_2-CH=CH-Ph$-\mathrm{CH_2-CH=CH-Ph}$. This asymmetric carbon center can exist in 2 distinct optical configurations: (R$R$) or (S$S$). Since the molecule is unsymmetrical, the two stereogenic units behave independently (n = 2$n = 2$): textTotal Stereoisomers = 2^2 = 4$$\text{Total Stereoisomers} = 2^2 = 4$$ ### Visual Mapping The structural tracking confirms the presence of these stereogenic sites:

### Pattern Recognition Always break the molecule down to count chiral centers and stereogenic double bonds independently. Since the molecule has asymmetric ends, you can safely use the simplified 2^n$2^n$ formula without worrying about meso configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula For a hydrocarbon with the molecular formula mathrmC_x mathrmH_y$\mathrm{C}_x \mathrm{H}_y$, the Double Bond Equivalent (DBE) or Degree of Unsaturation (DU) is given by: textDU = x + 1 - fracy2$$\text{DU} = x + 1 - \frac{y}{2}$$ ### Core Logic 1. Calculate the mass of carbon in 1 mole of the hydrocarbon: textMass of Carbon = 80text g cdot 90\% = 72text g$$\text{Mass of Carbon} = 80\text{ g} \cdot 90\% = 72\text{ g}$$ 2. Find the number of carbon atoms (x$x$): x = frac7212 = 6$$x = \frac{72}{12} = 6$$ 3. Find the mass and number of hydrogen atoms (y$y$): textMass of Hydrogen = 80text g - 72text g = 8text g$$\text{Mass of Hydrogen} = 80\text{ g} - 72\text{ g} = 8\text{ g}$$ y = frac81 = 8$$y = \frac{8}{1} = 8$$ Thus, the molecular formula of hydrocarbon (X) is mathrmC_6mathrmH_8$\mathrm{C}_{6}\mathrm{H}_{8}$. 4. Calculate the degree of unsaturation: textDU = 6 + 1 - frac82 = 7 - 4 = 3$$\text{DU} = 6 + 1 - \frac{8}{2} = 7 - 4 = 3$$ ### Pattern Recognition First, use the percentage composition and total molar mass to determine the exact number of carbon and hydrogen atoms. Once you have the molecular formula, plug it into the standard textDU = x + 1 - fracy2$\text{DU} = x + 1 - \frac{y}{2}$ equation to find the total number of rings and/or pi bonds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula \% text Bromine = fractextMolar Mass of BrtextMolar Mass of AgBr cdot fractextMass of AgBr formedtextMass of Organic Sample cdot 100$$\% \text{ Bromine} = \frac{\text{Molar Mass of Br}}{\text{Molar Mass of AgBr}} \cdot \frac{\text{Mass of AgBr formed}}{\text{Mass of Organic Sample}} \cdot 100$$ ### Core Logic 1. Calculate the molar mass of silver bromide (mathrmAgBr$\mathrm{AgBr}$): textMolar Mass of AgBr = 108 + 80 = 188text g/mol$$\text{Molar Mass of AgBr} = 108 + 80 = 188\text{ g/mol}$$ 2. Substitute the given values into the Carius quantitative formula: * textMass of AgBr = 0.15text g$\text{Mass of AgBr} = 0.15\text{ g}$ * textMass of sample = 0.25text g$\text{Mass of sample} = 0.25\text{ g}$ \% mathrmBr = frac80188 cdot frac0.150.25 cdot 100$$\% \mathrm{Br} = \frac{80}{188} \cdot \frac{0.15}{0.25} \cdot 100$$ \% mathrmBr = frac80188 cdot 0.6 cdot 100 = frac48188 cdot 100 = 25.5319\%$$\% \mathrm{Br} = \frac{80}{188} \cdot 0.6 \cdot 100 = \frac{48}{188} \cdot 100 = 25.5319\%$$ 3. Convert the percentage to match the requested output units (times 10^-1\%$\times 10^{-1}\%$): 25.5319\% = 255.319 cdot 10^-1\% approx 255 cdot 10^-1\%$$25.5319\% = 255.319 \cdot 10^{-1}\% \approx 255 \cdot 10^{-1}\%$$ Rounding to the nearest integer gives 255. ### Pattern Recognition Pay close attention to the final multiplier units requested in the blank (times 10^-1\%$\times 10^{-1}\%$). Always calculate the raw percentage first, then adjust the decimal place to match the required format. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Carbocations are stabilized by structural factors such as the inductive effect (+I$+I$), mesomeric effect (+M$+M$), and hyperconjugation. In Structure (2), the carbocation center is situated directly adjacent to a strong electronic donor methoxy group (-OCH_3$-OCH_3$). This configuration allows highly effective lone pair donation into the vacant p-orbital of the carbocation via the structural +M$+M$ mesomeric path, rendering it exceptionally stable. ### Pattern Recognition An adjacent heteroatom with a lone pair (O, N$O, N$) triggers dynamic back-bonding stability (+M$+M$), which fundamentally outweighs basic hyperconjugation trends. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques