Consider the following sequence of reactions:
mathrmCH_3-CH_2-CH_2-CH(Br)-CH_3 xrightarrowtextalcoholic KOH mathrmP text (Major Product) xrightarrowmathrmBr_2 mathrmQ$$\mathrm{CH_3-CH_2-CH_2-CH(Br)-CH_3} \xrightarrow{\text{alcoholic KOH}} \mathrm{P} \text{ (Major Product)} \xrightarrow{\mathrm{Br_2}} \mathrm{Q}$$
Consider the above sequence of reactions. 151~mathrmg$151~\mathrm{g}$ of 2-bromopentane is made to react. Yield of major product mathrmP$\mathrm{P}$ is 80\%$80\%$ whereas mathrmQ$\mathrm{Q}$ is 100\%$100\%$.
Mass of product mathrmQ$\mathrm{Q}$ obtained is _______ g.
Given molar mass in mathrmg~mol^-1$\mathrm{g~mol^{-1}}$ H: 1, C: 12, O: 16, Br: 80
Numerical Answer Type:
Enter a numerical valueAnswer: 184 to 184+4 marks
Solution & Explanation
### Related Formula
textActual Yield = textTheoretical Yield times \% text Yield$$\text{Actual Yield} = \text{Theoretical Yield} \times \% \text{ Yield}$$
### Core Logic
Let us break down each chemical reaction step:
- **Step 1**: 2-bromopentane undergoes dehydrohalogenation via an E2 mechanism using alcoholic mathrmKOH$\mathrm{KOH}$. According to Saytzeff's rule, the more substituted alkene is the major product. Thus, **pent-2-ene** is the major product mathrmP$\mathrm{P}$.
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
- **Step 2**: Pent-2-ene undergoes electrophilic bromination with liquid bromine (mathrmBr_2$\mathrm{Br_2}$) to give **2,3-dibromopentane** (product mathrmQ$\mathrm{Q}$):
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
### Step 1: Calculate Initial Moles of Reactant
Calculate the molar mass of 2-bromopentane (mathrmC_5H_11Br$\mathrm{C_5H_{11}Br}$):
textMolar mass = 5(12) + 11(1) + 80 = 60 + 11 + 80 = 151~mathrmg~mol^-1$$\text{Molar mass} = 5(12) + 11(1) + 80 = 60 + 11 + 80 = 151~\mathrm{g~mol^{-1}}$$textInitial moles = frac151~mathrmg151~mathrmg~mol^-1 = 1~mathrmmol$$\text{Initial moles} = \frac{151~\mathrm{g}}{151~\mathrm{g~mol^{-1}}} = 1~\mathrm{mol}$$
### Step 2: Calculate Moles of Intermediate P and Q
Since the yield of mathrmP$\mathrm{P}$ is 80\%$80\%$:
textMoles of P formed = 1 times 0.80 = 0.8~mathrmmol$$\text{Moles of P formed} = 1 \times 0.80 = 0.8~\mathrm{mol}$$
Since the conversion of mathrmP rightarrow mathrmQ$\mathrm{P} \rightarrow \mathrm{Q}$ has a yield of 100\%$100\%$, the mole count remains stoichiometric:
textMoles of Q formed = 0.8 times 1.00 = 0.8~mathrmmol$$\text{Moles of Q formed} = 0.8 \times 1.00 = 0.8~\mathrm{mol}$$
### Step 3: Calculate Mass of Q
Product mathrmQ$\mathrm{Q}$ is 2,3-dibromopentane (mathrmC_5H_10Br_2$\mathrm{C_5H_{10}Br_2}$).
Calculate its molar mass:
textMolar mass of Q = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~mathrmg~mol^-1$$\text{Molar mass of Q} = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~\mathrm{g~mol^{-1}}$$textMass of Q = 0.8 times 230 = 184~mathrmg$$\text{Mass of Q} = 0.8 \times 230 = 184~\mathrm{g}$$
### Pattern Recognition
Saytzeff vs Hofmann: Alcoholic mathrmKOH$\mathrm{KOH}$ is a small, non-bulky base, which selectively targets the internal secondary proton to yield the thermodynamic trans-alkene (pent-2-ene) as the major product rather than the terminal 1-alkene.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
In the following substitution reaction:
The structural layout depicts a 1,2-dibromo-4-nitrobenzene reacting with sodium ethoxide.
Product mathrmP$\mathrm{P}$ formed is:
A.
B.
C.
D.
Solution
### Related Formula
Nucleophilic Aromatic Substitution (S_NAr$S_NAr$) occurs via a Meisenheimer complex intermediate, where strong electron-withdrawing groups (-mathrmNO_2$-\mathrm{NO}_2$) activate positions strictly ortho and para to themselves.
### Core Logic
The reactant is 1,2-dibromo-4-nitrobenzene. Let us evaluate the two bromine positions relative to the nitro group:
* The bromine at C-1 is para to the strong activating -mathrmNO_2$-\mathrm{NO}_2$ group.
* The bromine at C-2 is meta to the -mathrmNO_2$-\mathrm{NO}_2$ group.
Since the para position facilitates effective negative charge delocalization onto the oxygen atoms of the nitro group during intermediate formation, the para-bromine undergoes substitution exclusively by the ethoxide ion (^-mathrmOC_2mathrmH_5$^-\mathrm{OC}_2\mathrm{H}_5$). This yields the final product shown below:
The structural layout depicts a 1,2-dibromo-4-nitrobenzene reacting with sodium ethoxide.
### Pattern Recognition
In aromatic pathways activated by -mathrmNO_2$-\mathrm{NO}_2$, substitution happens exclusively at positions ortho or para relative to the nitro flag; meta positions remain unactivated.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Haloalkanes react with KCN$KCN$ to form alkyl cyanides as a main product while with AgCN$AgCN$ form isocyanide as the main product.
Reason (R) : KCN$KCN$ and AgCN$AgCN$ both are highly ionic compounds.
In the light of the above statement, choose the most appropriate answer from the options given below:
A.text(A) is correct but (R) is not correct$\text{(A) is correct but (R) is not correct}$
B.textBoth (A) and (R) are correct but (R) is not the correct explanation of (A)$\text{Both (A) and (R) are correct but (R) is not the correct explanation of (A)}$
C.text(A) is not correct but (R) is correct$\text{(A) is not correct but (R) is correct}$
D.textBoth (A) and (R) are correct and (R) is the correct explanation of (A)$\text{Both (A) and (R) are correct and (R) is the correct explanation of (A)}$
Solution
### Core Logic
KCN$KCN$ is predominantly ionic and provides cyanide ions (CN^-$CN^-$) in solution. Although both carbon and nitrogen are in a position to donate electron pairs, the attack takes place mainly through carbon because C-C bond is more stable than C-N bond, forming alkyl cyanides (nitriles) as major product.
KCN + R-X rightarrow R-CN quad text(Major)$$KCN + R-X \rightarrow R-CN \quad \text{(Major)}$$
However, AgCN$AgCN$ is mainly covalent in nature and nitrogen is free to donate an electron pair forming isocyanide as the main product.
AgCN + R-X rightarrow R-NC quad text(Major)$$AgCN + R-X \rightarrow R-NC \quad \text{(Major)}$$
### Step 1: Final Conclusion
Assertion (A) is correct.
Reason (R) states both are highly ionic, which is incorrect because AgCN$AgCN$ is largely covalent.
Therefore, (A) is correct but (R) is not correct.
### Pattern Recognition
Ambidentate nucleophile shortcut: Alkali metal cyanides (KCN$KCN$, NaCN$NaCN$) are ionic rightarrow$\rightarrow$ attack from Carbon rightarrow$\rightarrow$ Cyanide. Heavy metal cyanides (AgCN$AgCN$) are covalent rightarrow$\rightarrow$ attack from Nitrogen rightarrow$\rightarrow$ Isocyanide.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Number of optical isomers possible for 2-chlorobutane
Numerical Answer.Answer: 2 to 2
Solution
### Core Logic
The structure of 2-chlorobutane is CH_3-CH(Cl)-CH_2-CH_3$CH_3-CH(Cl)-CH_2-CH_3$.
There is exactly one chiral center in this molecule, which is the carbon atom at position 2 (bonded to H$H$, Cl$Cl$, CH_3$CH_3$, and CH_2CH_3$CH_2CH_3$).
For a molecule with n$n$ distinct chiral centers and no plane of symmetry, the number of optical isomers (stereoisomers) is 2^n$2^n$.
### Step 1: Calculate Isomers
Number of chiral centers n = 1$n = 1$.
Total optical isomers = 2^1 = 2$2^1 = 2$ (one pair of enantiomers: the (R) and (S) configurations).
Optical Isomerism diagram for Q81 - JEE Main 2024 Morning
### Pattern Recognition
1 chiral center always gives 2 optical isomers (a pair of enantiomers).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is;
A. Retention occurs in S_N1$S_N1$ reaction and inversion occurs in S_N2$S_N2$ reaction.
B. Racemisation occurs in S_N1$S_N1$ reaction and retention occurs in S_N2$S_N2$ reaction.
C. Racemisation occurs in both S_N1$S_N1$ and S_N2$S_N2$ reactions.
D. Racemisation occurs in S_N1$S_N1$ reaction and inversion occurs in S_N2$S_N2$ reaction.
Solution
### Core Logic
In an textS_textN1$\text{S}_\text{N}1$ pathway, a planar carbocation intermediate is produced. Attack by the nucleophile can take place with equal probability from either side, resulting in complete/partial racemisation.
In an textS_textN2$\text{S}_\text{N}2$ pathway, the nucleophile attacks exclusively from the backside opposite the leaving group, causing an absolute structural inversion (Walden inversion).
### Pattern Recognition
S_N1 rightarrow$S_N1 \rightarrow$ planar intermediate carbocation rightarrow$\rightarrow$ Racemisation. S_N2 rightarrow$S_N2 \rightarrow$ direct backside launch rightarrow$\rightarrow$ Inversion.
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q70jee_main_2024_29_jan_morningPreparation of Haloarenes
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom.
Reason R : Phenols react with halogen acids violently.
In the light of the above statements, choose the most appropriate from the options given below:
A.textBoth A and R are true but R is NOT the correct explanation of A$\text{Both A and R are true but R is NOT the correct explanation of A}$
B.textA is false but R is true$\text{A is false but R is true}$
C.textA is true but R is false$\text{A is true but R is false}$
D.textBoth A and R are true and R is the correct explanation of A$\text{Both A and R are true and R is the correct explanation of A}$
Solution
### Core Logic
**Assertion (A):** In phenols, the C-O$C-O$ bond possesses partial double bond character due to resonance (the lone pair of oxygen delocalizes into the benzene ring). Because of this strong C-O$C-O$ bond, nucleophilic substitution reactions where a halide ion would replace the hydroxyl group do not occur under normal conditions. Thus, aryl halides cannot be prepared directly from phenols by reaction with HX$HX$. The statement is True.
**Reason (R):** Phenols do NOT react violently with halogen acids. In fact, they practically do not react with halogen acids (HX$HX$) to form aryl halides because the C-O$C-O$ bond is difficult to break. The statement is False.
### Step 1: Visualization
Preparation of Haloarenes diagram for Q70 - JEE Main 2024 Morning
Given reason is false.
### Step 2: Conclusion
Assertion (A) is correct but Reason (R) is false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Alcohols Phenols and Ethers
More Haloalkanes and Haloarenes Questions — jee_main_2025_02_april_evening
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