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

### 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$\mathrm{P}$.

- **Step 2**: Pent-2-ene undergoes electrophilic bromination with liquid bromine (mathrmBr_2$\mathrm{Br_2}$) to give **2,3-dibromopentane** (product mathrmQ$\mathrm{Q}$):

### 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