Organic Chemistry – Some Basic Principles and Techniques MCQs with Answers – Part 5 (Class 11 Chemistry)

Explanation: Row P is suitable because adding hydrogen across \( \mathrm{C=C} \) converts ethene into ethane, which is a reduction of the alkene. Row Q is suitable because ethanol is oxidised to ethanal by forming a carbonyl group. Row R is suitable because bromine is replaced by \( \mathrm{OH} \), so it is substitution. Row S is unsuitable because removal of water from ethanol to form ethene is elimination, not addition. The product pattern \( \mathrm{C-C \rightarrow C=C} \) with loss of a small molecule signals elimination.

Explanation: In Step P, bromine is replaced by \( \mathrm{OH} \), so the reaction is substitution. In Step Q, ethanol loses water and forms ethene, so the reaction is elimination. In Step R, hydrogen adds across the carbon-carbon double bond of ethene, so the reaction is addition. The same two-carbon skeleton appears throughout, but the reaction class changes according to what happens to bonds and groups. A sequence should be classified stepwise rather than by applying one label to the whole pathway.

Explanation: Reaction type is decided by the change in the organic substrate. In this reaction, chloroethane becomes ethanol because \( \mathrm{Cl} \) is replaced by \( \mathrm{OH} \). The carbon skeleton remains a two-carbon saturated chain. Formation of \( \mathrm{KCl} \) is a by-product result and does not make the organic reaction elimination. Elimination would normally remove atoms from the organic molecule and form a multiple bond.

Explanation: \( \textbf{Reaction idea:} \) One molecule of an alkene adds one molecule of \( \mathrm{Br_2} \) across one \( \mathrm{C=C} \) bond. \( \textbf{Molar mass of hydrocarbon:} \) \[ M(\mathrm{C_3H_6})=3(12)+6(1)=36+6=42\,\text{g mol}^{-1} \] \( \textbf{Moles of hydrocarbon used:} \) \[ n(\mathrm{C_3H_6})=\frac{28.0}{42}=0.6667\,\text{mol} \] \( \textbf{Mole ratio:} \) \[ 1\,\text{mol }\mathrm{C_3H_6}:1\,\text{mol }\mathrm{Br_2} \] \( \textbf{Moles of \( \mathrm{Br_2} \) required:} \) \[ n(\mathrm{Br_2})=0.6667\,\text{mol} \] \( \textbf{Molar mass of \( \mathrm{Br_2} \):} \) \[ M(\mathrm{Br_2})=2(80)=160\,\text{g mol}^{-1} \] \( \textbf{Mass of \( \mathrm{Br_2} \):} \) \[ m=0.6667\times160=106.7\,\text{g} \] \( \textbf{Final answer:} \) The required mass of \( \mathrm{Br_2} \) is \(106.7\,\text{g}\). The calculation uses both reaction stoichiometry and molar masses, so the bromine mass is not obtained by simply comparing formula subscripts.

Explanation: \( \textbf{Organic change:} \) \( \mathrm{C_2H_5Br} \) changes to \( \mathrm{C_2H_5OH} \) by substitution. \( \textbf{Mole relation:} \) One mole of \( \mathrm{C_2H_5Br} \) gives one mole of \( \mathrm{C_2H_5OH} \). \( \textbf{Molar mass of \( \mathrm{C_2H_5Br} \):} \) \[ 2(12)+5(1)+80=24+5+80=109\,\text{g mol}^{-1} \] \( \textbf{Moles of \( \mathrm{C_2H_5Br} \):} \) \[ n=\frac{10.9}{109}=0.100\,\text{mol} \] \( \textbf{Moles of ethanol formed:} \) \[ n(\mathrm{C_2H_5OH})=0.100\,\text{mol} \] \( \textbf{Molar mass of ethanol:} \) \[ M(\mathrm{C_2H_6O})=2(12)+6(1)+16=46\,\text{g mol}^{-1} \] \( \textbf{Mass of ethanol formed:} \) \[ m=0.100\times46=4.6\,\text{g} \] \( \textbf{Final answer:} \) The mass of ethanol formed is \(4.6\,\text{g}\). The product mass is lower because the heavy bromine atom is replaced by the lighter \( \mathrm{OH} \) group.

Explanation: Sublimation is used for solids that can pass directly from solid state to vapour state on heating. The vapour is then cooled and deposited as a solid. This technique is useful when the desired solid sublimes but the impurity does not, or when the impurity sublimes differently. It is a physical separation method and should not depend on chemical reaction with the impurity. The method is not suitable merely because a sample is solid; the subliming behaviour is the deciding feature.

Explanation: \( \mathrm{Naphthalene} \) is a sublimable organic solid. Common salt does not sublime under ordinary sublimation conditions. On heating the mixture, \( \mathrm{naphthalene} \) can change into vapour and then deposit as solid on a cooler surface. The salt remains behind as non-sublimed residue. This makes sublimation more appropriate than distillation, which is mainly used for liquid mixtures or volatile liquids.

Explanation: Crystallisation depends on different solubilities of the compound and impurities in a chosen solvent. The impure solid is usually dissolved in a minimum amount of hot solvent. On cooling, the desired compound crystallises out more purely, while many impurities remain in solution. The solvent should not normally react with the compound being purified. This method is especially useful for solids whose solubility changes appreciably with temperature.

Explanation: The given solubility pattern is ideal for crystallisation. The organic compound dissolves well in hot ethanol, so a hot saturated solution can be prepared. On cooling, the compound becomes much less soluble and separates as crystals. Since the coloured impurities remain in the mother liquor, they are removed when the crystals are filtered. Fractional distillation and steam distillation are not the first choice for purifying a non-volatile solid described through solvent solubility.

Explanation: \( \textbf{Mass of impure sample:} \) \(12.0\,\text{g}\). \( \textbf{Purity of desired compound in original sample:} \) \(80.0\%\). \( \textbf{Actual desired compound initially present:} \) \[ \frac{80.0}{100}\times12.0=9.60\,\text{g} \] \( \textbf{Mass of pure crystals recovered:} \) \(8.4\,\text{g}\). \( \textbf{Percentage recovery relation:} \) \[ \%\text{ recovery}=\frac{\text{mass recovered}}{\text{mass originally present}}\times100 \] \( \textbf{Substitution:} \) \[ \%\text{ recovery}=\frac{8.4}{9.60}\times100 \] \( \textbf{Calculation:} \) \[ \%\text{ recovery}=87.5\% \] \( \textbf{Final answer:} \) The percentage recovery is \(87.5\%\). The denominator is the amount of desired compound originally present, not the total impure sample mass.

Explanation: Simple distillation separates substances based on difference in volatility. It is useful when a volatile liquid can vaporise and then condense away from non-volatile impurities. It can also work when two liquids have a large difference in boiling points. If boiling points are very close, fractional distillation is more suitable because repeated vapour-condensation cycles are needed. Distillation is therefore a boiling-point-based purification method.

Explanation: Fractional distillation is used for separating miscible liquids with relatively close boiling points. The fractionating column provides repeated vaporisation and condensation steps. This improves separation because the vapour becomes richer in the more volatile component as it rises through the column. Simple distillation is less effective when boiling points are close. The choice between simple and fractional distillation depends on volatility difference, not on colour or chemical family alone.

Explanation: Row P is suitable because a volatile liquid can be separated from non-volatile impurities by simple distillation. Row Q is suitable because close-boiling miscible liquids require fractional distillation for better separation. Row R is suitable because sublimation works when one solid sublimes and the impurity does not. Row S is unsuitable because close boiling points are exactly the situation where simple distillation alone is usually inefficient. The fractionating column is used to improve separation between liquids of similar volatility.

Explanation: The two liquids are miscible, so they cannot be separated by simple layer separation. Their boiling points differ by only \(4^\circ\text{C}\), which is a small difference. Fractional distillation is designed for such close-boiling liquid mixtures. The fractionating column allows repeated condensation and vaporisation, improving enrichment of the lower-boiling component. Simple distillation would give poor separation because both liquids would tend to vaporise over a similar temperature range.

Explanation: During distillation of a pure liquid, the temperature remains nearly constant at the boiling point while liquid and vapour coexist. A horizontal region on the temperature-time graph indicates that the distilling vapour has a nearly constant composition. In a mixture, the temperature may change as composition changes during distillation. The graph does not by itself show formation of a new functional group. The plateau is linked to phase change at a characteristic boiling point.

Explanation: Steam distillation is used for substances that are volatile in steam and are usually immiscible with water. The method allows distillation at a temperature lower than the normal boiling point of the organic compound. This is useful for compounds that may decompose if heated strongly. The organic compound co-distils with steam and can later be separated from the water layer if immiscible. The method is not meant for completely non-volatile salts or ordinary crystallisation of solids.

Explanation: The liquid is immiscible with water and volatile in steam, so it can be carried over with steam. Since it decomposes near its normal boiling point, ordinary distillation may damage it. Steam distillation allows separation at a lower temperature because the mixture boils when the sum of vapour pressures reaches atmospheric pressure. This protects heat-sensitive compounds. Filtration would not separate a liquid mixture based on volatility.

Explanation: \( \textbf{Condition for steam distillation:} \) For immiscible liquids, the total vapour pressure is the sum of the individual vapour pressures. \( \textbf{Given vapour pressure of water:} \) \[ P_{\text{water}}=88\,\text{kPa} \] \( \textbf{Given vapour pressure of organic liquid:} \) \[ P_{\text{organic}}=13\,\text{kPa} \] \( \textbf{Total vapour pressure:} \) \[ P_{\text{total}}=88+13=101\,\text{kPa} \] \( \textbf{Atmospheric pressure:} \) \[ P_{\text{atm}}=101\,\text{kPa} \] \( \textbf{Boiling condition:} \) A liquid mixture boils when its total vapour pressure equals atmospheric pressure. \( \textbf{Final answer:} \) The mixture can boil under these conditions. The organic liquid need not reach \(101\,\text{kPa}\) vapour pressure by itself because water contributes to the total vapour pressure.

Explanation: Differential extraction uses two immiscible solvents, such as water and an organic solvent. The solute is more soluble in one solvent than the other, so it preferentially moves into that layer. The two layers can then be separated using a separating funnel. Repeated extraction can improve recovery of the solute. The method depends on distribution between two liquid phases, not on boiling point or sublimation.

Explanation: Since the organic compound is more soluble in ether than in water, it will preferentially move into the ether layer. Ether and water are immiscible, so two separate layers form. A separating funnel can be used to separate the ether layer from the aqueous layer. The ether can then be evaporated to recover the organic compound. This is an application of differential extraction based on solubility difference between immiscible solvents.