201. Which conversion is best achieved by the Finkelstein reaction?
ⓐ. \( R-Br \rightarrow R-I \)
ⓑ. \( R-Cl \rightarrow R-F \)
ⓒ. \( Ar-Cl \rightarrow Ar-I \)
ⓓ. \( R-I \rightarrow R-Cl \)
Correct Answer: \( R-Br \rightarrow R-I \)
Explanation: Finkelstein reaction is specifically used to convert alkyl chlorides or bromides into alkyl iodides. It is not a fluorination method, and it is not generally used for haloarenes. The halogen exchange occurs on an alkyl carbon and works well because iodide is a good nucleophile in acetone. So conversion of \( R-Br \) to \( R-I \) fits this reaction correctly.
202. Which statement correctly compares Swarts and Finkelstein reactions?
ⓐ. Swarts gives alkyl fluorides, while Finkelstein gives alkyl iodides.
ⓑ. Swarts gives alkyl iodides, while Finkelstein gives alkyl fluorides.
ⓒ. Both are used mainly to prepare aryl fluorides.
ⓓ. Both require arenediazonium salts as starting materials.
Correct Answer: Swarts gives alkyl fluorides, while Finkelstein gives alkyl iodides.
Explanation: Both Swarts and Finkelstein reactions are halogen-exchange reactions of alkyl halides, but they introduce different halogens. Swarts reaction introduces fluorine, usually using metallic fluorides. Finkelstein reaction introduces iodine, commonly using \( NaI \) in acetone. So they are related in concept but different in reagent and final product.
203. How many moles of \( NaCl \) are formed when \( 0.80 \) mol of \( R-Cl \) reacts completely in the Finkelstein reaction?
ⓐ. \( 0.40 \) mol
ⓑ. \( 1.60 \) mol
ⓒ. \( 0.80 \) mol
ⓓ. \( 0.20 \) mol
Correct Answer: \( 0.80 \) mol
Explanation: \( \textbf{Given:} \)
Amount of \( R-Cl = 0.80 \) mol
Reaction:
\[
R-Cl + NaI \xrightarrow{\text{acetone}} R-I + NaCl
\]
\( \textbf{Required:} \)
Moles of \( NaCl \) formed
\( \textbf{Relevant Principle:} \)
The balanced equation gives the mole ratio between \( R-Cl \) and \( NaCl \).
\( \textbf{Why this principle applies:} \)
One mole of alkyl chloride produces one mole of sodium chloride.
\( \textbf{Identify known values:} \)
\( 1 \) mol \( R-Cl \rightarrow 1 \) mol \( NaCl \)
\( \textbf{Substitution:} \)
\[
0.80 \times \frac{1}{1} = 0.80
\]
\( \textbf{Intermediate simplification:} \)
\[
0.80 \text{ mol } R\text{-}Cl \rightarrow 0.80 \text{ mol } NaCl
\]
\( \textbf{Final simplification:} \)
The amount of sodium chloride formed is \( 0.80 \) mol.
The answer is correctly expressed in moles.
\( \textbf{Final Answer:} \)
\[
0.80 \text{ mol}
\]
204. Which of the following is the correct product when \( CH_3CH_2Br \) is treated with \( NaI \) in acetone?
ⓐ. \( CH_3CH_2F \)
ⓑ. \( CH_3CH_2I \)
ⓒ. \( CH_3CH_2Cl \)
ⓓ. \( CH_2=CH_2 \)
Correct Answer: \( CH_3CH_2I \)
Explanation: Ethyl bromide undergoes halogen exchange with sodium iodide in acetone. The bromine atom is replaced by iodine to form ethyl iodide, \( CH_3CH_2I \). Since this is a Finkelstein reaction, the product is an alkyl iodide. No elimination or fluorination is expected under these standard conditions.
205. Which statement about halogen-exchange reactions of alkyl halides is correct?
ⓐ. Swarts reaction is used only for haloarenes.
ⓑ. Finkelstein reaction is mainly used to prepare alkyl fluorides.
ⓒ. In both reactions, the carbon skeleton changes during halogen exchange.
ⓓ. One halogen on an alkyl group is replaced by another.
Correct Answer: One halogen on an alkyl group is replaced by another.
Explanation: Swarts and Finkelstein reactions are both examples of halogen exchange in alkyl halides. In these reactions, the alkyl framework remains the same while one halogen attached to carbon is replaced by another halogen. Swarts is used mainly for fluorides, and Finkelstein for iodides. The essential feature is exchange of the halogen without altering the carbon chain.
206. Which statement best describes the physical appearance of most pure haloalkanes and haloarenes?
ⓐ. They are generally colourless compounds.
ⓑ. They are usually bright yellow solids.
ⓒ. They are commonly black crystalline substances.
ⓓ. They are always deep red liquids.
Correct Answer: They are generally colourless compounds.
Explanation: Most pure haloalkanes and haloarenes are colourless. Any strong colour seen in a sample is often due to impurities or decomposition products rather than the pure compound itself. This is a standard general property of many simple halogen derivatives of hydrocarbons. So colourlessness is the usual description for the pure substances.
207. Which statement is correct about the physical state of lower members of haloalkanes?
ⓐ. They are always solids at room temperature.
ⓑ. Many lower members are gases or volatile liquids.
ⓒ. They are all nonvolatile waxy substances.
ⓓ. They are always high-melting crystalline compounds.
Correct Answer: Many lower members are gases or volatile liquids.
Explanation: Lower haloalkanes have relatively small molecular masses, so the intermolecular forces between their molecules are not very strong. As a result, many of them exist as gases or volatile liquids at room temperature. With increase in molecular size, boiling point rises and heavier members become liquids or solids. This trend is similar to that seen in many homologous organic series.
208. As molecular mass increases in a homologous series of halogen derivatives, the physical state tends to change from:
ⓐ. solid to liquid to gas
ⓑ. liquid to gas to solid
ⓒ. gas to liquid to solid
ⓓ. gas to solid to liquid
Correct Answer: gas to liquid to solid
Explanation: Smaller members of the series have weaker intermolecular attractions and so may exist as gases. As molecular mass increases, van der Waals forces become stronger and the compounds tend to become liquids. On further increase in size, some members may exist as solids. Thus the usual progression is from gas to liquid to solid.
209. Many haloalkanes are described as having:
ⓐ. no odour at all
ⓑ. a sharp acidic or ammoniacal odour
ⓒ. an ammonia-like odour only
ⓓ. a sweet or characteristic odour
Correct Answer: a sweet or characteristic odour
Explanation: Many haloalkanes and related halogen derivatives are known for a sweet or characteristic smell. The exact odour can vary from compound to compound, but the description is often used as a general property. This is only a physical observation and does not imply safety. Many compounds with such odours can still be harmful or irritating.
210. Which of the following is most likely to be a solid at room temperature?
ⓐ. A higher molecular mass haloarene
ⓑ. A very low molecular mass haloalkane
ⓒ. Chloromethane only
ⓓ. Bromomethane only
Correct Answer: A higher molecular mass haloarene
Explanation: As molecular size and mass increase, intermolecular forces become stronger, which raises melting and boiling points. Because of this, higher members of halogen-derivative series are more likely to be solids. Very low molecular mass haloalkanes are usually gases or volatile liquids. So a higher molecular mass haloarene is the most likely solid among the choices.
211. Why are haloalkanes generally only sparingly soluble in water?
ⓐ. They react completely with water to form salts.
ⓑ. Their molecules are always symmetrical and nonpolar.
ⓒ. They ionise fully into halide ions and carbocations in water.
ⓓ. They cannot form strong hydrogen bonds with water molecules.
Correct Answer: They cannot form strong hydrogen bonds with water molecules.
Explanation: Water is a highly hydrogen-bonded solvent, and good solubility in water often requires strong interaction with water molecules. Haloalkanes do not form hydrogen bonds with water as effectively as alcohols do. Although they may have polar bonds, that alone is not sufficient to overcome the strong water-water attractions. Therefore their solubility in water is generally low.
212. Haloalkanes and haloarenes are usually more soluble in organic solvents than in water because:
ⓐ. organic solvents always convert them into ions
ⓑ. they form stronger hydrogen bonds with organic solvents than water does
ⓒ. their molecular nature is more compatible with organic solvents
ⓓ. water destroys the carbon-halogen bond immediately
Correct Answer: their molecular nature is more compatible with organic solvents
Explanation: Organic solvents usually dissolve substances of similar molecular character more effectively. Haloalkanes and haloarenes are covalent organic compounds, so they mix better with many organic solvents than with water. Their interactions with water are not strong enough to produce high solubility. This is why the phrase “like dissolves like” is useful here in a broad sense.
213. Which statement correctly compares the solubility of haloalkanes and alcohols in water?
ⓐ. Haloalkanes are usually more soluble because halogens are electronegative.
ⓑ. Alcohols are usually less soluble because oxygen reduces polarity.
ⓒ. Haloalkanes are usually more soluble because they are heavier.
ⓓ. Alcohols dissolve better because they hydrogen-bond with water.
Correct Answer: Alcohols dissolve better because they hydrogen-bond with water.
Explanation: Alcohols contain the \( -OH \) group, which can form hydrogen bonds with water molecules. This greatly improves their interaction with water and increases solubility, especially for lower alcohols. Haloalkanes, in contrast, do not form hydrogen bonds with water effectively. Therefore alcohols are generally more soluble in water than comparable haloalkanes.
214. Which statement about solubility of haloarenes in water is correct?
ⓐ. Haloarenes are highly soluble because the benzene ring is strongly hydrophilic.
ⓑ. Haloarenes are sparingly soluble because water interacts with them weakly.
ⓒ. Haloarenes dissolve well because the aromatic ring forms hydrogen bonds with water.
ⓓ. Haloarenes are fully miscible with water in all proportions.
Correct Answer: Haloarenes are sparingly soluble because water interacts with them weakly.
Explanation: Haloarenes contain a large hydrophobic aromatic ring and do not form strong hydrogen-bonding interactions with water. As a result, water cannot solvate them effectively. Their overall interaction with water is too weak to produce high solubility. So haloarenes are generally only sparingly soluble in water.
215. Which property pair is correctly matched for many lower haloalkanes?
ⓐ. highly water-soluble and odourless
ⓑ. volatile and often sweet-smelling
ⓒ. nonvolatile and strongly ionic
ⓓ. solid and completely miscible with water
Correct Answer: volatile and often sweet-smelling
Explanation: Lower haloalkanes often have low enough boiling points to be volatile. Many of them are also described as having a sweet or characteristic odour. They are not generally highly soluble in water, and they are covalent rather than ionic compounds. So volatility together with characteristic odour is the correct pairing.
216. For the series \( CH_3F \), \( CH_3Cl \), \( CH_3Br \), and \( CH_3I \), the boiling point generally increases in the order:
ⓐ. \( CH_3F < CH_3Cl < CH_3Br < CH_3I \)
ⓑ. \( CH_3I < CH_3Br < CH_3Cl < CH_3F \)
ⓒ. \( CH_3Cl < CH_3F < CH_3I < CH_3Br \)
ⓓ. \( CH_3Br < CH_3Cl < CH_3F < CH_3I \)
Correct Answer: \( CH_3F < CH_3Cl < CH_3Br < CH_3I \)
Explanation: As the halogen atom becomes heavier, the molecular mass of the haloalkane increases. Larger, heavier molecules experience stronger van der Waals forces. Because more energy is needed to overcome these intermolecular attractions, the boiling point rises. So methyl iodide has the highest boiling point among these compounds, while methyl fluoride has the lowest.
217. Which statement best explains why \( CH_3I \) has a higher boiling point than \( CH_3Cl \)?
ⓐ. The \( C-I \) bond is stronger than the \( C-Cl \) bond.
ⓑ. Greater mass gives stronger attractions.
ⓒ. \( CH_3I \) forms hydrogen bonds, while \( CH_3Cl \) does not.
ⓓ. \( CH_3I \) is ionic, while \( CH_3Cl \) is covalent.
Correct Answer: Greater mass gives stronger attractions.
Explanation: The main reason is the increase in molecular size and mass from chlorine to iodine. A larger electron cloud is more easily polarised, which strengthens van der Waals attractions between molecules. Hydrogen bonding is not the important factor here, and these compounds are covalent, not ionic. Therefore the heavier iodo compound boils at a higher temperature.
218. Which one of the following has the highest boiling point?
ⓐ. \( CH_3Cl \)
ⓑ. \( CH_3Br \)
ⓒ. \( CH_3I \)
ⓓ. \( CH_3F \)
Correct Answer: \( CH_3I \)
Explanation: Among methyl halides, boiling point rises with increase in the size and mass of the halogen atom. Iodine is the largest halogen in the series, so \( CH_3I \) has the strongest intermolecular attractions. More thermal energy is therefore required to separate its molecules into the vapour phase. This makes \( CH_3I \) the highest-boiling member among the options.
219. Among isomeric haloalkanes, branching usually causes the boiling point to:
ⓐ. increase sharply
ⓑ. remain exactly the same
ⓒ. become zero
ⓓ. decrease
Correct Answer: decrease
Explanation: Branching reduces the surface area available for close intermolecular contact. When molecules are more compact, van der Waals attractions between them become weaker. Because of this, branched isomers usually boil at lower temperatures than their straight-chain isomers. The same general trend is seen in many other organic compounds as well.
220. Which pair correctly compares the boiling points of the two isomeric alkyl bromides?
ⓐ. \( n \)-Butyl bromide has a lower boiling point than tert-butyl bromide.
ⓑ. \( n \)-Butyl bromide and tert-butyl bromide have the same boiling point.
ⓒ. tert-Butyl bromide has a higher boiling point than \( n \)-butyl bromide.
ⓓ. \( n \)-Butyl bromide has a higher boiling point than tert-butyl bromide.
Correct Answer: \( n \)-Butyl bromide has a higher boiling point than tert-butyl bromide.
Explanation: The straight-chain isomer \( n \)-butyl bromide has a larger effective surface area than the more compact tert-butyl bromide. This allows stronger intermolecular attractions between its molecules. As a result, more heat is required to separate them during boiling. Therefore the unbranched isomer has the higher boiling point.
