401. The harmful effect of freons on the ozone layer is mainly related to the release of:
ⓐ. hydrogen radicals
ⓑ. oxygen atoms
ⓒ. chlorine radicals
ⓓ. carbon dioxide molecules
Correct Answer: chlorine radicals
Explanation: In the upper atmosphere, freons can break down under the influence of high-energy radiation. This releases chlorine radicals, which participate in chain reactions that destroy ozone. Because one chlorine radical can destroy many ozone molecules, the environmental effect becomes serious. This is the main reason for restricting many chlorofluorocarbons.
402. DDT is prepared by the condensation of chloral with:
ⓐ. fluorobenzene in concentrated \( H_2SO_4 \)
ⓑ. bromobenzene in concentrated \( H_2SO_4 \)
ⓒ. chlorobenzene in concentrated \( H_2SO_4 \)
ⓓ. iodobenzene in concentrated \( H_2SO_4 \)
Correct Answer: chlorobenzene in concentrated \( H_2SO_4 \)
Explanation: DDT is formed by condensing chloral with chlorobenzene. The reaction is carried out in the presence of concentrated sulfuric acid, which promotes the condensation. This synthesis gives the well-known insecticide dichlorodiphenyltrichloroethane. The preparation is an important recognition point for DDT chemistry.
403. Which compound is chloral?
ⓐ. \( CCl_4 \)
ⓑ. \( CHCl_3 \)
ⓒ. \( CCl_3CH_2OH \)
ⓓ. \( CCl_3CHO \)
Correct Answer: \( CCl_3CHO \)
Explanation: Chloral is trichloroacetaldehyde, so its formula is \( CCl_3CHO \). It contains the aldehyde group along with three chlorine atoms on the adjacent carbon. This compound is used in the preparation of DDT by condensation with chlorobenzene. It should not be confused with chloroform or carbon tetrachloride.
404. Which statement correctly describes the ecological problem associated with DDT?
ⓐ. It decomposes too quickly to remain effective.
ⓑ. It is highly biodegradable and harmless in soil.
ⓒ. It rapidly evaporates and disappears from ecosystems.
ⓓ. It persists in nature and biomagnifies.
Correct Answer: It persists in nature and biomagnifies.
Explanation: DDT is non-biodegradable, so it remains in the environment for long periods. Once it enters a food chain, its concentration can increase from one trophic level to the next. This process is called biomagnification. Because of this persistence and accumulation, DDT became environmentally problematic despite its insecticidal efficiency.
405. Which pair is correctly matched?
ⓐ. chloroform — stored in dark bottles with a little ethanol
ⓑ. iodoform — common refrigerant in air conditioners
ⓒ. freon-\( 12 \) — common antiseptic yellow solid
ⓓ. DDT — obtained by hydrolysis of chlorobenzene under high pressure
Correct Answer: chloroform — stored in dark bottles with a little ethanol
Explanation: Chloroform is stored in dark bottles because light helps oxidise it to poisonous phosgene. A small amount of ethanol is often added to destroy any phosgene that may form. Iodoform is the yellow antiseptic solid, freon-\( 12 \) is a refrigerant, and DDT is an insecticide rather than a hydrolysis product. So only the first pair is correctly matched.
406. In a Grignard reagent such as \( RMgX \), the carbon atom bonded to magnesium behaves most like:
ⓐ. an electrophilic carbonyl carbon
ⓑ. a neutral aromatic carbon
ⓒ. a carbanion-like nucleophilic centre
ⓓ. a positively charged carbocation centre
Correct Answer: a carbanion-like nucleophilic centre
Explanation: In the \( C-Mg \) bond of a Grignard reagent, carbon is more electronegative than magnesium. As a result, electron density is drawn toward carbon, making it strongly nucleophilic in behavior. This is why Grignard reagents react readily with water, alcohols, carbonyl compounds, and other electrophilic species. The carbon does not behave as an electron-poor centre here.
407. Which reagent is commonly used to reduce a haloalkane into the corresponding alkane?
ⓐ. \( LiAlH_4 \)
ⓑ. \( NaBH_4 \)
ⓒ. \( K_2Cr_2O_7 \)
ⓓ. \( AgNO_3 \)
Correct Answer: \( LiAlH_4 \)
Explanation: Reduction of a haloalkane means replacement of halogen by hydrogen. \( LiAlH_4 \) is a strong reducing agent that can carry out this dehalogenation. The other listed reagents are not the usual choice for converting haloalkanes into alkanes. So \( LiAlH_4 \) is the appropriate reagent.
408. What is the main organic product when bromoethane is reduced with \( LiAlH_4 \)?
ⓐ. \( CH_2=CH_2 \)
ⓑ. \( CH_3CH_2OH \)
ⓒ. \( CH_3CH_2NH_2 \)
ⓓ. \( CH_3CH_3 \)
Correct Answer: \( CH_3CH_3 \)
Explanation: Reduction of bromoethane replaces bromine by hydrogen without changing the carbon skeleton. Since bromoethane has two carbon atoms, the product is the corresponding two-carbon alkane. That alkane is ethane, \( CH_3CH_3 \). This is an example of dehalogenation by reduction.
409. Which equation correctly represents hydrolysis of a Grignard reagent?
ⓐ. \( R-MgX + H_2O \rightarrow R-OH + MgXH \)
ⓑ. \( R-MgX + H_2O \rightarrow RH + Mg(OH)X \)
ⓒ. \( R-MgX + H_2O \rightarrow R-X + Mg(OH)_2 \)
ⓓ. \( R-MgX + H_2O \rightarrow R-R + Mg(OH)X \)
Correct Answer: \( R-MgX + H_2O \rightarrow RH + Mg(OH)X \)
Explanation: \( \textbf{Given:} \)
A Grignard reagent, \( R-MgX \), reacts with water.
\( \textbf{Required:} \)
Correct reaction equation
\( \textbf{Relevant Principle:} \)
Grignard reagents are destroyed by compounds containing acidic hydrogen, including water.
\( \textbf{Why this principle applies:} \)
The carbon of \( R-MgX \) behaves as a carbanion-like centre and takes a proton from water.
\( \textbf{Identify products:} \)
The organic product becomes \( RH \), and the inorganic product is \( Mg(OH)X \).
\( \textbf{Balanced equation:} \)
\[
R-MgX + H_2O \rightarrow RH + Mg(OH)X
\]
\( \textbf{Final simplification:} \)
This is the standard hydrolysis equation of a Grignard reagent.
All formulas and symbols are written correctly.
\( \textbf{Final Answer:} \)
\[
R-MgX + H_2O \rightarrow RH + Mg(OH)X
\]
410. How many moles of propane are formed when \( 0.75 \) mol of \( C_3H_7MgBr \) is completely hydrolysed?
ⓐ. \( 0.75 \) mol
ⓑ. \( 1.50 \) mol
ⓒ. \( 0.375 \) mol
ⓓ. \( 0.25 \) mol
Correct Answer: \( 0.75 \) mol
Explanation: \( \textbf{Given:} \)
Amount of \( C_3H_7MgBr = 0.75 \) mol
Reaction:
\[
C_3H_7MgBr + H_2O \rightarrow C_3H_8 + Mg(OH)Br
\]
\( \textbf{Required:} \)
Moles of propane formed
\( \textbf{Relevant Principle:} \)
One mole of a Grignard reagent gives one mole of the corresponding hydrocarbon on hydrolysis.
\( \textbf{Why this principle applies:} \)
The organic group \( C_3H_7- \) simply takes a proton from water to become \( C_3H_8 \).
\( \textbf{Identify known ratio:} \)
\( 1 \) mol \( C_3H_7MgBr \rightarrow 1 \) mol \( C_3H_8 \)
\( \textbf{Substitution:} \)
\[
0.75 \times \frac{1}{1} = 0.75
\]
\( \textbf{Final simplification:} \)
The amount of propane formed is \( 0.75 \) mol.
The answer is in moles and matches the stoichiometric ratio.
\( \textbf{Final Answer:} \)
\[
0.75 \text{ mol}
\]
411. On treatment with zinc dust, vicinal dihalides generally form:
ⓐ. alcohols
ⓑ. aldehydes
ⓒ. nitriles
ⓓ. alkenes
Correct Answer: alkenes
Explanation: Zinc removes the two halogen atoms from adjacent carbon atoms of a vicinal dihalide. This dehalogenation creates a double bond between those carbons. As a result, a vicinal dihalide is converted into an alkene. This is a useful recognition reaction for adjacent dihalides.
412. Which equation correctly represents dehalogenation of \( 1,2 \)-dibromoethane with zinc?
ⓐ. \( BrCH_2CH_2Br + Zn \rightarrow CH_3CH_3 + ZnBr_2 \)
ⓑ. \( BrCH_2CH_2Br + Zn \rightarrow CH_3CH_2Br + ZnBr \)
ⓒ. \( BrCH_2CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2 \)
ⓓ. \( BrCH_2CH_2Br + Zn \rightarrow HC\equiv CH + ZnBr_2 + H_2 \)
Correct Answer: \( BrCH_2CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2 \)
Explanation: \( \textbf{Given:} \)
Substrate: \( 1,2 \)-dibromoethane, \( BrCH_2CH_2Br \)
Reagent: zinc dust
\( \textbf{Required:} \)
Correct reaction equation
\( \textbf{Relevant Principle:} \)
Vicinal dihalides undergo dehalogenation with zinc to form alkenes.
\( \textbf{Why this principle applies:} \)
The two bromine atoms are attached to adjacent carbons, so zinc can remove them and produce a double bond.
\( \textbf{Identify the organic product:} \)
Removal of both bromine atoms gives ethene, \( CH_2=CH_2 \).
\( \textbf{Identify the inorganic product:} \)
The bromine atoms combine with zinc to give \( ZnBr_2 \).
\( \textbf{Final simplification:} \)
So the correct equation is:
\[
BrCH_2CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2
\]
All formulas are correctly balanced and written.
\( \textbf{Final Answer:} \)
\[
BrCH_2CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2
\]
413. Which statement about geminal and vicinal dihalides is correct?
ⓐ. Both can give alkynes on strong-base dehydrohalogenation.
ⓑ. Both always give alcohols only with strong base.
ⓒ. Only vicinal dihalides can form alkenes, while geminal dihalides never eliminate.
ⓓ. Neither can undergo elimination because they contain two halogen atoms.
Correct Answer: Both can give alkynes on strong-base dehydrohalogenation.
Explanation: Geminal and vicinal dihalides can both lose hydrogen halide in elimination reactions. If dehydrohalogenation occurs twice under sufficiently strong basic conditions, an alkyne can be formed. This makes dihalides useful precursors in alkyne synthesis at the recognition level. So the important idea is that both types can lead to alkynes through successive elimination.
414. Sulfonation of chlorobenzene gives mainly:
ⓐ. a mixture of \( o \)- and \( p \)-chlorobenzenesulfonic acids
ⓑ. a mixture of \( m \)- and \( p \)-chlorobenzenesulfonic acids
ⓒ. only \( p \)-chlorobenzenesulfonic acid
ⓓ. only \( m \)-chlorobenzenesulfonic acid
Correct Answer: a mixture of \( o \)- and \( p \)-chlorobenzenesulfonic acids
Explanation: Chlorine is deactivating but ortho- and para-directing in electrophilic substitution. Therefore sulfonation of chlorobenzene does not occur mainly at the meta position. Instead, the incoming sulfonic acid group enters chiefly at the ortho and para positions. This gives a mixture of the ortho and para isomers.
415. Which reaction is generally not favored by chlorobenzene under ordinary Lewis acid-catalyzed conditions?
ⓐ. electrophilic nitration
ⓑ. electrophilic sulfonation
ⓒ. Lewis acid halogenation
ⓓ. Friedel–Crafts alkylation
Correct Answer: Friedel–Crafts alkylation
Explanation: Chlorobenzene can undergo nitration, sulfonation, and halogenation, though it reacts more slowly than benzene. However, Friedel–Crafts alkylation is generally difficult with chlorobenzene because the halogen deactivates the ring. The halogen can also interact with the Lewis acid catalyst, which further reduces the tendency for the reaction. So Friedel–Crafts alkylation is not generally favored for chlorobenzene under ordinary conditions.
416. Warming ethanol or acetone with bleaching powder is a known laboratory method for preparing:
ⓐ. carbon tetrachloride
ⓑ. chloroform
ⓒ. iodoform
ⓓ. dichloromethane
Correct Answer: chloroform
Explanation: Chloroform can be prepared in the laboratory from ethanol or acetone by reaction with bleaching powder. This is a classic preparation method associated with trichloromethane. It is an important recognition point for polyhalogen compounds. So ethanol or acetone with bleaching powder leads to chloroform.
417. Iodoform is obtained in the haloform reaction when ethanol or acetone is treated with:
ⓐ. \( Br_2 \) and water
ⓑ. concentrated \( HCl \)
ⓒ. \( I_2 \) and alkali
ⓓ. \( NaCl \) and dilute acid
Correct Answer: \( I_2 \) and alkali
Explanation: Iodoform formation is associated with the haloform reaction, where iodine and alkali are used. Ethanol and acetone both can give iodoform under suitable conditions because they generate the required methyl carbonyl-related system during the reaction. The product formed is the yellow solid \( CHI_3 \). This preparation route is a standard identification point for iodoform chemistry.
418. In the reduction of a haloalkane \( R-X \) to the corresponding alkane \( R-H \), what replaces the halogen atom in the organic product?
ⓐ. Hydrogen
ⓑ. Hydroxyl group
ⓒ. Cyano group
ⓓ. Magnesium halide group
Correct Answer: Hydrogen
Explanation: Reduction of a haloalkane removes the halogen atom from the organic molecule and replaces it with hydrogen. The carbon skeleton remains the same, so \( R-X \) becomes \( R-H \). This conversion is different from hydrolysis, where halogen is replaced by \( -OH \), and from cyanide substitution, where halogen is replaced by \( CN \). Suitable reducing agents or catalytic hydrogenation conditions can bring about this dehalogenation. The defining organic change is replacement of halogen by hydrogen.
419. What is the organic product when ethylmagnesium bromide is hydrolysed with water?
ⓐ. ethane
ⓑ. ethanol
ⓒ. ethene
ⓓ. bromoethane
Correct Answer: ethane
Explanation: A Grignard reagent behaves as if the carbon bonded to magnesium carries negative character. On hydrolysis, that carbon simply takes a proton from water. Therefore ethylmagnesium bromide, \( C_2H_5MgBr \), gives the corresponding hydrocarbon, ethane. This is why Grignard reagents must be protected from moisture.
420. Successive dehydrohalogenation of \( 1,2 \)-dibromoethane with a strong base gives:
ⓐ. ethane
ⓑ. ethene
ⓒ. ethanol
ⓓ. ethyne
Correct Answer: ethyne
Explanation: \( 1,2 \)-Dibromoethane is a vicinal dihalide, so it can lose hydrogen bromide in elimination reactions. One elimination first gives a bromoalkene, and a second elimination removes another molecule of \( HBr \). The final product is an alkyne. Therefore the product formed is ethyne.
