101. A Wurtz reaction uses \( \mathrm{C_2H_5Br} \) as the only alkyl halide. The hydrocarbon product has how many carbon atoms?
ⓐ. \(2\)
ⓑ. \(3\)
ⓒ. \(4\)
ⓓ. \(5\)
Correct Answer: \(4\)
Explanation: \( \textbf{Alkyl group present:} \) \( \mathrm{C_2H_5Br} \) contains the ethyl group, \( \mathrm{C_2H_5-} \).
\( \textbf{Reaction type:} \) Sodium in dry ether couples two identical alkyl groups.
\( \textbf{Carbon count per group:} \) Each ethyl group has \(2\) carbon atoms.
\( \textbf{Coupled product skeleton:} \)
\[
\mathrm{C_2H_5-C_2H_5}
\]
\( \textbf{Total carbon atoms:} \)
\[
2+2=4
\]
\( \textbf{Product formula:} \) The alkane is \( \mathrm{C_4H_{10}} \).
\( \textbf{Final answer:} \) The product has \(4\) carbon atoms.
The carbon count doubles because the same alkyl group appears on both sides of the new carbon-carbon bond.
102. The side product formed along with \( \mathrm{R-R} \) in the Wurtz reaction \( \mathrm{2R-X+2Na\rightarrow R-R+2NaX} \) is
ⓐ. \( \mathrm{HX} \)
ⓑ. \( \mathrm{NaX} \)
ⓒ. \( \mathrm{CO_2} \)
ⓓ. \( \mathrm{H_2O} \)
Correct Answer: \( \mathrm{NaX} \)
Explanation: In the Wurtz reaction, sodium removes the halogen from the alkyl halide during coupling. The two alkyl groups form the new carbon-carbon bond in \( \mathrm{R-R} \). The halogen combines with sodium to form \( \mathrm{NaX} \). For example, chloromethane gives \( \mathrm{NaCl} \) as the inorganic side product. Formation of \( \mathrm{CO_2} \) belongs to decarboxylation reactions, not Wurtz coupling.
103. A reaction planner wants \( \mathrm{CH_3CH_2CH_2CH_3} \) as the main product by Wurtz reaction. The most suitable single alkyl halide is
ⓐ. \( \mathrm{CH_3Cl} \)
ⓑ. \( \mathrm{C_3H_7Cl} \)
ⓒ. a mixture of \( \mathrm{CH_3Br} \) and \( \mathrm{C_2H_5Br} \)
ⓓ. \( \mathrm{C_2H_5Br} \)
Correct Answer: \( \mathrm{C_2H_5Br} \)
Explanation: \( \textbf{Desired product:} \) \( \mathrm{CH_3CH_2CH_2CH_3} \) is butane, \( \mathrm{C_4H_{10}} \).
\( \textbf{Wurtz logic:} \) A single alkyl halide gives a symmetrical product \( \mathrm{R-R} \).
\( \textbf{Split the product symmetrically:} \)
\[
\mathrm{CH_3CH_2-CH_2CH_3}
\]
\( \textbf{Each half:} \) Each alkyl group is ethyl, \( \mathrm{C_2H_5-} \).
\( \textbf{Suitable alkyl halide:} \) An ethyl halide such as \( \mathrm{C_2H_5Br} \) can provide this group.
\( \textbf{Reaction representation:} \)
\[
\mathrm{2C_2H_5Br+2Na\rightarrow C_4H_{10}+2NaBr}
\]
\( \textbf{Final answer:} \) The suitable single alkyl halide is \( \mathrm{C_2H_5Br} \).
Using \( \mathrm{C_3H_7Cl} \) would couple two propyl groups and give a six-carbon alkane.
104. The best reason dry ether is specified in Wurtz reaction is that
ⓐ. sodium coupling needs a dry, non-aqueous medium
ⓑ. water is needed to dissolve sodium completely
ⓒ. ether supplies the carbon atoms of the product
ⓓ. ether oxidises the alkyl halide to a carboxylate
Correct Answer: sodium coupling needs a dry, non-aqueous medium
Explanation: Wurtz reaction uses sodium metal, so an aqueous medium is unsuitable. Water would react with sodium and interfere with the intended coupling of alkyl halides. Dry ether acts as a non-aqueous solvent in which the coupling can occur. It does not supply the alkyl groups that form the product; those come from the alkyl halide. The word “dry” is therefore part of the reaction condition, not a decorative detail.
105. Soda-lime decarboxylation of a sodium carboxylate is mainly used to prepare an alkane by
ⓐ. adding \( \mathrm{H_2} \) across a double bond
ⓑ. removing the carboxyl carbon and forming \( \mathrm{R-H} \)
ⓒ. coupling two alkyl halides with sodium in dry ether
ⓓ. adding bromine across a carbon-carbon multiple bond
Correct Answer: removing the carboxyl carbon and forming \( \mathrm{R-H} \)
Explanation: Soda-lime decarboxylation uses a sodium salt of a carboxylic acid with \( \mathrm{NaOH/CaO} \) and heat. The carboxylate carbon is removed from the organic molecule, and the remaining alkyl part becomes an alkane. A general representation is \( \mathrm{RCOONa+NaOH\rightarrow R-H+Na_2CO_3} \), with \( \mathrm{CaO} \) helping maintain the reaction medium. This is different from Wurtz reaction, where two alkyl groups couple. The most important carbon-count feature is that the alkane has one carbon atom fewer than the carboxylate salt.
106. Heating sodium acetate with soda lime gives
ⓐ. methane, \( \mathrm{CH_4} \)
ⓑ. ethane, \( \mathrm{C_2H_6} \)
ⓒ. ethene, \( \mathrm{C_2H_4} \)
ⓓ. propane, \( \mathrm{C_3H_8} \)
Correct Answer: methane, \( \mathrm{CH_4} \)
Explanation: \( \textbf{Starting salt:} \) Sodium acetate is \( \mathrm{CH_3COONa} \).
\( \textbf{Reagent condition:} \) Soda lime means \( \mathrm{NaOH/CaO} \) with heat.
\( \textbf{Reaction type:} \) Decarboxylation removes the carboxyl carbon.
\( \textbf{General pattern:} \)
\[
\mathrm{RCOONa+NaOH\rightarrow R-H+Na_2CO_3}
\]
\( \textbf{Identify \(R\):} \) In \( \mathrm{CH_3COONa} \), \( \mathrm{R=CH_3} \).
\( \textbf{Organic product:} \)
\[
\mathrm{CH_3-H=CH_4}
\]
\( \textbf{Balanced representation:} \)
\[
\mathrm{CH_3COONa+NaOH\rightarrow CH_4+Na_2CO_3}
\]
\( \textbf{Final answer:} \) The alkane formed is methane, \( \mathrm{CH_4} \).
The carbonyl carbon of the carboxylate does not remain in the hydrocarbon product.
107. The alkane formed by soda-lime decarboxylation of sodium propionate, \( \mathrm{CH_3CH_2COONa} \), is
ⓐ. methane, \( \mathrm{CH_4} \)
ⓑ. propane, \( \mathrm{C_3H_8} \)
ⓒ. ethane, \( \mathrm{C_2H_6} \)
ⓓ. butane, \( \mathrm{C_4H_{10}} \)
Correct Answer: ethane, \( \mathrm{C_2H_6} \)
Explanation: \( \textbf{Starting salt:} \) \( \mathrm{CH_3CH_2COONa} \) is sodium propionate.
\( \textbf{Total carbon count in salt:} \) It contains \(3\) carbon atoms.
\( \textbf{Decarboxylation rule:} \) The carboxyl carbon is removed during soda-lime decarboxylation.
\( \textbf{Remaining alkyl group:} \) \( \mathrm{CH_3CH_2-} \).
\( \textbf{Hydrogen replacement:} \) The remaining group becomes \( \mathrm{CH_3CH_2H} \).
\( \textbf{Product formula:} \)
\[
\mathrm{CH_3CH_2COONa+NaOH\rightarrow CH_3CH_3+Na_2CO_3}
\]
\( \textbf{Product name:} \) \( \mathrm{CH_3CH_3} \) is ethane.
\( \textbf{Final answer:} \) Sodium propionate gives ethane.
Counting all three carbons of the salt in the alkane would wrongly keep the carbon lost as carbonate.
108. In the reaction \( \mathrm{RCOONa+NaOH\rightarrow R-H+Na_2CO_3} \), the role of \( \mathrm{CaO} \) in soda lime is best described as
ⓐ. the source of the alkyl group \( \mathrm{R} \)
ⓑ. the carbon source for the alkane product
ⓒ. a Lewis acid catalyst like \( \mathrm{FeBr_3} \)
ⓓ. a dry component of soda lime
Correct Answer: a dry component of soda lime
Explanation: Soda lime is commonly represented as \( \mathrm{NaOH/CaO} \). \( \mathrm{NaOH} \) participates in the decarboxylation of the sodium carboxylate, while \( \mathrm{CaO} \) helps provide a dry, high-temperature solid medium. The alkyl group \( \mathrm{R} \) comes from the carboxylate salt, not from \( \mathrm{CaO} \). \( \mathrm{CaO} \) does not supply carbon atoms to the alkane product. It should not be confused with Lewis acid catalysts used in aromatic halogenation.
109. A reaction scheme changes \( \mathrm{C_4H_9COONa} \) into an alkane by soda-lime decarboxylation. The alkane formed contains
ⓐ. \(3\) carbon atoms
ⓑ. \(5\) carbon atoms
ⓒ. \(4\) carbon atoms
ⓓ. \(6\) carbon atoms
Correct Answer: \(4\) carbon atoms
Explanation: \( \textbf{Starting salt:} \) \( \mathrm{C_4H_9COONa} \) has an alkyl group \( \mathrm{C_4H_9-} \) attached to \( \mathrm{-COONa} \).
\( \textbf{Reaction type:} \) Soda-lime decarboxylation removes the carboxylate carbon.
\( \textbf{General pattern:} \)
\[
\mathrm{RCOONa\rightarrow R-H}
\]
\( \textbf{Identify \(R\):} \) \( \mathrm{R=C_4H_9} \).
\( \textbf{Product:} \) \( \mathrm{C_4H_{10}} \).
\( \textbf{Carbon count in product:} \) \(4\) carbon atoms.
\( \textbf{Carbon count comparison:} \) The salt has \(5\) carbons total, but the alkane has one fewer.
\( \textbf{Final answer:} \) The alkane formed contains \(4\) carbon atoms.
The \( \mathrm{-COONa} \) carbon is counted in the salt but not in the alkane.
110. A table of soda-lime decarboxylation products is given below.
| Row | Sodium carboxylate | Claimed alkane product |
|---|
| P | \( \mathrm{HCOONa} \) | \( \mathrm{H_2} \) |
| Q | \( \mathrm{CH_3COONa} \) | \( \mathrm{CH_4} \) |
| R | \( \mathrm{CH_3CH_2COONa} \) | \( \mathrm{C_2H_6} \) |
| S | \( \mathrm{C_3H_7COONa} \) | \( \mathrm{C_3H_8} \) |
The row that is least suitable as an alkane-product entry is
ⓐ. Q
ⓑ. P
ⓒ. R
ⓓ. S
Correct Answer: P
Explanation: Sodium acetate, \( \mathrm{CH_3COONa} \), gives methane because \( \mathrm{R=CH_3} \). Sodium propionate gives ethane because the \( \mathrm{CH_3CH_2-} \) group remains. \( \mathrm{C_3H_7COONa} \) gives propane by the same one-carbon-loss rule. Sodium formate, \( \mathrm{HCOONa} \), does not give an alkane in the same carbon-chain sense because there is no alkyl group \( \mathrm{R} \) attached to the carboxylate carbon. The general \( \mathrm{RCOONa\rightarrow R-H} \) pattern is most directly applied when \( \mathrm{R} \) is an alkyl group.
111. Assertion: Soda-lime decarboxylation of sodium acetate gives methane.
Reason: In soda-lime decarboxylation, the carbon atom of the \( \mathrm{-COONa} \) group is not retained in the alkane.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Sodium acetate has the structure \( \mathrm{CH_3COONa} \). During soda-lime decarboxylation, the carboxylate carbon is removed into the inorganic carbonate product. The remaining \( \mathrm{CH_3-} \) group becomes \( \mathrm{CH_4} \). Thus the product has one carbon atom, not two. The Reason explains the carbon-count change that leads specifically from sodium acetate to methane.
112. A reaction comparison is shown below.
| Reaction route | Typical carbon-count effect |
|---|
| P. \( \mathrm{R-X} \) reduction | carbon skeleton retained |
| Q. Wurtz reaction with one alkyl halide | alkyl group count effectively doubles |
| R. Soda-lime decarboxylation | alkane has one carbon fewer than the carboxylate salt |
The comparison supports the conclusion that
ⓐ. decarboxylation involves loss of the carboxyl carbon
ⓑ. all three routes have the same carbon-count effect
ⓒ. Wurtz reaction always gives one carbon fewer
ⓓ. alkyl halide reduction always doubles the chain
Correct Answer: decarboxylation involves loss of the carboxyl carbon
Explanation: The three preparation routes differ in their carbon-count behaviour. Reduction of an alkyl halide generally retains the carbon skeleton while replacing halogen by hydrogen. Wurtz reaction couples alkyl groups, so a single alkyl halide often gives a symmetrical higher alkane. Soda-lime decarboxylation removes the carboxylate carbon, giving an alkane with one fewer carbon atom than the salt. Recognising this carbon-count pattern helps select the correct product without confusing nearby preparation methods.
113. A student predicts that \( \mathrm{CH_3CH_2COONa} \) gives propane on heating with soda lime because the salt has three carbon atoms. The better prediction is ethane because
ⓐ. soda lime adds one carbon atom to the salt
ⓑ. sodium propionate contains no carbon atoms
ⓒ. all carboxylates become alkenes in soda lime
ⓓ. the carboxyl carbon is lost
Correct Answer: the carboxyl carbon is lost
Explanation: Sodium propionate, \( \mathrm{CH_3CH_2COONa} \), contains three carbon atoms in total. In soda-lime decarboxylation, the \( \mathrm{-COONa} \) carbon is not retained in the hydrocarbon product. The remaining \( \mathrm{CH_3CH_2-} \) group becomes \( \mathrm{CH_3CH_3} \), which is ethane. The reaction does not add carbon atoms and does not form an alkene as the normal product. The product carbon count should be taken from \( \mathrm{R} \), not from the whole \( \mathrm{RCOONa} \) formula.
114. Kolbe electrolysis prepares alkanes by electrolysing sodium or potassium salts of carboxylic acids, mainly through
ⓐ. hydration of a double bond at the cathode
ⓑ. direct addition of \( \mathrm{H_2} \) to benzene
ⓒ. substitution of hydrogen by nitro group
ⓓ. coupling of alkyl radicals formed at the anode
Correct Answer: coupling of alkyl radicals formed at the anode
Explanation: In Kolbe electrolysis, carboxylate ions are oxidised at the anode. They lose carbon dioxide and form alkyl radicals. Two such radicals can couple to form a new carbon-carbon bond and produce an alkane. The method is therefore different from soda-lime decarboxylation, where the remaining group becomes \( \mathrm{R-H} \). Kolbe electrolysis is best understood as decarboxylation followed by radical coupling.
115. During Kolbe electrolysis of sodium acetate, the main alkane product is
ⓐ. methane, \( \mathrm{CH_4} \)
ⓑ. ethane, \( \mathrm{C_2H_6} \)
ⓒ. propane, \( \mathrm{C_3H_8} \)
ⓓ. butane, \( \mathrm{C_4H_{10}} \)
Correct Answer: ethane, \( \mathrm{C_2H_6} \)
Explanation: \( \textbf{Starting salt:} \) Sodium acetate is \( \mathrm{CH_3COONa} \).
\( \textbf{Reactive ion:} \) The carboxylate ion is \( \mathrm{CH_3COO^-} \).
\( \textbf{At the anode:} \) The carboxylate ion loses \( \mathrm{CO_2} \) and forms a methyl radical, \( \mathrm{CH_3\cdot} \).
\( \textbf{Radical coupling:} \) Two methyl radicals combine.
\[
\mathrm{CH_3\cdot+\cdot CH_3\rightarrow CH_3CH_3}
\]
\( \textbf{Product:} \) \( \mathrm{CH_3CH_3} \) is ethane.
\( \textbf{Carbon-count check:} \) Each acetate ion contributes one \( \mathrm{CH_3} \) group to the product.
\( \textbf{Final answer:} \) Kolbe electrolysis of sodium acetate gives ethane.
This differs from soda-lime decarboxylation of sodium acetate, which gives methane.
116. The anode step in Kolbe electrolysis is most closely represented by
ⓐ. \( \mathrm{RCOO^- \rightarrow R\cdot + CO_2 + e^-} \)
ⓑ. \( \mathrm{R-X+H_2\rightarrow R-H+HX} \)
ⓒ. \( \mathrm{R-H\rightarrow R-X} \)
ⓓ. \( \mathrm{RCH=CH_2+H_2\rightarrow RCH_2CH_3} \)
Correct Answer: \( \mathrm{RCOO^- \rightarrow R\cdot + CO_2 + e^-} \)
Explanation: Kolbe electrolysis involves oxidation of carboxylate ions at the anode. The carboxylate ion loses an electron and releases \( \mathrm{CO_2} \), forming an alkyl radical \( \mathrm{R\cdot} \). These radicals then couple to form \( \mathrm{R-R} \). The other options describe reduction, substitution, or hydrogenation patterns rather than the Kolbe anode process. The appearance of \( \mathrm{R\cdot} \) and \( \mathrm{CO_2} \) is the key mechanistic clue.
117. Electrolysis of a single carboxylate salt \( \mathrm{C_2H_5COONa} \) by the Kolbe method most directly gives an alkane with
ⓐ. \(2\) carbon atoms
ⓑ. \(3\) carbon atoms
ⓒ. \(4\) carbon atoms
ⓓ. \(6\) carbon atoms
Correct Answer: \(4\) carbon atoms
Explanation: \( \textbf{Starting salt:} \) \( \mathrm{C_2H_5COONa} \) contains the alkyl group \( \mathrm{C_2H_5-} \).
\( \textbf{Kolbe event:} \) The carboxylate part loses \( \mathrm{CO_2} \).
\( \textbf{Radical formed:} \) \( \mathrm{C_2H_5\cdot} \).
\( \textbf{Coupling step:} \) Two identical ethyl radicals combine.
\[
\mathrm{C_2H_5\cdot+\cdot C_2H_5\rightarrow C_2H_5-C_2H_5}
\]
\( \textbf{Carbon count:} \)
\[
2+2=4
\]
\( \textbf{Product formula:} \) \( \mathrm{C_4H_{10}} \).
\( \textbf{Final answer:} \) The alkane has \(4\) carbon atoms.
The carboxyl carbon is lost before radical coupling, so it is not counted in the final alkane.
118. A compact comparison of sodium acetate reactions is given below.
Route P: \( \mathrm{CH_3COONa} \) is heated with soda lime. Route Q: \( \mathrm{CH_3COONa} \) is electrolysed under Kolbe conditions.
The hydrocarbon products of Route P and Route Q are respectively
ⓐ. \( \mathrm{C_2H_6} \) and \( \mathrm{CH_4} \)
ⓑ. \( \mathrm{CH_4} \) and \( \mathrm{C_2H_6} \)
ⓒ. \( \mathrm{C_2H_4} \) and \( \mathrm{C_2H_2} \)
ⓓ. \( \mathrm{C_3H_8} \) and \( \mathrm{C_4H_{10}} \)
Correct Answer: \( \mathrm{CH_4} \) and \( \mathrm{C_2H_6} \)
Explanation: In soda-lime decarboxylation, \( \mathrm{CH_3COONa} \) loses the carboxyl carbon and the \( \mathrm{CH_3-} \) group becomes \( \mathrm{CH_4} \). In Kolbe electrolysis, the \( \mathrm{CH_3COO^-} \) ion forms \( \mathrm{CH_3\cdot} \) radicals after loss of \( \mathrm{CO_2} \). Two methyl radicals then couple to give \( \mathrm{CH_3CH_3} \), which is ethane. The same salt can therefore give different alkanes under different conditions. The condition decides whether the remaining group is protonated or coupled.
119. A table lists possible Kolbe electrolysis outcomes for single carboxylate salts.
| Row | Salt | Alkyl radical after \( \mathrm{CO_2} \) loss | Expected coupled alkane |
|---|
| P | \( \mathrm{CH_3COONa} \) | \( \mathrm{CH_3\cdot} \) | \( \mathrm{C_2H_6} \) |
| Q | \( \mathrm{C_2H_5COONa} \) | \( \mathrm{C_2H_5\cdot} \) | \( \mathrm{C_4H_{10}} \) |
| R | \( \mathrm{C_3H_7COONa} \) | \( \mathrm{C_3H_7\cdot} \) | \( \mathrm{C_6H_{14}} \) |
| S | \( \mathrm{C_2H_5COONa} \) | \( \mathrm{C_2H_5\cdot} \) | \( \mathrm{C_3H_8} \) |
The row needing correction is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: In Kolbe electrolysis, the alkyl group left after loss of \( \mathrm{CO_2} \) forms a radical. Two identical radicals from a single salt couple to form \( \mathrm{R-R} \). Acetate gives methyl radicals and then ethane, so row P is suitable. Propionate gives ethyl radicals and then butane, so row Q is suitable. Butyrate gives propyl radicals and then a six-carbon alkane, so row R is suitable. Row S is wrong because two ethyl radicals cannot couple to give propane; their coupling gives butane.
120. Kolbe electrolysis of a single sodium carboxylate \( \mathrm{RCOONa} \) usually gives a symmetrical alkane because
ⓐ. the carboxylate carbon remains between two \( \mathrm{R} \) groups
ⓑ. sodium adds to the hydrocarbon chain as a substituent
ⓒ. two identical \( \mathrm{R\cdot} \) radicals couple
ⓓ. water supplies one carbon atom to each radical
Correct Answer: two identical \( \mathrm{R\cdot} \) radicals couple
Explanation: A single carboxylate salt gives one kind of alkyl radical after decarboxylation at the anode. Since the same \( \mathrm{R\cdot} \) radical is produced repeatedly, coupling gives \( \mathrm{R-R} \). This product is symmetrical when both sides of the new carbon-carbon bond are the same. The carboxyl carbon is released as \( \mathrm{CO_2} \), so it does not remain in the alkane. Sodium is part of the salt and electrolyte system, not a substituent in the hydrocarbon product.
