Solution: Amide reduction replaces C=O by CH₂ while retaining the C-N and N-methyl connections.

A. Correct. Amide reduction replaces C=O by CH₂ while retaining the C-N and N-methyl connections.

B. Incorrect. This adds an unsupported methyl.

C. Incorrect. That describes cleavage rather than amide-to-amine reduction.

D. Incorrect. Excess hydride reduces beyond an imine.

E. Incorrect. Reduction of this secondary amide retains its N-methyl group; this nitrile has the wrong connectivity and carbon count.

Solution: Grignard adds to CN carbon, and hydrolysis converts its C=N-derived bond to C=O.

A. Incorrect. That remains remote in the ethyl group.

B. Incorrect. That becomes the methyl substituent.

C. Incorrect. Water supplies no carbon.

D. Correct. Grignard adds to CN carbon, and hydrolysis converts its C=N-derived bond to C=O.

E. Incorrect. Bromide supplies none.

Solution: Carboxylic acid reduction gives the primary alcohol without deleting the carboxyl carbon.

A. Incorrect. That replaces the side-chain carbon by oxygen.

B. Incorrect. This adds one unsupported carbon.

C. Incorrect. That would delete the side-chain carbon.

D. Incorrect. No methyl donor is present.

E. Correct. Carboxylic acid reduction gives the primary alcohol without deleting the carboxyl carbon.

Solution: Basic ester hydrolysis gives acetate; neutral acetic acid requires a separate acidic workup.

A. Incorrect. Ester saponification is not reduction to an aldehyde.

B. Incorrect. Ethanol comes from the ester ethoxy group, not the acyl fragment.

C. Correct. Basic ester hydrolysis gives acetate; neutral acetic acid requires a separate acidic workup.

D. Incorrect. The carboxyl group is predominantly deprotonated in strongly basic solution.

E. Incorrect. Aqueous hydroxide does not convert the ester into an acyl chloride.

Solution: The ring contains carbonyl C, three CH₂ carbons and the alcohol O.

A. Incorrect. No extra atoms enter the intramolecular ring.

B. Incorrect. This counts the tether incompletely.

C. Incorrect. There are only three methylene carbons in the tether.

D. Correct. The ring contains carbonyl C, three CH₂ carbons and the alcohol O.

E. Incorrect. This omits the ring oxygen or carbonyl carbon.

Solution: The acyl carbonyl becomes CH₂OH; the methoxy fragment becomes methanol.

A. Incorrect. The fragments do not couple.

B. Incorrect. This loses the carbonyl carbon.

C. Incorrect. Excess LiAlH₄ reduces the aldehyde further.

D. Incorrect. That is hydrolysis rather than complete reduction.

E. Correct. The acyl carbonyl becomes CH₂OH; the methoxy fragment becomes methanol.

Solution: Conversion to CH₃CH₂COCl replaces the acid OH function without changing carbon count.

A. Incorrect. No cleavage is required.

B. Correct. Conversion to CH₃CH₂COCl replaces the acid OH function without changing carbon count.

C. Incorrect. SOCl₂ contains no carbon.

D. Incorrect. The carbonyl carbon remains.

E. Incorrect. An acyl chloride is not an acid dimer.

Solution: Fischer esterification is reversible; lowering product-water activity favors ester formation.

A. Incorrect. A catalyst changes rate, not equilibrium constant.

B. Incorrect. That favors hydrolysis.

C. Correct. Fischer esterification is reversible; lowering product-water activity favors ester formation.

D. Incorrect. That gives carboxylate rather than acid-catalyzed esterification.

E. Incorrect. Removing a reactant opposes ester formation.

Solution: Acyl substitution gives CH₃CONHCH₂CH₃ and acid that otherwise protonates the nucleophile.

A. Incorrect. Amine contains no oxygen.

B. Correct. Acyl substitution gives CH₃CONHCH₂CH₃ and acid that otherwise protonates the nucleophile.

C. Incorrect. The acyl framework remains.

D. Incorrect. Ordinary acylation gives an amide.

E. Incorrect. Chloride has no carbon.

Solution: An ester undergoes addition-elimination to ketone, then a second Grignard addition.

A. Correct. An ester undergoes addition-elimination to ketone, then a second Grignard addition.

B. Incorrect. Only two additions occur at an ordinary ester carbonyl.

C. Incorrect. This stops at the ketone intermediate.

D. Incorrect. Grignard carbon is incorporated.

E. Incorrect. That would exceed the target carbon's valence.

Solution: Amide hydrolysis yields acetic acid; activation replaces acid OH by Cl.

A. Incorrect. No carbonyl reduction occurs.

B. Incorrect. The sequence is not amide dehydration.

C. Incorrect. No carbon nucleophile is added.

D. Incorrect. The N-methyl fragment leaves during hydrolysis rather than becoming acyl carbon.

E. Correct. Amide hydrolysis yields acetic acid; activation replaces acid OH by Cl.

Solution: Hydrolysis opens the anhydride bridge to a dicarboxylic acid.

A. Incorrect. That would cleave the central C-C bond incorrectly.

B. Correct. Hydrolysis opens the anhydride bridge to a dicarboxylic acid.

C. Incorrect. Hydrolysis does not remove both acid functions.

D. Incorrect. That requires reduction.

E. Incorrect. One carbonyl carbon is lost.

Solution: Reduction turns the original acetyl group into an N-ethyl substituent; reacylation adds a new acetyl group.

A. Incorrect. Both original N-C connections survive reduction.

B. Incorrect. The reduced acyl fragment is ethyl.

C. Incorrect. No phenyl group is supplied.

D. Correct. Reduction turns the original acetyl group into an N-ethyl substituent; reacylation adds a new acetyl group.

E. Incorrect. The secondary amine already has two carbon substituents.

Solution: One organic group replaces chloride, giving a ketone.

A. Correct. One organic group replaces chloride, giving a ketone.

B. Incorrect. That is the common excess-Grignard overaddition endpoint.

C. Incorrect. That would require methoxide rather than a carbon nucleophile.

D. Incorrect. No nitrogen reagent is present.

E. Incorrect. This is hydride reduction, not cuprate coupling.

Solution: Cyanide adds one carbon to the two-carbon chain; hydrolysis gives propanoic acid.

A. Incorrect. This omits cyanide carbon.

B. Incorrect. That is not the acid endpoint.

C. Correct. Cyanide adds one carbon to the two-carbon chain; hydrolysis gives propanoic acid.

D. Incorrect. That is an alcohol formula.

E. Incorrect. This adds an unsupported extra carbon.

Solution: Reduction of the prochiral ketone gives equal enantiomers absent chiral influence.

A. Incorrect. Only one center is created.

B. Correct. Reduction of the prochiral ketone gives equal enantiomers absent chiral influence.

C. Incorrect. Starting nitrile geometry does not provide chirality.

D. Incorrect. Negative charge does not impose an enantiotopic-face preference.

E. Incorrect. The carbonyl oxygen becomes OH.

Solution: The amine's ethyl group remains attached to nitrogen.

A. Incorrect. Oxygen cannot supply an ethyl group.

B. Incorrect. No carbon splitting occurs.

C. Incorrect. Water has no carbon.

D. Correct. The amine's ethyl group remains attached to nitrogen.

E. Incorrect. It supplies no carbon.

Solution: Methanol oxygen attacks acyl carbon, and chloride leaves after collapse.

A. Incorrect. The ring does not undergo chlorination.

B. Incorrect. Pyridine acts as base here, not the intended retained amide group.

C. Incorrect. The neutral ester's methoxy O bonds to acyl carbon.

D. Incorrect. That would be carbon alkylation of the ring.

E. Correct. Methanol oxygen attacks acyl carbon, and chloride leaves after collapse.

Solution: Collapse of the first tetrahedral intermediate expels the alkoxy group and exposes a ketone.

A. Incorrect. The carbon substituent on the acyl carbon is retained.

B. Incorrect. They are incorporated in the tertiary alcohol.

C. Incorrect. It remains the central alcohol carbon.

D. Incorrect. No aromatic ring is present.

E. Correct. Collapse of the first tetrahedral intermediate expels the alkoxy group and exposes a ketone.

Solution: Reduction gives CH₃CH₂N(CH₃)₂; nitrogen has three carbon substituents.

A. Incorrect. Two N-methyl groups remain.

B. Correct. Reduction gives CH₃CH₂N(CH₃)₂; nitrogen has three carbon substituents.

C. Incorrect. No fourth carbon substituent is added.

D. Incorrect. There are three carbon groups on N.

E. Incorrect. The amide is reduced rather than dehydrated.

Solution: Hydrolysis releases the amine-derived nitrogen fragment; the acid-derived chain is retained.

A. Incorrect. The acyl chloride retains a carbonyl oxygen.

B. Incorrect. It remains in the chain.

C. Correct. Hydrolysis releases the amine-derived nitrogen fragment; the acid-derived chain is retained.

D. Incorrect. It remains in the acyl chloride.

E. Incorrect. It remains next to C=O.

Solution: Ester enolate acylation gives ethyl acetoacetate, with one ketone and one ester carbonyl.

A. Incorrect. Claisen condensation does not reduce the ester to butanol.

B. Incorrect. No nitrogen reagent is present to form an amide.

C. Correct. Ester enolate acylation gives ethyl acetoacetate, with one ketone and one ester carbonyl.

D. Incorrect. This is unchanged ethyl acetate, not the condensation product.

E. Incorrect. This hydroxy aldehyde has neither the required ester nor the beta-keto ester framework.

Solution: The original amide carbonyl was reduced to CH₂; the new acetyl carbonyl is introduced by acylation.

A. Correct. The original amide carbonyl was reduced to CH₂; the new acetyl carbonyl is introduced by acylation.

B. Incorrect. That carbon is now part of a propyl group.

C. Incorrect. It stays a methyl substituent.

D. Incorrect. Elements are conserved.

E. Incorrect. Et3N serves as base and is not the intended acyl donor.

Solution: Ethyl substitution gives HO₂C-CH(Et)-CO₂H, then decarboxylation leaves Et-CH₂CO₂H.

A. Incorrect. This omits the malonate methylene carbon.

B. Incorrect. The ethyl substituent was lost incorrectly.

C. Incorrect. Malonic synthesis gives an acid, not a methyl ketone.

D. Incorrect. One extra carbon is retained incorrectly.

E. Correct. Ethyl substitution gives HO₂C-CH(Et)-CO₂H, then decarboxylation leaves Et-CH₂CO₂H.

Solution: The benzamide carbonyl stays attached to the ring through hydrolysis and activation.

A. Incorrect. No reduction to CH₂ occurs.

B. Incorrect. This moves nitrogen into a different connection.

C. Incorrect. That would remove the carbonyl carbon.

D. Incorrect. The stated route is not dehydration.

E. Correct. The benzamide carbonyl stays attached to the ring through hydrolysis and activation.

Solution: The final N has benzyl, methyl and acyl substituents with no N-H remaining.

A. Correct. The final N has benzyl, methyl and acyl substituents with no N-H remaining.

B. Incorrect. Neutral amide N normally has three sigma bonds.

C. Incorrect. The ring remains.

D. Incorrect. The displayed secondary amine has one.

E. Incorrect. It is retained.

Solution: Mono-methylation followed by beta-keto-acid decarboxylation gives a four-carbon methyl ketone.

A. Incorrect. This is an acid rather than the methyl-ketone endpoint.

B. Incorrect. This loses too many carbons and changes the endpoint class.

C. Incorrect. This requires ethyl rather than methyl alkylation.

D. Incorrect. This omits the introduced methyl carbon.

E. Correct. Mono-methylation followed by beta-keto-acid decarboxylation gives a four-carbon methyl ketone.

Solution: One amine attacks and another accepts the acid generated.

A. Incorrect. Neither follows the substitution balance.

B. Incorrect. One acyl group makes one amide.

C. Correct. One amine attacks and another accepts the acid generated.

D. Incorrect. Methylamine is not methanol.

E. Incorrect. That would require reduction after amide formation.

Solution: Two methyl additions to the ester give a tertiary alcohol after workup.

A. Incorrect. The ketone intermediate reacts further with excess Grignard.

B. Incorrect. This deletes the acyl carbon.

C. Correct. Two methyl additions to the ester give a tertiary alcohol after workup.

D. Incorrect. That is hydride reduction.

E. Incorrect. This omits the second addition.

Solution: Both terminal CO₂H groups become acyl chlorides, then N-methyl amides.

A. Incorrect. Both activated ends are supplied with excess amine.

B. Correct. Both terminal CO₂H groups become acyl chlorides, then N-methyl amides.

C. Incorrect. No carbonyl reduction occurs.

D. Incorrect. Only two acyl groups are present.

E. Incorrect. No new carbonyl carbons are added by methylamine.

Solution: OH replacement, cyanide C-C coupling and hydrolysis give CH₃CH₂CO₂H.

A. Incorrect. No carbon donor or acid-forming stage appears.

B. Incorrect. This returns an alcohol without homologation.

C. Incorrect. This does not supply cyanide carbon.

D. Correct. OH replacement, cyanide C-C coupling and hydrolysis give CH₃CH₂CO₂H.

E. Incorrect. That does not homologate the skeleton.

Solution: C=O becomes CH₂, giving CH₃CH₂NHCH₃ without cleaving either N-C bond.

A. Incorrect. No phenyl source is supplied.

B. Correct. C=O becomes CH₂, giving CH₃CH₂NHCH₃ without cleaving either N-C bond.

C. Incorrect. That is not the reduction pathway.

D. Incorrect. Amine reduction removes oxygen rather than making N-O.

E. Incorrect. The acyl carbon is retained.

Solution: The three-carbon acyl fragment becomes propanol, then cyanation adds one carbon.

A. Correct. The three-carbon acyl fragment becomes propanol, then cyanation adds one carbon.

B. Incorrect. This loses the original carbonyl carbon.

C. Incorrect. This adds two carbons instead of one.

D. Incorrect. That tracks a different fragment and invents a coupling.

E. Incorrect. The carbonyl was reduced before substitution.

Solution: The three-carbon alcohol framework gains labeled cyanide carbon that becomes the carboxyl carbon.

A. Incorrect. Hydrolysis retains nitrile carbon.

B. Incorrect. The isotope remains carbon.

C. Incorrect. Both count and placement are wrong.

D. Correct. The three-carbon alcohol framework gains labeled cyanide carbon that becomes the carboxyl carbon.

E. Incorrect. Only one new carbon enters.

Solution: The former carbonyl carbon is now CH₂, and both N-C bonds remain.

A. Correct. The former carbonyl carbon is now CH₂, and both N-C bonds remain.

B. Incorrect. This loses one carbon.

C. Incorrect. This removes nitrogen incorrectly for amide reduction.

D. Incorrect. The imine-like stage is reduced further.

E. Incorrect. That is unreduced starting amide.

Solution: CN replaces Br and adds one carbon to the benzylic chain.

A. Incorrect. That came from the original ester carbonyl.

B. Correct. CN replaces Br and adds one carbon to the benzylic chain.

C. Incorrect. They were present from the start.

D. Incorrect. Br is the leaving group.

E. Incorrect. It was removed during bromide formation.

Solution: Both retain nitrile carbon but produce different heteroatom functionality.

A. Correct. Both retain nitrile carbon but produce different heteroatom functionality.

B. Incorrect. Water adds none.

C. Incorrect. Elements are not converted.

D. Incorrect. The carbon is retained.

E. Incorrect. LiAlH₄ gives an amine.

Solution: Hydrolysis and hydride reduction preserve the C-C bond made by Grignard addition.

A. Incorrect. The ring came from the starting material.

B. Incorrect. Oxygen comes from hydrolysis water.

C. Incorrect. It supplies H, not carbon.

D. Correct. Hydrolysis and hydride reduction preserve the C-C bond made by Grignard addition.

E. Incorrect. It is retained.

Solution: The sequence gives propanol then propanal while retaining the acyl three-carbon chain.

A. Incorrect. That gives a tertiary alcohol with added carbons.

B. Incorrect. That produces an amide route.

C. Incorrect. An ordinary ester is not reduced by these mild conditions.

D. Correct. The sequence gives propanol then propanal while retaining the acyl three-carbon chain.

E. Incorrect. That is hydrolysis, not reduction to alcohol.

Solution: Alcohol oxygen replaces the leaving group; a nitrogen donor is required for the amide.

A. Incorrect. No dehydration of a nitrogen-containing precursor occurs.

B. Incorrect. No hydride is supplied.

C. Correct. Alcohol oxygen replaces the leaving group; a nitrogen donor is required for the amide.

D. Incorrect. They give different functional groups.

E. Incorrect. Acyl substitution retains it.