Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option C (Aldehyde) is correct because the Wacker method is used to convert alkenes into corresponding aldehydes.

Option A is correct because diborane ($B_2H_6$) readily reduces carboxylic acids to primary alcohols.

Option B is correct because the Lucas test is specific to alcohols and is used to differentiate between primary, secondary, and tertiary alcohols, not aldehydes.

Option D (Clemmensen reduction) is correct because the reaction described is the Clemmensen reduction, which reduces a ketone to an alkane using zinc amalgam and concentrated HCl.

Option A is correct because the conversion of benzaldehyde to 3-phenylprop-2-en-1-oic acid (cinnamic acid) using acetic anhydride in the presence of a base is the Perkin condensation.

The correct answer is B because both A and R are true, but R is not the correct explanation of A. The acidity is due to the -I effect of chlorine, not the +R effect.

Option A ($\mathrm{C}{\mathrm{H}_3} - \mathrm{COOH}$) is correct because carboxylic acids are already in a high oxidation state and are generally resistant to further oxidation.

Option B (Formic acid) is correct because formic acid, when heated with ${{\mathrm{P}_2}{\mathrm{O}_5}}$, dehydrates to produce carbon monoxide, which burns with a pale blue flame.

Option D (Assertion (A) and Reason (R) both are false) is correct because formic acid is not the strongest monocarboxylic acid, and trifluoroacetic acid is stronger than formic acid.

Option C (10% acetic acid) is correct because pyroligneous acid, obtained from the destructive distillation of wood, contains approximately 10% acetic acid.

Option C is correct. The assertion is correct because electron-withdrawing groups increase acidity, but the reason is incorrect because chlorine has a -I effect, not a +M effect.

Option B (${Cl}_3 {C} . {COOH}>{Cl}_2 {CH} \cdot {COOH}> Cl. {CH}_2 {COOH}>{CH}_3 {COOH}$) is correct because the acidity increases with the number of chlorine atoms due to the electron-withdrawing inductive effect (-I effect) of chlorine.

The correct answer is A (Both statement I and II are correct) because aldehydes can be oxidized to carboxylic acids, and carboxylic acids can be reduced to alcohols using $LiAlH_4$.

Option C ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{COCH}_3$) is correct because both compounds give a positive iodoform test.

Option B (It is reducing agent) is correct because formic acid can reduce Tollen’s reagent, Fehling solution, mercuric chloride and $\mathrm{KMnO}_4$ etc.

The correct answer is D (All of the above) because the formation of A is a nucleophilic addition, B is optically active, and the intermediate in step I is a hydroxy alkoxide ion.

Option D (It does not reduce Tollens' reagent) is correct because formic acid (HCOOH) *does* reduce Tollens' reagent due to the presence of an aldehyde-like group.

Option C (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CHO}$) is correct because benzaldehyde undergoes Cannizzaro's reaction, reduces Schiff's reagent, but does not reduce Fehling's solution.

Option C (Schiff’s base) is correct because imine derivatives of aldehydes and ketones are known as Schiff's bases.

Option D (Hexanal) is correct because as the size of the aldehyde molecule increases, the odor becomes less pungent and more fragrant.

Option A (C6H5CHO) is correct because benzaldehyde (C6H5CHO) lacks alpha-hydrogens and undergoes Cannizzaro's reaction with 50% NaOH to give alcohol and acid.

Option D (Ester) is correct because a silver salt of a fatty acid reacts with an alkyl halide to form an ester in the Hunsdiecker reaction.

Option C is correct because the electrophile involved in the Reimer-Tiemann reaction is dichlorocarbene (:${CC{l_2}}$).

Option A is correct because the acidic strength decreases as the alkyl chain length increases due to the +I inductive effect, so ethanoic acid is the strongest.

Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the reason for benzaldehyde's lower reactivity is due to resonance stabilization, not solely the hybridization of its carbon atoms.

Option D (Urotropine) is correct because the reaction of ammonia and formaldehyde produces hexamethylenetetramine, also known as urotropine.

Option A (I > II > III) is correct because the rate of decarboxylation depends on the stability of the carbanion formed, which is influenced by the -I effect of the halogens.

Option D (All of these) is correct because $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ and ${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COOH}$ differ in physical state, solubility, and flame characteristics.

Option C ($\mathrm{p - }$ rosaniline hydrochloride) is correct because Magenta is $\mathrm{p - }$ rosaniline hydrochloride, a dye that is decolorized by ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}$ to give Schiff’s reagent.

Option A (3 – Pentanone) is correct because 3-pentanone (a ketone) forms phenyl hydrazone, gives negative Tollen's and iodoform tests, and yields n-pentane upon reduction.