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JEE Advanced Aldehydes, Ketones and Carboxylic Acids — practice questions — page 4

Explore 122 free Aldehydes, Ketones and Carboxylic Acids questions in JEE Advanced Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Aldehydes, Ketones and Carboxylic Acids questions and exercises with answers

Showing questions 91–120 of 122.

**Reimer–Tiemann reaction** of phenol with $CHCl_3$ and aq. $NaOH$, followed by hydrolysis, givesRead the answer and worked solution →The reactive electrophile in the Reimer–Tiemann reaction isRead the answer and worked solution →A **cross aldol** condensation between benzaldehyde and acetone (excess acetone) gives, as the major product,Read the answer and worked solution →In the **Etard reaction**, toluene is oxidised to benzaldehyde byRead the answer and worked solution →**Clemmensen reduction** converts $>C=O$ to $>CH_2$ usingRead the answer and worked solution →**Sandmeyer's reaction** converts $ArN_2^+X^-$ to $ArCl$ usingRead the answer and worked solution →Sandmeyer's reaction proceeds via which type of intermediate?Read the answer and worked solution →The **Gattermann reaction** (not Gatterman–Koch) converts $ArN_2^+X^-$ to $ArX$ usingRead the answer and worked solution →The Wurtz–Fittig reaction between $C_6H_5Br$ and $CH_3CH_2Br$ with Na in dry ether givesRead the answer and worked solution →The **Balz–Schiemann reaction** converts $ArN_2^+$ to $ArF$ via which intermediate?Read the answer and worked solution →Friedel–Crafts alkylation of benzene with $1$-chloropropane and $AlCl_3$ gives, as the major product,Read the answer and worked solution →A **Grignard reagent** has the general formulaRead the answer and worked solution →**Stephen's reduction** converts a nitrile $RCN$ into an aldehyde usingRead the answer and worked solution →In the **cumene process** for industrial phenol production, the by-product formed alongside phenol isRead the answer and worked solution →The **Swarts reaction** converts an alkyl chloride to an alkyl fluoride usingRead the answer and worked solution →**Cannizzaro's reaction** is given by aldehydes that haveRead the answer and worked solution →**Diazotisation** of a primary aromatic amine uses $NaNO_2 + HCl$ atRead the answer and worked solution →Which reaction is being described?Read the answer and worked solution →The reaction of $CH_3MgBr$ with formaldehyde ($HCHO$), followed by acidic workup, givesRead the answer and worked solution →**Lucas reagent** is a mixture ofRead the answer and worked solution →In **Wurtz's reaction**, two alkyl halides couple in the presence ofRead the answer and worked solution →Self-aldol condensation of acetaldehyde, after dehydration, givesRead the answer and worked solution →**Rosenmund's reduction** converts an acyl chloride to an aldehyde usingRead the answer and worked solution →The **Hell–Volhard–Zelinsky (HVZ) reaction** converts a carboxylic acid (with $\alpha$-H) to an $\alpha$-halo acid usingRead the answer and worked solution →Identify the correct conversion: $CH_3CN \xrightarrow{?} CH_3CHO$ (one step).Read the answer and worked solution →In **Dow's process**, chlorobenzene is converted to phenol by treatment withRead the answer and worked solution →**Gatterman–Koch reaction** formylates benzene usingRead the answer and worked solution →Among $1$-butanol, $2$-butanol and $tert$-butanol, the order of reactivity in the Lucas test isRead the answer and worked solution →Match each named organic reaction to its description.Read the answer and worked solution →The product of the Hunsdiecker reaction on $CH_3CH_2COOAg$ with $Br_2/CCl_4$ isRead the answer and worked solution →

Worked practice questions

**Reimer–Tiemann reaction** of phenol with $CHCl_3$ and aq. $NaOH$, followed by hydrolysis, gives

  1. Salicylaldehyde
  2. Benzoic acid
  3. Benzaldehyde
  4. Salicylic acid
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Correct answer: A. Salicylaldehyde

Reimer–Tiemann installs a $-CHO$ group ortho to phenol: $C_6H_5OH + CHCl_3 + NaOH \to o\text{-hydroxybenzaldehyde}$. The reactive intermediate is dichlorocarbene $:CCl_2$.

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The reactive electrophile in the Reimer–Tiemann reaction is

  1. $Cl^-$
  2. $OH^-$
  3. Dichlorocarbene $:CCl_2$
  4. $CHCl_3$ itself
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Correct answer: C. Dichlorocarbene $:CCl_2$

$CHCl_3 + OH^- \to :CCl_2 + Cl^- + H_2O$. The carbene attacks the ortho carbon of phenoxide. Subsequent hydrolysis of the geminal dichloride gives $-CHO$.

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A **cross aldol** condensation between benzaldehyde and acetone (excess acetone) gives, as the major product,

  1. Acetophenone
  2. 4-Phenyl-3-buten-2-one
  3. Dibenzalacetone (only)
  4. Cinnamaldehyde
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Correct answer: B. 4-Phenyl-3-buten-2-one

Acetone has $\alpha$-H, benzaldehyde does not (so benzaldehyde cannot form an enolate — only acetone can). Excess acetone gives mono-condensation product $C_6H_5CH=CHCOCH_3$ (4-phenyl-3-buten-2-one). With excess benzaldehyde, both methyls of acetone react to give dibenzalacetone.

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