Aldehydes, Ketones and Carboxylic Acids
The total number of carboxylic acid groups is the product P is _________.
Explanation:
The reaction involved is
The number of carboxylic acid groups in the product is 2.
After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are)
The major product H of the given reaction sequence is
$C{H_3} - C{H_2} - CO - C{H_3}\buildrel {\overline C N} \over \longrightarrow G\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Heat}^{95\% \,{H_2}S{O_4}}} H$
The compound that undergoes decarboxylation most readily under mild condition is
The compound K is
The carboxyl functional group ($-$COOH) is present in
The number of aldol reactions that occur in the given transformation are
The number of optically active products obtained from the complete ozonolysis of the given compound is ______.
With reference to the scheme given, which of the given statement(s) about T, U, V and W is(are) correct?
Identify the binary mixtures that can be separated into individual compounds, by differential extraction, as shown in the given scheme.
The structure of compound P is
The structure of the compound Q is
In the scheme given below, the total number of intra molecular aldol condensation products formed from Y is ____________.
Explanation:
There is only one product formed in intramolecular aldol condensation, which is explained as follows:
Amongst the following, the total number of compounds soluble in aqueous NaOH is _________.
Explanation:
Aromatic alcohols and carboxylic acids forms salt with NaOH, will dissolve in aqueous NaOH :
Benzylic alcohol is less acidic than water and hence does dissolve in aqueous NaOH.
2PhCHO $\buildrel {\mathop {:OH}\limits^{\left( - \right)} } \over \longrightarrow $ PhCH2OH + $PhC\mathop O\limits^{..} $2(-)
the slowest step is :
Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) |  |
(P) | Nucleophilic substitution |
| (B) |  |
(Q) | Elimination |
| (C) |  |
(R) | Nucleophilic addition |
| (D) |  |
(S) | Esterification with acetic anhydride |
| (T) | Dehydrogenation |
The structure of the carbonyl compound P is
The structures of the products Q and R, respectively, are
The structure of the product S is
In the following reaction sequence, the correct structure of E, F and G are :
The structure of the product I is :
The structure of compounds J and K, respectively, are:
The structure of product L is :
Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is :
Match the compounds/ions in Column I with their properties/reactions in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $\mathrm{C_6H_5CHO}$ | (P) | gives precipitate with 2, 4-dinitrophenylhydrazine |
| (B) | $\mathrm{CH_3C\equiv CH}$ | (Q) | gives precipitate with $\mathrm{AgNO_3}$ |
| (C) | $\mathrm{CN^-}$ | (R) | is a nucleophile |
| (D) | $\mathrm{I^-}$ | (S) | is involved in cyanohydrin formation |
Statement 1 : p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.
Statement 2 : o-Hydroxybenzoic acid has intramolecular hydrogen bonding.
(A) HCHO
(B) CH3COCH3
(C) PhCOCH3
(D) PhCOPh
Which of the following reactants on reaction with conc. NaOH followed by acidification gives the following lactone as the only product?
The conversion of compound B to lactone is shown below:
The conversion of compound C to lactone is shown below:
Step 2:
Step 3:
Note: Option B is not possible as it on reaction with conc. NaOH undergoes intermolecular cannizaro's reaction with possibility of forming pthalic acid as one of the products.
$ \text { Step 2: } $
$ \text { Step 3: } $
Two different oximes are formed.
Three different oximes are formed.
Two oximes are optically active.
All oximes are optically active.
The oxime formed is optically inactive as there is plane of symmetry present in it.
An oxime will be obtained which shows two enantiomers based on the position of bulky group, $\mathrm{CH}_3 \mathrm{CH}_2$ with respect to the OH group attached to N atom of the oxime.
The two enantiomers are optically active as they lack a plane of symmetry and exist as $R$ and S isomers.
