Aldehydes, Ketones and Carboxylic Acids

$ \begin{aligned} &\begin{aligned} & \text { Degree of unsaturation }=(8+1)-\left(\frac{8}{2}\right) \\ & =9-4=5 \end{aligned}\\ &\text { In question it is mentioned that oxidation under drastic condition, therefore it must be ketone } \end{aligned} $

$ \begin{aligned} &\text { }\\B.\,\,\, &\mathrm{CH}_3 \mathrm{COOH} \xrightarrow[\text { (ii) } \mathrm{H}_2 \text {, catalyst }]{\text { (i) } \mathrm{CH}_3 \mathrm{OH}, \mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \end{aligned} $

$ \text { So, correct Answer is : Acetate, } \mathrm{CH}_3 \mathrm{COO}^{-} $

$ \text { Therefore, compound } 2 \text { cannot be obtained by aldol condensation of given compounds. } $

Esters are reduced to aldehydes with DIBAL-H

Electron donating group decreases the acidity of carboxylic acids.

So correct order is

$\mathrm{HCOOH}>\mathrm{CH}_3 \mathrm{COOH}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCOOH}>\left(\mathrm{CH}_3\right)_3 \mathrm{CCOOH}$

Functional groups present in product are alcohol and ketone.

Rosenmund reaction

It is an example of nucleophilic addition reaction

$\bullet$ Gabriel phthalimide synthesis is used for preparation of aliphatic primary amines.

$\bullet$ Kolbe synthesis with phenol gives salicylic acid

$\bullet$ Williamson synthesis gives ether on reaction of alkyl halide and alcoxide

$\bullet$ Etard reaction gives benzaldehyde from benzene

$\bullet$ The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses due to weak molecular association in aldehydes and ketones arising out of the dipole - dipole interaction.

$\bullet$ Alcohols involved intermolecular hydrogen bonding, because of which the boiling point of aldehydes and ketones are lower than the alcohols of similar molecular masses.

Here Y is RCOO^$-$Mg⁺X

(a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)

The given reaction is the reductive ozonolysis of an alkene. The alkene will be

Cross Aldol condensation reaction:

Both reactants contain $\alpha$-Hydrogens, so multiple products are possible which are as follows:

The given reaction (as per the options) is crossed Cannizzaro reaction. A simple Cannizzaro reaction is shown by those aldehydes in which $\alpha$-hydrogen is absent. These aldehydes in the presence of a strong base like $\mathrm{NaOH}$ show disproportionation, i.e. simultaneous oxidation and reduction of the same molecule. Thus, the result includes an acid alongwith an alcohol. In crossed Cannizzaro reaction however, molecules of two different can participate, out of which the one more prone to nucleophilic attack will show oxidation, while the other show reduction. Here, out of the given two substrates, $\mathrm{HCHO}$ is more prone to nucleophilic attack hence, this will undergo oxidation while $\mathrm{CH}_3 \mathrm{CHO}$ will undergo reduction. Hence, the possible products are $\mathrm{HCOO}^{-} \mathrm{Na}^{+}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$.

Note Same combination of substrates give crossed aldol as well, in the presence of $\mathrm{NaOH}$ to form

But as, it is not among the options as the product, hence we are considering option (c) as the most appropriate option.

Assertion:

Anhydrides are more reactive than esters for nucleophilic substitution.

Reason:

$ RCOO^- $ is a better leaving group than $ \mathrm{R}-\mathrm{O}^{-} $.

Explanation:

In nucleophilic acyl substitution reactions, the reactivity of the compounds depends on the leaving group's quality. A good leaving group is one that can stabilize the negative charge well when it departs from the molecule.

Anhydrides contain the $ RCOO^- $ group, while esters contain the $ \mathrm{R}-\mathrm{O}^{-} $ group. The $ RCOO^- $ group is more stable due to resonance stabilization, which disperses the negative charge over two oxygen atoms. This makes $ RCOO^- $ a better leaving group compared to $ \mathrm{R}-\mathrm{O}^{-} $, which does not have such extensive resonance stabilization. Therefore, anhydrides are more reactive in nucleophilic substitution reactions than esters.

Conclusion:

Both the assertion and the reason are true, and the reason provides the correct explanation for the assertion.

Correct Option: A

The ease of nucleophilic substitution depends upon the nature of leaving group. When the leaving tendency of a group in a compound is high, then the compound is more reactive towards nucleophilic substitution. The nucleophilic acyl substitution is completed in two steps as shown below.

The order of leaving tendency is $\mathrm{Cl}^{-} > \mathrm{RCOO}^{-} > \mathrm{RO}^{-} > \mathrm{NH}_2^{-}$ and therefore, the order of reactivity of acyl compound is as

Necessary condition for aldol condensation is the presence of atleast one $\alpha$-H atom. Thus,

among the given options,

do not undergoes aldol condensation as it does not contains any $\alpha$-hydrogen atom.