Carboxylic Acids and Its Derivatives

The complete reaction sequence is as follows

$ \text { Thus } X \text { and } Y \text { are } \mathrm{CH}_3 \mathrm{Cl} \text { and } \mathrm{CH}_3 \mathrm{COCl} \text {. } $

The complete reaction sequence is as follows

Thus $A$ and $B$ respectively are $\mathrm{Na} / \mathrm{Hg}$, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and DIBAL- $\mathrm{H}_2 \mathrm{H}_2 \mathrm{O}$.

Carboxylic acid having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give $\alpha$-halocarboxylic acids. This reaction is called Hell-Volhard Zelinsky reaction. Thus among the given compounds,

$ \text { will undergo this reaction is } $

The complete reaction involved is

The correct structures of succinic acid $(x)$ and malonic acid $(y)$ respectively are

$ \text { The complete reaction sequence is as follows } $

$ \text { The complete reaction sequence is as follows } $

$ \text { The complete reaction sequence is as follows } $

$ \text { The complete reaction is as follows, } $

$ \begin{aligned} & X=\mathrm{KMnO}_4 / \mathrm{H}^{+}, Y=\mathrm{SOCl}_2 \\ & Z=\text { Acetophenone } \end{aligned} $

$ \text { The complete reaction is as follows, } $

Among the given options, most acidic carboxylic acid is HCOOH .

This is due to the sability of its conjugate base, the formate ion $\left[\mathrm{HCOO}^{-}\right)$.

The complete reaction sequence is as follows

Thus, the correct decreasing order of their acid strength is II $>$ I $>$ III.

The complete reaction is as follows,

$ \underset{\text { Formic add }}{ \mathrm { HCOOH }} \xrightarrow[\Delta]{\mathrm{H}_{2} \mathrm{SO}_{4}} \underset{\text { (Product } X \text { ) }}{\mathrm{H}_{2} \mathrm{O}}+\underset{\text { (Product } Y \text { ) }}{\mathrm{CO}} $

CO has 3 bonds, $ 1 \sigma $ bond and $ 2 \pi $ bonds.

$ \mathrm{H}_{2} \mathrm{O} $ has 2 bonds, both are $ \sigma $ bonds. Hence, $ X=2,0 $ and $ Y=1,2 $

The complete reaction sequence is as follows,

The complete reaction sequence is as follows

$\mathrm{NH}_2 \mathrm{CONH}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{(X)}{\left[\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3\right]} \rightleftharpoons 2 \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}+\underset{(Y)}{\mathrm{CO}_2} $

So, the hybridisation of carbon in compound $X$ and $Y$ are $s p^2$ and $s p$ respectively.

The complete reaction is as follows.

The reaction provided in option (d) results in the formation of a primary alcohol.

The specific reaction is as follows:

$ \mathrm{CH}_3 \mathrm{COOH} \xrightarrow[\text{(ii) } \mathrm{H}_2 \mathrm{O}]{\text{(i) } \mathrm{LiAlH}_4 \text{ /ether }} \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} $

This procedure leads to the production of ethanol (a primary alcohol), not acetaldehyde.

Compound $ Y $, which is the desired product, is acetaldehyde $\left(\mathrm{CH}_3 \mathrm{CHO}\right)$. Therefore, the reaction outlined in option (d) cannot be used to produce compound $ Y $.

Both A and R are correct and R is the correct explanation of A.

$ \text { The reaction involved is as follows. } $

The complete reaction is as follows,

The complete reaction is as follows,

When glucose is oxidised with nitric acid, saccharic acid is formed.

$ \text { The conversion of } X \text { to } Y \text { is called as carbylamine reaction. } $

When ether is reduced with DIBAL-H, aldehyde is formed

When ether is reacted with excess $\mathrm{H}_2$, then alcohol is formed.

Hence, $X=\mathrm{H}_2 /$ catalyst, $Y=$ DIBAL-H (ii) $\mathrm{H}_2 \mathrm{O}$.

When 2-methylhexane is passed over $\mathrm{Cr}_2 \mathrm{O}_3$ at 773 K , toluene is formed as $A$.

$A$ reacts with $\mathrm{KMnO}_4 / \mathrm{OH}^{-}$followed by acidic hydrolysis to give Bi.e. benzoic acid.

This benzoic acid undergoes nitration in presence of mixture of conc. $\mathrm{H}_2 \mathrm{SO}_4$ and conc. $\mathrm{HNO}_3$ to give $m$-nitrobenzoic acid $(C)$.

Smaller the $\mathrm{p} K_a$ stronger the acid. The correct order of $\mathrm{p} K_a$ is

Electron vithdrawing group increases acidic strength. So, $\mathrm{CF}_3 \mathrm{COOH}$ has least $\mathrm{pK} K_a$ while $\mathrm{NCCH}_2 \mathrm{COOH}$ has highest $\mathrm{p} K_a$ value.

(i), (ii) 2-hexanone undergoes haloform reaction to form bromoform and a carboxylate ion. This carboxylate ion undergoes hydrolysis to give pentanoic acid.

(iii), (iv) Pentanoic acid react with bromine In presence of red-phosphorus to give $\alpha$-bromocarboxylic acid. This reaction is Hell-Volhard-Zelinsky reaction.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} \xrightarrow[\text { (iv) } \mathrm{H}_2 \mathrm{O}]{\text { (iii) Red } \mathrm{Br}_2} $

Phenol are weaker acid than carboxylic acids. So, phenolic -OH group does not react with $\mathrm{NaHCO}_3$. Carboxylic acids react with $\mathrm{NaHCO}_3$ to give brisk effervescences of carbon dioxide gas.

Hence, compounds III and IV only react with $\mathrm{NaHCO}_3$.

$n$-alcohol on reaction with hydrohalide gives alkyl halide. Further, reaction of alkyl halide with KCN followed by hydrolysis leads to formation of acid which have 1 carbon atom more than parent compound. The reaction involved is as follows

The reagent that can reduce the carboxylic acid to corresponding alcohol is diborane (disassociated to $\mathrm{BH}_3$ in ether or THF solution)

$\mathrm{NaBH}_4$ being weak reducing agent does not reduce carboxylic acids but reduces aldehyde and ketones.

The reaction of aldehydes and ketones with zinc amalgam ( $\mathrm{Zn} / \mathrm{Hg}$ alloy) in concentrated hydrochloric acid, reduces the aldehyde or ketone to a hydrocarbon, is called Clemmensen reduction. It does not reduce carboxylic acid.

$\mathrm{H}_2$ over $\mathrm{Pd} / \mathrm{C}$ will reduce alkenes and alkynes all the way to alkanes.

Carboxylic acid can be tested by treating with sodium hydrogen carbonate. The brisk effervescences of $\mathrm{CO}_2$ will confirm the test. Further by passing it through lime water, the milky solution too confirms the test.

$\mathrm{Ar}-\mathrm{COOH}+\mathrm{NaHCO}_3 \longrightarrow \mathrm{ArCOO}^{\ominus} \mathrm{Na}^{\oplus} +\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}$

Styrene reacts with HBr in presence of peroxide to add Br at less substituted carbon. This reaction follows anti-Markownikoff rule.

Product $(X)$ reacts with Mg in dry ether to form Grignard reagent which on treatment with $\mathrm{CO}_2$ followed by hydrolysis gives carboxylic acid.

$Y$ on treatment with $\mathrm{PCl}_5$ form acyl chloride. This acyl chloride on reduction with $\mathrm{H}_2$; $\mathrm{Pd}-\mathrm{BaSO}_4$ gives aldehyde and this reaction is known as "Rosenmund reaction."

Hence, Z is

Carboxylic acids are the most acidic compounds among the organic compounds. Due to presence of electron withdrawing group i.e. chlorine and fluorine, $\mathrm{FCH}_2 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOOH}$ are more acidic, than $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COOH}$. Fluorine is more electron withdrawing group than chlorine.

Therefore, $\mathrm{\mathop {FC{H_2}COOH}\limits_{Fluoro\,ethanoic\,acid} > \mathop {C{H_3}C{H_2}CHClCOOH}\limits_{2 - chlorobu\tan oic\,acid} > \mathop {{C_6}{H_5}COOH}\limits_{Benzoic\,acid} > \mathop {C{H_3}COOH}\limits_{Acetic\,acid}}$

$ \text { In the given reaction, the final product is as follows : } $

Here, $-\mathrm{CH}_3$ will make bond with -COO of ring. Hence, we get the product as shown. Hence, option (d) is the correct answer.

$ \text { In the given reaction, the final product is as follows : } $

In step (i), we get oxidation product and in step (ii), when the product is treated with $\mathrm{Br}_2$ in presence of $\mathrm{FeBr}_3$ and $\mathrm{H}_3 \mathrm{O}^{+}$, we get meta-substituted substituted bromo-product as shown in the reaction. Hence, option (d) is the correct answer.

$ \text { The product }(R) \text { in the given reaction sequence is } $

$ \text { Which after using sodalime gives, } $

$ \text { Hence, option (b) is the correct answer. } $

$ \text { In the above } P, Q \text { and } R \text { are respectively } $