General Introduction of Carboxylic Acids and Their Derivatives Questions in English

(B) The structures of the given acids are:
$(A)$ Carbamic acid: $NH_2COOH$ (contains $-COOH$ group)
$(B)$ Barbituric acid: $C_4H_4N_2O_3$ (a cyclic amide,does not contain $-COOH$ group)
$(C)$ Lactic acid: $CH_3CH(OH)COOH$ (contains $-COOH$ group)
$(D)$ Succinic acid: $HOOCCH_2CH_2COOH$ (contains $-COOH$ group)
Therefore,barbituric acid is the correct answer as it is a pyrimidine derivative and not a carboxylic acid.

(C) dicarboxylic acid contains two carboxylic acid $(-COOH)$ functional groups.
Valeric acid $(CH_3(CH_2)_3COOH)$,Caproic acid $(CH_3(CH_2)_4COOH)$,and Butyric acid $(CH_3(CH_2)_2COOH)$ are monocarboxylic acids.
Glutaric acid has the structure $HOOC-(CH_2)_3-COOH$,which contains two $-COOH$ groups,making it a dicarboxylic acid.

(B) dicarboxylic acid contains two carboxyl $(-COOH)$ groups.
Malonic acid is $HOOC-CH_2-COOH$.
Glutaric acid is $HOOC-(CH_2)_3-COOH$.
Succinic acid is $HOOC-(CH_2)_2-COOH$.
Caproic acid is a monocarboxylic acid with the formula $CH_3(CH_2)_4COOH$,containing only one $-COOH$ group.
Therefore,the correct answer is $B$.

(A) mono carboxylic acid contains only one $-COOH$ group in its structure.
$1$. Phthalic acid $(C_6H_4(COOH)_2)$ contains two $-COOH$ groups attached to the benzene ring at ortho positions,making it a dicarboxylic acid.
$2$. Salicylic acid $(C_6H_4(OH)(COOH))$ contains one $-COOH$ group and one $-OH$ group,making it a mono carboxylic acid.
$3$. $o$-toluic acid $(C_6H_4(CH_3)(COOH))$ contains one $-COOH$ group,making it a mono carboxylic acid.
$4$. Benzoic acid $(C_6H_5COOH)$ contains one $-COOH$ group,making it a mono carboxylic acid.
Therefore,Phthalic acid is not a mono carboxylic acid.

(D) Monocarboxylic acids contain only one $-COOH$ group in their structure.
$A$. Malonic acid $(HOOC-CH_2-COOH)$ is a dicarboxylic acid,while propionic acid $(CH_3-CH_2-COOH)$ is a monocarboxylic acid.
$B$. Valeric acid $(CH_3(CH_2)_3COOH)$ is a monocarboxylic acid,while succinic acid $(HOOC-(CH_2)_2-COOH)$ is a dicarboxylic acid.
$C$. Acetic acid $(CH_3COOH)$ is a monocarboxylic acid,while adipic acid $(HOOC-(CH_2)_4-COOH)$ is a dicarboxylic acid.
$D$. Butyric acid $(CH_3(CH_2)_2COOH)$ and caproic acid $(CH_3(CH_2)_4COOH)$ both contain only one $-COOH$ group,so they are both monocarboxylic acids.

(C) Phthalic acid is benzene-$1,2$-dicarboxylic acid.
Its chemical structure consists of a benzene ring with two carboxylic acid $(-COOH)$ groups attached at the ortho positions.
Therefore,the number of $-COOH$ groups present in phthalic acid is $2$.

(A) An amide is an organic compound that contains the functional group $-CONH_2$.
In the given options,$CH_3CONH_2$ (acetamide) contains the $-CONH_2$ group,therefore it is an amide.
$CH_3CN$ is a nitrile,$CH_3NH_2$ is an amine,and $CH_3NO_2$ is a nitro compound.

(C) dicarboxylic acid contains two carboxyl $(-COOH)$ groups.
$1$. Valeric acid: $CH_3(CH_2)_3COOH$ (Monocarboxylic acid)
$2$. Caproic acid: $CH_3(CH_2)_4COOH$ (Monocarboxylic acid)
$3$. Glutaric acid: $HOOC(CH_2)_3COOH$ (Dicarboxylic acid)
$4$. Butyric acid: $CH_3(CH_2)_2COOH$ (Monocarboxylic acid)
Therefore,Glutaric acid is a dicarboxylic acid.

(B) The chemical formula for acetyl chloride is $CH_3COCl$.
Among the given options:
$A$ is acetamide $(CH_3CONH_2)$.
$B$ is acetyl chloride $(CH_3COCl)$.
$C$ is acetic anhydride $((CH_3CO)_2O)$.
$D$ is trichloroacetic acid $(CCl_3COOH)$.
Therefore,the correct option is $B$.

(D) tricarboxylic acid is an organic compound that contains three carboxyl groups $(-COOH)$.
$1$. Valeric acid $(CH_3(CH_2)_3COOH)$ is a monocarboxylic acid.
$2$. Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid.
$3$. Caproic acid $(CH_3(CH_2)_4COOH)$ is a monocarboxylic acid.
$4$. Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ is a tricarboxylic acid.
Therefore,the correct option is $D$.

(D) tricarboxylic acid is an organic compound that contains three carboxylic acid $(-COOH)$ functional groups.
$1$. Propionic acid $(CH_3CH_2COOH)$ is a monocarboxylic acid.
$2$. Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid.
$3$. Malonic acid $(HOOC-CH_2-COOH)$ is a dicarboxylic acid.
$4$. Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ contains three $-COOH$ groups,making it a tricarboxylic acid.
Therefore,the correct option is $D$.

(C) The compound formed by attaching two $-COOH$ groups to adjacent carbon atoms (ortho position) of a benzene ring is known as benzene$-1,2-$dicarboxylic acid,which is commonly called Phthalic acid.

(A) dicarboxylic acid is an organic compound containing two carboxylic acid functional groups $(-COOH)$.
Glutaric acid $(HOOC-(CH_2)_3-COOH)$ and Malonic acid $(HOOC-CH_2-COOH)$ are both dicarboxylic acids.
Valeric acid,Caproic acid,and Propionic acid are monocarboxylic acids.
Therefore,the correct pair is Glutaric acid and Malonic acid.

(D) Glycolic acid,also known as hydroxyacetic acid,has the chemical formula $HO-CH_2-COOH$.
It consists of a carboxylic acid group $(-COOH)$ attached to a hydroxymethyl group $(-CH_2OH)$.

(A) Methyl salicylate is an ester formed from salicylic acid ($2$-hydroxybenzoic acid) and methanol. \\ The $IUPAC$ name for the ester of salicylic acid with methanol is Methyl $2$-hydroxybenzoate. \\ Therefore,the correct option is $A$.

(D) $Methyl$ salicylate is also known as oil of wintergreen.
It is an organic ester with the chemical structure $C_6H_4(OH)(COOCH_3)$.

(B) Vinegar is a dilute solution of acetic acid $(CH_3 COOH)$.
Butter contains butyric acid $(CH_3 CH_2 CH_2 COOH)$,which is a four-carbon carboxylic acid.

(D) The main constituent of vinegar is acetic acid $\left( CH_3COOH \right)$.
It contains $5-8 \%$ acetic acid by volume.
Vinegar is mainly used in culinary arts.

(D) Salol is phenyl salicylate. Its chemical structure consists of a benzene ring substituted with a hydroxyl group $(-OH)$ at the ortho position and an ester group ($-COOC_6H_5$,which is a type of $-COOR$) at the adjacent position.
Therefore,the functional groups present in salol are $-OH$ (hydroxyl) and $-COOR$ (ester).

(A) Analyzing the given compounds:
$I$: $(CH_3CO)_2O$ is acetic anhydride.
$II$: $CH_3COOH$ is acetic acid (a carboxylic acid).
$III$: $PhOH$ is phenol.
$IV$: $CH_3COCHO$ is methylglyoxal (a dicarbonyl compound).
Therefore,the correct statement is that compound $I$ is an anhydride.