Mix Examples of 8-1.Organic Chemistry : Classification and Nomenclature of Organic compounds Questions in English

(D) Tetraethyl lead $(Pb(C_2H_5)_4)$ is added to gasoline as an antiknock agent.
It prevents the premature ignition of the fuel-air mixture in the engine,thereby increasing the antiknock property of the fuel.

(D) Glycerol is not used as an antiseptic agent. It is primarily used in the manufacture of nitroglycerin (explosive),in cosmetics and soaps to prevent drying,and as an antifreeze in automobile radiators.

(D) The enol content is highest in acetyl acetone $(CH_3COCH_2COCH_3)$ because it forms an intramolecular hydrogen-bonded stable six-membered ring in its enolic form.
Additionally,the enol form is stabilized by conjugation between the $C=C$ double bond and the $C=O$ carbonyl group.
The percentage of the enolic form in acetyl acetone is approximately $76\%$,which is significantly higher than in simple ketones like acetone or acetophenone.

(D) The reaction between ammonium chloride $(NH_4Cl)$ and potassium cyanate $(KCNO)$ is a classic synthesis of urea.
The chemical equation is:
$NH_4Cl + KCNO \rightarrow NH_4CNO + KCl$
Upon heating,ammonium cyanate $(NH_4CNO)$ undergoes rearrangement to form urea $(H_2NCONH_2)$:
$NH_4CNO \xrightarrow{\Delta} H_2NCONH_2$
Thus,the final product is urea $(H_2NCONH_2)$.

(D) The structure of allyl isocyanide is $CH_2=CH-CH_2-N\rightleftharpoons C$.
Counting the bonds:
- $C-H$ bonds: $5$
- $C-C$ bonds: $2$
- $C-N$ bond: $1$
- $N-C$ bond: $1$
Total sigma bonds = $5 + 2 + 1 + 1 = 9$.
- $C=C$ bond: $1$ pi bond
- $N\rightleftharpoons C$ bond: $2$ pi bonds
Total pi bonds = $1 + 2 = 3$.
- Non-bonding electrons: The nitrogen atom has $1$ lone pair ($2$ electrons) and the terminal carbon atom has $1$ lone pair ($2$ electrons). However,in the coordinate bond representation $N\rightleftharpoons C$,the terminal carbon has $2$ non-bonding electrons.
Thus,it has $9$ sigma bonds,$3$ pi bonds,and $2$ non-bonding electrons.

(C) To identify the presence of $1^o, 2^o, 3^o$ and $4^o$ carbon atoms,let us analyze the structure of $2,3,3$-trimethylpentane:
The structure is $CH_3-CH(CH_3)-C(CH_3)_2-CH_2-CH_3$.
$1^o$ carbons are terminal methyl groups (there are $5$ such carbons).
$2^o$ carbon is the $CH_2$ group at position $4$.
$3^o$ carbon is the $CH$ group at position $2$.
$4^o$ carbon is the central carbon at position $3$ which is bonded to four other carbons.
Thus,$2,3,3$-trimethylpentane contains all four types of carbon atoms.

(A) The keto form of acetone is $CH_3COCH_3$. The enolic form is obtained by the migration of an $\alpha$-hydrogen to the oxygen atom,resulting in $CH_3-C(OH)=CH_2$.
In $CH_3-C(OH)=CH_2$:
$1$. Total $\sigma$ bonds: $3$ (in $CH_3$) + $1$ ($C$-$C$) + $1$ ($C$-$O$) + $1$ ($O$-$H$) + $2$ ($C$-$H$ in $CH_2$) + $1$ ($C$=$C$) = $9\sigma$ bonds.
$2$. Total $\pi$ bonds: $1$ (in $C=C$ double bond) = $1\pi$ bond.
$3$. Lone pairs: Oxygen has $2$ lone pairs in the keto form,but in the enolic form,the oxygen atom involved in the $C-OH$ group retains $2$ lone pairs. However,looking at the structure $CH_3-C(OH)=CH_2$,the oxygen atom has $2$ lone pairs. Re-evaluating the options,the standard representation for this specific question often considers the lone pairs on the oxygen atom. Given the structure,there are $2$ lone pairs on the oxygen atom.

(A) The reaction between $4$-methylbenzenesulfonic acid $(CH_3-C_6H_4-SO_3H)$ and sodium acetate $(CH_3COONa)$ is an acid-base reaction. The sulfonic acid is a stronger acid than acetic acid,so it protonates the acetate ion to form acetic acid $(CH_3COOH)$ and the corresponding sodium sulfonate salt $(CH_3-C_6H_4-SO_3Na)$.

(B) . Friedrich $W$öhler synthesized the first organic compound,urea,in the laboratory in $1828$ by heating ammonium cyanate $(NH_4OCN)$.

(C) The correct answer is $(C)$ $CH_3COCH_2COCH_3$.
In $CH_3COCH_2COCH_3$ (acetylacetone),the enol form is significantly stabilized by intramolecular hydrogen bonding and resonance,forming a stable six-membered ring. This makes the enol content much higher compared to simple aldehydes or ketones.

(A) secondary carbon atom is a carbon atom bonded to exactly two other carbon atoms.
$1$. Benzene $(C_6H_6)$: All $6$ carbon atoms are bonded to two other carbon atoms within the ring. Thus,it has $6$ secondary carbon atoms.
$2$. $o$-Xylene $(C_8H_{10})$: The ring has $6$ carbons. Two carbons are bonded to methyl groups (tertiary carbons),and the other $4$ carbons are bonded to two other ring carbons (secondary carbons). Thus,it has $4$ secondary carbon atoms.
$3$. Toluene $(C_7H_8)$: The ring has $6$ carbons. One carbon is bonded to a methyl group (tertiary carbon),and the other $5$ carbons are bonded to two other ring carbons (secondary carbons). Thus,it has $5$ secondary carbon atoms.
Therefore,the number of secondary carbon atoms are $6, 4, 5$ respectively.

(A) The degree of unsaturation $(DU)$ is calculated using the formula: $DU = C + 1 - \frac{H}{2} - \frac{X}{2} + \frac{N}{2}$,where $C$ is the number of carbon atoms,$H$ is the number of hydrogen atoms,$X$ is the number of halogen atoms,and $N$ is the number of nitrogen atoms.
Given the molecular formula $C_{20}H_{22}O_2$:
$C = 20$,$H = 22$,$X = 0$,$N = 0$.
Substituting these values into the formula:
$DU = 20 + 1 - \frac{22}{2} = 21 - 11 = 10$.
Wait,checking the provided options,if the compound is a specific structure like a steroid or complex polycyclic system,let us re-evaluate. If the question implies a specific structure not provided,we assume the formula $C_{20}H_{22}O_2$. Based on the calculation,$DU = 10$. However,if the intended answer is $10$ and it is not in the options,there might be a typo in the question's formula. Given the options provided,if we assume the question refers to a specific structure where $DU = 10$,but the options are $17-21$,the question is likely flawed. Assuming the intended answer is $10$ based on the formula $C_{20}H_{22}O_2$.

(D) Let us analyze the given chemical structures:
$1$. Structure $P$ is $OHC-CH_2-CH=CH-CH_2-CHO$,which is a dialdehyde (hex$-3-$enedial).
$2$. Structure $Q$ is $HO-CH_2-CH_2-CH_2-CH_2-CH_2-CH_2-OH$,which is a diol (hexane$-1,6-$diol).
$3$. Structure $R$ is $HO-CH_2-CH_2-CH=CH-CH_2-CHO$,which contains both a hydroxyl group $(-OH)$ and an aldehyde group $(-CHO)$,making it a hydroxy aldehyde.
Since statements $(B)$ and $(C)$ are both correct,the correct option is $(D)$.

(D) $1$. The molecular formula $C_3H_6O$ corresponds to the degree of unsaturation $U = 3 - (6/2) + 1 = 1$. This indicates the presence of one double bond or one ring.
$2$. The compound does not react with $2,4-$Dinitrophenylhydrazine ($2,4-DNP$), which means it does not contain a carbonyl group (aldehyde or ketone).
$3$. The compound does not react with metallic sodium, which means it does not contain an active hydrogen atom (like in alcohols or phenols).
$4$. Given the formula $C_3H_6O$, the structure $CH_2=CH-O-CH_3$ (methyl vinyl ether) fits all criteria: it has no carbonyl group, no hydroxyl group, and satisfies the molecular formula.

(B) $1$. $n$-butane $(CH_3CH_2CH_2CH_3)$ and isobutane $(CH_3CH(CH_3)CH_3)$ have the same molecular formula $(C_4H_{10})$ but different carbon skeleton structures,so they are chain isomers. Thus,option $A$ is correct.
$2$. Homologues are members of a series having the same functional group and similar chemical properties,differing by a $-CH_2-$ unit. $n$-butane and isobutane are isomers,not homologues. Thus,option $B$ is incorrect.
$3$. The $IUPAC$ name for $CHCl_3$ is $trichloromethane$ (or $1, 1, 1-$trichloromethane). Thus,option $C$ is correct.
$4$. $n$-pentane and neopentane have the same molecular formula $(C_5H_{12})$ but different carbon chains,so they are chain isomers. Thus,option $D$ is correct.

(D) $1$. Analyze the structure of the given compound.
$2$. The structure contains a $C=O$ group attached to two carbon atoms,which is a $Ketone$ group.
$3$. It contains a $-CONH-$ group,which is an $Amide$ group.
$4$. It contains a $-COO-$ group within a ring,which is an $Ester$ group.
$5$. There is no $Ether$ group ($-O-$ linkage between two carbon atoms) present in the structure.
$6$. Therefore,the functional group $NOT$ present is $Ether$.

(C) The degree of unsaturation (DoU) is calculated by counting the number of rings and $\pi$ bonds in the structure.
$1$. Rings: There are $3$ rings (one benzene ring and two saturated six-membered rings).
$2$. $\pi$ bonds: There is $1$ triple bond ($2 \pi$ bonds),$2$ double bonds ($2 \pi$ bonds),and $1$ carbonyl group ($1 \pi$ bond).
Total $\pi$ bonds $= 2 + 2 + 1 = 5$.
Total DoU $= \text{Number of rings} + \text{Number of } \pi \text{ bonds} = 3 + 5 = 8$.
Wait,re-evaluating the structure: The benzene ring contributes $4$ ($1$ ring + $3$ $\pi$ bonds). The other two rings contribute $2$. The triple bond contributes $2$. The double bond contributes $1$. The carbonyl group contributes $1$. Total $= 4 + 2 + 2 + 1 + 1 = 10$.
Therefore,the correct option is $C$.

(D) The formula for amoxycillin is $C_{16}H_{19}N_3O_5S$.
Double bond equivalent $(DBE)$ or degree of unsaturation can be calculated using the formula: $DBE = C + 1 - \frac{H}{2} - \frac{X}{2} + \frac{N}{2}$.
Here,$C = 16, H = 19, N = 3, X = 0$.
$DBE = 16 + 1 - \frac{19}{2} + \frac{3}{2} = 17 - 9.5 + 1.5 = 9$.
Alternatively,by counting: $1$ benzene ring ($4$ $DBE$),$1$ amide group ($1$ $DBE$),$1$ $\beta$-lactam ring ($1$ $DBE$),$1$ thiazolidine ring ($1$ $DBE$),$1$ carboxylic acid $C=O$ ($1$ $DBE$),and $1$ amide $C=O$ ($1$ $DBE$). Total = $9$.

(B) The structure of cyclohexa-$1,3$-diene is a six-membered ring with two double bonds at positions $1$ and $3$.
$1$. Vinylic hydrogen atoms are those attached directly to the carbon atoms involved in a double bond ($sp^2$ hybridized carbons).
In cyclohexa-$1,3$-diene,there are $4$ such carbon atoms,each having one hydrogen atom attached. Thus,there are $4$ vinylic hydrogen atoms.
$2$. Allylic hydrogen atoms are those attached to the $sp^3$ hybridized carbon atoms adjacent to the double bond.
In cyclohexa-$1,3$-diene,there are $2$ such $sp^3$ hybridized carbon atoms (at positions $5$ and $6$). Each of these carbons is bonded to $2$ hydrogen atoms. Thus,there are $2 \times 2 = 4$ allylic hydrogen atoms.
Therefore,the molecule contains $4$ vinylic and $4$ allylic hydrogen atoms.

(C) The reaction shows that $(A)$ reacts with $1 \ mole$ of $H_2$ in the presence of $Pt$ to form a saturated product.
This indicates that $(A)$ contains one double bond.
The structure of $(A)$ consists of two cyclobutane rings connected by a single bond,which accounts for $2$ rings $(DBE = 2)$.
Additionally,there is one double bond in the structure,which adds $1$ to the $DBE$.
Therefore,the total $DBE = 2 \text{ (rings)} + 1 \text{ (double bond)} = 3$.

(B) $1$. Analyze compound $X$: It decolourises bromine (indicates unsaturation),reacts with $Na$ metal (indicates $-OH$ group),and reacts with chromic acid (indicates $1^{\circ}$ or $2^{\circ}$ alcohol). It shows no reaction with Lucas reagent,which is characteristic of a $1^{\circ}$ alcohol. Thus,$X$ is $CH_2=CH-CH_2-CH_2OH$ (but$-3-$en$-1-$ol).
$2$. Analyze compound $Y$: It shows no reaction with any of the reagents. Given the formula $C_4H_8O$,$Y$ is a saturated cyclic ether,$tetrahydrofuran$ $(THF)$.
$3$. Therefore,$X$ is $CH_2=CH-CH_2-CH_2OH$ and $Y$ is $tetrahydrofuran$.

(B) To have a single set of structurally equivalent hydrogen atoms,all hydrogen atoms in the molecule must be chemically identical due to symmetry.
$I$ (Benzene): All $6$ hydrogen atoms are equivalent due to the resonance-stabilized hexagonal symmetry.
$IV$ (Triafulvalene derivative/isomer): The structure shown is $1,3,5$-trimethylenecyclopropane (or similar isomer). In this highly symmetric structure,all hydrogen atoms attached to the terminal methylene groups are equivalent.
$II$,$III$,and $V$ possess different types of hydrogen atoms (e.g.,bridgehead,vinylic,or methylene protons in different environments).
Therefore,$I$ and $IV$ have a single set of structurally equivalent hydrogen atoms.

(A) The molecule is $CH_3-CH(Br)-CH(Br)-CH_3$ ($2$,$3$-dibromobutane).
$1.$ $A$ secondary carbon ($2^\circ$ $C$) is a carbon atom bonded to two other carbon atoms. In this molecule,the two central carbons (C2 and C3) are each bonded to two other carbons (one methyl group and one other central carbon). Thus,there are $2$ secondary carbons.
$2.$ Secondary hydrogens are the hydrogen atoms attached to secondary carbons. Each of the two secondary carbons is bonded to one hydrogen atom. Therefore,the total number of secondary hydrogens is $1 + 1 = 2$.
Thus,the number of secondary carbons and secondary hydrogens are $2$ and $2$ respectively.

(C) Homologous series is a group of organic compounds having the same functional group and similar chemical properties,in which successive members differ by a $CH_2$ group.
Because the molecular mass increases as we move up the series,physical properties like boiling point,melting point,and density change gradually.
Therefore,homologues have different physical properties such as boiling point,while they share the same general formula,functional group,and type of atoms.

(C) To identify the number of functional groups in the given molecule,we analyze the structure:
$1$. Hydroxyl group $(-OH)$: Attached to the cyclohexene ring.
$2$. Alkene group $(C=C)$: Present in the cyclohexene ring.
$3$. Primary amine group $(-NH_2)$: Attached to the alpha carbon of the amide.
$4$. Amide group $(-CONH-)$: Linking the two parts of the molecule.
$5$. Secondary amine $(-NH-)$: Part of the four-membered ring (azetidine derivative).
$6$. Thioether (sulfide) group $(-S-)$: Part of the thiazolidine ring.
$7$. Carboxylic acid group $(-COOH)$: Attached to the thiazolidine ring.
There are $7$ distinct functional groups present in the molecule.

(D) All the given statements are correct.
$1.$ $C_3H_7-$ has $2$ isomers: $n$-propyl and isopropyl.
$2.$ $C_4H_9-$ has $4$ isomers: $n$-butyl,$sec$-butyl,isobutyl,and $tert$-butyl.
$3.$ $C_5H_{11}-$ has $8$ isomers: $n$-pentyl,$2$-pentyl,$3$-pentyl,$2$-methylbutyl,$3$-methylbutyl,$2,2$-dimethylpropyl,$2$-methylbutan-$2$-yl,and $3$-methylbutan-$2$-yl.

(D) Let us analyze each option:
$A$: The structure shown is a substituted cyclohexanol,which is a homocyclic compound,not heterocyclic.
$B$: The structure is a substituted cyclohexane,which contains multiple types of carbon environments $(1^o, 2^o, 3^o)$.
$C$: The structure is a substituted tetrahydrofuran,which is a heterocyclic compound because it contains an oxygen atom in the ring. The label says 'Homocyclic',which is incorrect.
$D$: The structure is $1,3,5$-trimethylcyclohexane. Let us count the types of carbons:
- $1^o$ carbons: $3$ (the three methyl groups attached to the ring).
- $2^o$ carbons: $3$ (the $CH_2$ groups in the ring).
- $3^o$ carbons: $3$ (the $CH$ groups in the ring to which methyl groups are attached).
Since the number of $1^o, 2^o$,and $3^o$ carbons are all equal ($3$ each),this is the correctly matched option.

(B) The structure of propane$-1, 2, 3-$tricarboxylic acid is $HOOC-CH_2-CH(COOH)-CH_2-COOH$.
Counting the bonds:
$1$. There are $3$ carboxylic acid groups $(-COOH)$,each containing one $C=O$ double bond. Thus,there are $3\pi$ bonds.
$2$. Counting the $\sigma$ bonds:
- $C-C$ bonds: $2$ (in the backbone) $+ 3$ (to carboxyl groups) $= 5 \sigma$ bonds.
- $C-H$ bonds: $2$ (on $C_1$) $+ 1$ (on $C_2$) $+ 2$ (on $C_3$) $= 5 \sigma$ bonds.
- $C=O$ bonds: $3 \sigma$ bonds.
- $C-O$ bonds: $3 \sigma$ bonds.
- $O-H$ bonds: $3 \sigma$ bonds.
Total $\sigma$ bonds $= 5 + 5 + 3 + 3 + 3 = 19$.
Therefore,the molecule has $19\sigma$ and $3\pi$ bonds.

(A) The enolic form of acetone $(CH_3COCH_3)$ is propen$-2-$ol,which has the structure $CH_2=C(OH)CH_3$.
Counting the bonds:
- $C-H$ bonds: $3$ (on $CH_2$) + $3$ (on $CH_3$) = $6 \sigma$ bonds.
- $C-C$ bond: $1 \sigma$ bond.
- $C-O$ bond: $1 \sigma$ bond.
- $O-H$ bond: $1 \sigma$ bond.
- Total $\sigma$ bonds = $6 + 1 + 1 + 1 = 9 \sigma$ bonds.
- $C=C$ bond: $1 \pi$ bond.
- Total $\pi$ bonds = $1 \pi$ bond.
- Lone pairs: The oxygen atom in the $-OH$ group has $2$ lone pairs.

(B) In $CH_3COCH_2COCH_3$ (acetylacetone),the enol form is highly stabilized by resonance and intramolecular hydrogen bonding. The enol content is approximately $76\%$.
The enol structure is: $CH_3-C(OH)=CH-CO-CH_3$.

(N/A) Organic chemistry is the study of carbon-containing compounds and is essential for several reasons:
$1$. It is the foundation of life,as all living organisms are composed of organic molecules like proteins,carbohydrates,lipids,and nucleic acids ($DNA$ and $RNA$).
$2$. It is crucial in the pharmaceutical industry for the synthesis of life-saving drugs,antibiotics,and vaccines.
$3$. It plays a vital role in the development of materials such as plastics,polymers,synthetic fibers (like nylon and polyester),and rubbers.
$4$. It is fundamental to the petrochemical industry,which provides fuels,lubricants,and various industrial chemicals.
$5$. It is essential in agriculture for the production of fertilizers,pesticides,and herbicides that enhance crop yield.

(N/A) The three-dimensional $(3-D)$ structure of organic molecules can be represented on paper using specific conventions:
$(a)$ Wedge-and-dash formula:
- Solid wedge $(\blacktriangle)$: Used to indicate a bond projecting out of the plane of the paper towards the observer.
- Dashed wedge $(\dots)$: Used to depict a bond projecting out of the plane of the paper away from the observer.
- Normal line $(-)$: Used to represent bonds lying in the plane of the paper.
Example: The $3-D$ representation of a methane $(CH_4)$ molecule is shown in the provided image.
$(b)$ Molecular Models: These are physical devices used for better visualization of the $3-D$ shapes of organic molecules.
$(i)$ Framework Model: Represents only the bonds connecting the atoms,ignoring the size of the atoms.
$(ii)$ Ball and Stick Model: Represents atoms as balls and bonds as sticks.
$(iii)$ Space Filling Model: Emphasizes the surface area of the molecule by representing atoms as spheres.

(N/A) In the 3D representation of organic molecules,the solid wedge is used to indicate a bond projecting out of the plane of the paper towards the observer.
The dashed wedge is used to depict a bond projecting out of the plane of the paper and away from the observer.

(B) The space filling model emphasizes the relative size of each atom based on its van der Waals radius. It conveys the volume occupied by each atom in the molecule.

(A) $(i) \text{ True}, (ii) \text{ True}, (iii) \text{ True}, (iv) \text{ False}$.
Explanation:
$(i)$ $F$. Wohler synthesized urea from ammonium cyanate,which was the first organic compound synthesized in the laboratory.
$(ii)$ Berzelius proposed the Vital Force Theory,suggesting a mysterious force in living organisms.
$(iii)$ Wohler's synthesis in $1828$ actually disproved the Vital Force Theory,but the theory was widely accepted before his experiment.
$(iv)$ Kolbe synthesized acetic acid,but Berthelot synthesized methane,so the statement is technically true in its individual parts,but often evaluated as false in specific textbook contexts if the synthesis order or context is misinterpreted. However,based on historical facts,$(iv)$ is True. Given the provided solution format,we clarify: $(i) T, (ii) T, (iii) T, (iv) T$.

(B) $(i) - \text{False}$. The large number of organic compounds is primarily due to the unique property of carbon known as catenation and its ability to form multiple bonds,not just its small size.
$(ii) - \text{True}$. Catenation is the ability of carbon atoms to link with each other to form long chains and rings,which is the main reason for the existence of a vast number of organic compounds.
$(iii) - \text{False}$. While tetravalency is a necessary condition for forming complex structures,it is not the primary reason for the vast number of organic compounds; catenation is the dominant factor.

(A) $(1)$ Equal (Both have $13$ $\sigma$-bonds and $0$ $\pi$-bonds).
$(2)$ Heteroatom (like $O, N, S$) in the cyclic ring.
$(3)$ Homocyclic (or carbocyclic),Heterocyclic.
$(4)$ Space-filling model.

(C) The enol content is significantly increased when the enol form is aromatic.
Cyclohexane$-1,3,5-$trione (phloroglucinol) can tautomerize to form benzene$-1,3,5-$triol.
The enol form,benzene$-1,3,5-$triol,is aromatic,which provides extra stability due to resonance energy.
Therefore,cyclohexane$-1,3,5-$trione has the highest enol content among the given options.

(A) Step $1$: Analyze the structures in List-$I$.
$(A)$ is Tetrahydrofuran,which is a non-planar heterocyclic compound $(III)$.
$(B)$ is a bicyclo compound $(IV)$.
$(C)$ is a spiro compound $(I)$.
$(D)$ is Furan,which is an aromatic compound $(II)$.
Step $2$: Match the findings.
$A-III, B-IV, C-I, D-II$.
Therefore,the correct option is $A$.

(B) Vanillin is an organic compound with the molecular formula $C_8H_8O_3$. It is a phenolic aldehyde containing aldehyde,hydroxyl,and ether functional groups.
Structure analysis:
$1$. Total number of oxygen atoms = $3$ (one in $-CHO$,one in $-OH$,and one in $-OCH_3$).
$2$. Total number of $\pi$-bonds:
- $3$ $\pi$-bonds in the benzene ring.
- $1$ $\pi$-bond in the $C=O$ group of the aldehyde.
- Total $\pi$-bonds = $4$.
$3$. Total number of $\pi$-electrons = $4 \times 2 = 8$.
Sum of oxygen atoms and $\pi$-electrons = $3 + 8 = 11$.

(A) To determine the least acidic compound,we compare the stability of their conjugate bases:
$(A)$ $m$-Ethoxyphenol: The conjugate base is a phenoxide ion,which is resonance-stabilized by the benzene ring.
$(B)$ Benzenesulfonic acid: The conjugate base is a sulfonate ion $(PhSO_3^-)$,which is highly resonance-stabilized by three oxygen atoms.
$(C)$ Butanoic acid: The conjugate base is a carboxylate ion $(CH_3CH_2CH_2COO^-)$,which is resonance-stabilized by two oxygen atoms.
$(D)$ Ethyl propiolate $(EtO_2C-C \equiv CH)$: The acidic proton is on the $sp$-hybridized carbon. The conjugate base is $EtO_2C-C \equiv C^-$. Although the negative charge is on an $sp$-hybridized carbon (which is more electronegative than $sp^2$ or $sp^3$),it is not resonance-stabilized by oxygen atoms like the others.
Comparing the acidity,sulfonic acids are the strongest,followed by carboxylic acids,then phenols,and finally terminal alkynes. Therefore,ethyl propiolate is the least acidic.

(B) Neopentane $(CH_3)_4C$ and Isopentane $(CH_3)_2CHCH_2CH_3$ are chain isomers,not position isomers. So,$(A)$ is incorrect.
$(B)$ Isopentane $(CH_3)_2CHCH_2CH_3$ has three primary carbons (at the ends of the branches). So,$(B)$ is correct.
$(C)$ Isobutylene $(CH_3)_2C=CH_2$ has one primary carbon (in the $CH_2$ group) and two primary carbons (in the $CH_3$ groups). It has no secondary carbon. So,$(C)$ is correct.
$(D)$ Isobutane $(CH_3)_3CH$ has three primary carbons. So,$(D)$ is incorrect.
Therefore,the correct pairs are $(B)$ and $(C)$.

(B) Two successive members of a homologous series differ by one $-CH_2-$ (methylene) unit,which corresponds to a molar mass difference of $(12 + 2 \times 1) = 14 \ g \ mol^{-1}$.
Given,molar mass of the first member $= 46 \ g \ mol^{-1}$.
The second member will have a molar mass of $46 + 14 = 60 \ g \ mol^{-1}$.
The third member will have a molar mass of $60 + 14 = 74 \ g \ mol^{-1}$.
Therefore,the molar mass of the third member is $74 \ g \ mol^{-1}$.

(A) The acidity of a compound depends on the stability of its conjugate base. Carboxylic acids $(R-COOH)$ are significantly more acidic than alcohols $(R-OH)$ due to the resonance stabilization of the carboxylate ion $(R-COO^-)$.
Among the carboxylic acids,the acidity is influenced by the inductive effect of the carbon atom attached to the carboxyl group. The hybridization of the carbon atom attached to the $-COOH$ group determines its electron-withdrawing ability:
$1$. In $(i)$ $HC \equiv C-COOH$,the carbon is $sp$ hybridized ($50$% s-character),which is the most electronegative and exerts the strongest $-I$ effect,stabilizing the carboxylate ion the most.
$2$. In $(ii)$ $CH_2=CH-COOH$,the carbon is $sp^2$ hybridized ($33$% s-character),exerting a moderate $-I$ effect.
$3$. In $(iii)$ $CH_3-CH_2-COOH$,the carbon is $sp^3$ hybridized ($25$% s-character),exerting the weakest $-I$ effect among the acids.
$4$. $(iv)$ $CH_3-CH_2-OH$ is an alcohol and is the least acidic.
Therefore,the correct order of acid strength is $i > ii > iii > iv$.

(D) The degree of unsaturation (or double bond equivalent) is calculated using the formula: $U = C + 1 - \frac{H + X - N}{2}$.
For a stable organic compound,the degree of unsaturation must be an integer ($0$,$1$,$2$,...).
For $C_4H_{10}O_4$: $U = 4 + 1 - \frac{10}{2} = 0$.
For $C_4H_8O_4$: $U = 4 + 1 - \frac{8}{2} = 1$.
For $C_4H_7ClO_4$: $U = 4 + 1 - \frac{7+1}{2} = 1$.
For $C_4H_9O_4$: $U = 4 + 1 - \frac{9}{2} = 0.5$.
Since the degree of unsaturation cannot be $0.5$ for a stable organic molecule,$C_4H_9O_4$ does not represent a valid organic compound.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. Acetic acid $(CH_3COOH)$ is the most acidic because its conjugate base (acetate ion,$CH_3COO^-$) is stabilized by resonance with two equivalent oxygen atoms.
$2$. $p-$Nitrophenol is the next most acidic because the phenoxide ion is stabilized by resonance,and the $-NO_2$ group at the para position exerts a strong $-I$ and $-M$ effect,further stabilizing the negative charge.
$3$. Ethanol $(CH_3CH_2OH)$ is less acidic than phenols because the ethoxide ion $(CH_3CH_2O^-)$ is destabilized by the $+I$ effect of the ethyl group.
$4$. Acetylene $(HC \equiv CH)$ is the least acidic among these as the negative charge resides on an $sp$ hybridized carbon atom,which is less electronegative than oxygen.
Thus,the correct order of acidity is: $\text{acetic acid} > p-\text{nitrophenol} > \text{ethanol} > \text{acetylene}$.

(A) Statement-$I$: Compound $(X)$ contains a carboxylic acid group $(-COOH)$,which is acidic enough to react with $NaHCO_3$ to evolve $CO_2$ gas,thus it dissolves. It also has two chiral centers (marked with $*$),so Statement-$I$ is true.
Statement-$II$: Compound $(Y)$ is $CH_3-CH=C=O$. The structure is $CH_3-CH=C=O$. The first carbon $(-CH_3)$ is $sp^3$,the second carbon $(-CH=)$ is $sp^2$,the third carbon $(=C=)$ is $sp$,and the fourth carbon $(=O)$ is not a carbon atom. Thus,it has one $sp^3$,one $sp^2$,and one $sp$ hybridized carbon. Therefore,Statement-$II$ is false.