Structural isomerism Questions in English

When two or more compounds have the same molecular formula but possess different functional groups,they are called functional isomers,and this phenomenon is known as functional group isomerism.
Example: The molecular formula $C_3H_6O$ represents both an aldehyde and a ketone.
$1$. Propanal $(CH_3CH_2CHO)$: Contains an aldehyde functional group $(-CHO)$.
$2$. Propanone $(CH_3COCH_3)$: Contains a ketone functional group $(>C=O)$.

(N/A) Metamerism is a type of structural isomerism that arises due to the presence of different alkyl chains on either side of the same functional group in a molecule.
Example: The molecular formula $C_4H_{10}O$ (ether) exhibits metamerism.
$(i)$ Methoxypropane: $CH_3-O-CH_2-CH_2-CH_3$
$(ii)$ Ethoxyethane: $CH_3-CH_2-O-CH_2-CH_3$

(N/A) Structural isomers: Compounds having the same molecular formula but different structural arrangements are known as structural isomers.
This phenomenon is known as structural isomerism.
Chain isomers: If two or more compounds have the same molecular formula but different carbon chain structures,such compounds are known as chain isomers,and this phenomenon is known as chain isomerism. Alkanes having more than three carbon atoms exhibit chain isomerism.
Example-$1$: Butane $(C_4H_{10})$ has $2$ structural isomers:
$(i)$ $n$-butane: $CH_3-CH_2-CH_2-CH_3$ (continuous chain).
$(ii)$ $2$-methylpropane (isobutane): $CH_3-CH(CH_3)-CH_3$ (branched chain).
Example-$2$: Pentane $(C_5H_{12})$ has $3$ isomers:
$(i)$ $n$-pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$ (straight chain).
$(ii)$ Isopentane ($2$-methylbutane): $CH_3-CH(CH_3)-CH_2-CH_3$ (branched chain).
$(iii)$ Neopentane ($2,2$-dimethylpropane): $CH_3-C(CH_3)_2-CH_3$ (branched chain).

(N/A) Methane $(CH_{4})$,ethane $(C_{2}H_{6})$,and propane $(C_{3}H_{8})$ do not exhibit structural isomerism as they have only one possible structure.

Name of Hydrocarbon	Number of Isomers
Butane $(C_{4}H_{10})$	$2$
Pentane $(C_{5}H_{12})$	$3$
Hexane $(C_{6}H_{14})$	$5$
Heptane $(C_{7}H_{16})$	$9$
Decane $(C_{10}H_{22})$	$75$

(N/A) The molecular formula $C_4H_6$ corresponds to an alkyne (general formula $C_nH_{2n-2}$).
$(a)$ Position Isomers:
$CH_3-CH_2-C \equiv CH$ (But$-1-$yne)
$CH_3-C \equiv C-CH_3$ (But$-2-$yne)
$(b)$ Chain Isomers:
There are no chain isomers for $C_4H_6$ alkynes as the carbon chain cannot be branched without violating valency.
$(c)$ Functional Isomers (Dienes):
$CH_2=C=CH-CH_3$ (Buta$-1,2-$diene)
$CH_2=CH-CH=CH_2$ (Buta$-1,3-$diene)
$(d)$ Cyclic Isomers:
$1$-Methylcyclopropene,$3$-Methylcyclopropene,and Cyclobutene.

(N/A) The molecular formula $C_6H_{14}$ corresponds to hexane. The chain isomers that contain only a methyl group as a substituent are as follows:

Structure	Name
$(i) \ CH_3-CH(CH_3)-CH_2-CH_2-CH_3$	$2\text{-methylpentane}$
$(ii) \ CH_3-CH_2-CH(CH_3)-CH_2-CH_3$	$3\text{-methylpentane}$

(N/A) The molecular formula $C_6H_{14}$ corresponds to an alkane. Chain isomers with two methyl groups are:
$(i)$ $CH_3-C(CH_3)_2-CH_2-CH_3$: The longest chain has $4$ carbon atoms,and there are two methyl groups at the $2^{nd}$ position. Its $IUPAC$ name is $2,2-Dimethylbutane$.
(ii) $CH_3-CH(CH_3)-CH(CH_3)-CH_3$: The longest chain has $4$ carbon atoms,and there are methyl groups at the $2^{nd}$ and $3^{rd}$ positions. Its $IUPAC$ name is $2,3-Dimethylbutane$.

(N/A) Isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural arrangements of atoms.
For example,$n$-butane $(CH_3-CH_2-CH_2-CH_3)$ undergoes isomerization in the presence of anhydrous $AlCl_3$ and $HCl$ gas to form isobutane $(CH_3-CH(CH_3)-CH_3)$.
This occurs because the branched-chain isomer is thermodynamically more stable than the straight-chain isomer.

(N/A) Isomerism: Compounds having the same molecular formula but different physical or chemical properties due to different arrangements of atoms are called isomers,and the phenomenon is called isomerism.
$(b)$ Position isomerism: It occurs when the position of a functional group or substituent changes along the same carbon chain (e.g.,$CH_3-CH_2-CH_2-OH$ and $CH_3-CH(OH)-CH_3$).
$(c)$ Chain isomerism: It arises due to the difference in the arrangement of the carbon skeleton (e.g.,$n$-butane and isobutane).
$(d)$ Functional group isomerism: It occurs when compounds have the same molecular formula but contain different functional groups (e.g.,ethanol $CH_3CH_2OH$ and dimethyl ether $CH_3OCH_3$).
$(e)$ Metamerism: It arises due to the different distribution of carbon atoms on either side of a polyvalent functional group (e.g.,diethyl ether $CH_3CH_2-O-CH_2CH_3$ and methyl propyl ether $CH_3-O-CH_2CH_2CH_3$).
$(f)$ Optical isomerism: It is a type of stereoisomerism where isomers are non-superimposable mirror images of each other,known as enantiomers,which rotate plane-polarized light.

(A) Functional group isomers are compounds that have the same molecular formula but different functional groups.
$I, II, III, IV$ are alcohols (functional group $-OH$).
$V, VI, VII$ are ethers (functional group $-O-$).
Any pair consisting of one alcohol and one ether will be functional group isomers (e.g.,$I$ and $V$).
$V, VI, VII$ are ethers and exhibit metamery due to different alkyl groups attached to the oxygen atom.

(D) Position isomers have the same molecular formula and same carbon skeleton but differ in the position of the functional group or substituent on the carbon chain.
- Pair $(I)$ and $(II)$ are position isomers ($Butan-1-ol$ and $Butan-2-ol$).
- Pair $(III)$ and $(IV)$ are position isomers ($2-Methylpropan-2-ol$ and $2-Methylpropan-1-ol$).
- Pair $(VI)$ and $(VII)$ are position isomers ($1-Methoxypropane$ and $2-Methoxypropane$).

(A) Position isomers have the same molecular formula and same functional group but differ in the position of the functional group on the parent chain.
- In $(I)$ $CH_3-CH_2-CH_2-CH_2-OH$ (butan$-1-$ol) and $(II)$ $CH_3-CH_2-CH(OH)-CH_3$ (butan$-2-$ol),the position of the $-OH$ group differs on the same butane chain.
- In $(VI)$ $CH_3-O-CH_2-CH_2-CH_3$ ($1$-methoxypropane) and $(VII)$ $CH_3-O-CH(CH_3)-CH_3$ ($2$-methoxypropane),the position of the $-OCH_3$ group differs on the same propane chain.
- Therefore,$(I), (II)$ and $(VI), (VII)$ are pairs of position isomers.

(B) Chain isomerism occurs when compounds have the same molecular formula but different carbon chain structures (e.g.,straight chain vs. branched chain).
$1$. $I$ $(CH_3-CH_2-CH_2-CH_2-OH)$ is a straight-chain alcohol (butan$-1-$ol).
$2$. $IV$ $((CH_3)_2CH-CH_2-OH)$ is a branched-chain alcohol ($2$-methylpropan$-1-$ol).
Both $I$ and $IV$ have the same molecular formula $(C_4H_{10}O)$ but differ in the arrangement of the carbon chain. Therefore,they represent chain isomerism.

(C) $Metamerism$ is a type of structural isomerism that occurs when different alkyl groups are attached to the same polyvalent functional group (such as $-O-$,$-S-$,$-NH-$,or $-CO-$).
In the given pair,the alkyl groups attached to the sulfur atom are different (methyl and $n$-propyl in the first,methyl and isopropyl in the second). Therefore,they are metamers.

(B) The keto-enol tautomerism in acetone $[CH_{3}COCH_{3}]$ results in a very small amount of enol form $(< 0.1 \%)$ because the enol form is less stable than the keto form.
In acetylacetone $[CH_{3}COCH_{2}COCH_{3}]$,the enol form is significantly more stable due to the formation of a six-membered ring stabilized by intramolecular hydrogen bonding.
Therefore,both the assertion $A$ and the reason $R$ are true,and $R$ provides the correct explanation for $A$.

(D) The molecular formula $C_3H_6O$ represents compounds such as propanal $(CH_3CH_2CHO)$ and propanone $(CH_3COCH_3)$.
Since these compounds have the same molecular formula but different functional groups (aldehyde and ketone),they exhibit functional group isomerism.

(D) Metamerism is shown by compounds having a polyvalent functional group (like ether,thioether,or amine) where the alkyl groups attached to the functional group differ.
$C_4H_{10}O$ can represent diethyl ether $(CH_3CH_2-O-CH_2CH_3)$ and methyl propyl ether $(CH_3-O-CH_2CH_2CH_3)$.
Since the alkyl groups attached to the oxygen atom are different in these two isomers,they are metamers.
Therefore,the correct option is $D$.

(B) Metamerism arises due to the difference in the nature of the alkyl groups attached to the same polyvalent functional group (like $-O-$,$-CO-$,$-NH-$,etc.).
In the pair $CH_3-CH_2-CO-CH_2-CH_3$ (pentan$-3-$one) and $CH_3-CO-CH_2-CH_2-CH_3$ (pentan$-2-$one),the ketone functional group $(-CO-)$ is attached to different alkyl groups (ethyl-ethyl vs methyl-propyl). Thus,they are metamers.

(A) Tautomerism requires the presence of an $\alpha$-hydrogen atom adjacent to a carbonyl group.
Compounds with an active methylene group $(-CH_2-)$ flanked by two electron-withdrawing carbonyl groups (like in $\beta$-keto esters) undergo tautomerization most readily due to the stabilization of the resulting enol form by resonance and intramolecular hydrogen bonding.
In $CH_3COCH_2CO_2C_2H_5$ (ethyl acetoacetate),the methylene group is between two carbonyl groups,making it highly acidic and prone to tautomerization.
Therefore,the correct option is $A$.

(D) The first compound is $hex-1-yne$ (an alkyne).
The second compound is $hexa-1,3-diene$ (a conjugated diene).
Since they possess different functional groups (triple bond vs. two double bonds),they are classified as functional group isomers.

(B) The given compounds are $CH_3-CH=CH-CH_3$ (but-$2$-ene) and $CH_3-CH_2-CH=CH_2$ (but-$1$-ene).
Both compounds have the same molecular formula $C_4H_8$ but differ in the position of the double bond in the carbon chain.
Therefore,they are positional isomers.

(B) . The molecular formula $C_3H_4$ corresponds to a degree of unsaturation of $2$. The possible structural isomers are:
$1$. Propyne: $CH_3-C\equiv CH$
$2$. Cyclopropene: $A$ three-membered ring with one double bond.

(C) The molecular formulas for the given compounds are as follows:
acetone: $C_3H_6O$
propanol: $C_3H_8O$
methyl acetate: $C_3H_6O_2$
propionic acid: $C_3H_6O_2$
Since the molecular formulas for both propionic acid $(CH_3CH_2COOH)$ and methyl acetate $(CH_3COOCH_3)$ are the same $(C_3H_6O_2)$ but they possess different functional groups,they are functional isomers.

(D) Metamers are isomers that have the same molecular formula and the same functional group,but differ in the nature of the alkyl or aryl groups attached to either side of the functional group.
The given compound $(X)$ is $N$-phenylcyclohexanecarboxamide,which has a cyclohexyl group on one side of the amide functional group and a phenyl group on the other.
To form a metamer,we need to change the distribution of the carbon atoms attached to the amide nitrogen and carbonyl carbon while keeping the total number of carbon atoms and the functional group the same.
Compound $(D)$,$N$-cyclohexylbenzamide,is a metamer of $(X)$ because it has the same molecular formula and the same amide functional group,but the phenyl group is now attached to the carbonyl carbon and the cyclohexyl group is attached to the nitrogen atom.

(B) The molecular formula $C_4H_6$ corresponds to a degree of unsaturation (double bond equivalent) of $4 - (6/2) + 1 = 2$.
For cyclic isomers,we consider structures with one ring and one double bond,or two rings.
The possible cyclic isomers are:
$1$. Cyclobutene
$2$. $1-$Methylcyclopropene
$3$. $3-$Methylcyclopropene
$4$. Methylenecyclopropane
$5$. Bicyclobutane
Thus,there are a total of $5$ cyclic isomers.

(C) To determine if the molecules are identical,we assign $IUPAC$ names to each Newman projection:
$P$: The structure corresponds to $2,3,3$-trimethylpentan-$2$-ol.
$Q$: The structure corresponds to $3$-ethyl-$2$-methylpentan-$2$-ol.
$R$: The structure corresponds to $3$-ethyl-$2$-methylpentan-$2$-ol.
$S$: The structure corresponds to $3$-ethyl-$2$-methylpentan-$3$-ol.
Comparing the $IUPAC$ names,$Q$ and $R$ have the same name ($3$-ethyl-$2$-methylpentan-$2$-ol),which means they represent the same molecule. Thus,the correct option is $C$.

(B) Statement $-I \rightarrow$ True
$CH_3CH=CHCH_3$ and $cyclobutane$ both have the molecular formula $C_4H_8$. They are ring-chain isomers,which is a type of structural isomerism.
Statement $-II \rightarrow$ True
$CH_3CH_2CH_2CH_2NH_2$ ($1^{\circ}$ amine) and $(CH_3CH_2)_2NH$ ($2^{\circ}$ amine) have different functional groups (primary amine vs secondary amine). Therefore,they are functional group isomers.

(D) . Pentan$-3-$one $(CH_3CH_2COCH_2CH_3)$ and Pentan$-2-$one $(CH_3COCH_2CH_2CH_3)$ have different alkyl groups attached to the carbonyl group,so they are metamers. (Correct)
$B$. Butanenitrile $(CH_3CH_2CH_2CN)$ and Butaneisonitrile $(CH_3CH_2CH_2NC)$ have different functional groups (nitrile vs isonitrile),so they are functional isomers. (Correct)
$C$. Butan$-1-$ol $(CH_3CH_2CH_2CH_2OH)$ and Butan$-2-$ol $(CH_3CH_2CH(OH)CH_3)$ differ only in the position of the $-OH$ group,so they are position isomers. (Correct)
$D$. Butan$-1-$amine $(CH_3CH_2CH_2CH_2NH_2)$ is a primary amine,while $N$-methylpropan$-1-$amine $(CH_3CH_2CH_2NHCH_3)$ is a secondary amine. They are functional isomers,not homologs. (Incorrect)
Therefore,statements $A, B,$ and $C$ are correct.

(A) Metamerism arises due to the difference in the nature of the alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,or $-CO-$).
In diethyl ether $(CH_3CH_2-O-CH_2CH_3)$,the oxygen atom is attached to two ethyl groups.
In $n$-propyl methyl ether $(CH_3CH_2CH_2-O-CH_3)$,the oxygen atom is attached to a methyl group and an $n$-propyl group.
Since the alkyl groups attached to the oxygen atom differ,these compounds exhibit metamerism.

(C) Let's analyze each pair:
$(A)$ $CH_2=CH-OCH_3$ (methoxyethene) and oxetane $(C_3H_6O)$ are ring-chain isomers because one is an open-chain compound and the other is a cyclic compound with the same molecular formula.
$(B)$ $o$-methylbenzyl alcohol $(C_8H_{10}O)$ and $2,3$-dimethylphenol $(C_8H_{10}O)$ are functional isomers because they contain different functional groups (alcohol vs. phenol).
$(C)$ Butan-$2$-ol and butan-$1$-ol are position isomers because the position of the $-OH$ group changes,not the carbon chain length. Therefore,calling them chain isomers is incorrect.
$(D)$ Pent-$1$-ene and $2$-methylbut-$1$-ene are chain isomers because the parent carbon chain length changes from $5$ to $4$ carbons.

(C) Step $1$: Analyze $(i)$. The compounds are $2-chloro-2-methylbutane$ and $2-chloro-2-methylpropane$. They differ by a $-CH_2-$ group,so they are Homologues $(P)$.
Step $2$: Analyze $(ii)$. The compounds are $ethoxybenzene$ and $1-methoxy-2-methylbenzene$. These are Metamers $(R)$ as the alkyl group attached to the oxygen atom changes.
Step $3$: Analyze $(iii)$. The compounds are $methoxybenzene$ $(ether)$ and $2-methylphenol$ $(alcohol)$. These have different functional groups,so they are Functional group isomers $(Q)$.
Step $4$: Analyze $(iv)$. The compounds are $2-methylpropan-1-ol$ and $2-methylpropan-2-ol$. The position of the $-OH$ group changes,so they are Position isomers $(S)$.
Therefore,the correct match is $i-P, ii-R, iii-Q, iv-S$.

(C) The given structures are Newman projections of substituted ethanes. Let us convert them to their $IUPAC$ names:
$I$: $CH_3-CH(D)-CH_2(T)$ ($1$-tritio$-2-$deutero-propane)
$II$: $CH_3-CH(T)-CH_2(D)$ ($1$-deutero$-2-$tritio-propane)
$III$: $CH_3-CH_2-CH(DT)$ ($1$-deutero$-1-$tritio-propane)
Comparing the structures:
$I$ and $II$ are positional isomers because the positions of $D$ and $T$ atoms on the carbon chain are different.
$II$ and $III$ are positional isomers because the positions of $D$ and $T$ atoms are different.
$I$ and $III$ are positional isomers because the positions of $D$ and $T$ atoms are different.
None of the given options correctly describe the relationship. However,based on the provided options,if we re-evaluate the structures as isomers of the same molecular formula,they are all structural isomers. Given the standard nature of such questions,if $I$,$II$,and $III$ represent different substitution patterns on the same propane backbone,they are positional isomers.

(C) The general formula for both ethers and alcohols is $C_n H_{2n+2} O$.
In ethers,the functional group is an ether linkage $(-O-)$,while in alcohols,the functional group is a hydroxyl group $(-OH)$.
Since they possess different functional groups despite having the same molecular formula,they are classified as functional isomers.

(B) Functional isomers are compounds that have the same molecular formula but different functional groups.
In option $B$,Butan-$1$-ol $(C_4H_{10}O)$ is an alcohol,while $1$-Methoxypropane $(C_4H_{10}O)$ is an ether.
Since they have the same molecular formula but different functional groups ($-OH$ vs $-O-$),they are functional isomers.

(C) $Pentan-2-ol$ $(C_5H_{12}O)$ is an alcohol.
Functional isomers are compounds with the same molecular formula but different functional groups.
$Ethoxypropane$ $(C_2H_5-O-C_3H_7)$ has the molecular formula $C_5H_{12}O$ and belongs to the ether functional group.
Therefore,$ethoxypropane$ is a functional isomer of $pentan-2-ol$.
$Pentan-1-ol$ and $pentan-3-ol$ are positional isomers,while $pentan-2-one$ is a ketone with the formula $C_5H_{10}O$.

(C) The correct answer is $C$.
In the Newman projection formula,the molecule is viewed along the $C-C$ bond axis.
The carbon atom nearer to the observer is represented by a point,and the rear carbon atom (further from the observer) is represented by a circle around that point.

(C) $Ethoxyethane$ $(CH_3-CH_2-O-CH_2-CH_3)$ and $methoxypropane$ $(CH_3-O-CH_2-CH_2-CH_3)$ have the same functional group (ether) but have a different distribution of carbon atoms attached to the ethereal oxygen atom. Thus,this pair of compounds exhibits metamerism.

(D) Position isomerism occurs when the position of a functional group or a double bond changes while the carbon skeleton remains the same.
In $CH_3-CH_2-CH=CH_2$ (but$-1-$ene),the double bond is at the $C-1$ position.
In $CH_3-CH=CH-CH_3$ (but$-2-$ene),the double bond is at the $C-2$ position.
Since the parent carbon chain is the same and only the position of the double bond differs,they are position isomers.

(D) Metamerism arises due to the presence of different alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,etc.).
In $Ethoxyethane$ $(CH_3CH_2-O-CH_2CH_3)$,the oxygen atom is attached to two ethyl groups.
In $1-Methoxypropane$ $(CH_3-O-CH_2CH_2CH_3)$,the oxygen atom is attached to a methyl group and a propyl group.
Since the alkyl groups attached to the oxygen atom are different,these compounds are metamers.

(C) Position isomers are compounds that have the same molecular formula but differ in the position of the functional group or multiple bond on the carbon chain.
In $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$,the double bond is at the $1^{st}$ position.
In $but-2-ene$ $(CH_3-CH=CH-CH_3)$,the double bond is at the $2^{nd}$ position.
Since the position of the double bond is different while the carbon skeleton remains the same,they are position isomers.

(C) The molecular formula for both methoxyethane $(CH_3-O-CH_2-CH_3)$ and propan-$1$-ol $(CH_3-CH_2-CH_2-OH)$ is $C_3H_8O$.
Methoxyethane is an ether,while propan-$1$-ol is an alcohol.
Since they have the same molecular formula but different functional groups,they exhibit functional group isomerism.

(B) Metamerism arises due to the difference in the nature of alkyl groups attached to the same polyvalent functional group (in this case,the ether oxygen atom,$-O-$).
In $CH_3-CH_2-O-CH_2-CH_3$ (ethoxy ethane),the alkyl groups attached to the oxygen are two ethyl groups $(-C_2H_5)$.
In $CH_3-O-CH_2-CH_2-CH_3$ (methoxy propane),the alkyl groups attached to the oxygen are a methyl group $(-CH_3)$ and a propyl group $(-C_3H_7)$.
Since the distribution of alkyl groups around the oxygen atom is different,these compounds are metamers.

(A) The molecular formula $C_{4}H_{10}O$ corresponds to saturated acyclic compounds (alcohols and ethers). The structural isomers are as follows:
Alcohols:
$1$. $CH_{3}CH_{2}CH_{2}CH_{2}OH$ ($n$-butanol)
$2$. $CH_{3}CH_{2}CH(OH)CH_{3}$ (butan-$2$-ol)
$3$. $(CH_{3})_{2}CHCH_{2}OH$ (isobutanol)
$4$. $(CH_{3})_{3}COH$ (tert-butanol)
Ethers:
$5$. $CH_{3}OCH_{2}CH_{2}CH_{3}$ (methyl propyl ether)
$6$. $CH_{3}OCH(CH_{3})_{2}$ (methyl isopropyl ether)
$7$. $CH_{3}CH_{2}OCH_{2}CH_{3}$ (diethyl ether)
There are a total of $7$ structural isomers.

(C) Pentane has the molecular formula $C_5H_{12}$. Isomers must have the same molecular formula but different structural arrangements.
$1.$ $n$-pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$ $(C_5H_{12})$
$2.$ $2,2$-dimethylpropane: $CH_3-C(CH_3)_2-CH_3$ $(C_5H_{12})$
$3.$ $2$-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_3$ $(C_5H_{12})$
$4.$ $2,3$-dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ $(C_6H_{14})$
Since $2,3$-dimethylbutane has $6$ carbon atoms and the formula $C_6H_{14}$,it is an isomer of hexane,not pentane. Thus,option $C$ is correct.

(C) Metamers are isomers that have the same molecular formula but differ in the distribution of alkyl groups on either side of the functional group (in this case,the ether oxygen atom).
For the molecular formula $C_{4}H_{10}O$,the possible ether metamers are:
$i$) $CH_{3}CH_{2}-O-CH_{2}CH_{3}$ (Diethyl ether)
$ii$) $CH_{3}-O-CH_{2}CH_{2}CH_{3}$ (Methyl propyl ether)
$iii$) $CH_{3}-O-CH(CH_{3})_{2}$ (Methyl isopropyl ether)
Thus,there are $3$ possible metamers.

(A) $1$. The reaction sequence is as follows:
$CH_3CHO \xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} CH_3-CH(OH)-CH_3 (P)$
$CH_3-CH(OH)-CH_3 \xrightarrow[\Delta]{conc. H_2SO_4} CH_3-CH=CH_2 (Q)$
$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) B_2H_6} CH_3-CH_2-CH_2OH (R)$
$2$. Compound $P$ is propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ and compound $R$ is propan$-1-$ol $(CH_3-CH_2-CH_2OH)$.
$3$. Both compounds have the same molecular formula $(C_3H_8O)$ and the same functional group $(-OH)$,but the position of the hydroxyl group is different (carbon-$2$ in $P$ and carbon-$1$ in $R$).
$4$. Therefore,$P$ and $R$ are position isomers.

(D) The reaction is between isobutylmagnesium bromide and methylamine. The Grignard reagent acts as a base and abstracts the acidic proton from the amine to form an alkane.
$ (CH_3)_2CHCH_2MgBr + CH_3NH_2 \rightarrow (CH_3)_2CHCH_3 + Mg(Br)NHCH_3 $
The organic compound $X$ formed is isobutane,which is $2$-methylpropane,$(CH_3)_2CHCH_3$.
The molecular formula of $X$ is $C_4H_{10}$.
The isomers of $C_4H_{10}$ are:
$1$. $n$-butane: $CH_3-CH_2-CH_2-CH_3$
$2$. Isobutane ($2$-methylpropane): $CH_3-CH(CH_3)-CH_3$
Thus,there are $2$ possible isomers for the organic compound $X$.

(C) Hexane is an alkane with the molecular formula $C_6H_{14}$. Isomers of hexane must also have the molecular formula $C_6H_{14}$.
$A$. $3$-methylpentane $(C_6H_{14})$ is an isomer.
$B$. $n$-hexane $(C_6H_{14})$ is an isomer.
$C$. Ethylcyclobutane has a ring structure,and its molecular formula is $C_6H_{12}$. Therefore,it is not an isomer of hexane.
$D$. $2$-methylpentane $(C_6H_{14})$ is an isomer.
Thus,the compound that is not an isomer of hexane is ethylcyclobutane.

(B) $Propanone$ $(CH_3COCH_3)$ and $propanal$ $(CH_3CH_2CHO)$ have the same molecular formula,$C_3H_6O$.
$Propanone$ contains a ketone functional group $(-CO-)$,whereas $propanal$ contains an aldehyde functional group $(-CHO)$.
Since they possess the same molecular formula but different functional groups,they are classified as functional isomers.

(A) Functional isomers are compounds that have the same molecular formula but possess different functional groups.
In option $A$,both $C_{2}H_{5}OC_{2}H_{5}$ (diethyl ether) and $C_{3}H_{7}OCH_{3}$ (methyl propyl ether) contain the same functional group,which is an ether $(-O-)$. Therefore,they are metamers,not functional isomers.
In option $B$,$CH_{3}CH_{2}OH$ (alcohol) and $CH_{3}OCH_{3}$ (ether) are functional isomers.
In option $C$,$CH_{3}CH_{2}NO_{2}$ (nitroalkane) and $H_{2}NCH_{2}COOH$ (amino acid) represent different functional groups.
In option $D$,$CH_{3}COOH$ (carboxylic acid) and $HCOOCH_{3}$ (ester) are functional isomers.