Structural isomerism Questions in English

(A) The molecular formula $C_3H_7Cl$ corresponds to a propyl chloride derivative.
There are two possible structural isomers based on the position of the chlorine atom:
$1$. $CH_3-CH_2-CH_2-Cl$ ($1$-chloropropane)
$2$. $CH_3-CH(Cl)-CH_3$ ($2$-chloropropane)
Therefore,the total number of structural isomers is $2$.

(A) Esters have the general formula $R-COOR'$.
Functional isomers are compounds that have the same molecular formula but different functional groups.
Carboxylic acids (alkanoic acids) have the general formula $R-COOH$,which is a functional isomer of an ester.
For example,methyl methanoate ($HCOOCH_3$,$C_2H_4O_2$) and ethanoic acid ($CH_3COOH$,$C_2H_4O_2$) are functional isomers.
Therefore,alkanoic acids are functional isomers of esters.

(A) The molecular formula $C_2BrClF_4$ represents a derivative of ethane where hydrogen atoms are replaced by halogens.
To find the isomers,we consider the arrangement of atoms around the $C-C$ bond.
For $C_2BrClF_4$,the possible structural isomers are based on the distribution of atoms on the two carbon atoms:
$1$. $CF_3-CFBrCl$ (one isomer)
$2$. $CF_2Cl-CF_2Br$ (one isomer)
For $CF_3-CFBrCl$,the carbon atom attached to $Br$ and $Cl$ is chiral,leading to $2$ enantiomers ($d$ and $l$ forms).
For $CF_2Cl-CF_2Br$,there are no chiral centers,but it can exhibit rotational isomerism (conformations),though typically in structural isomerism counting,we look for distinct connectivity.
However,considering stereoisomerism:
$CF_3-CFBrCl$ has $2$ enantiomers.
$CF_2Cl-CF_2Br$ has no stereoisomers.
Total isomers = $2$ (enantiomers) + $1$ (structural isomer) = $3$.

(D) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom attached to an $sp^3$ hybridized carbon atom adjacent to a functional group like the nitro group $(-NO_2)$.
In $RCH_2NO_2$,the carbon atom attached to the $-NO_2$ group has two $\alpha$-hydrogen atoms,which are acidic due to the electron-withdrawing effect of the nitro group.
Therefore,$RCH_2NO_2$ can exhibit nitro-aci tautomerism.

(B) $n$-butyl alcohol is $CH_3-CH_2-CH_2-CH_2-OH$ (a straight chain).
Isobutyl alcohol is $(CH_3)_2CH-CH_2-OH$ (a branched chain).
Since the carbon skeleton differs between the two isomers,they exhibit chain isomerism.

(B) pseudo-asymmetric carbon atom is a carbon atom bonded to four groups,where two of the groups are enantiomeric (mirror images of each other) and the other two are different.
In the compound $CH_3-CH(Br)-CH(OH)-CH(Br)-CH_3$,the central carbon atom is bonded to $-H$,$-OH$,and two identical groups $-CH(Br)CH_3$.
However,if the two $-CH(Br)CH_3$ groups have different configurations (one $R$ and one $S$),the central carbon becomes a pseudo-asymmetric center.

(C) Diethylamine is a secondary amine with the formula $(C_2H_5)_2NH$.
It is a functional isomer of primary amines like $n$-butylamine or $2$-butanamine,as they all share the same molecular formula $C_4H_{11}N$ but have different functional groups (secondary amine vs primary amine).
Therefore,diethylamine and $2$-butanamine are functional isomers.

(A) Tautomerism is a special type of functional isomerism where a chemical compound exists in two or more interconvertible forms that differ in the position of at least one hydrogen atom.
For a compound to exhibit tautomerism,it must possess an $\alpha$-hydrogen atom adjacent to a polar group (like a carbonyl group,$C=O$).
In $2-$pentanone $(CH_3COCH_2CH_2CH_3)$,there are $\alpha$-hydrogen atoms present on the carbon adjacent to the carbonyl group.
Therefore,it can undergo keto-enol tautomerism.
Phenol,$2-$butene,and lactic acid do not exhibit this specific type of keto-enol tautomerism under standard conditions.

(D) Metamerism is observed in polyvalent functional groups where the alkyl groups attached to the central atom differ.
Examples include ethers $(R-O-R')$,ketones $(R-CO-R')$,and tertiary amines $(R-N(R')-R'')$.
Since all the given options (ether,ketone,and tertiary amine) are polyvalent functional groups,they all exhibit metamerism.

(A) The given compounds are $CH_3CH_2CH_2OCH_3$ (methoxypropane) and $CH_3CH_2OCH_2CH_3$ (ethoxyethane).
Both are ethers with the same molecular formula $C_4H_{10}O$.
Metamerism arises due to the difference in the distribution of carbon atoms on either side of the functional group $(-O-)$.
In $CH_3CH_2CH_2OCH_3$,the alkyl groups attached to oxygen are methyl and propyl.
In $CH_3CH_2OCH_2CH_3$,the alkyl groups attached to oxygen are ethyl and ethyl.
Since the distribution of alkyl groups around the functional group is different,they are metamers.

(C) Methoxy methane $(CH_3-O-CH_3)$ is an ether with the molecular formula $C_2H_6O$.
Ethanol $(CH_3-CH_2-OH)$ is an alcohol with the same molecular formula $C_2H_6O$.
Since both compounds have the same molecular formula but contain different functional groups (ether group $-O-$ and hydroxyl group $-OH$),they are classified as functional isomers.

(C) The structural isomers of $C_6H_{14}$ (hexane) are as follows:
$1$. $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$ ($n$-hexane)
$2$. $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$ ($2$-methylpentane)
$3$. $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3$-methylpentane)
$4$. $CH_3-C(CH_3)_2-CH_2-CH_3$ ($2,2$-dimethylbutane)
$5$. $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ ($2,3$-dimethylbutane)
Thus,there are $5$ structural isomers.

(A) The first compound is $n$-butanenitrile $(CH_3-CH_2-CH_2-CN)$,which has a straight chain of $4$ carbon atoms.
The second compound is $2$-methylpropanenitrile $(CH_3-CH(CN)-CH_3)$,which has a branched chain of $3$ carbon atoms in the main chain.
Since the carbon skeleton (the arrangement of carbon atoms) is different in the two isomers,they exhibit chain isomerism.

(C) Position isomerism requires a minimum of $3$ carbon atoms in an alkane chain (e.g.,$1$-chloropropane and $2$-chloropropane). However,for simple substituted alkanes like dichloroethane,$2$ carbon atoms are sufficient to show position isomerism (e.g.,$1,1$-dichloroethane and $1,2$-dichloroethane).

(B) The molecular formula $C_5H_{10}$ corresponds to a degree of unsaturation (double bond equivalent) of $1$.
For cyclic isomers,the possible structures are:
$1$. Cyclopentane
$2$. Methylcyclobutane
$3$. Ethylcyclopropane
$4$. $1,1$-Dimethylcyclopropane
$5$. $1,2$-Dimethylcyclopropane (cis and trans are stereoisomers,so we count this as one structural isomer)
Thus,there are $5$ structural cyclic isomers.

(B) The first compound is $hexan-3-one$ $(CH_3-CH_2-CO-CH_2-CH_2-CH_3)$.
The second compound is $3,3-dimethylbutan-2-one$ $(CH_3-CO-C(CH_3)_3)$.
Both compounds have the same molecular formula $C_6H_{12}O$.
They differ in the position of the functional group (ketone) relative to the carbon chain and the branching of the alkyl chain.
Since they have the same functional group but different carbon skeleton structures,they are chain isomers. However,among the given options,they are best described as metamers because the alkyl groups attached to the carbonyl group are different ($ethyl/propyl$ vs $methyl/tert-butyl$).
Therefore,the correct classification is $Metamers$.

(B) carbon atom is chiral if it is bonded to four different groups or atoms.
Structure $(I)$: $C_2H_5-CH(CH_3)-C_3H_7$. The central carbon is bonded to $-H, -CH_3, -C_2H_5, -C_3H_7$. All four groups are different,so it is chiral.
Structure $(II)$: $C_2H_5-CO-CH(CH_3)-C_2H_5$. The carbon atom in the $CH(CH_3)$ group is bonded to $-H, -CH_3, -C_2H_5, -CO-C_2H_5$. All four groups are different,so it is chiral.
Structure $(III)$: $CH_3^+$. This is a planar carbocation with three hydrogens,not chiral.
Structure $(IV)$: $C_2H_5-CH(CH_3)-C_2H_5$. The central carbon is bonded to $-H, -CH_3, -C_2H_5, -C_2H_5$. Two groups are identical,so it is achiral.
Therefore,only $I$ and $II$ are chiral.

(D) The molecular formula $C_4H_{10}O$ corresponds to alcohols and ethers. The structural isomers are as follows:
Alcohols:
$1$. $CH_3CH_2CH_2CH_2OH$ (Butan$-1-$ol)
$2$. $CH_3CH_2CH(OH)CH_3$ (Butan$-2-$ol)
$3$. $(CH_3)_2CHCH_2OH$ ($2$-Methylpropan$-1-$ol)
$4$. $(CH_3)_3COH$ ($2$-Methylpropan$-2-$ol)
Ethers:
$5$. $CH_3OCH_2CH_2CH_3$ ($1$-Methoxypropane)
$6$. $CH_3OCH(CH_3)_2$ ($2$-Methoxypropane)
$7$. $CH_3CH_2OCH_2CH_3$ (Ethoxyethane)
Total structural isomers = $4$ (alcohols) + $3$ (ethers) = $7$.

(C) Functional group isomerism occurs when compounds have the same molecular formula but different functional groups.
$1$. Alcohols can show functional group isomerism with ethers (e.g.,$CH_3CH_2OH$ and $CH_3OCH_3$).
$2$. Aldehydes can show functional group isomerism with ketones (e.g.,$CH_3CH_2CHO$ and $CH_3COCH_3$).
$3$. Cyanides (nitriles) can show functional group isomerism with isocyanides (e.g.,$R-CN$ and $R-NC$).
$4$. Alkyl halides $(R-X)$ do not possess any other functional group isomer that can be formed with the same molecular formula,as they lack the structural diversity required for such isomerism.
Therefore,alkyl halides do not exhibit functional group isomerism.

(C) Ethyl acetate $(CH_3COOCH_2CH_3)$ has the molecular formula $C_4H_8O_2$.
Functional isomers must have the same molecular formula but different functional groups.
$4$-hydroxybutanone $(C_4H_8O_2)$,$3$-hydroxybutanal $(C_4H_8O_2)$,and isobutyric acid $(C_4H_8O_2)$ all have the same molecular formula $C_4H_8O_2$.
However,butane-$2,3$-diol has the molecular formula $C_4H_{10}O_2$,which is different from ethyl acetate.
Therefore,butane-$2,3$-diol is not an isomer of ethyl acetate.

(C) The first compound is benzoic acid $(C_6H_5COOH)$,which contains a carboxylic acid functional group.
The second compound is $2$-hydroxybenzaldehyde $(o-HOC_6H_4CHO)$,which contains both a hydroxyl $(-OH)$ and an aldehyde $(-CHO)$ functional group.
Since the two compounds have different functional groups,they are classified as functional isomers.

(A) The first compound is benzoic acid $(C_6H_5COOH)$,which contains a carboxylic acid functional group $(-COOH)$.
The second compound is $3-hydroxybenzaldehyde$,which contains both an aldehyde $(-CHO)$ and a hydroxyl $(-OH)$ functional group.
Since the two compounds possess different functional groups,they exhibit functional group isomerism.

(C) Chain isomerism requires a minimum of $4$ carbon atoms for alkanes $(C_4H_{10})$,but for alkynes $(C_nH_{2n-2})$,a minimum of $5$ carbon atoms is required to show chain isomerism.
$C_4H_6$ corresponds to an alkyne $(C_4H_{2(4)-2} = C_4H_6)$,which cannot form a branched chain isomer with only $4$ carbon atoms.

(B) Metamerism arises due to the difference in the alkyl chains attached to the same polyvalent functional group.
In option $A$,the functional group is $-CO-NH-$. In the first compound,the groups attached are $C_6H_5$ and $CH_3$. In the second compound,the groups attached are $CH_3$ and $C_6H_5$. These are identical,not metamers.
In option $B$,the functional group is $-CO-O-CO-$. In the first compound,the groups attached are $C_2H_5$ and $C_2H_5$. In the second compound,the groups attached are $CH_3$ and $C_3H_7$. Since the alkyl groups attached to the central functional group differ,they are metamers.
Therefore,only option $B$ represents a pair of metamers.

(D) Tautomerism requires the presence of an $\alpha$-hydrogen atom adjacent to a polar group like $C=O$,$C=N$,or $N=O$ to allow for proton migration.
$CH_3NO_2$,$CH_3CH_2NO_2$,and $C_6H_5CH=CH-OH$ (an enol) can exhibit tautomerism.
$CH_3CH_2OH$ (ethanol) is a saturated alcohol and does not contain the necessary functional group or $\alpha$-hydrogen arrangement to exhibit tautomerism.

(C) Position isomerism occurs when the same functional group is attached to different positions on the same carbon skeleton.
Structure $(1)$ is $o$-chloroanisole (or $1$-chloro-$2$-methoxybenzene).
Structure $(4)$ is $p$-chloroanisole (or $1$-chloro-$4$-methoxybenzene).
Since both $(1)$ and $(4)$ have the same molecular formula and the same functional groups attached to different positions on the benzene ring,they are position isomers.
Structures $(2)$ and $(3)$ are epoxides with different carbon chain lengths,making them chain isomers,not position isomers.

(B) The molecular formula of diethyl ether $(C_2H_5-O-C_2H_5)$ is $C_4H_{10}O$.
To be an isomer,a substance must have the same molecular formula $(C_4H_{10}O)$.
$1$. Butan$-1-$ol $(CH_3CH_2CH_2CH_2OH)$ has the formula $C_4H_{10}O$.
$2$. Butanone $(CH_3COCH_2CH_3)$ has the formula $C_4H_8O$.
$3$. $2-$Methylpropan$-2-$ol $((CH_3)_3COH)$ has the formula $C_4H_{10}O$.
$4$. n-Propyl methyl ether $(CH_3CH_2CH_2OCH_3)$ has the formula $C_4H_{10}O$.
Butanone has a different molecular formula,so it is not an isomer of diethyl ether.

(A) Compound $A$ is $CH_3OOC-CH(OH)-CH(OH)-COOH$.
Compound $B$ is $HOOC-CH(OH)-CH(OH)-COOCH_3$.
These two structures are identical because they represent the same molecule viewed from different orientations (rotating $B$ by $180^{\circ}$ in the plane of the paper gives $A$).
Therefore,$A$ and $B$ are identical.

(A) Acetic acid $(CH_3COOH)$ and methyl formate $(HCOOCH_3)$ have the same molecular formula,$C_2H_4O_2$.
Acetic acid contains a carboxylic acid functional group $(-COOH)$.
Methyl formate contains an ester functional group $(-COOCH_3)$.
Since they possess different functional groups,they exhibit functional isomerism.

(D) Tautomerism requires the presence of an $\alpha$-hydrogen atom adjacent to a carbonyl group $(-C=O)$.
In $1,2$-cyclohexanedione,the carbon atoms adjacent to the carbonyl groups possess $\alpha$-hydrogen atoms,which can migrate to the oxygen atom to form an enol tautomer.
Structure $A$ is an enol itself,but it does not exhibit keto-enol tautomerism in the same way as the diones.
Structure $B$ ($1,4$-cyclohexanedione) has $\alpha$-hydrogens and can exhibit tautomerism.
Structure $D$ ($1,2$-cyclohexanedione) also has $\alpha$-hydrogens and is a classic example of a compound exhibiting tautomerism.
However,in the context of standard chemistry problems of this type,$1,2$-cyclohexanedione is the most commonly cited example for keto-enol tautomerism among these choices.

(D) The given compounds are $CH_3-CH_2-CH_2-COOH$ (butanoic acid) and $CH_3-CH(CH_3)-COOH$ ($2$-methylpropanoic acid).
Both compounds have the same molecular formula $(C_4H_8O_2)$ but differ in the arrangement of the carbon chain (straight chain vs. branched chain).
Therefore,they are examples of chain isomerism.

(A) For a compound to exhibit keto-enol tautomerism,it must possess at least one $\alpha$-hydrogen atom adjacent to the carbonyl group.
$A$. $p$-Benzoquinone: It has no $\alpha$-hydrogen atoms available for tautomerization.
$B$. Benzaldehyde oxime: It can exhibit tautomerism (nitroso-oxime tautomerism).
$C$. Cyclohexane$-1,3,5-$trione: It has $\alpha$-hydrogen atoms and exhibits keto-enol tautomerism.
$D$. Cyclohexane$-1,4-$dione: It has $\alpha$-hydrogen atoms and can exhibit keto-enol tautomerism.
Therefore,$p$-Benzoquinone does not exhibit tautomerism.

(A) is the correct answer.
Propanal $(CH_3CH_2CHO)$ is an aldehyde,while propanone $(CH_3COCH_3)$ is a ketone.
Both compounds have the same molecular formula,$C_3H_6O$,but contain different functional groups (aldehyde and ketone).
Therefore,they exhibit functional group isomerism.

(B) Ethanol $(C_2H_5OH)$ and dimethyl ether $(CH_3OCH_3)$ both have the same molecular formula,$C_2H_6O$.
Since they have the same molecular formula but different structural arrangements (functional groups),they are functional isomers of each other.
Therefore,the correct option is $(B)$.

(A) Tautomerism in carbonyl compounds requires the presence of an $\alpha$-hydrogen atom to form an enol. Additionally,the formation of the enol must not violate $Bredt's$ $Rule$,which states that a double bond cannot be placed at the bridgehead of a bridged bicyclic system unless the ring is large enough.
$I$: This is bicyclo[$2.2$.$1$]heptan$-2-$one. The $\alpha$-hydrogen is present,but forming the enol would place a double bond at the bridgehead,violating $Bredt's$ $Rule$.
$II$: This is $3,3-$diphenylbicyclo[$2.2$.$1$]heptan$-2-$one. Similar to $I$,the enol form would violate $Bredt's$ $Rule$.
$III$: This is bicyclo[$2.2$.$1$]heptan$-2-$one (specifically the isomer where the carbonyl is at the $2-$position). The enol form can be formed by removing an $\alpha$-hydrogen from the $C_3$ position,which does not violate $Bredt's$ $Rule$ as the double bond is not at the bridgehead.
Therefore,only molecule $III$ can exhibit tautomerism.

(C) Tautomerism in carbonyl compounds generally requires the presence of an $\alpha$-hydrogen atom.
$I$: This compound has an $\alpha$-hydrogen atom adjacent to the carbonyl group,so it can exhibit keto-enol tautomerism.
$II$: This compound also possesses an $\alpha$-hydrogen atom adjacent to the carbonyl group,allowing it to exhibit keto-enol tautomerism.
$III$: Although this compound lacks an $\alpha$-hydrogen,it can exhibit $\gamma$-tautomerism (also known as $p$-tautomerism) because it has a conjugated system where a $\gamma$-hydrogen can participate in the tautomeric shift.
Therefore,all three compounds ($I, II$,and $III$) can exhibit some form of tautomerism.

(D) Compound $(III)$ is the most stable because it is a conjugated enol and is further stabilized by intramolecular hydrogen bonding,which forms a stable six-membered ring.
Compound $(II)$ is the keto form,which is generally more stable than a non-conjugated enol.
Compound $(I)$ is a non-conjugated enol,making it the least stable.
Therefore,the stability order is $III > II > I$.

(D) To determine which compound is not an isomer,we calculate the Degree of Unsaturation $(DU)$ for each:
$1$. Cyclooctene: $C_8H_{14}$,$DU = 8 - 14/2 + 1 = 2$.
$2$. Bicyclobutyl: $C_8H_{14}$,$DU = 8 - 14/2 + 1 = 2$.
$3$. Bicyclo[$3.3.0$]octane: $C_8H_{14}$,$DU = 8 - 14/2 + 1 = 2$.
$4$. Tricyclo[$3.3.0.0^{2,6}$]octane: $C_8H_{12}$,$DU = 8 - 12/2 + 1 = 3$.
Since the fourth compound has a different molecular formula and a different $DU$ value,it is not an isomer of the others.

(A) The molecular formula $C_5H_{12}$ corresponds to an alkane (pentane).
The possible structural isomers are:
$1.$ $CH_3-CH_2-CH_2-CH_2-CH_3$ ($n$-pentane)
$2.$ $CH_3-CH(CH_3)-CH_2-CH_3$ ($2$-methylbutane or isopentane)
$3.$ $CH_3-C(CH_3)_2-CH_3$ ($2,2$-dimethylpropane or neopentane)
Therefore,there are a total of $3$ structural isomers.

(D) To determine the relationship between the two compounds,we assign the $R/S$ configuration to each chiral center.
For the first compound (left): The chiral centers are at the second and third carbons. Assigning priorities based on Cahn-Ingold-Prelog rules,the configuration is $(2S, 3S)$.
For the second compound (right): Assigning priorities,the configuration is $(2S, 3R)$.
Since the configurations are $(2S, 3S)$ and $(2S, 3R)$,they are non-mirror image stereoisomers.
Therefore,the two compounds are diastereomers.

(B) chiral centre is a carbon atom bonded to four different groups. In a Fischer projection of a chain with three chiral carbons (like the ones shown,which are derivatives of pentane$-2,3,4-$triol or similar),the central carbon $(C-3)$ is the middle intersection point.
Looking at the structures:
- In structure $A$,$C-3$ is bonded to $H$,$Br$,the top group $(-CH(OH)CH_3)$,and the bottom group $(-CH(Br)CH_3)$. Since the top and bottom groups are different,$C-3$ is chiral.
- In structure $B$,$C-3$ is bonded to $H$,$OH$,the top group $(-CH(Br)CH_3)$,and the bottom group $(-CH(Br)CH_3)$. Here,the top and bottom groups are identical $(-CH(Br)CH_3)$. Therefore,the central carbon $C-3$ is bonded to two identical groups and is not a chiral centre.
- In structures $C$ and $D$,the groups attached to $C-3$ are different,making it chiral.
Thus,in structure $B$,the central carbon is not chiral.

(D) For a compound to show keto-enol tautomerism,it must have at least one $\alpha$-hydrogen atom adjacent to the carbonyl group $(C=O)$.
$(A)$ The structure shown in the image (cyclobut$-2-$ene$-1,2-$dione) has no $\alpha$-hydrogen atoms.
$(B)$ The structure shown in the image (camphor derivative) does not show tautomerism due to Bredt's rule,which prevents the formation of a double bond at the bridgehead position.
$(C)$ $Ph-CO-Ph$ (Benzophenone): Both $\alpha$-carbons are part of benzene rings and have no hydrogens.
$(D)$ $Ph-CO-CH_3$ (Acetophenone): It has three $\alpha$-hydrogens on the methyl $(-CH_3)$ group. Therefore,it can show tautomerism.

(D) The first compound is $2$-chlorobutan$-1$-ol $(CH_2(OH)-CHCl-CH_2-CH_3)$.
The second compound is $3$-chlorobutan$-1$-ol $(CH_2(OH)-CH_2-CHCl-CH_3)$.
Both compounds have the same molecular formula $(C_4H_9ClO)$ but differ in the position of the chlorine atom on the carbon chain.
Therefore,they are positional isomers,which is a type of structural isomerism.

(B) Let us analyze each option:
$A$. The given compounds are $N$-methylbutan$-1-$amine and $N$-methylpropan$-2-$amine. These are position isomers,not chain isomers.
$B$. The first compound is $5$-oxohexanoyl chloride and the second is $4$-methyl-$5$-oxopentanoyl chloride. These are chain isomers because the carbon skeleton of the main chain is different.
$C$. Metamers require the same functional group attached to different alkyl groups. Here,the groups attached to the carbonyl group are identical in both structures,just flipped,so they are the same compound,not metamers.
$D$. In ethane$-1,2-$diol,the gauche form is more stable than the anti form due to intramolecular hydrogen bonding.
Therefore,option $B$ is the correct pair.

(A) Cyanides $(R-C \equiv N)$ and isocyanides $(R-N \equiv C)$ are functional isomers of each other.
They exhibit tautomerism,which is a special type of functional isomerism where the isomers exist in dynamic equilibrium with each other.
Therefore,cyanides exist in tautomeric form.

(A) The molecular formula $C_4H_{10}O$ corresponds to the general formula $C_nH_{2n+2}O$,which indicates either an alcohol or an ether.
There are $4$ alcohols: $n$-butanol,butan-$2$-ol,$2$-methylpropan-$1$-ol,and $2$-methylpropan-$2$-ol.
There are $3$ ethers: diethyl ether,$1$-methoxypropane,and $2$-methoxypropane.
Thus,there are a total of $4 + 3 = 7$ structural isomers.

(C) The given structures are secondary amines with the same molecular formula but different alkyl groups attached to the nitrogen atom.
Structure $1$ has an isopropyl group attached to the nitrogen atom.
Structure $2$ has an ethyl group and a methyl group attached to the nitrogen atom.
Since the distribution of alkyl groups around the polyvalent functional group (nitrogen) is different,these are metamers.
Therefore,the correct type of isomerism is Metamerism.