Structural isomerism Questions in English

(C) Chain isomerism occurs when compounds have the same molecular formula but different carbon skeletons (different branching).
In option $A$,$2$-methylbutane and $n$-pentane are chain isomers.
In option $B$,$2,2$-dimethylpropane and $2$-methylbutane are chain isomers.
In option $D$,$n$-butane and $2$-methylpropane are chain isomers.
However,the question asks for an example of chain isomerism. All options $A$,$B$,and $D$ represent chain isomers. Given the structure in the image (which represents $2$-methylbutane and $2,2$-dimethylpropane),option $C$ is the most accurate representation of the provided image.

(C) The molecular formula $C_2H_2Br_2$ has three structural and geometrical isomers:
$(1)$ $cis-1,2-dibromoethene$
$(2)$ $trans-1,2-dibromoethene$
$(3)$ $1,1-dibromoethene$
Thus,the total number of isomers is $3$.

(A) Compounds that possess the same molecular formula but differ in their structural arrangement are known as isomers. Specifically,when the difference is in the structural formula,they are referred to as structural isomers.

(A) $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$ differ in the arrangement of their carbon chain (straight chain vs. branched chain).
Therefore,they are examples of chain isomers.
Correct option is $A$.

(B) Position isomers have the same molecular formula and the same carbon skeleton but differ in the position of the functional group or multiple bond.
In $CH_3-CH_2-CH=CH_2$ (But-$1$-ene) and $CH_3-CH=CH-CH_3$ (But-$2$-ene),the position of the double bond is different ($1$ and $2$),while the carbon chain remains the same.
Therefore,they are position isomers.

(C) . Metamerism is a special type of isomerism shown by compounds containing polyvalent functional groups like ethers $(-O-)$,ketones $(-CO-)$,and secondary amines $(-NH-)$. Alkenes do not contain such polyvalent functional groups,hence they do not exhibit metamerism.

(D) The molecular formula $C_{4}H_{10}O$ corresponds to the general formula $C_{n}H_{2n+2}O$,which indicates either an alcohol or an ether.
There are $4$ isomeric alcohols:
$1.$ $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (butan-$1$-ol)
$2.$ $CH_{3}CH_{2}CH(OH)CH_{3}$ (butan-$2$-ol)
$3.$ $(CH_{3})_{2}CHCH_{2}OH$ ($2$-methylpropan-$1$-ol)
$4.$ $(CH_{3})_{3}COH$ ($2$-methylpropan-$2$-ol)
There are $3$ isomeric ethers:
$5.$ $CH_{3}OCH_{2}CH_{2}CH_{3}$ ($1$-methoxypropane)
$6.$ $CH_{3}OCH(CH_{3})_{2}$ ($2$-methoxypropane)
$7.$ $CH_{3}CH_{2}OCH_{2}CH_{3}$ (ethoxyethane)
Thus,a total of $7$ structural isomers are possible.

(A) $n$-propyl alcohol is $CH_3-CH_2-CH_2-OH$ and isopropyl alcohol is $CH_3-CH(OH)-CH_3$.
These two compounds have the same molecular formula $(C_3H_8O)$ but differ in the position of the functional group ($-OH$ group).
Therefore,they are position isomers.

(A) The isomerism exhibited by alkyl cyanide $(R-CN)$ and alkyl isocyanide $(R-NC)$ is functional isomerism.
In these compounds,the functional group itself is different (cyanide group vs. isocyanide group),which is the definition of functional isomerism.

(D) Diethyl ether has the molecular formula $C_4H_{10}O$.
$n$-propyl methyl ether $(CH_3-O-CH_2-CH_2-CH_3)$,butan-$1$-ol $(CH_3-CH_2-CH_2-CH_2-OH)$,and $2$-methylpropan-$2$-ol $((CH_3)_3C-OH)$ all have the molecular formula $C_4H_{10}O$,making them isomers of diethyl ether.
Butanone $(CH_3-CO-CH_2-CH_3)$ has the molecular formula $C_4H_8O$.
Since the molecular formulas are different,butanone is not an isomer of diethyl ether.
Therefore,the correct option is $(D)$.

(B) The given compounds are $CH_3-O-CH_2CH_3$ (methoxyethane,an ether) and $CH_3-CH_2-CH_2OH$ (propan$-1-$ol,an alcohol).
Since these compounds possess the same molecular formula $(C_3H_8O)$ but contain different functional groups (ether and alcohol),they are examples of functional isomerism.

(C) Isomers are compounds with the same molecular formula but different structural arrangements.
$A$. Ethyl ethanoate $(CH_3COOCH_2CH_3)$ and methyl propanoate $(CH_3CH_2COOCH_3)$ are functional isomers (esters) with the same molecular formula $C_4H_8O_2$.
$B$. Butanone $(CH_3COCH_2CH_3)$ and butanal $(CH_3CH_2CH_2CHO)$ are functional isomers with the same molecular formula $C_4H_8O$.
$C$. Ethoxy propane $(CH_3CH_2OCH_2CH_2CH_3)$ and propoxy ethane $(CH_3CH_2CH_2OCH_2CH_3)$ are the same compound,just named differently. They are not isomers.
$D$. Methoxy methane $(CH_3OCH_3)$ and ethanol $(CH_3CH_2OH)$ are functional isomers with the same molecular formula $C_2H_6O$.

(A) Functional isomers are compounds that have the same molecular formula but different functional groups.
$A$. Acetone $(CH_3COCH_3)$ is a ketone,and propionaldehyde $(CH_3CH_2CHO)$ is an aldehyde. Both have the molecular formula $C_3H_6O$,but they contain different functional groups,hence they are functional isomers.
$B$. Diethyl ether and methyl propyl ether are metamers.
$C$. Butane and isobutane are chain isomers.
$D$. $1-$butene and $2-$butene are position isomers.

(A) The molecular formula $C_5H_{10}$ corresponds to a degree of unsaturation of $1$. This allows for either one double bond or one ring.
Structural isomers with one double bond (alkenes):
$1$. $CH_3-CH_2-CH_2-CH=CH_2$ (pent$-1-$ene)
$2$. $CH_3-CH_2-CH=CH-CH_3$ (pent$-2-$ene)
$3$. $CH_2=C(CH_3)-CH_2-CH_3$ ($2-$methylbut$-1-$ene)
$4$. $CH_3-CH=C(CH_3)-CH_3$ ($2-$methylbut$-2-$ene)
$5$. $CH_2=CH-CH(CH_3)_2$ ($3-$methylbut$-1-$ene)
Structural isomers with one ring (cycloalkanes):
$6$. Cyclopentane
$7$. Methylcyclobutane
$8$. $1,1-$Dimethylcyclopropane
$9$. $1,2-$Dimethylcyclopropane (cis and trans are stereoisomers,but structural isomers include $1,1-$ and $1,2-$)
$10$. Ethylcyclopropane
Total structural isomers = $10$.

(B) $C_6H_5C \equiv N$ (benzonitrile) and $C_6H_5N \equiv C$ (phenyl isocyanide) contain different functional groups (nitrile vs isocyanide).
Therefore,they exhibit functional isomerism.

(B) $R-C \equiv N$ (Cyanide) and $R-N \rightleftharpoons C$ (Isocyanide) contain different functional groups,hence they are functional isomers.

(C) Chain isomers have the same molecular formula but different carbon skeletons.
Neo-pentane $(2,2-dimethylpropane)$ and isopentane $(2-methylbutane)$ both have the molecular formula $C_5H_{12}$ but different arrangements of the carbon chain.
$1.$ $CH_3-CHBr_2$ and $CH_2Br-CH_2Br$ are position isomers.
$2.$ $1-propanol$ and $2-propanol$ are position isomers.
$3.$ Diethyl ether and methyl-$n$-propyl ether are metamers.

(D) $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$ and neopentane $(C(CH_3)_4)$ have the same molecular formula $(C_5H_{12})$.
They differ in the arrangement of the carbon chain (branching).
Therefore,they are chain isomers.

(B) $n$-butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ has a straight carbon chain of $4$ carbons.
Isobutyl alcohol $((CH_3)_2CHCH_2OH)$ has a branched carbon chain of $3$ carbons in the main chain.
Since they differ in the arrangement of the carbon skeleton (branching),they are chain isomers.

(A) $Methyl$ $acetate$ $(CH_3COOCH_3)$ has the molecular formula $C_3H_6O_2$ and belongs to the ester functional group.
$Propionic$ $acid$ $(CH_3CH_2COOH)$ also has the molecular formula $C_3H_6O_2$ and belongs to the carboxylic acid functional group.
Since they have the same molecular formula but different functional groups,they are functional isomers.

(A) Propionic acid (propanoic acid) has the molecular formula $C_3H_6O_2$.
Isomers are compounds with the same molecular formula but different structural arrangements.
For $HCOOC_2H_5$ (ethyl formate): Molecular formula is $C_3H_6O_2$.
For $CH_3COOCH_3$ (methyl acetate): Molecular formula is $C_3H_6O_2$.
Both compounds have the same molecular formula as propionic acid,therefore they are isomers of propionic acid.
Thus,option $A$ is correct.

(A) The molecular formula for ethyl alcohol $(C_2H_5OH)$ is $C_2H_6O$.
Functional isomers are compounds that have the same molecular formula but different functional groups.
$CH_3OCH_3$ (dimethyl ether) also has the molecular formula $C_2H_6O$.
Since ethyl alcohol contains a hydroxyl group $(-OH)$ and dimethyl ether contains an ether group $(-O-)$,they are functional isomers.

(C) The given compounds are ethers with the general formula $R-O-R'$.
Metamerism is a type of structural isomerism that arises due to the different distribution of carbon atoms (alkyl groups) on either side of the functional group (in this case,the ether oxygen atom).
In the provided structures,the alkyl groups attached to the oxygen atom are different in each case,which is the characteristic feature of metamerism.

(A) compound whose molecules are superimposable on their mirror images,even though they contain asymmetric carbon atoms,is called a meso compound.
Due to the presence of a plane of symmetry,these compounds are optically inactive despite containing chiral carbons.

(A) . Diethyl ketone $(CH_3CH_2COCH_2CH_3)$ and methyl propyl ketone $(CH_3COCH_2CH_2CH_3)$ are functional isomers (both are ketones),but specifically,they are metamers because the alkyl groups attached to the carbonyl group are different. They are not position isomers. Thus,this statement is wrong.
$B$. $2-$chloropentane and $1-$chloropentane differ in the position of the chlorine atom on the carbon chain,so they are position isomers. This is correct.
$C$. $n-$butane (straight chain) and $2-$methylpropane (branched chain) are chain isomers. This is correct.
$D$. Acetone $(CH_3COCH_3)$ is a ketone,while propionaldehyde $(CH_3CH_2CHO)$ is an aldehyde. They have the same molecular formula $(C_3H_6O)$ but different functional groups,so they are functional isomers. This is correct.

(C) Dimethyl ether $(CH_3-O-CH_3)$ and ethyl alcohol $(C_2H_5OH)$ have the same molecular formula $(C_2H_6O)$ but different functional groups (ether and alcohol,respectively).
Therefore,they are functional isomers.

(A) Compound $A$ is methyl hydrogen tartrate (where the top group is $COOCH_3$ and the bottom is $COOH$).
Compound $B$ is also methyl hydrogen tartrate (where the top group is $COOH$ and the bottom is $COOCH_3$).
By rotating compound $B$ by $180^{\circ}$ in the plane of the paper,the $COOH$ group moves to the top and the $COOCH_3$ group moves to the bottom,while the $H$ and $OH$ groups on the chiral carbons remain in their relative positions.
After rotation,the structure of $B$ becomes identical to the structure of $A$.
Therefore,$A$ and $B$ are identical.

(D) The correct answer is $(d)$.
Ethyl acetoacetate $(CH_3COCH_2COOC_2H_5)$ exhibits keto-enol tautomerism.
It exists as an equilibrium mixture of the keto form $(CH_3COCH_2COOC_2H_5)$ and the enol form $(CH_3C(OH)=CHCOOC_2H_5)$.

(B) meso compound is an optically inactive molecule that contains chiral centers but also possesses an internal plane of symmetry or a center of inversion.
$2, 3-$Dichlorobutane $(CH_3-CHCl-CHCl-CH_3)$ has two identical chiral centers at $C2$ and $C3$.
In its eclipsed conformation,it possesses a plane of symmetry,making it a meso isomer.
Therefore,the correct option is $(B)$.

(D) The structural isomers $C_2H_5OH$ (ethanol) and $CH_3OCH_3$ (dimethyl ether) have the same molecular formula,$C_2H_6O$,and therefore the same molar mass $(M = 46.07 \ g/mol)$.
According to the ideal gas law,$PV = nRT = (m/M)RT$,which can be rearranged to $d = (PM)/(RT)$,where $d$ is the density.
Since both compounds have the same molar mass $(M)$,they will have the same gaseous density $(d)$ at the same temperature $(T)$ and pressure $(P)$.
Properties like boiling point,vapour pressure,and heat of vaporization depend on intermolecular forces.
Ethanol exhibits strong hydrogen bonding,whereas dimethyl ether does not,leading to significant differences in these physical properties.

(C) $Nitroalkanes$ containing at least one $\alpha-hydrogen$ atom exhibit tautomerism.
In $nitroethane$ $(CH_3CH_2NO_2)$,the $\alpha-hydrogen$ atoms are acidic due to the electron-withdrawing nature of the $-NO_2$ group.
These atoms can migrate to the oxygen atom of the nitro group to form an aci-nitro form (also known as the nitronic acid form),as shown below:
$H_3C-CH_2-N(=O)-O \leftrightarrow H_3C-CH=N(OH)-O$

(B) The given compounds are $CH_3CH_2CH = CH_2$ (but$-1-$ene) and $CH_3CH = CHCH_3$ (but$-2-$ene).
Both compounds have the same molecular formula $(C_4H_8)$ and the same carbon chain length.
The only difference is the position of the double bond in the carbon chain.
Therefore,they exhibit position isomerism.

(D) The stability of substituted cyclohexanes depends on steric hindrance and intramolecular interactions.
In $1,2-$cyclohexanediol,the $OH$ groups are closer,leading to higher steric repulsion.
In $1,3-$cyclohexanediol,the $OH$ groups are further apart,reducing steric repulsion.
Between $cis$ and $trans$ isomers,the $trans-1,3-$cyclohexanediol allows both $OH$ groups to occupy equatorial positions in the chair conformation,which significantly minimizes steric strain compared to other isomers.
Therefore,$trans-1,3-$cyclohexanediol is the most stable compound.

(D) Diphenyl methane has two equivalent phenyl rings attached to a central $CH_2$ group.
When one hydrogen atom is replaced by a chlorine atom,the possible positions are:
$1$. Ortho-position on the phenyl ring.
$2$. Meta-position on the phenyl ring.
$3$. Para-position on the phenyl ring.
$4$. The central $CH_2$ group (benzylic position).
Thus,there are $4$ distinct structural isomers possible,as shown in the provided image.

(A) The molecular formula $C_4H_{10}$ corresponds to an alkane.
There are two structural isomers for this formula:
$1$. $n$-Butane: $CH_3-CH_2-CH_2-CH_3$
$2$. Isobutane ($2$-methylpropane): $CH_3-CH(CH_3)-CH_3$
Therefore,the total number of isomers is $2$.

(B) The structural isomers for the molecular formula $C_2H_3Cl_2Br$ are determined by the different arrangements of the halogen atoms on the ethane backbone:
$1$. $CH_3-C(Cl)_2Br$: $1$-bromo-$1,1$-dichloroethane
$2$. $Br-CH_2-CHCl_2$: $1$-bromo-$2,2$-dichloroethane
$3$. $Cl-CH_2-CHClBr$: $1$-bromo-$1,2$-dichloroethane
Thus,there are $3$ possible structural isomers.

(D) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom attached to a carbon atom adjacent to a functional group that can undergo tautomeric rearrangement.
In the case of nitro compounds $(RCH_2NO_2)$,the $\alpha$-hydrogen atom is acidic and can migrate to the oxygen atom of the nitro group to form an aci-nitro form.
The equilibrium is represented as:
$R-CH_2-N^+(O)O^- \rightleftharpoons R-CH=N^+(OH)O^-$
Since $(CH_3)_3CNO$,$(CH_3)_2NH$,and $R_3CNO_2$ lack an $\alpha$-hydrogen atom,they do not exhibit tautomerism.
Therefore,the correct option is $D$.

(A, C, D) Tautomerism requires the presence of an $\alpha$-hydrogen atom adjacent to a carbonyl group or a similar system that allows for the migration of a proton between two polyvalent atoms.
$(a)$ $C_6H_5-CH=CH-OH$ (enol) can tautomerize to $C_6H_5-CH_2-CHO$ (phenylacetaldehyde).
$(b)$ $p$-Benzoquinone does not have any $\alpha$-hydrogen atoms,so it cannot exhibit tautomerism.
$(c)$ Cyclohex$-2-$ene$-1,4-$dione has $\alpha$-hydrogen atoms at the $C-5$ and $C-6$ positions,allowing it to exhibit tautomerism.
$(d)$ Cyclohexane$-1,2-$dione has $\alpha$-hydrogen atoms,allowing it to exhibit tautomerism.
Therefore,compounds $(a)$,$(c)$,and $(d)$ exhibit tautomerism.

(C) ${C_4}{H_{10}}O$ has $7$ structural isomers,consisting of $4$ alcohols and $3$ ethers.
Alcohols:
$1. \text{ } CH_3-CH_2-CH_2-CH_2-OH \text{ } (n-\text{butyl alcohol})$
$2. \text{ } CH_3-CH_2-CH(OH)-CH_3 \text{ } (\text{sec-butyl alcohol})$
$3. \text{ } (CH_3)_2CH-CH_2-OH \text{ } (\text{isobutyl alcohol})$
$4. \text{ } (CH_3)_3C-OH \text{ } (\text{tert-butyl alcohol})$
Ethers:
$5. \text{ } CH_3-CH_2-O-CH_2-CH_3 \text{ } (\text{diethyl ether})$
$6. \text{ } CH_3-O-CH_2-CH_2-CH_3 \text{ } (\text{methyl propyl ether})$
$7. \text{ } CH_3-O-CH(CH_3)_2 \text{ } (\text{methyl isopropyl ether})$

(B) Metamerism is shown by compounds having the same molecular formula but different alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,$-CO-$).
$C_2H_5-S-C_2H_5$ (Diethyl thioether) can form metamers like $CH_3-S-C_3H_7$ (Methyl propyl thioether).
Since the sulfur atom is flanked by different alkyl groups in these isomers,it exhibits metamerism.
Therefore,the correct option is $B$.

(B) $1.$ $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$ $(n-hexane)$
$2.$ $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$ $(2-methylpentane)$
$3.$ $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ $(3-methylpentane)$
$4.$ $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ $(2,3-dimethylbutane)$
$5.$ $CH_3-C(CH_3)_2-CH_2-CH_3$ $(2,2-dimethylbutane)$
There are a total of $5$ structural isomers for $C_6H_{14}$.

(A) The molecular formula $C_3H_7Cl$ corresponds to a saturated alkyl halide.
Two structural isomers are possible for this formula:
$1$. $CH_3-CH_2-CH_2Cl$ ($1$-chloropropane)
$2$. $CH_3-CHCl-CH_3$ ($2$-chloropropane)
Thus,the total number of structural isomers is $2$.

(A) The compound with the molecular formula $C_3H_7Cl$ has two structural isomers: $1$-chloropropane $(CH_3CH_2CH_2Cl)$ and $2$-chloropropane $(CH_3CHClCH_3)$.
These isomers differ in the position of the chlorine atom on the propane chain,representing position isomerism.

(B) Isomers are compounds with the same molecular formula but different structural arrangements.
Ethyl alcohol $(C_2H_5OH)$ and dimethyl ether $(CH_3-O-CH_3)$ both have the molecular formula $C_2H_6O$.
Therefore,they are functional isomers.

(D) The molecular formula of diethyl ether $(C_2H_5-O-C_2H_5)$ is $C_4H_{10}O$.
Isomers are compounds with the same molecular formula but different structural arrangements.
$n-$propyl methyl ether $(CH_3-O-CH_2-CH_2-CH_3)$ has the formula $C_4H_{10}O$.
Butan$-1-$ol $(CH_3-CH_2-CH_2-CH_2-OH)$ has the formula $C_4H_{10}O$.
$2-$methylpropan$-2-$ol $(CH_3-C(OH)(CH_3)-CH_3)$ has the formula $C_4H_{10}O$.
Butanone $(CH_3-CO-CH_2-CH_3)$ has the molecular formula $C_4H_8O$.
Therefore,butanone is not an isomer of diethyl ether.

(A) Metamerism is a type of structural isomerism exhibited by organic compounds.
Compounds exhibiting metamerism have the same molecular formula and the same functional group,but they differ in the nature of the alkyl groups attached to the polyvalent functional group (such as $-O-$,$-S-$,$-NH-$,etc.).
In the given compounds,$CH_3-O-C_3H_7$ (methyl propyl ether) and $C_2H_5-O-C_2H_5$ (diethyl ether),both contain the ether functional group $(-O-)$. However,the alkyl groups attached to the oxygen atom are different ($CH_3$ and $C_3H_7$ in the first,$C_2H_5$ and $C_2H_5$ in the second).
Therefore,these compounds exhibit metamerism.

(C) $(i)$ Metamers have different alkyl groups attached to the same functional group.
$(ii)$ Functional isomers have different functional groups.
$(iii)$ Homologous series are represented by the same general formula and two successive homologs differ by $CH_2$.
$(iv)$ Position isomers differ in the position of a double bond,triple bond,or functional group.
$CH_3-O-CH_3$ (dimethyl ether) and $CH_3CH_2OH$ (ethyl alcohol) have the same molecular formula $(C_2H_6O)$ but different functional groups (ether and alcohol,respectively).
$\therefore$ They are functional isomers.

(B) The molecular formula $C_4H_{10}O$ represents ethers with the following structures:
$1$. $CH_3CH_2OCH_2CH_3$ (Diethyl ether)
$2$. $CH_3OCH_2CH_2CH_3$ (Methyl propyl ether)
$3$. $CH_3OCH(CH_3)_2$ (Methyl isopropyl ether)
Metamers are isomers that have the same molecular formula but differ in the distribution of carbon atoms on either side of the functional group $(-O-)$.
Comparing these structures:
- Structure $1$ and $2$ are metamers (distribution $2,2$ vs $1,3$).
- Structure $1$ and $3$ are metamers (distribution $2,2$ vs $1,3$ branched).
- Structure $2$ and $3$ are chain isomers,not metamers.
Thus,there are $3$ ethers that exhibit metamerism.

(C) Phthalic acid is the ortho-isomer ($1, 2-$benzene dicarboxylic acid).
$1, 4-$benzene dicarboxylic acid (terephthalic acid) is a structural isomer of phthalic acid because both have the same molecular formula $(C_8H_6O_4)$ but different positions of the carboxylic acid groups on the benzene ring.

(D) $C_2H_5OH$ and $CH_3OCH_3$ are structural isomers with the same molecular formula $C_2H_6O$.
Since they have the same molecular mass,they will have the same vapor density at the same temperature and pressure.