Structural isomerism Questions in English

(A) The molecular formula for both acetone $(CH_3COCH_3)$ and propanal $(CH_3CH_2CHO)$ is $C_3H_6O$.
Acetone contains a ketonic functional group $(>C=O)$,while propanal contains an aldehydic functional group $(-CHO)$.
Since they have the same molecular formula but different functional groups,they are classified as functional isomers.

(C) Methoxy methane $(CH_3-O-CH_3)$ and ethanol $(CH_3-CH_2-OH)$ have the same molecular formula $(C_2H_6O)$ but different functional groups (ether and alcohol).
Therefore,they are functional isomers.

(C) $C_2H_5OH$ shows functional isomerism with dimethyl ether.
$\text{CH}_3\text{CH}_2\text{OH} \text{ (Ethanol)} \rightleftharpoons \text{CH}_3\text{OCH}_3 \text{ (Dimethyl ether)}$

(C) Option $(A)$: $n$-Butane and iso-butane are chain isomers,not functional isomers.
Option $(B)$: Dimethyl ether $(CH_3OCH_3)$ and ethanol $(CH_3CH_2OH)$ are functional isomers because they contain different functional groups (ether and alcohol).
Option $(C)$: Propan-$1$-ol $(CH_3CH_2CH_2OH)$ and propan-$2$-ol $(CH_3CH(OH)CH_3)$ are position isomers because the position of the hydroxyl $(-OH)$ group changes along the carbon chain.
Option $(D)$: Ethanoic acid $(CH_3COOH)$ and methyl methanoate $(HCOOCH_3)$ are functional isomers because they contain different functional groups (carboxylic acid and ester).

(D) The molecule is $2,3-$dibromobutane$-1,4-$diol. The structure is $HOCH_2-CH(Br)-CH(Br)-CH_2OH$.
An asymmetric carbon (chiral center) is a carbon atom bonded to four different groups.
In the given molecule,the carbons at the $2$nd and $3$rd positions are bonded to $-H$,$-Br$,$-CH_2OH$,and the other $-CH(Br)CH_2OH$ group.
Since each of these two carbons is bonded to four distinct groups,they are asymmetric carbons.
Thus,there are $2$ asymmetric carbons.

(A) $C_5H_{12}$ is an alkane. It can have the following structures:
$(a)$ $n-$pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$
$(b)$ $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$
$(c)$ $2,2-$dimethylpropane $(CH_3-C(CH_3)_2-CH_3)$. It is also known as neopentane.
Therefore,there are $3$ isomers of $C_5H_{12}$.

(D) Functional isomers are compounds that have the same molecular formula but different functional groups.
In option $D$,$CH_3CH_2CH_2OH$ is an alcohol (functional group $-OH$),while $CH_3-O-CH_2CH_3$ is an ether (functional group $-O-$).
Both compounds have the molecular formula $C_3H_8O$,but they belong to different functional group classes,making them functional isomers.

(C) The structural isomers for $C_4H_9Br$ are:
$1$. $CH_3CH_2CH_2CH_2Br$ ($1$-bromobutane)
$2$. $CH_3CH_2CH(Br)CH_3$ ($2$-bromobutane)
$3$. $(CH_3)_2CHCH_2Br$ ($1$-bromo-$2$-methylpropane)
$4$. $(CH_3)_3CBr$ ($2$-bromo-$2$-methylpropane)
Among these,$2$-bromobutane $(CH_3CH_2CH(Br)CH_3)$ has a chiral center at the $C_2$ position,meaning it exists as a pair of enantiomers ($R$ and $S$ configurations).
Therefore,the total number of isomers including stereoisomers is $4$ structural isomers + $1$ additional enantiomer = $5$ isomers.

(B) Metamerism arises due to the different distribution of carbon atoms on either side of a polyvalent (bivalent or trivalent) functional group.
Examples of functional groups that show metamerism include ethers $(-O-)$,ketones $(-CO-)$,and secondary amines $(-NH-)$.

(B) Functional isomerism occurs when compounds have the same molecular formula but different functional groups.
$C_2H_5CO_2H$ (propanoic acid) and $CH_3CO_2CH_3$ (methyl acetate) both have the molecular formula $C_3H_6O_2$.
Propanoic acid contains a carboxylic acid group $(-COOH)$,while methyl acetate contains an ester group $(-COOCH_3)$.
Therefore,they are functional isomers.

(D) Functional isomers are compounds that have the same molecular formula but possess different functional groups.
In option $(D)$,the first compound is $CH_3CH_2CH_2OH$ (propan$-1-$ol),which contains an alcohol $(-OH)$ functional group.
The second compound is $CH_3OCH_2CH_3$ (methoxyethane),which contains an ether $(-O-)$ functional group.
Since they have the same molecular formula $(C_3H_8O)$ but different functional groups,they are functional isomers.

(C) molecule is asymmetric if it lacks any element of symmetry,such as a plane of symmetry or a center of inversion.
$A$: $cis-1-bromo-3-chlorocyclobutane$ has a plane of symmetry passing through $C1$ and $C3$.
$B$: $1-bromo-2,2-dichlorocyclopropane$ has a plane of symmetry passing through the $C1-Br$ bond and the $C2$ atom.
$C$: $trans-1-bromo-3-chlorocyclobutane$ lacks both a plane of symmetry and a center of symmetry,making it asymmetric.
$D$: $1-bromo-2,2-dichlorocyclopropane$ (as shown in the image) also possesses a plane of symmetry.
Therefore,the correct option is $C$.

(B) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom adjacent to a carbonyl group $(C=O)$.
$A$. $(Me_3CCO)_3CH$ has an $\alpha$-hydrogen atom on the central carbon,which is flanked by three carbonyl groups,making it highly acidic and capable of tautomerism.
$B$. Cyclohex$-2-$ene$-1,4-$dione has $\alpha$-hydrogen atoms at the $C-5$ and $C-6$ positions,allowing it to exhibit keto-enol tautomerism.
$C$. Cyclohexa$-2,5-$diene$-1,4-$dione (p-benzoquinone) has no $\alpha$-hydrogen atoms,so it cannot exhibit tautomerism.
$D$. Cyclohexane$-1,2-$dione has $\alpha$-hydrogen atoms at the $C-3$ and $C-6$ positions,allowing it to exhibit keto-enol tautomerism.
Given the options provided in the image,compounds $(a)$,$(b)$,and $(d)$ all possess $\alpha$-hydrogens and can exhibit tautomerism. However,in many standard contexts for this specific question,$(b)$ is often highlighted for its specific structural features.

(B) The compound $MeCOCH_{2}CO_{2}Et$ is ethyl acetoacetate,a $\beta$-keto ester.
It can form an enol by the migration of an $\alpha$-hydrogen to the carbonyl oxygen.
The most stable enol form is the one that is stabilized by intramolecular hydrogen bonding and conjugation.
In $MeC(OH)=CHCO_{2}Et$,the enol double bond is conjugated with the ester carbonyl group,and it forms a stable six-membered ring through intramolecular hydrogen bonding between the hydroxyl group and the ester carbonyl oxygen.
Therefore,$MeC(OH)=CHCO_{2}Et$ is the most contributing tautomeric enol form.

(A) Keto-enol tautomerism requires the presence of at least one $\alpha$-hydrogen atom adjacent to the carbonyl group $(C=O)$.
In $C_6H_5COC_6H_5$ (benzophenone),the carbonyl carbon is attached to two phenyl groups $(C_6H_5)$.
There is no $\alpha$-carbon with an $\alpha$-hydrogen atom available in this molecule.
Therefore,it cannot exhibit keto-enol tautomerism.

(C) The reaction is $C_3H_8 \xrightarrow{-2H, +2Cl} C_3H_6Cl_2$.
The structural isomers of $C_3H_6Cl_2$ are:
$1,1$-dichloropropane
$2,2$-dichloropropane
$1,2$-dichloropropane
$1,3$-dichloropropane
Among these,$1,2$-dichloropropane contains a chiral carbon atom (the $C-2$ atom),which means it exists as a pair of enantiomers (optical isomers).
Therefore,the total number of isomers is $4$ structural isomers + $1$ additional enantiomer = $5$ isomers.

(C) To determine the relationship between the two structures,we assign the $R/S$ configuration to the chiral centers.
For the first structure $(I)$:
The top chiral center has $-OH$ $(1)$,$-CH_2OH$ $(2)$,$-CH_3$ $(3)$,and $-H$ $(4)$. With $-H$ on the horizontal,the configuration is $R$.
The bottom chiral center has $-OH$ $(1)$,$-CH(OH)CH_3$ $(2)$,$-CH_2OH$ $(3)$,and $-H$ $(4)$. The configuration is $R$.
For the second structure $(II)$:
Rotating the Fischer projection by $180^{\circ}$ in the plane of the paper does not change the configuration of the chiral centers.
After a $180^{\circ}$ rotation of structure $(II)$,it becomes identical to structure $(I)$.
Therefore,both structures represent the same molecule.

(D) Metamerism arises due to the difference in the nature of alkyl groups attached to the same polyvalent functional group (like $-O-$,$-S-$,$-NH-$,$-CO-$).
In option $(D)$,both molecules are ethers with the same molecular formula $C_{4}H_{10}O$.
$CH_{3}CH_{2}OCH_{2}CH_{3}$ is diethyl ether,where the oxygen atom is attached to two ethyl groups.
$CH_{3}OCH_{2}CH_{2}CH_{3}$ is methyl propyl ether,where the oxygen atom is attached to one methyl group and one propyl group.
Since the distribution of alkyl groups around the oxygen atom is different,they are metamers.