Nomenclature of organic compounds Questions in English

(C) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has higher priority than the ether group $(-OCH_3)$.
$2$. Select the longest carbon chain containing the principal functional group: The chain has $3$ carbon atoms,so the parent alkane is propane.
$3$. Number the chain starting from the aldehyde carbon as $C-1$: The aldehyde carbon is $C-1$,the alpha carbon is $C-2$,and the terminal methyl carbon is $C-3$.
$4$. Identify substituents: There is a methoxy group $(-OCH_3)$ attached to the $C-2$ position.
$5$. Combine the parts: The name is $2-$methoxypropanal.

(A) The given compound is an ester: $CH_2=C(CH_3)-COOCH_3$.
$1$. Identify the ester group: The alkyl group attached to the oxygen is methyl $(-CH_3)$.
$2$. Identify the parent chain: The acid part is $CH_2=C(CH_3)-COOH$. The longest carbon chain containing the double bond and the carboxyl group has $3$ carbons,so the parent alkane is propane.
$3$. Numbering: The carbon of the carboxyl group is $C-1$. The double bond starts at $C-2$.
$4$. Substituents: There is a methyl group at $C-2$.
$5$. Naming: The parent chain is prop$-2-$enoic acid. As an ester,it becomes methyl $2-$methylprop$-2-$enoate.

(A) The given compound is $CH_3-CH=CH-COOH$.
$1$. Identify the longest carbon chain containing the functional group (carboxylic acid). The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain starting from the carboxylic acid carbon as $C-1$.
$3$. The double bond starts at $C-2$.
$4$. Thus,the name is but$-2-$ene$-1-$oic acid.

(A) The given compound is $CH_3-CH(OH)-COOH$.
$1$. Identify the principal functional group,which is the carboxylic acid group $(-COOH)$.
$2$. Number the carbon chain starting from the carbon of the $-COOH$ group as $C-1$.
$3$. The chain has $3$ carbons,so the parent alkane is propane.
$4$. The $-OH$ group is present at the $C-2$ position,so it is named as $2-$hydroxy.
$5$. Combining these,the $IUPAC$ name is $2-$hydroxypropanoic acid.

(A) The given compound is $HOOC-CH(OH)-CH(OH)-COOH$.
$1$. Identify the longest carbon chain containing the functional groups. The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. The functional group is carboxylic acid $(-COOH)$,which is present at positions $1$ and $4$.
$3$. The hydroxyl group $(-OH)$ is present at positions $2$ and $3$.
$4$. Combining these,the $IUPAC$ name is $2, 3-$ dihydroxybutane $-1, 4-$ dioic acid.

(A) The structure is $CH_3-CH(CH_3)-CO-COOH$.
$1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has higher priority than the ketone group $(>C=O)$.
$2$. Number the carbon chain starting from the carboxylic acid carbon as $C-1$.
$3$. The chain is $C_1(OOH)-C_2(=O)-C_3(H)(CH_3)-C_4(H_3)$.
$4$. At $C-2$,there is an oxo group. At $C-3$,there is a methyl group.
$5$. The parent chain has $4$ carbons,so it is a butanoic acid derivative.
$6$. The $IUPAC$ name is $3-$methyl$-2-$oxobutanoic acid.

(D) The given compound is $NC-CH_2-CH_2-COOH$.
In $IUPAC$ nomenclature,the carboxylic acid group $(-COOH)$ is given priority over the nitrile group $(-CN)$.
The carbon atom of the $-COOH$ group is assigned position $1$.
The chain consists of $3$ carbons: $C_1$ (in $-COOH$),$C_2$,and $C_3$ (attached to $-CN$).
Thus,the substituent $-CN$ is at position $3$,and the parent chain is propanoic acid.
The correct $IUPAC$ name is $3-$cyanopropanoic acid.

(B) $1$. Identify the longest carbon chain containing the principal functional groups. The longest chain containing the three carboxylic acid groups has $3$ carbons.
$2$. Number the chain such that the substituents get the lowest possible locants. Numbering from either end gives the central carbon the position $2$.
$3$. The structure is a propane derivative with three carboxylic acid groups at positions $1, 2,$ and $3$,and a hydroxyl group at position $2$.
$4$. The correct $IUPAC$ name is $2-$hydroxypropane$-1,2,3-$tricarboxylic acid (commonly known as citric acid).

(A) The given structure is $CH_3-C(COOC_2H_5)=CH-CH_2-COOH$.
The principal functional group is the carboxylic acid $(-COOH)$,so the numbering starts from its carbon atom.
The longest carbon chain containing the principal functional group and the double bond consists of $5$ carbon atoms.
The double bond is located at position $3$ and the ethoxycarbonyl group $(-COOC_2H_5)$ is attached at position $4$.
Therefore,the $IUPAC$ name is $4-$ethoxycarbonylpent$-3-$enoic acid.

(A) The structure is $CH_3-CH(CH_3)-CO-CONHBr$.
$1$. The parent chain is a four-carbon amide chain (butanamide).
$2$. There is a keto group at the $C-2$ position.
$3$. There is a methyl group at the $C-3$ position.
$4$. There is a bromine atom attached to the nitrogen atom of the amide group,denoted as $N$-Bromo.
$5$. Combining these,the $IUPAC$ name is $N$-Bromo$-2-$keto$-3-$methylbutanamide.

(A) In the given structure $Cl-CH_2-C(CH_3)=CH-CH_2OH$,the principal functional group is the alcohol $(-OH)$.
The carbon chain is numbered starting from the carbon attached to the $-OH$ group to give it the lowest possible number $(1)$.
Numbering:
$C1: -CH_2OH$
$C2: =CH-$
$C3: -C(CH_3)=$
$C4: -CH_2Cl$
The parent chain has $4$ carbons (but),a double bond at $C2$ (en),and an alcohol at $C1$ (ol).
Substituents are chloro at $C4$ and methyl at $C3$.
Alphabetically,chloro comes before methyl.
Thus,the $IUPAC$ name is $4-Chloro-3-methylbut-2-en-1-ol$.

(A) $1$. Identify the principal functional group: The amide group $(-CONH_2)$ has higher priority than the ketone group $(-CO-)$ and the halide $(-Br)$. Thus,the parent chain is an amide.
$2$. Number the chain starting from the amide carbon as $C-1$: $C-1$ is the carbon of $-CONH_2$,$C-2$ is the carbon attached to the amide,$C-3$ is the ketone carbon,$C-4$ is the next carbon,and $C-5$ is the terminal carbon.
$3$. The structure is $Br-CH_2-CH(CONH_2)-CO-CH_2-CH_3$. The longest chain containing the amide and ketone is a $5-$carbon chain (pentanamide).
$4$. At $C-2$,there is a substituent group: $Br-CH_2-$ which is a bromomethyl group.
$5$. At $C-3$,there is a ketone group,which is named as an oxo substituent.
$6$. Combining these,the name is $2-$(Bromomethyl)$-3-$oxopentanamide.

(B) The given structure is $(CH_3)_3C-CH_2-CH_2-Cl$.
$1$. Identify the longest carbon chain containing the functional group (chlorine). The longest chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain starting from the end closer to the substituent (chlorine). Thus,$C-1$ is attached to $Cl$,$C-2$ is $CH_2$,$C-3$ is $C(CH_3)_2$,and $C-4$ is $CH_3$.
$3$. The substituent $Cl$ is at position $1$,and two methyl groups are at position $3$.
$4$. Combining these,the $IUPAC$ name is $1-$chloro$-3, 3-$dimethylbutane.

(A) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has the highest priority,so the parent chain must include the carbon of the $-CHO$ group and it is numbered as $C-1$.
$2$. Select the longest carbon chain containing the principal functional group: The longest chain containing the $-CHO$ group has $5$ carbon atoms,so the parent alkane is pentane.
$3$. Number the chain: The aldehyde carbon is $C-1$. The chain is numbered as $C-1$ (aldehyde),$C-2$ $(-OH)$,$C-3$ $(-OH)$,$C-4$ $(-CH_3)$,and $C-5$ (terminal methyl).
$4$. Identify substituents: There are two hydroxyl groups $(-OH)$ at positions $2$ and $3$,and a methyl group $(-CH_3)$ at position $4$.
$5$. Combine the parts: The name is $2, 3-$dihydroxy$-4-$methylpentanal.

(C) $1$. Identify the principal functional group: The ketone group $(-CO-)$ has higher priority than the chloro group $(-Cl)$.
$2$. Select the longest carbon chain containing the principal functional group: The chain has $5$ carbon atoms,so the parent alkane is pentane.
$3$. Number the chain to give the principal functional group the lowest possible locant: Numbering from the right gives the ketone group position $2$.
$4$. Identify substituents: There is a methyl group at position $3$ and a chloro group at position $5$.
$5$. Combine the parts: $5-$Chloro$-3-$methylpentan$-2-$one.

(B) In the given compound $CH_3-CH(OH)-CH_2-C(=O)-CH_3$,there are two functional groups: ketone $(>C=O)$ and alcohol $(-OH)$.
According to $IUPAC$ priority rules,the ketone group has higher priority than the alcohol group.
Therefore,the principal functional group is the ketone,and the alcohol group is treated as a substituent (hydroxy).
Numbering starts from the right side to give the ketone group the lowest possible number $(2)$.
The parent chain has $5$ carbon atoms (pentane).
Thus,the name is $4-$hydroxypentan$-2-$one.

(A) The given compound is $CH_3-CH(Cl)-CH(Br)-CH(OH)-CH_3$.
The principal functional group is the hydroxyl group $(-OH)$.
According to $IUPAC$ rules,the carbon chain is numbered from the end that gives the principal functional group the lowest possible locant.
Numbering from the right side gives the $-OH$ group the position $2$.
The substituents are $Bromo$ at position $3$ and $Chloro$ at position $4$.
Arranging the substituents alphabetically,the name is $3-Bromo-4-chloropentan-2-ol$.

(C) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$: The chain has $5$ carbons,so the parent alkane is pentane.
$2$. Number the chain to give the lowest locants to the functional group and substituents: Numbering from the right side gives $-OH$ at $C_3$,$Cl$ at $C_1$ and $C_2$,and $Br$ at $C_3$.
$3$. The structure is $CH_2(Cl)-CH(Cl)-C(Br)(OH)-CH_2-CH_3$.
$4$. Alphabetical order for substituents: Bromo $(B)$ comes before Chloro $(C)$.
$5$. Combining these,the name is $3-$Bromo$-1,2-$dichloropentan$-3-$ol.

(A) The structure is $CH_3-CH(OH)-CH_2-O-CH_2-CH_3$.
$1$. Identify the longest carbon chain containing the principal functional group (alcohol). The chain has $3$ carbons,so the parent alkane is propane.
$2$. Number the chain starting from the end closer to the hydroxyl group $(-OH)$: $C1$ is $CH_2(OC_2H_5)$,$C2$ is $CH(OH)$,and $C3$ is $CH_3$.
$3$. The substituent is an ethoxy group $(-OCH_2CH_3)$ at position $1$.
$4$. The hydroxyl group is at position $2$.
$5$. Combining these,the $IUPAC$ name is $1-$ethoxypropan$-2-$ol.

(A) The structure consists of a benzene ring with an $-NH_2$ group and a $-Br$ group at the para position.
According to $IUPAC$ nomenclature,the $-NH_2$ group attached to the benzene ring is named as benzenamine (or aniline).
The $-NH_2$ group is given priority,so the carbon attached to it is position $1$.
The $-Br$ group is at position $4$.
Therefore,the correct $IUPAC$ name is $4-$bromo benzenamine.

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain must include this carbon as $C-1$.
$2$. Number the chain: Starting from the carboxylic acid carbon,the chain is numbered to give the lowest possible locants to the double bonds and substituents. The longest chain containing the $-COOH$ group and the double bonds is a heptenoic acid derivative.
$3$. Identify substituents and unsaturation: There is a methyl group at position $2$ and another methyl group at position $6$. There are double bonds at positions $2$ and $5$.
$4$. Construct the name: The parent chain is heptane,with two double bonds (dienoic acid) and two methyl groups. The correct name is $2,6-$dimethylhepta$-2,5-$dienoic acid.
$5$. Conclusion: The correct option is $A$.

(B) $1$. Identify the principal functional group: The aldehyde group $(-CHO)$ has the highest priority and must be assigned the lowest possible number,so it is carbon $1$.
$2$. Select the longest carbon chain containing the principal functional group and the double bond: The chain has $5$ carbons,so the parent alkane is pentane.
$3$. Number the chain: Starting from the aldehyde carbon as $C-1$,the double bond starts at $C-4$ and the methyl group is at $C-3$.
$4$. Combine the parts: The parent is $pent-4-enal$. With a methyl group at position $3$,the name is $3-methylpent-4-enal$.

(C) The given structure is a benzene ring with two hydroxyl $(-OH)$ groups attached at adjacent positions (ortho positions).
According to $IUPAC$ nomenclature rules for polyhydric phenols,the parent chain is benzene and the hydroxyl groups are indicated by the suffix '$-diol$'.
Numbering the ring starting from one of the hydroxyl groups gives the positions $1$ and $2$.
Therefore,the correct $IUPAC$ name is $Benzene-1, 2-diol$.

(B) The given structure is a benzene ring substituted with a carboxylic acid group $(-COOH)$ and a hydroxyl group $(-OH)$ at the ortho position relative to each other.
According to $IUPAC$ nomenclature rules,the carboxylic acid group has higher priority than the hydroxyl group.
Therefore,the parent chain is benzoic acid.
The hydroxyl group is treated as a substituent at the $2-$position.
Thus,the correct $IUPAC$ name is $2-$hydroxybenzoic acid.

(A) The given structure is $Ph-CH=CH-COOH$.
In this molecule,the principal functional group is the carboxylic acid $(-COOH)$,which is assigned position $1$.
The longest carbon chain containing the double bond and the carboxylic acid group consists of $3$ carbon atoms,hence the parent alkane is $propane$ (prop-).
The double bond starts at position $2$,so it is named $prop-2-enoic$ acid.
The phenyl group $(Ph-)$ is attached to the $3^{rd}$ carbon atom.
Therefore,the $IUPAC$ name is $3-phenylprop-2-enoic$ acid.

(C) The given structure is $C_6H_5CH_2Cl$.
According to $IUPAC$ nomenclature rules,the parent chain is methane,which is substituted by a phenyl group and a chlorine atom.
Therefore,the $IUPAC$ name is chlorophenylmethane.
Chloromethylbenzene is a common name (benzyl chloride) but is often accepted in some contexts; however,chlorophenylmethane is the systematic $IUPAC$ name.
Since both names are frequently used to describe this compound in various textbooks,option $(c)$ is the most appropriate choice.

(C) $1$. Identify the longest carbon chain containing the double bond. The chain has $4$ carbon atoms,so the parent alkane is butane,and with the double bond,it is but$-1-$ene.
$2$. Number the chain starting from the end closest to the double bond to give it the lowest possible locant. Thus,$C_1=CH_2$,$C_2=CH$,$C_3=C(CH_3)_2$,and $C_4=CH_3$.
$3$. There are two methyl groups attached at the $C_3$ position.
$4$. Combining these,the name is $3,3-$Dimethylbut$-1-$ene.

(B) To determine the $IUPAC$ name,first identify the longest carbon chain containing the double bond.
The chain $CH_2 = CH - CH(CH_3) - CH_3$ has $4$ carbon atoms,so the parent alkane is butane.
Numbering the chain from the end closer to the double bond gives the double bond position at $C-1$.
The methyl group is attached at the $C-3$ position.
Therefore,the correct name is $3-$methylbut$-1-$ene.

(C) The compound $C_6H_5COCl$ consists of a benzene ring attached to an acyl chloride group $(-COCl)$.
According to $IUPAC$ nomenclature rules for acid chlorides,the suffix used is "oyl chloride".
When the $-COCl$ group is attached to a benzene ring,the parent name is derived from benzoic acid,and the acid chloride is named as "Benzoyl chloride".

(B) To identify the compounds containing both an aldehyde $(-CHO)$ and an ether $(-O-)$ functional group,let's analyze each structure:
$(i)$ This structure contains an ether linkage $(-O-)$ and an aldehyde group $(-CHO)$. Thus,it is both an aldehyde and an ether.
$(ii)$ This structure contains a ketone group $(C=O)$ and an ether group $(-O-)$.
$(iii)$ This structure contains two ketone groups $(C=O)$.
$(iv)$ This structure contains an ether linkage within a cyclic ring and an aldehyde group $(-CHO)$. Thus,it is both an aldehyde and an ether.
$(v)$ This structure contains an aldehyde group $(-CHO)$ and an ester group $(-COOCH_3)$.
Therefore,compounds $(i)$ and $(iv)$ contain both an aldehyde and an ether functional group.

(C) According to $IUPAC$ nomenclature rules for cyclic compounds containing a double bond,the double bond is given priority and assigned positions $1$ and $2$.
For compound $(I)$,the bromine is at position $6$.
For compound $(II)$,the bromine is at position $3$.
For compound $(III)$,the bromine is at position $1$.
For compound $(IV)$,the bromine is at position $5$.
Sum of positions $= 6 + 3 + 1 + 5 = 15$.

(A) Let us determine the $IUPAC$ names for each structure:
$1$. $pentan-1-ol$
$2$. $pentan-2-ol$
$3$. $pentan-3-ol$
$4$. $2-methylbutan-1-ol$
$5$. $2,2-dimethylpropan-1-ol$
$6$. $3-methylbutan-1-ol$
$7$. $3-methylbutan-2-ol$
$8$. $2-methylbutan-2-ol$
All the given compounds have distinct $IUPAC$ names. Therefore,the number of compounds with the same $IUPAC$ name is $0$.

(C) To determine the number of substituents,we first identify the parent chain ($P$.$C$.) which is the longest carbon chain that maximizes the number of substituents attached to it.
By analyzing the structure,we select the longest chain such that the branches attached to it are maximized.
As shown in the solution image,there are $5$ groups attached to the parent chain,which are treated as substituents: one $Br$-containing group,one $Cl$-containing group,one ethyl group,and two methyl groups.
Therefore,the total number of substituents is $5$.

(B) $1$. In cyclic alkenes,the double bond is assigned positions $1$ and $2$.
$2$. Numbering is done to give the lowest possible locants to the substituents.
$3$. Starting from the methyl group at position $1$ and moving towards the double bond,the ethyl group is at position $5$.
$4$. Alphabetical order is followed for naming substituents: $E$ (ethyl) comes before $M$ (methyl).
$5$. Thus,the correct $IUPAC$ name is $5-$ ethyl $-1-$ methylcyclohexene.

(D) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbon atoms,so the parent alkane is hexane. Since there is a double bond,it is a hexene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from right to left gives the double bond the position $2$ $(C2=C3)$.
$3$. Identify the substituents: There is an ethyl group at position $4$ and a methyl group at position $3$.
$4$. Combine these to get the name: $4-$ethyl$-3-$methylhex$-2-$ene.

(B) $1$. Identify the parent chain: The parent chain is a cyclohexane ring.
$2$. Numbering: Number the ring to give the lowest possible locants to the substituents.
$3$. If we start numbering from the carbon with two methyl groups as $1$,the ethyl group gets position $2$. This gives the locant set $(1, 1, 2)$.
$4$. Alphabetical order: Ethyl comes before methyl.
$5$. Therefore,the name is $2-$ethyl$-1,1-$dimethylcyclohexane.

(B) The longest carbon chain contains $6$ carbon atoms,so the parent alkane is hexane.
There are two substituents: an ethyl group and a methyl group.
Numbering the chain from either end gives the locants $3$ and $4$.
According to $IUPAC$ rules,when two different substituents are present at equivalent positions,the substituent that comes first in alphabetical order (ethyl) is assigned the lower number.
Therefore,the correct name is $3-ethyl-4-methylhexane$.

(D) The given compound is a three-membered cyclic ether containing an oxygen atom.
According to $IUPAC$ nomenclature rules for epoxides,the oxygen atom is treated as an epoxy substituent on the alkane chain.
The parent chain is propane ($3$ carbons).
The oxygen atom is bonded to carbon $1$ and carbon $2$.
Therefore,the correct $IUPAC$ name is $1, 2-$ epoxy propane.

(A) $1.$ The longest carbon chain containing both multiple bonds has $7$ carbon atoms,so the parent name is hept.
$2.$ Numbering the chain from left to right gives the multiple bonds the lowest locants ($2$ and $4$). The double bond is at $C2$ and the triple bond is at $C4$.
$3.$ Since the higher priority groups ($CH_3$ and the alkyne group) are on opposite sides of the double bond,the configuration is $(E)$.
$4.$ Therefore,the $IUPAC$ name is $(E)-2-\text{hepten}-4-\text{yne}$.

(D) To determine the $IUPAC$ name,follow these steps:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $5$ carbons,so the parent alkane is pentene.
$2$. Number the chain starting from the end that gives the double bond the lowest possible locant. Numbering from left to right gives the double bond at position $2$.
$3$. Identify substituents: There is a methyl group at position $3$ and a bromo group at position $4$.
$4$. Combine these: The name is $4-$bromo$-3-$methylpent$-2-$ene.

(A) The principal functional group is $-COOH$ (carboxylic acid),which is assigned the lowest locant $(1)$.
Numbering the parent chain starting from the carboxylic acid carbon:
$C^5H_3-C^4H(CH_3)-C^3H(OH)-C^2H_2-C^1OOH$
There is a methyl group at position $4$ and a hydroxy group at position $3$.
According to $IUPAC$ nomenclature rules,substituents are listed in alphabetical order.
'Hydroxy' precedes 'methyl'.
Therefore,the correct name is $3$-hydroxy-$4$-methylpentanoic acid.

(D) To name the given compound according to $IUPAC$ rules,we identify the parent chain as the benzene ring.
We assign the lowest possible locants to the substituents.
Here,the substituents are a methyl group $(-CH_3)$,a chloro group $(-Cl)$,and a nitro group $(-NO_2)$.
Assigning position $1$ to the methyl group,the chloro group is at position $2$ and the nitro group is at position $4$.
Following alphabetical order for substituents (chloro,methyl,nitro),the name is $2-$chloro$-1-$methyl$-4-$nitrobenzene.

(A) $1$. Identify the longest carbon chain containing both the double and triple bonds. The chain has $7$ carbons,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the multiple bonds. Starting from the double bond end,the double bond is at $C-1$ and the triple bond is at $C-6$. If we start from the other end,the triple bond is at $C-1$ and the double bond is at $C-6$. According to $IUPAC$ rules,when both double and triple bonds are present,the double bond gets priority for the lowest number if they are at equivalent positions. Thus,the chain is numbered from the alkene end.
$3$. The substituents are two methyl groups at positions $3$ and $5$,and a propyl group at position $4$.
$4$. Combining these,the name is $3, 5-$dimethyl$-4-$propylhept$-1-$en$-6-$yne.

(D) Let us analyze each structure and its common name:
$(A)$ The structure shown is $CH_3-CH(CH_3)-CH_2-CH_3$,which is $2$-methylbutane. Its common name is Isopentane. This is correct.
$(B)$ The structure shown is $CH_3-C(CH_3)_2-CH_2-CH_3$,which is $2,2$-dimethylbutane. Its common name is Neohexane. This is correct.
$(C)$ The structure shown is $CH_3-C(CH_3)_2-CH_2-CH(CH_3)-CH_3$,which is $2,2,4$-trimethylpentane. Its common name is Isooctane. This is correct.
$(D)$ The structure shown is $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$,which is $3$-methylpentane. The common name Isohexane refers to $2$-methylpentane $(CH_3-CH(CH_3)-CH_2-CH_2-CH_3)$. Therefore,the label Isohexane for $3$-methylpentane is incorrect.

(A) The structure is $CH_2=CH-CH_2-COOH$.
According to $IUPAC$ nomenclature rules,the principal functional group (carboxylic acid,$-COOH$) is assigned the lowest possible number,which is $1$.
The carbon chain consists of $4$ carbon atoms,so the parent alkane is butane.
The double bond is present at the $3^{rd}$ carbon position.
Therefore,the correct $IUPAC$ name is but$-3-$enoic acid.

(D) The correct answer is $(d)$. The name $3-$amino$-2-$hydroxybutanedioic acid is incorrect.
According to $IUPAC$ rules,when two different substituents are present at equivalent positions,the one that comes first in alphabetical order is given the lower locant.
In the structure $HOOC-CH(OH)-CH(NH_2)-COOH$,numbering from either side gives locants $2$ and $3$.
Since 'amino' comes before 'hydroxy' alphabetically,the amino group should be at $C2$ and the hydroxy group at $C3$.
Therefore,the correct $IUPAC$ name is $2-$amino$-3-$hydroxybutanedioic acid.

(D) The given structure is $2$-acetoxybenzoic acid,which is commonly known as $Aspirin$.
It is also referred to as $o-$acetoxybenzoic acid because the acetoxy group $(-OCOCH_3)$ is at the ortho position relative to the carboxylic acid group $(-COOH)$.
It is also known as acetylsalicylic acid because it is the acetylated derivative of salicylic acid ($o$-hydroxybenzoic acid).
Therefore,all the given names are correct for the compound.

(B) The given structure is an ester,$CH_2=CH-O-CO-CH_3$.
$1$. Identify the alkyl group attached to the oxygen atom: $CH_2=CH-$ is the ethenyl (or vinyl) group.
$2$. Identify the parent carboxylic acid part: The acyl group is $CH_3-CO-$,which is derived from ethanoic acid.
$3$. The $IUPAC$ name for esters is written as 'alkyl alkanoate'.
$4$. Therefore,the name is ethenyl ethanoate.

(A) The given compound is $CH_3-CH=CH-C \equiv N$.
$1$. The $IUPAC$ name is $But-2-enenitrile$.
$2$. The common name is $Crotononitrile$.
$3$. $Allyl$ $cyanide$ refers to the structure $H_2C=CH-CH_2-CN$.
Therefore,$Allyl$ $cyanide$ is the invalid name for the given compound.

(A) The given compound is $CH_3-CO-CH_2-C(CH_3)_2-CN$.
According to $IUPAC$ priority rules,the nitrile group $(-CN)$ has higher priority than the ketone group $(>C=O)$.
Therefore,the carbon of the $-CN$ group is assigned number $1$.
The numbering of the chain is: $C(5)H_3-C(4)(=O)-C(3)H_2-C(2)(CH_3)_2-C(1)N$.
Substituents: Two methyl groups at $C2$ ($2,2-$ dimethyl) and an oxo group at $C4$ ($4-$ oxo).
Parent chain: Pentanenitrile.
Thus,the $IUPAC$ name is $2,2-$ Dimethyl $-4-$ oxopentanenitrile.