Nomenclature of organic compounds Questions in English

(A) In $Br-CH_2-CH=CH_2$,the double bond is given priority over the halogen substituent for numbering.
The numbering of the carbon chain should start from the end containing the double bond:
$CH_2(1)=CH(2)-CH_2(3)-Br$
Therefore,the correct $IUPAC$ name is $3-$Bromoprop$-1-$ene.
Thus,option $(A)$ is not according to the $IUPAC$ system.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,making it a hexene derivative.
$2$. Number the chain starting from the end closest to the double bond. The double bond starts at $C-1$.
$3$. The substituents are an ethyl group at $C-3$ and a propyl group at $C-4$.
$4$. Combining these,the correct $IUPAC$ name is $4-$ethyl$-3-$propylhex$-1-$ene.

$(A)$ The longest carbon chain contains $5$ carbon atoms with a $C=C$ bond starting at the second carbon atom, so it is a $2$-pentene derivative.
The substituents are a chlorine atom $(Cl)$ at position $2$ and an iodine atom $(I)$ at position $3$.
Looking at the geometry across the $C=C$ bond, the higher priority groups (based on atomic number, $Cl$ vs $CH_3$ on one side, and $I$ vs $CH_2CH_3$ on the other) are on opposite sides of the double bond, which corresponds to the $E$ configuration, often referred to as $trans$ in simple cases where similar groups are on opposite sides.
Specifically, the $Cl$ and the $CH_2CH_3$ group are on opposite sides of the double bond, and the $CH_3$ and $I$ are on opposite sides. Thus, the configuration is $trans-2-chloro-3-iodo-2-pentene$.

(B) The given compound is $CH_3-CH=CH-C\equiv CH$.
According to $IUPAC$ rules,the numbering of the carbon chain starts from the end that gives the lowest possible locants to the multiple bonds.
If there is a choice between a double bond and a triple bond,the double bond is given priority for the lower number.
Here,the double bond starts at $C-3$ and the triple bond starts at $C-1$.
Thus,the correct name is pent$-3-$en$-1-$yne.

(C) The structure is $CH \equiv C-CH=CH_2$.
According to $IUPAC$ rules for naming hydrocarbons containing both double and triple bonds,if both are at the same position from either end,the double bond is given priority in numbering.
Numbering from right to left: $CH_2(1)=CH(2)-C(3) \equiv CH(4)$.
The double bond is at position $1$ and the triple bond is at position $3$.
Therefore,the name is but$-1-$ene$-3-$yne.

(C) $1$. Identify the principal functional group: The compound contains an acyl chloride group $(-COCl)$,which takes priority in numbering.
$2$. Select the longest carbon chain containing the functional group: The longest chain starting from the carbonyl carbon has $5$ carbons,making it a pentanoyl chloride derivative.
$3$. Number the chain: The carbonyl carbon of the $-COCl$ group is assigned position $1$.
$4$. Identify substituents: There are methyl groups at positions $2$ and $3$.
$5$. Combine the parts: The name is $2,3-$dimethylpentanoyl chloride.

(A) $1$. In cyclic compounds containing a double bond,the double bond is given priority for numbering.
$2$. The carbon atoms of the double bond are assigned positions $1$ and $2$ such that the substituents get the lowest possible locants.
$3$. Starting from the carbon attached to the chlorine atom as position $1$,we move towards the double bond to assign position $2$ to the other carbon of the double bond.
$4$. This gives the bromine atom at position $3$.
$5$. Thus,the correct name is $3-$bromo$-1-$chlorocyclohexene.

(A) $1$. Identify the longest carbon chain in the given structure. The longest chain contains $7$ carbon atoms,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the substituents. Numbering from right to left gives the substituents at positions $3$ and $4$.
$3$. At position $3$,there is an ethyl group $(-CH_2CH_3)$.
$4$. At position $4$,there are two methyl groups $(-CH_3)$.
$5$. Combining these,the $IUPAC$ name is $3-$ethyl$-4,4-$dimethylheptane.

(A) In the $IUPAC$ system of nomenclature,the priority order for functional groups is determined by specific rules.
According to the $IUPAC$ priority table,the carboxylic acid group $(-COOH)$ has the highest priority among the given groups.
The sulfonic acid group $(-SO_3H)$ follows,then the amide group $(-CONH_2)$,and finally the aldehyde group $(-CHO)$.
Therefore,the correct decreasing order is: $-COOH > -SO_3H > -CONH_2 > -CHO$.

(A) The structure of neopentane is $CH_3-C(CH_3)_2-CH_3$.
To determine the $IUPAC$ name,we identify the longest carbon chain,which contains $3$ carbon atoms,making it a propane derivative.
There are two methyl groups attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2,2-$dimethylpropane.

(D) $6-$ Methylbicyclo $[3.2.1]$oct$-2-$ene is the correct $IUPAC$ name.
$(a)$ $1, 2, 3-$ Trihydroxypropane is a common name; the $IUPAC$ name is Propane-$1,2,3$-triol.
$(b)$ For $CH_3-CH(C_2H_5)-CH(CH_3)-CH_3$,the longest chain has $5$ carbons,so the correct name is $2,3-$Dimethylpentane.
$(c)$ For $CH_3-CH(COOH)-COOH$,the correct $IUPAC$ name is Methylpropanedioic acid.

(B) $1.$ In the given compound $CH_3-CH_2-O-C(=O)-CH_2-COOH$,there are two functional groups: carboxylic acid $(-COOH)$ and ester $(-COOR)$.
$2.$ The carboxylic acid group has higher priority than the ester group according to $IUPAC$ rules.
$3.$ The parent chain is the longest carbon chain containing the principal functional group $(-COOH)$,which is ethanoic acid $(CH_3-COOH)$.
$4.$ The ester group $-COOCH_2CH_3$ is treated as a substituent and named as 'ethoxycarbonyl'.
$5.$ This substituent is attached to the $2$nd carbon of the ethanoic acid chain.
$6.$ Therefore,the correct $IUPAC$ name is Ethoxycarbonyl ethanoic acid.

(D) The given structure is $CH_3CH_2-NH-CHO$.
This compound is an $N$-substituted amide.
The parent amide chain contains one carbon atom,which is derived from methanoic acid,so the parent name is methanamide.
The nitrogen atom is substituted with an ethyl group $(CH_3CH_2-)$.
Therefore,the $IUPAC$ name is $N$-ethyl methanamide.

(C) $1$. Identify the principal functional group: The ester group $(-COOC_2H_5)$ has higher priority than the aldehyde group $(-CHO)$. Thus,the parent chain is a cyclopentanecarboxylate ester.
$2$. Numbering: The carbon attached to the ester group is assigned position $1$. The aldehyde group $(-CHO)$ is then at position $2$.
$3$. Naming: The substituent is the formyl group at position $2$. The ester part is ethyl. Combining these,the name is Ethyl $2-$formylcyclopentanecarboxylate.
$4$. Therefore,the correct option is $C$.

(B) $1$. Identify the principal functional group: The carboxylic acid $(-COOH)$ group has the highest priority,so the parent chain is a butanoic acid derivative.
$2$. Number the parent chain: Start numbering from the carbon of the $-COOH$ group as $C-1$. The chain is $C1-C2-C3(CH_3)-C4$.
$3$. Identify the substituent: At position $C-4$,there is a cyclohexenyl ring. Number the ring starting from the point of attachment to the main chain as $1'$. The double bond is at $2'$ and the hydroxyl group is at $4'$. Thus,the substituent is $4'-$hydroxycyclohex$-2'-$enyl.
$4$. Combine the parts: The compound is $4-(4'-$hydroxycyclohex$-2'-$enyl)$-3-$methylbutanoic acid.

(C) The given structure is a bicyclic compound.
To name it,we first identify the bridgehead carbons,which are carbons $1$ and $5$.
The number of carbons in the three bridges are $3$ (between $1$ and $5$ via $2,3,4$),$1$ (between $1$ and $5$ via $6$),and $1$ (between $1$ and $5$ via $7$).
Thus,the bicyclo system is bicyclo$[3.1.1]$heptane.
Numbering starts from a bridgehead carbon,proceeds along the longest bridge to the other bridgehead,then along the next longest bridge,and finally the shortest bridge.
The double bond is at position $2$ and there are methyl groups at positions $2$ and $6,6$.
Therefore,the correct $IUPAC$ name is $2,6,6-$trimethylbicyclo$[3.1.1]$hept$-2-$ene.

(B) The given structure is $CH_3-CH_2-C(=CH_2)-CH(CH_3)-CH_3$.
$1.$ The longest carbon chain containing the double bond is selected as the parent chain. Here,it is a $4-$carbon chain (butene).
$2.$ Numbering starts from the end closer to the double bond,so the double bond is at position $1$.
$3.$ There is an ethyl group at position $2$ and a methyl group at position $3$.
$4.$ According to $IUPAC$ rules,substituents are listed in alphabetical order (ethyl before methyl).
Therefore,the name is $2-$ethyl$-3-$methyl$-1-$butene.

(A) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$. The chain is a butane chain with the $-OH$ group at position $2$ and the $-Br$ group at position $3$. The parent name is $3-$bromo$-2-$butanol.
$2$. Assign priorities to the substituents at each chiral center using Cahn-Ingold-Prelog $(CIP)$ rules.
$3$. At $C-2$: The priorities are $-OH (1) > -CH(Br)CH_3 (2) > -CH_3 (3) > -H (4)$. The $-H$ is on a dashed bond (pointing away). The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,so the configuration is $2R$.
$4$. At $C-3$: The priorities are $-Br (1) > -CH(OH)CH_3 (2) > -CH_3 (3) > -H (4)$. The $-H$ is on a wedged bond (pointing towards). The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise,which would be $S$,but since $-H$ is pointing towards us,we reverse it to $R$. Thus,the configuration is $3R$.
$5$. The complete name is $(2R, 3R)-3-$bromo$-2-$butanol.

(D) The double bond equivalent $(DBE)$ or degree of unsaturation is calculated as: $DBE = \text{Number of rings} + \text{Number of double bonds} + 2 \times \text{Number of triple bonds}$.
In the given structure:
$1$. There are $3$ rings (two benzene rings and one central six-membered ring).
$2$. Each benzene ring has $3$ double bonds,so $2 \times 3 = 6$ double bonds.
$3$. The $-COOH$ group contains $1$ $C=O$ double bond.
$4$. The phenyl group $(-Ph)$ attached to the structure is a benzene ring,which contributes $1$ ring and $3$ double bonds.
Total rings = $3$ (from the main fused system) $+ 1$ (from the phenyl group) $= 4$.
Total double bonds = $6$ (from the two fused benzene rings) $+ 1$ (from the $C=O$ group) $+ 3$ (from the phenyl group) $= 10$.
$DBE = 4 + 10 = 14$. Wait,let's re-evaluate the structure carefully.
The structure is a phenalene derivative. The core is phenalene ($3$ rings,$10$ $\pi$ electrons). The phenyl group adds $1$ ring and $3$ double bonds. The $-COOH$ group adds $1$ double bond.
Total rings = $3 + 1 = 4$.
Total double bonds = $5$ (in phenalene) $+ 3$ (in phenyl) $+ 1$ (in $COOH$) $= 9$.
$DBE = 4 + 9 = 13$.
Thus,the correct option is $D$.

(A) $1$. Identify the principal functional group: The ester group $(-COOCH_3)$ is the principal functional group,so the parent chain is a cyclohexane ring with the suffix 'carboxylate'.
$2$. Numbering the ring: The carbon atom of the cyclohexane ring attached to the ester group is assigned position $1$.
$3$. Substituent placement: To give the lowest locants to the substituents,we number the ring clockwise. The ethyl group is at position $2$ and the bromo group is at position $5$.
$4$. Alphabetical order: 'Bromo' comes before 'ethyl'.
$5$. Final name: Methyl $5$-bromo-$2$-ethylcyclohexanecarboxylate.

(B) The compound is $CH_3-CH=CH-CH=CH-C \equiv C-CH_3$.
$1$. The longest carbon chain contains $8$ carbon atoms,so the root name is $octa$.
$2$. Numbering the chain from the left gives the double bonds positions at $2$ and $4$,and the triple bond position at $6$.
$3$. Numbering from the right gives the triple bond position at $2$ and the double bonds at $4$ and $6$.
$4$. According to $IUPAC$ rules,when both double and triple bonds are present,the lowest locant set is preferred. Comparing the sets $(2, 4, 6)$ and $(2, 4, 6)$,we apply the rule that double bonds get priority over triple bonds for the lower locant if there is a tie in the set.
$5$. Thus,the correct name is $Octa-2,4-diene-6-yne$.

(B) The principal functional group is the carboxylic acid $(-COOH)$.
The longest carbon chain containing the $-COOH$ group and one of the $-CHO$ groups has $4$ carbons (butanoic acid).
The carbon of the terminal $-CHO$ group is $C4$,and it is designated as "oxo".
The other $-CHO$ group at $C3$ is treated as a substituent and named "formyl".
Thus,the name is $3-$formyl$-4-$oxobutanoic acid.

(B) The priority order of functional groups is $-CN > -CHO > >C=O$.
Therefore,the principal functional group is nitrile $(-CN)$.
The longest carbon chain containing the $-CN$ group is selected as the parent chain,and numbering starts from the carbon of the $-CN$ group.
The $-CHO$ group at position $2$ is treated as a substituent and named 'formyl',and the keto group at position $3$ is named 'oxo'.
The parent chain has $5$ carbons,so it is 'pentanenitrile'.
The correct $IUPAC$ name is $2-$formyl$-3-$oxopentanenitrile.

(B) According to $IUPAC$ nomenclature rules,the priority of functional groups determines the suffix and the numbering of the carbon chain.
$1$-chloro-$1$-ethoxypropane $(CH_3CH_2CH(Cl)OCH_2CH_3)$ is a valid name where the chloro and ethoxy groups are treated as substituents.
$1$-ethoxy-$2$-propanol $(CH_3CH(OH)CH_2OCH_2CH_3)$ is a valid name where the hydroxyl group $(-OH)$ has higher priority than the ethoxy group $(-OCH_2CH_3)$,thus it is named as an alcohol.
$1$-ethoxy-$1$-propanamine $(CH_3CH_2CH(NH_2)OCH_2CH_3)$ is a valid name where the amine group $(-NH_2)$ has higher priority than the ethoxy group,thus it is named as an amine.
$1$-amino-$1$-ethoxypropane is incorrect because the amine group $(-NH_2)$ has higher priority than the ethoxy group $(-OCH_2CH_3)$. Therefore,the parent chain should be named as an amine (propanamine),not as a propane derivative with an amino substituent.

(C) The principal functional group is the ester $(-COOCH_3)$,so the suffix is 'oate' and the prefix is 'Methyl'.
The longest carbon chain containing the ester group has $4$ carbons,making it a 'butanoate'.
There is an acetyl group $(-COCH_3)$ at the $2^{nd}$ carbon and an oxo group $(=O)$ at the $3^{rd}$ carbon.
Therefore,the $IUPAC$ name is Methyl-$2$-acetyl-$3$-oxobutanoate.

(C) The structure of $\beta$-ethoxy-$\alpha$-hydroxy propionic acid is $CH_3CH_2OCH_2CH(OH)COOH$.
$1$. Identify the longest carbon chain containing the carboxylic acid group: The chain has $3$ carbons,so the parent alkane is propane,and the acid is propanoic acid.
$2$. Number the chain starting from the carboxylic acid carbon as $C-1$.
$3$. The hydroxy group $(-OH)$ is at the $C-2$ position,and the ethoxy group $(-OCH_2CH_3)$ is at the $C-3$ position.
$4$. Combining these,the $IUPAC$ name is $3-$Ethoxy$-2-$hydroxy propanoic acid.

(C) The structure is $CH_3-CO-CH(CH_3)-CH(CHO)-CH(CH_3)_2$.
According to $IUPAC$ rules,the aldehyde group $(-CHO)$ has higher priority than the ketone group $(>C=O)$.
Thus,the parent chain is the longest chain containing the $-CHO$ group,which consists of $5$ carbons.
The carbon of the $-CHO$ group is numbered as $C1$.
The substituents are: an isopropyl group at $C2$,a methyl group at $C3$,and an oxo group at $C4$.
Combining these,the name is $2-Isopropyl-3-methyl-4-oxopentanal$.

(B) The $IUPAC$ name is determined by:
$1.$ Identifying the principal functional group: Carboxylic acid $(-COOH)$,which is assigned $C1$.
$2.$ The longest chain containing the double bond and $-COOH$ has $5$ carbons: $pent-2-enoic acid$.
$3.$ Substituents are $3-amino$,$4-chloro$,and $2-methyl$.
$4.$ Arranging substituents alphabetically: $3-amino-4-chloro-2-methyl-2-pentenoic acid$.

(C) The principal functional group is the carboxylic acid $(-COOH)$.
The parent chain must include both $-COOH$ groups to maximize the number of principal functional groups,making it a $4-$carbon chain,i.e.,butane$-1, 4-$dioic acid.
The substituents are an amino group $(-NH_2)$ at $C2$ and a formyl group $(-CHO)$ at $C3$.
According to $IUPAC$ rules,the chain is numbered to give the alphabetically prior substituent (amino) the lower locant.
Thus,the amino group is at $C2$ and the formyl group is at $C3$.
The correct $IUPAC$ name is $2-$amino$-3-$formylbutane$-1, 4-$dioic acid.

(D) The structure provided for iso-octane is $2,2,4$-trimethylpentane. While this is the universally accepted common name for this compound in fuel chemistry (octane rating),it does not follow the systematic "iso" prefix rule. The "iso" prefix typically denotes a single methyl branch on the second carbon of an otherwise straight chain,which for an octane would be $2$-methylheptane. Therefore,compared to the other standard common names,this is considered an exception or a mismatch in strict nomenclature terms.

(D) The given compound is $CH_2(OH)-CH(OH)-CH_2(OH)$.
The longest carbon chain contains $3$ carbon atoms,so the parent alkane is propane.
There are three hydroxyl $(-OH)$ groups at positions $1$,$2$,and $3$.
According to $IUPAC$ rules,the suffix $'-ol'$ is used for alcohols. For three groups,it becomes $'-triol'$.
Therefore,the $IUPAC$ name is Propane$-1,2,3-$triol.

(D) According to the $IUPAC$ priority order of functional groups,the priority sequence is: $-CHO > >C=O > -OH > -C \equiv C^{-}$.
Therefore,the aldehyde group $(-CHO)$ is regarded as the principal functional group among the given options.

(C) The structure $2-$ethyl$-3-$pentyne is drawn by placing an ethyl group at position $2$ of a $5-$carbon chain containing a triple bond at position $3$.
By expanding the structure,we identify the longest carbon chain containing the triple bond.
The longest chain consists of $6$ carbon atoms.
Numbering the chain to give the lowest possible locant to the triple bond,we start from the right side.
The triple bond is at position $2$ and a methyl group is at position $4$.
Therefore,the correct $IUPAC$ name is $4-$methylhex$-2-$yne.

(C) $1$. Identify the longest carbon chain containing the double bond. The chain has $5$ carbons,so the parent alkane is pentene.
$2$. Number the chain to give the double bond the lowest possible locant. Numbering from the left gives the double bond at position $2$ $(CH_3-C(Cl)=C(I)-CH_2CH_3)$.
$3$. The substituents are chloro at position $2$ and iodo at position $3$. The name is $2-$chloro$-3-$iodo$-2-$pentene.
$4$. Determine the $E/Z$ configuration using Cahn-Ingold-Prelog priority rules.
- At $C-2$: $Cl$ (atomic number $17$) has higher priority than $CH_3$ (atomic number $6$).
- At $C-3$: $I$ (atomic number $53$) has higher priority than $CH_2CH_3$ (atomic number $6$).
- The high-priority groups ($Cl$ and $I$) are on the same side of the double bond,which corresponds to the $Z$ configuration.
$5$. Therefore,the correct name is $Z-2-$chloro$-3-$iodo$-2-$pentene.

(B) The compound $CH_3-C \equiv C-CH(CH_3)-COOH$ has $5$ carbon atoms in the parent chain including the functional group.
The triple bond is at the $3^{rd}$ position.
Therefore,the correct $IUPAC$ name is $2-$Methylpent$-3-$ynoic acid.
The given name $2-$Methylhex$-3-$enoic acid is incorrect because it indicates $6$ carbons (hex) and a double bond (ene).

(C) $1$. Identify the principal functional group,which is the hydroxyl group $(-OH)$. The parent chain is cyclohexane,so the suffix is $-ol$.
$2$. Number the ring starting from the carbon attached to the $-OH$ group as $1$.
$3$. Continue numbering to give the substituents the lowest possible locants. Moving clockwise,the two methyl groups are at position $3$.
$4$. Thus,the compound is $3,3-$ dimethylcyclohexan$-1-$ol,which is commonly written as $3,3-$ dimethyl $-1-$ cyclohexanol.

(B) The parent chain consists of $3$ carbon atoms including the carboxylic acid group,so it is propanoic acid.
Numbering starts from the $-COOH$ carbon as $C1$.
There is a chloro group at $C2$ and an $N,N-$diethylamino group at $C3$.
Alphabetically,'chloro' comes before 'diethylamino'.
Thus,the correct $IUPAC$ name is $2-$chloro$-3-(N,N-$diethylamino$)$propanoic acid.

(A) $1$. Identify the longest carbon chain: The chain contains two carbon atoms,so the parent alkane is ethane.
$2$. Numbering the chain: Numbering starts from the carbon attached to the halogens to give the lowest locants. Thus,the carbon with $Br$,$Cl$,and $I$ is $C-2$,and the $CF_3$ group is on $C-1$.
$3$. Alphabetical order: The substituents are bromo,chloro,fluoro,and iodo. Alphabetically,they are bromo,chloro,fluoro,and iodo.
$4$. Combining the name: The substituents are at position $2$ (bromo,chloro,iodo) and position $1$ (trifluoro). The correct $IUPAC$ name is $2-$bromo$-2-$chloro$-2-$iodo$-1,1,1-$trifluoroethane.

(B) The given structure is $CH_3-CH_2-C(=CH_2)-COOH$.
$1.$ The principal functional group is the carboxylic acid $(-COOH)$,which is assigned position $1$.
$2.$ The parent chain must contain the principal functional group and the double bond.
$3.$ The longest chain containing both the $-COOH$ group and the $C=C$ bond has $3$ carbon atoms (propanoic acid backbone).
$4.$ There is an ethyl group $(-CH_2-CH_3)$ attached to the $2^{nd}$ carbon atom.
$5.$ Therefore,the $IUPAC$ name is $2-$ethylprop$-2-$enoic acid.

(A) The given structure is $(CH_3)_3C-C\equiv C-CH(CH_3)_2$.
First,identify the longest carbon chain containing the triple bond.
The structure is $CH_3-C(CH_3)_2-C\equiv C-CH(CH_3)_2$.
The longest chain has $6$ carbons,so the parent alkane is hexane.
Numbering from the left,the triple bond starts at position $3$.
There are methyl groups at positions $2$ and $5$.
Thus,the $IUPAC$ name is $2,5-$dimethylhex$-3-$yne.

(A) The given structure is $[CH_3CH(CH_3)]_2C(CH_2CH_3)C(CH_3)C(CH_2CH_3)_2$.
First,identify the longest carbon chain containing the double bond.
The longest chain consists of $7$ carbons,making it a heptene derivative.
Numbering the chain to give the double bond the lowest possible locant,we start from the right side.
The substituents are identified at their respective positions: $3,5-$diethyl,$4,6-$dimethyl,and $5-(1-$methylethyl$)$.
Thus,the correct $IUPAC$ name is $3,5-$diethyl$-4,6-$dimethyl$-5-(1-$methylethyl$)-3-$heptene.

(B) The structure of acetyl acetone is $CH_3-CO-CH_2-CO-CH_3$.
The parent chain contains $5$ carbon atoms,so the root name is pentane.
There are two ketone groups at the $2^{nd}$ and $4^{th}$ positions.
Therefore,the $IUPAC$ name is pentane-$2,4-$dione.

(A) The parent hydrocarbon contains $3$ carbon atoms.
Hence,its name is propane. Each carbon atom has a hydroxyl group attached to it.
The suffix $-triol$ is used to represent the presence of $3$ hydroxyl groups.
Therefore,the correct $IUPAC$ name of glycerine is Propane-$1,2,3$-triol.

(D) According to $IUPAC$ rules,when three or more identical carbon-containing functional groups (like $-CN$) are attached to a parent chain,the carbons of these groups are not included in the parent chain.
In such cases,the suffix 'carbonitrile' is used for the $-CN$ group.
For the given structure,the parent chain is propane,and the three $-CN$ groups are attached at positions $1, 2,$ and $3$.
Therefore,the correct $IUPAC$ name is Propane$-1,2,3-$tricarbonitrile.

(B) $1$. Identify the longest carbon chain containing the double bond and the functional groups $(-OH)$. The structure is $CH_3-CH_2-C(OH)=C(OH)-CH_3$.
$2$. The longest chain has $5$ carbons,so the parent alkane is pentane.
$3$. Numbering should be done to give the lowest possible locants to the functional groups $(-OH)$. Numbering from the right gives the $-OH$ groups at positions $2$ and $3$.
$4$. The double bond is at position $2$.
$5$. Combining these,the name is $Pent-2-ene-2,3-diol$.

(A) The given structure is $Br-CH_2-CH(CONH_2)-CO-CH_2-CH_2-CH_3$.
In this structure,the principal functional group is the amide $(-CONH_2)$.
The longest carbon chain containing the amide carbon consists of $6$ carbons,which corresponds to the parent chain $hexanamide$.
Numbering starts from the amide carbon as $C1$.
At the $C2$ position,there is a bromomethyl group $(-CH_2Br)$.
At the $C3$ position,there is an oxo group $(=O)$.
Combining these,the $IUPAC$ name is $2-(\text{bromomethyl})-3-\text{oxohexanamide}$.

(D) $1$. Identify the longest carbon chain containing the functional group $(-OH)$ and the double bond $(C=C)$. The chain has $5$ carbon atoms,so the parent alkane is pentene.
$2$. Number the chain starting from the carbon attached to the $-OH$ group to give it the lowest possible locant $(C-1)$.
$3$. The double bond starts at $C-2$. The methyl group is attached to $C-4$.
$4$. Combining these,the name is $4-$methylpent-$2-$en-$1-$ol.

(B) $1$. Identify the longest carbon chain containing the principal functional groups. Here,the longest chain containing the three carboxylic acid groups is a propane chain.
$2$. Number the chain such that the substituents get the lowest possible locants. Numbering from either end gives the same result: $1, 2, 3$.
$3$. The principal functional group is carboxylic acid $(-COOH)$,which is given the suffix 'oic acid' or 'tricarboxylic acid' when three are present.
$4$. The hydroxyl group $(-OH)$ is at position $2$.
$5$. Combining these,the name is $2-$ hydroxy propane $-1,2,3-$ tricarboxylic acid.

(C) The given compound is a three-membered cyclic ether containing an oxygen atom,which is known as an epoxide or oxirane.
In $IUPAC$ nomenclature,such compounds are named as epoxy derivatives of the corresponding alkane.
The parent chain is propane ($3$ carbons).
The oxygen atom is bonded to carbon $1$ and carbon $2$.
Therefore,the $IUPAC$ name is $1,2-$epoxypropane.