Nomenclature of organic compounds Questions in English

(B) The given compound is $CH_3-C(=O)-O-C(=O)-CH_2-OCH_3$.
It contains an anhydride group $(-CO-O-CO-)$ and an ether group $(-O-)$.
According to $IUPAC$ priority rules,the anhydride group has a higher priority than the ether group.
Therefore,the principal functional group is the anhydride.

(C) According to $IUPAC$ rules,when three or more identical carbon-containing functional groups (such as $-CN, -COOH, -CHO$) are directly attached to the main chain,the carbon atoms of these functional groups are not included in the parent chain.
In such cases,the suffix "-carbonitrile" is used for the $-CN$ group.
Here,the parent chain consists of $3$ carbon atoms (propane),and the three $-CN$ groups are attached at positions $1, 2,$ and $3$.
Therefore,the $IUPAC$ name is propane $-1, 2, 3-$ tricarbonitrile.

(B) $1$. Identify the principal functional group: The compound contains an acyl chloride group $(-COCl)$,which is the principal functional group. The carbon atom of the $-COCl$ group is assigned position $1$.
$2$. Select the longest carbon chain containing the principal functional group: The longest chain starting from the $-COCl$ carbon has $5$ carbon atoms,so the parent alkane is pentane. The suffix for acyl chloride is 'oyl chloride',so the parent name is pentanoyl chloride.
$3$. Number the chain: Starting from the $-COCl$ carbon as $C-1$,the methyl groups are at positions $2$ and $3$.
$4$. Name the substituents: There are two methyl groups at positions $2$ and $3$. Thus,the prefix is $2,3-$dimethyl.
$5$. Combine the parts: The complete $IUPAC$ name is $2,3-$dimethylpentanoyl chloride.

(D) The structure of Tartaric acid is $HOOC-CH(OH)-CH(OH)-COOH$.
To determine the $IUPAC$ name:
$1$. Identify the longest carbon chain containing the principal functional groups (carboxylic acid groups). Here,the chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain from either end to give the lowest possible locants to the substituents. Numbering from either end gives the $COOH$ groups at positions $1$ and $4$,and the $OH$ groups at positions $2$ and $3$.
$3$. The parent chain is butane,with two carboxylic acid groups at positions $1$ and $4$,making it a butane $-1, 4-$ dioic acid.
$4$. There are two hydroxyl groups at positions $2$ and $3$,so it is $2, 3-$ dihydroxy.
Combining these,the $IUPAC$ name is $2, 3-$ dihydroxybutane $-1, 4-$ dioic acid.

(B) According to the $CIP$ (Cahn-Ingold-Prelog) sequence rule,priority is assigned based on the atomic number of the atom directly attached to the chiral center.
$1$. Comparing the atoms directly attached: $-OH$ (Oxygen,$Z=8$) has the highest priority.
$2$. For the carbon-containing groups: $-COOH$,$-CHO$,and $-CH_2OH$,we look at the atoms attached to the first carbon:
- In $-COOH$,the carbon is bonded to $(O, O, O)$ (counting the double bond as two oxygens).
- In $-CHO$,the carbon is bonded to $(O, O, H)$.
- In $-CH_2OH$,the carbon is bonded to $(O, H, H)$.
Comparing these sets,the priority order is $-COOH > -CHO > -CH_2OH$.
Combining these,the overall order of decreasing priority is $-OH > -COOH > -CHO > -CH_2OH$.

(C) The given structure is $CH_2=CH-CH_2-CH(OCH_3)-CH_2NO_2$.
In this molecule,the principal functional group is the alkene (double bond).
According to $IUPAC$ rules,the numbering of the parent carbon chain starts from the end closer to the double bond to give it the lowest possible locant.
Numbering: $C1(H_2)=C2(H)-C3(H_2)-C4(H)(OCH_3)-C5(H_2)NO_2$.
The substituents are a methoxy group at $C4$ and a nitro group at $C5$.
Alphabetically,'methoxy' comes before 'nitro'.
Therefore,the correct name is $4$-methoxy-$5$-nitro-$1$-pentene.

(B) $1$. In the given cyclic compound,the double bond is the principal functional group. According to $IUPAC$ rules,the double bond must be assigned the lowest possible locants,i.e.,$1$ and $2$.
$2$. We number the ring starting from the double bond such that the substituent (methyl group) gets the lowest possible number.
$3$. If we number clockwise starting from the top carbon of the double bond,the methyl group is at position $4$.
$4$. Thus,the correct $IUPAC$ name is $4-$methylcyclohex$-1-$ene.

(C) To determine the structure of $3-$Methyl$-2-$cyclohexenol,follow these steps:
$1$. The parent chain is a cyclohexane ring with an alcohol group $(-OH)$ and a double bond $(-ene)$.
$2$. The suffix $-ol$ indicates the $-OH$ group is the principal functional group,so it is assigned position $1$.
$3$. The double bond starts at position $2$,making it a $2-$cyclohexenol derivative.
$4$. $A$ methyl group $(-CH_3)$ is attached at position $3$.
$5$. Combining these,the $-OH$ is at $C-1$,the double bond is between $C-2$ and $C-3$,and the methyl group is at $C-3$.
$6$. Looking at the options,image $830-c350$ shows the $-OH$ at $C-1$,the double bond between $C-2$ and $C-3$,and the methyl group at $C-3$.

(A) The given structure is $HOOC-CH_2-CH(COOH)-CH_2-COOH$.
According to $IUPAC$ rules,when three or more identical functional groups are present,the carbon atoms of these groups are not included in the main chain.
Here,the main chain is a propane chain with three carboxylic acid groups attached at positions $1, 2,$ and $3$.
Therefore,the correct $IUPAC$ name is Propane$-1, 2, 3-$tricarboxylic acid.

(B) The given structure is $CH_3-CH(CH_2-CH_3)-CHO$.
$1.$ The principal functional group is the aldehyde group $(-CHO)$.
$2.$ The longest carbon chain containing the $-CHO$ group has $4$ carbon atoms,so the parent alkane is butane.
$3.$ Numbering starts from the carbon of the $-CHO$ group.
$4.$ There is a methyl group $(-CH_3)$ at the $2^{nd}$ position.
$5.$ Therefore,the $IUPAC$ name is $2-methylbutanal$.

(A) The structure of caffeic acid consists of a benzene ring substituted with two hydroxyl groups at positions $3$ and $4$,and a propenoic acid side chain at position $1$.
$1$. The principal functional group is the carboxylic acid,so the parent chain is a propenoic acid derivative.
$2$. The benzene ring is attached at the $3$-position of the propenoic acid chain.
$3$. The benzene ring has hydroxyl groups at positions $3$ and $4$ relative to the attachment point of the propenoic acid chain.
$4$. Combining these,the $IUPAC$ name is $3-(3,4-\text{dihydroxyphenyl})\text{prop-2-enoic acid}$.

(D) $1$. Identify the principal functional group: The ester group $(-COOCH_2CH_3)$ is the principal functional group,so the suffix is 'carboxylate'.
$2$. Number the ring: The carbon attached to the ester group is assigned position $1$. To give the double bond and substituents the lowest possible locants,we number the cyclopentene ring clockwise.
$3$. Position $1$: $-COOCH_2CH_3$ (ethyl ester group).
$4$. Position $2$: The double bond starts here.
$5$. Position $5$: The $-Br$ substituent is at this position.
$6$. The methyl group is at position $2$ (attached to the same carbon as the double bond).
$7$. Combining these,the name is Ethyl $5-$bromo$-2-$methylcyclopent$-1-$enecarboxylate.

(C) The given structure consists of a benzene ring substituted with a dimethylamino group $(-N(CH_3)_2)$ and a nitroso group $(-N=O)$.
In $IUPAC$ nomenclature,the parent compound is aniline $(C_6H_5NH_2)$.
The dimethylamino group is attached to the nitrogen atom of aniline,making it $N,N-$dimethylaniline.
The nitroso group $(-N=O)$ is at the para position relative to the dimethylamino group.
Therefore,the correct $IUPAC$ name is $4-$nitroso$-N,N-$dimethylaniline.

(C) Phthalic acid is benzene-$1,2$-dicarboxylic acid.
When it is heated,it undergoes dehydration to form phthalic anhydride.
The $IUPAC$ name for the anhydride derived from benzene-$1,2$-dicarboxylic acid is benzene-$1,2$-dicarboxylic anhydride.

(A) $1$. Identify the principal functional group or the longest carbon chain containing the double bond. The chain is a butene derivative.
$2$. Number the chain starting from the end closest to the double bond to give it the lowest possible locant. The double bond starts at $C-1$.
$3$. The substituent at $C-1$ is a $4-\text{bromophenyl}$ group.
$4$. The substituent at $C-3$ is two methyl groups $(3,3-\text{dimethyl})$.
$5$. Combining these,the name is $1-(4-\text{bromophenyl})-3,3-\text{dimethylbut}-1-\text{ene}$.

(A) $1$. Identify the principal functional group/ring: The cyclopropene ring is the parent chain because it contains the double bond. The cyclopentyl group is a substituent attached at position $3$ of the cyclopropene ring.
$2$. Numbering the substituent: The cyclopentyl ring is numbered starting from the point of attachment to the cyclopropene ring as $1'$.
$3$. Substituents on the cyclopentyl ring: The bromine and chlorine atoms are at positions $3$ and $4$ of the cyclopentyl ring. According to alphabetical order,'bromo' comes before 'chloro'.
$4$. Assigning locants: To give the lowest possible locants to the substituents,we number the cyclopentyl ring such that bromo is at $3$ and chloro is at $4$ (or vice versa). Since $3-\text{bromo}-4-\text{chloro}$ is alphabetically preferred over $3-\text{chloro}-4-\text{bromo}$,we choose the former.
$5$. Final Name: The compound is $3-(3-\text{bromo}-4-\text{chlorocyclopentyl})\text{cyclopropene}$.

(B) To determine the correct structure of $5-$bromo$-2-$cyclopropylcyclohex$-2-$enol,we follow the $IUPAC$ nomenclature rules:
$1$. The principal functional group is the alcohol $(-OH)$,which gets the lowest possible number. Thus,the carbon attached to $-OH$ is $C-1$.
$2$. The parent chain is a cyclohexene ring. The double bond (alkene) is given priority over the halide substituent. The numbering should proceed to give the double bond the lowest possible locants,starting from $C-1$ (the alcohol carbon).
$3$. If we number the ring such that the double bond starts at $C-2$,the cyclopropyl group is at $C-2$ and the bromine is at $C-5$.
$4$. This matches the name $5-$bromo$-2-$cyclopropylcyclohex$-2-$enol.
$5$. Looking at the options,the structure in option $B$ shows the $-OH$ group at $C-1$,the double bond between $C-2$ and $C-3$,the cyclopropyl group at $C-2$,and the bromine atom at $C-5$.

(A) The given compound is $2-methylanthracen-9(10H)-one$.
To find the number of $C-C$ sigma bonds,we count the bonds between carbon atoms in the structure.
In the three fused benzene rings,there are $14$ carbons forming the skeleton.
Counting the $C-C$ sigma bonds:
$1$. There are $15$ $C-C$ bonds within the fused ring system.
$2$. There is $1$ $C-C$ bond connecting the methyl group to the ring.
Total $C-C$ sigma bonds = $15 + 1 = 16$.

(A) To determine the priority,we look at the atoms attached to the chiral center:
$1$. Compare the first atom: All are $C$ atoms.
$2$. Compare the next set of atoms attached to these $C$ atoms:
- $-CH_2Br$: $C$ is attached to $(Br, H, H)$.
- $-CH_2CH_2Br$: $C$ is attached to $(C, H, H)$.
- $-CH(CH_3)_2$: $C$ is attached to $(C, C, H)$.
- $-C(CH_3)_3$: $C$ is attached to $(C, C, C)$.
$3$. Comparing the atomic number of the first point of difference: $Br$ $(35)$ has the highest atomic number,so $-CH_2Br$ is priority $1$.
$4$. Next,compare the remaining groups. $-CH_2CH_2Br$ has $C$ attached to $(C, H, H)$,while $-CH(CH_3)_2$ has $C$ attached to $(C, C, H)$. Since $(C, C, H)$ has higher priority than $(C, H, H)$,$-CH(CH_3)_2$ is priority $2$,$-CH_2CH_2Br$ is priority $3$,and $-C(CH_3)_3$ is priority $4$.
Thus,the order is $C > A > B > D$.

(B) According to the $CIP$ (Cahn-Ingold-Prelog) priority rules,priority is assigned based on the atomic number of the atom directly attached to the chiral center.
$1$. The atoms attached are: $S$ (atomic number $16$) for $-SCH_3$,$N$ (atomic number $7$) for $-NO_2$,$C$ (atomic number $6$) for $-C \equiv CH$,and $C$ (atomic number $6$) for $-CH_2C_6H_5$.
$2$. Comparing these,$S$ has the highest atomic number,so $-SCH_3$ $(I)$ has the highest priority.
$3$. Next is $N$ in $-NO_2$ $(II)$.
$4$. Comparing the two carbon-based groups: $-C \equiv CH$ $(III)$ is treated as being bonded to three carbons (due to triple bond),whereas $-CH_2C_6H_5$ $(IV)$ is bonded to one carbon and two hydrogens. Thus,$-C \equiv CH$ has higher priority than $-CH_2C_6H_5$.
$5$. The decreasing order of priority is $I > II > III > IV$.
$6$. Therefore,the increasing order of priority is $IV < III < II < I$.

(A) According to the $CIP$ priority rules,we compare the atoms attached to the first carbon atom of the group.
$1: -CH(CH_3)_2$ is attached to $C$ (bonded to $C, C, H$).
$2: -CH_2Br$ is attached to $C$ (bonded to $Br, H, H$).
$3: -CH_2CH_2Br$ is attached to $C$ (bonded to $C, H, H$).
Comparing the first point of difference:
Group $2$ has $Br$ (atomic number $35$) attached to the first carbon,which has the highest priority.
Group $3$ has $C$ (atomic number $6$) attached to the first carbon.
Group $1$ has $C$ (atomic number $6$) attached to the first carbon,but the branching at the second carbon makes it lower priority than the linear chain in group $3$ when comparing subsequent atoms.
Thus,the order of decreasing priority is $2 > 3 > 1$.

(C) The given compound is a carboxylic acid derivative. The principal functional group is the carboxylic acid group $(-COOH)$.
Substituents are the groups attached to the main carbon chain that are not part of the principal functional group.
In the structure,we identify the following substituents attached to the carbon chain:
$1$. $A$ chloro group $(-Cl)$ at the end of the chain.
$2$. $A$ bromo group $(-Br)$.
$3$. An iodo group $(-I)$.
Therefore,there are a total of $3$ substituents present in the compound.

(B) $1$. The given compound is $CH_2=CH-CH_2-CH_2-C \equiv CH$.
$2$. The longest carbon chain contains $6$ carbon atoms,so the parent alkane is hexane.
$3$. According to $IUPAC$ rules,when both double and triple bonds are present,numbering is done such that the double bond gets the lower number if there is a choice.
$4$. Numbering from the left: $C_1=C_2-C_3-C_4-C_5 \equiv C_6$. The double bond is at position $1$ and the triple bond is at position $5$.
$5$. The name is constructed as $Hex-1-en-5-yne$.

(B) $1$. Identify the parent chain: The parent chain is a cyclohexane ring.
$2$. Identify substituents: There are two methyl groups $(-CH_3)$ and one ethoxy group $(-OCH_2CH_3)$.
$3$. Numbering the ring: According to $IUPAC$ rules,numbering should give the lowest possible locants to substituents. If we start at the carbon with the two methyl groups as $1$,the substituents are at positions $1, 1,$ and $2$. If we start at the carbon with the ethoxy group as $1$,the substituents are at $1, 2, 2$. Comparing the sets $(1, 1, 2)$ and $(1, 2, 2)$,the set $(1, 1, 2)$ is preferred.
$4$. Alphabetical order: Ethoxy comes before methyl. Therefore,the name is $2-$ ethoxy $-1, 1-$ dimethylcyclohexane.

(A) The structure is an aldehyde with an $8-$carbon chain.
Numbering starts from the aldehyde carbon as $C-1$.
The double bonds are at positions $2$ and $6$.
Methyl groups are at positions $3$ and $7$.
Thus,the correct $IUPAC$ name is $3, 7-$dimethyloct$-2, 6-$dienal.

(A) The structure consists of a benzene ring substituted with an ester group $(-COOC_2H_5)$ and an acyl chloride group $(-COCl)$.
According to $IUPAC$ priority rules,the ester group $(-COOR)$ has higher priority than the acyl chloride group $(-COCl)$.
Therefore,the parent chain is based on the ester,making it a benzoate derivative.
The ester is an ethyl ester,and the $-COCl$ group is attached at the ortho position (position $2$) relative to the ester group.
The $-COCl$ group is named as a 'chlorocarbonyl' substituent.
Thus,the correct $IUPAC$ name is ethyl $2$-(chlorocarbonyl)benzoate.

(A) $1$. Identify the longest carbon chain containing the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentene.
$2$. Number the chain to give the double bond the lowest possible locant. Numbering from left to right gives the double bond at position $2$.
$3$. Substituents are $Cl$ at position $2$ and $I$ at position $3$.
$4$. Determine the geometry: The groups of higher priority on each carbon of the double bond ($Cl$ on $C2$ and $CH_2CH_3$ on $C3$) are on opposite sides of the double bond,indicating $trans$ geometry.
$5$. Combining these,the name is $trans-2$-chloro-$3$-iodo-$2$-pentene.

(A) $1$. Identify the principal functional group. The amide group $(-CONH_2)$ has higher priority than the aldehyde group $(-CHO)$. Thus,the parent chain is an amide.
$2$. Number the chain starting from the amide carbon as $C-1$. The chain is $6$ carbons long.
$3$. The aldehyde group at $C-6$ is treated as a substituent and named as 'oxo'.
$4$. There is a double bond starting at $C-3$.
$5$. There is a methyl group at $C-2$.
$6$. Combining these,the name is $2-$methyl$-6-$oxohex$-3-$enamide.

(A) The given compound is $CH_3CH_2CH(C_2H_5)COOCH_3$.
First,identify the functional group,which is an ester $(-COOCH_3)$.
The alkyl group attached to the oxygen of the ester is methyl.
Next,identify the longest carbon chain containing the ester group. The chain starts from the carbonyl carbon of the ester.
The chain is $CH_3-CH_2-CH(C_2H_5)-COO-CH_3$.
The main chain has $4$ carbons: $C_1$ (carbonyl),$C_2$ (with ethyl group),$C_3$,and $C_4$.
At position $2$,there is an ethyl group $(-C_2H_5)$.
Thus,the name is methyl $2-$ethylbutanoate.

(C) The structure is $CH_3-CH=CH-COOH$.
The longest carbon chain contains $4$ carbon atoms,so the parent alkane is butane.
The carboxylic acid group $(-COOH)$ is assigned position $1$.
The double bond starts at carbon $2$.
Therefore,the name is But$-2-$en$-1-$oic acid,commonly known as crotonic acid.

(B) $1$. Identify the longest carbon chain containing the double bond. The chain has $4$ carbons,so the parent alkane is butane. With the double bond at position $2$,it is $but-2-ene$.
$2$. Identify the substituents. There is a methoxy group $(-OCH_3)$ attached to the second carbon.
$3$. Combining these,the $IUPAC$ name is $2-methoxybut-2-ene$.

(A) The given structure is $CH_2=C(OCH_3)-COOC_2H_5$.
The parent chain contains $3$ carbon atoms,so the root word is $prop$.
There is a double bond at the $2^{nd}$ position,so the suffix is $-2-$enoate.
The substituent $-OCH_3$ (methoxy group) is attached to the $2^{nd}$ carbon.
The ester group $-COOC_2H_5$ is named as Ethyl.
Combining these,the $IUPAC$ name is Ethyl $2-$methoxyprop$-2-$enoate.

(C) $1$. Identify the longest carbon chain containing the principal functional group (ketone). The chain has $5$ carbons,so the parent alkane is $pentane$ and the suffix is $-2-\text{one}$.
$2$. Number the chain from the end closer to the ketone group: $CH_3(1)-CO(2)-CH(3)(C_2H_5)-CH(4)(CH_3)_2(5)$.
$3$. At position $3$,there is an ethyl group $(-C_2H_5)$.
$4$. At position $4$,there is a methyl group $(-CH_3)$.
$5$. Combining these,the name is $3-\text{Ethyl}-4-\text{methylpentan}-2-\text{one}$.

(A) $1$. Identify the principal functional group: The compound contains an acyl bromide group $(-COBr)$,which is the principal functional group. The carbon atom of the $-COBr$ group is assigned position $1$.
$2$. Select the longest carbon chain containing the principal functional group: The longest chain containing the $-COBr$ group has $4$ carbon atoms,so the parent alkane is butane. The suffix for acyl bromide is 'oyl bromide'. Thus,the parent name is 'butanoyl bromide'.
$3$. Number the chain: Starting from the $-COBr$ carbon as $C-1$,the methyl group $(-CH_3)$ is at the $C-3$ position.
$4$. Combine the parts: The name is $3-$methylbutanoyl bromide.

(A) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$. The chain has $3$ carbons,so the parent alkane is propane.
$2$. Number the chain starting from the carbon attached to the $-OH$ group to give it the lowest possible locant. Thus,$C-1$ is the carbon attached to $-OH$,$C-2$ is the middle carbon,and $C-3$ is the terminal methyl group.
$3$. The substituent $-C_6H_5$ (phenyl group) is attached to the $C-2$ position.
$4$. Combining these,the name is $2-$phenylpropan$-1-$ol.

(D) $1$. Identify the longest carbon chain: The longest chain contains $7$ carbon atoms,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the substituents the lowest possible locants. Numbering from left to right gives substituents at positions $2, 2, 3, 4$. Numbering from right to left gives substituents at positions $4, 5, 5, 6$.
$3$. The substituents are: $2, 2-$dimethyl,$3-$methyl,and $4-$iodo.
$4$. Combining these,the name is $4-$iodo$-2,2,3-$trimethylheptane. However,checking the provided options,the structure $CH_3-C(CH_3)_2-CH(I)-CH(CH_3)-CH_2-CH_3$ corresponds to a $7$-carbon chain. If we re-evaluate the chain length,the longest chain is $7$ carbons. The correct name is $4-$iodo$-2,2,3-$trimethylheptane. Given the options,$D$ is the closest match if we consider a $6$-carbon chain interpretation,but strictly,the $IUPAC$ name is $4-$iodo$-2,2,3-$trimethylheptane.

(A) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$. The chain has $6$ carbons,so the parent alkane is hexane.
$2$. Number the chain starting from the end closest to the $-OH$ group. The $-OH$ is at $C-2$.
$3$. Identify substituents: a chloro group at $C-4$ and two methyl groups at $C-2$ and $C-3$.
$4$. Combine these to get the name: $4-$chloro$-2,3-$dimethylhexan$-2-$ol.

(C) The given structure is $C_6H_5-CH(CH_3)-CHO$.
$1$. Identify the principal functional group,which is the aldehyde $(-CHO)$ group.
$2$. Select the longest carbon chain containing the aldehyde group. The chain has $3$ carbon atoms,so the parent alkane is propane.
$3$. Number the chain starting from the aldehyde carbon as $C-1$.
$4$. The phenyl group $(-C_6H_5)$ is attached to the $C-2$ position.
$5$. Therefore,the $IUPAC$ name is $2-$phenylpropanal.

(D) $1$. Identify the longest carbon chain containing the functional group. The given structure is $CH_3-CO-CH(CH_3)_2$.
$2$. The longest chain has $4$ carbon atoms,so the parent alkane is butane.
$3$. The functional group is a ketone $(-CO-)$,which gets the suffix $-one$.
$4$. Number the chain from the end that gives the ketone group the lowest possible locant. Numbering from left to right,the ketone is at position $2$.
$5$. There is a methyl group $(-CH_3)$ attached at position $3$.
$6$. Combining these,the name is $3-$methylbutan$-2-$one.

(A) The given structure is an acid anhydride formed by the condensation of two different carboxylic acids: ethanoic acid $(CH_3COOH)$ and propanoic acid $(CH_3CH_2COOH)$.
According to $IUPAC$ nomenclature,acid anhydrides are named by listing the names of the parent carboxylic acids in alphabetical order,followed by the word 'anhydride'.
Since 'Ethanoic' comes before 'Propanoic' alphabetically,the correct name is Ethanoic propanoic anhydride.

(B) To determine the $IUPAC$ name,first identify the longest carbon chain containing the functional groups $(-OH)$.
The structure is $CH_3-CH(OH)-CH(C_2H_5)-CH_2-OH$.
The longest chain containing both $-OH$ groups has $4$ carbons (butane).
The numbering starts from the end closer to the substituents/functional groups.
Numbering from right to left: $C_1$ is $CH_2(OH)$,$C_2$ is $CH(C_2H_5)$,$C_3$ is $CH(OH)$,$C_4$ is $CH_3$.
At $C_2$,there is an ethyl group $(-C_2H_5)$.
At $C_1$ and $C_3$,there are hydroxyl groups $(-OH)$.
Thus,the name is $2-$ethylbutane$-1, 3-$diol.

(C) The given structure is $CH_3-CO-CO-CH_3$.
Numbering the carbon chain from either end gives the keto groups at positions $2$ and $3$.
Since there are $4$ carbon atoms in the parent chain,the prefix is $butane$.
Since there are two ketone groups,the suffix is $-dione$.
Therefore,the $IUPAC$ name is $butane-2, 3-dione$.

(B) The structure is $CH_2=CH-CH=CH_2$.
There are $4$ carbon atoms in the longest chain,so the parent alkane is butane.
There are two double bonds at positions $1$ and $3$.
Therefore,the $IUPAC$ name is $Buta-1,3-diene$.

(A) The given compound is $HOOC-CH_2-CH_2-CH_2-COOH$.
There are $5$ carbon atoms in the longest chain including the two carboxyl groups.
The suffix for a dicarboxylic acid is -dioic acid.
Therefore,the $IUPAC$ name is pentane-$1,5$-dioic acid.

(D) The structure is $OHC-CH_2-CH=CH-CHO$.
The longest carbon chain containing both aldehyde groups has $5$ carbon atoms.
Numbering starts from the end that gives the double bond the lowest possible locant.
Numbering from left to right: $C_1(CHO)-C_2(H_2)-C_3(H)=C_4(H)-C_5(CHO)$. The double bond is at $C_3$.
Numbering from right to left: $C_1(CHO)-C_2(H)=C_3(H)-C_4(H_2)-C_5(CHO)$. The double bond is at $C_2$.
Since $2 < 3$,the correct numbering is from right to left.
Thus,the $IUPAC$ name is $Pent-2-ene-1,5-dial$.

(D) The structure is $NC-CH_2-CH_2-CN$.
The longest carbon chain contains $4$ carbon atoms.
The functional group is nitrile $(-CN)$.
Since the carbon of the nitrile group is included in the parent chain,the suffix used is 'nitrile'.
The numbering starts from one of the nitrile carbons,giving the positions $1$ and $4$.
Thus,the $IUPAC$ name is Butane $-1, 4-$ dinitrile.

(A) $1$. Identify the longest carbon chain containing the functional groups. The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain from the end that gives the lowest locants to the substituents and functional groups. Numbering from left to right gives the amine groups at positions $1$ and $4$,and the methyl group at position $3$.
$3$. The structure is $H_2N(1)-CH_2(2)-CH(CH_3)(3)-CH_2(4)-NH_2$. Wait,let's re-number: $H_2N-CH_2(1)-CH(CH_3)(2)-CH_2(3)-CH_2(4)-NH_2$. The methyl group is at position $2$. Thus,the name is $2-$methylbutane$-1,4-$diamine.

(B) The given compound is $CH_2(Cl)-CH(CH_2Cl)-CH_2Cl$.
To find the $IUPAC$ name,we identify the longest carbon chain containing the principal functional groups.
The longest chain has $3$ carbon atoms,making it a propane derivative.
Numbering the chain from either end,the chlorine atoms are at positions $1, 2,$ and $3$.
Specifically,there are chlorine atoms at $C-1$ and $C-3$,and a chloromethyl group $(-CH_2Cl)$ at $C-2$.
Thus,the name is $1, 3-$dichloro$-2-$(chloromethyl)propane.

(A) $1$. Identify the longest carbon chain containing the functional groups (hydroxyl groups). The chain has $6$ carbon atoms,so the parent alkane is hexane.
$2$. Number the chain from the end that gives the lowest locants to the substituents and functional groups. Numbering from left to right gives hydroxyl groups at positions $2$ and $4$,and a methyl group at position $5$. Numbering from right to left gives hydroxyl groups at positions $3$ and $5$,and methyl groups at positions $2, 2, 4$.
$3$. According to the lowest locant rule for functional groups,numbering from left to right is preferred ($2, 4$ vs $3, 5$).
$4$. The substituents are at position $5$ (methyl). Wait,let's re-evaluate: The structure is $CH_3-CH(OH)-CH(CH_3)-CH(OH)-C(CH_3)_2-CH_3$. The longest chain is $6$ carbons. Numbering from left: $C1(CH_3)-C2(CHOH)-C3(CHCH_3)-C4(CHOH)-C5(C(CH_3)_2)-C6(CH_3)$. This gives substituents at $3, 5, 5$ and diol at $2, 4$. The name is $3, 5, 5-$ trimethylhexane $-2, 4-$ diol.

(A) The given compound is $HOOC-CH=CH-COOH$.
$1$. The longest carbon chain containing the double bond and the two carboxylic acid groups has $4$ carbon atoms.
$2$. The parent alkane is butane,and since there is a double bond at position $2$,it is named but$-2-$ene.
$3$. The two carboxylic acid groups are at positions $1$ and $4$.
$4$. Therefore,the $IUPAC$ name is $But-2-ene-1,4-dioic \text{ acid}$.