Nomenclature of organic compounds Questions in English

(C) Phloroglucinol is a benzene ring substituted with three hydroxyl $(-OH)$ groups at the $1$,$3$,and $5$ positions.
According to $IUPAC$ nomenclature,the parent hydrocarbon is benzene,and the substituents are indicated by their positions followed by the suffix 'triol' for three hydroxyl groups.
Therefore,the $IUPAC$ name is benzene-$1,3,5$-triol.

(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 at position $3$,it is a hex$-3-$ene derivative.
$2$. Number the chain from the end that gives the double bond the lowest possible number. Numbering from right to left gives the double bond position $3$.
$3$. Identify substituents: There is a chloro group $(-Cl)$ at position $3$ and a methyl group $(-CH_3)$ at position $4$.
$4$. Combine these to get the $IUPAC$ name: $3-$Chloro$-4-$methylhex$-3-$ene.

(B) $1$. Identify the longest carbon chain. The longest chain contains $8$ carbon atoms,so the parent alkane is octane.
$2$. Number the chain from the end that gives the substituents the lowest possible locants. Numbering from left to right gives substituents at positions $4, 3, 6, 6$. Numbering from right to left gives substituents at positions $3, 3, 6, 5$. Comparing the sets $(3, 3, 5, 6)$ and $(3, 4, 6, 6)$,the set $(3, 3, 5, 6)$ is lower.
$3$. The substituents are: bromo at position $5$,and methyl groups at positions $3, 3, 6$.
$4$. Combining these,the name is $5-$bromo$-3, 3, 6-$trimethyloctane.

(D) primary carbon atom is a carbon atom that is bonded to only one other carbon atom. In the structure of $3-$Ethyl$-2,4-$dimethylheptane,the terminal methyl $(-CH_3)$ groups are primary carbons.
Looking at the structure:
$1$. The $C-2$ position has one methyl group attached.
$2$. The $C-3$ position has an ethyl group attached,which ends in a methyl group.
$3$. The $C-4$ position has a methyl group attached.
$4$. The main chain heptane has two terminal methyl groups.
Counting these:
- Two methyls at $C-2$ (one is part of the chain,one is a substituent).
- One methyl at the end of the ethyl group at $C-3$.
- One methyl at $C-4$.
- Two terminal methyls of the heptane chain.
Total primary carbon atoms = $6$.

(B) The structure of isobutyl bromide is $(CH_3)_2CH-CH_2Br$.
To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group $(-Br)$.
The longest chain has $3$ carbon atoms,so the parent alkane is propane.
Numbering the chain starting from the carbon attached to the bromine atom gives the $1-$position to the bromine.
There is a methyl group at the $2-$position.
Thus,the $IUPAC$ name is $1-$bromo$-2-$methyl propane.

(A) The structure of acrolein is $CH_2=CH-CHO$.
It contains a carbon chain of $3$ carbon atoms with an aldehyde group $(-CHO)$ at position $1$ and a double bond starting at position $2$.
Therefore,the $IUPAC$ name is prop$-2-$enal.

(C) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclopentanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the lowest possible locants to the substituents ($-Cl$ and $-CH_3$),we number the ring in the direction that assigns the lowest numbers to the substituents. Numbering clockwise gives the $-CH_3$ at position $2$ and the $-Cl$ at position $4$. Numbering counter-clockwise gives the $-Cl$ at position $3$ and the $-CH_3$ at position $5$.
$3$. Compare locant sets: The set $(2, 4)$ is lower than $(3, 5)$. Therefore,we number clockwise starting from the carbon with the $-OH$ group.
$4$. Assemble the name: The substituents are $4-$chloro and $2-$methyl. Alphabetical order dictates that chloro comes before methyl. Thus,the $IUPAC$ name is $4-$chloro$-2-$methylcyclopentanol.

(C) The structure of $3-$bromopropene is $CH_2=CH-CH_2Br$.
Counting the atoms in the structure:
There are $3$ carbon atoms $(C_3)$.
There are $5$ hydrogen atoms $(H_5)$.
There is $1$ bromine atom $(Br)$.
Therefore,the molecular formula is $C_3H_5Br$.

(D) The structure of isobutyl chloride is $(CH_3)_2CH-CH_2Cl$.
To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group.
The longest chain has $3$ carbon atoms (propane).
There is a methyl group at the $2^{nd}$ position and a chlorine atom at the $1^{st}$ position.
Therefore,the $IUPAC$ name is $1-$chloro$-2-$methylpropane.

(C) The $IUPAC$ name $1-$iodo$-2,3-$dimethylpentane indicates a pentane chain ($5$ carbons) with an iodine atom at position $1$,and methyl groups at positions $2$ and $3$.
In the bond line notation,the carbon chain is represented by a zigzag line.
Starting from the end with the iodine atom (position $1$),the next carbon (position $2$) has a methyl group attached,and the third carbon (position $3$) also has a methyl group attached.
Analyzing the given options:
Option $A$ shows iodine at position $2$.
Option $B$ shows iodine at position $2$.
Option $C$ shows the iodine at the end of the chain (position $1$),with methyl groups at positions $2$ and $3$.
Option $D$ shows iodine at position $3$.
Therefore,the correct structure is represented by option $C$.

(C) $1$. Identify the parent chain: The compound consists of two cyclobutane rings connected by a single bond,forming a $1, 1'-$bicyclobutane system.
$2$. Numbering: Number the rings such that the substituents get the lowest possible locants. The connection point is $1$ and $1'$.
$3$. Substituents: There is a bromine atom at position $3$ and a chlorine atom at position $3'$.
$4$. Alphabetical order: $B$ (bromo) comes before $C$ (chloro).
$5$. Final Name: $3-$bromo$-3'-$chloro$-1, 1'-$bicyclobutane.

(B) $1$. Identify the longest carbon chain containing the functional group (aldehyde group,$-CHO$).
$2$. The longest chain has $5$ carbon atoms,so the parent alkane is pentane,and the suffix is $-al$ (pentanal).
$3$. Number the chain starting from the carbon of the $-CHO$ group as $C-1$.
$4$. The substituents are two methyl groups at positions $2$ and $3$.
$5$. Therefore,the $IUPAC$ name is $2, 3-$dimethylpentanal.

(C) The structural formula of crotonaldehyde is $CH_3-CH=CH-CHO$.
Numbering the carbon chain starting from the aldehyde group $(-CHO)$ as carbon-$1$,we get:
$CH_3(4)-CH(3)=CH(2)-CHO(1)$.
Since there are $4$ carbon atoms,the parent alkane is butane. The double bond is at position $2$,and the aldehyde group is at position $1$. Therefore,the $IUPAC$ name is but-$2$-en-$1$-al.

(C) $1$. Identify the longest carbon chain containing the ether functional group. The structure is $CH_3-CH(OC_3H_7)-C_3H_7$.
$2$. The longest chain has $5$ carbon atoms,which is a pentane chain.
$3$. The alkoxy group $-OC_3H_7$ (propoxy group) is attached to the second carbon atom of the pentane chain.
$4$. Numbering the chain from the end closer to the substituent gives the position $2$ for the propoxy group.
$5$. Therefore,the $IUPAC$ name is $2-$propoxy pentane.

(A) $1$. Identify the longest carbon chain containing the functional group. The longest chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. The functional group is an ether group $(-OC_2H_5)$,which is named as an alkoxy group (ethoxy).
$3$. Number the chain from the end that gives the lowest locant to the substituent. Numbering from left to right,the ethoxy group is at position $2$.
$4$. Therefore,the $IUPAC$ name is $2-$ethoxy pentane.

(D) The given compound is $HO-CH_2-CH_2-CH_2-CH(CH_3)-CH(CH_3)_2$.
This is a chain of $6$ carbon atoms in the main backbone with a hydroxyl group at the first carbon and methyl substituents at the $4^{th}$ and $5^{th}$ positions.
Counting from the $OH$ group:
$C_1$ is $CH_2OH$
$C_2$ is $CH_2$
$C_3$ is $CH_2$
$C_4$ is $CH(CH_3)$
$C_5$ is $CH(CH_3)_2$
Looking at the options,image $232466-$d represents a $6$-carbon chain with an $OH$ group at the end and two methyl groups at the $4^{th}$ and $5^{th}$ positions relative to the $OH$ group.

(B) The structure of $2-$chloro$-3,4,5-$trimethylhexane is $CH_3-CHCl-CH(CH_3)-CH(CH_3)-CH(CH_3)-CH_3$.
Let us analyze each carbon atom for chirality (a carbon atom bonded to four different groups):
$C2$: Bonded to $-H, -Cl, -CH_3, -CH(CH_3)CH(CH_3)CH_2CH_3$. This is chiral.
$C3$: Bonded to $-H, -CH_3, -CH(CH_3)CH_2CH_3, -CHClCH_3$. This is chiral.
$C4$: Bonded to $-H, -CH_3, -CH(CH_3)CH_3, -CH(CH_3)CHClCH_3$. This is chiral.
$C5$: Bonded to $-H, -CH_3, -CH_3, -CH(CH_3)CH(CH_3)CHClCH_3$. This is not chiral because it is bonded to two identical $-CH_3$ groups.
Thus,there are $3$ chiral carbon atoms $(C2, C3, C4)$.

(C) In the $Cahn-Ingold-Prelog$ $(CIP)$ priority rules,phantom atoms are used to represent multiple bonds by expanding them into single bonds.
- Acetaldehyde $(CH_3CHO)$ contains a $C=O$ double bond,which requires phantom atoms.
- Methyl cyanide $(CH_3CN)$ contains a $C \equiv N$ triple bond,which requires phantom atoms.
- Propionic acid $(CH_3CH_2COOH)$ contains a $C=O$ double bond,which requires phantom atoms.
- $n$-propyl alcohol $(CH_3CH_2CH_2OH)$ contains only single bonds between all atoms. Therefore,it does not require any phantom atoms.

(A) According to the Cahn-Ingold-Prelog $(CIP)$ priority rules,phantom atoms are used to account for multiple bonds (double or triple bonds) by expanding them into single bonds.
In $CH_{3}COOH$ (acetic acid),the carbonyl group $(C=O)$ contains a double bond between carbon and oxygen.
To assign priorities,the $C=O$ bond is expanded by treating the carbon as bonded to two oxygens (one real,one phantom) and the oxygen as bonded to two carbons (one real,one phantom).
Since the other options ($CH_{3}CH_{2}Cl$,$CH_{3}CH_{2}NH_{2}$,and $CH_{3}CH_{2}CH_{2}OH$) contain only single bonds,they do not require the use of phantom atoms for priority assignment.
Therefore,$CH_{3}COOH$ is the correct answer.

(D) 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.
Comparing the atoms attached to the chiral center: $N$ (in $-NO_2$,$-NH_2$,$-CN$) has an atomic number of $7$,while $O$ (in $-OH$) has an atomic number of $8$.
Since the atomic number of $O$ $(8)$ is greater than that of $N$ $(7)$,the $-OH$ group receives the highest priority among the given options.

(B) According to the Cahn-Ingold-Prelog $(CIP)$ priority rules,priority is assigned based on the atomic number of the atoms directly attached to the chiral center. If the atoms are the same,we look at the next atoms in the chain.
$1$. Compare the first atoms: All groups $(CONH_2, COCH_3, CHO, CH_2OH)$ are attached via a carbon atom.
$2$. Look at the atoms attached to these carbons:
- $CONH_2$ (Carbon is bonded to $O, O, N$)
- $COCH_3$ (Carbon is bonded to $O, O, C$)
- $CHO$ (Carbon is bonded to $O, O, H$)
- $CH_2OH$ (Carbon is bonded to $O, H, H$)
$3$. Comparing the atoms attached to the carbonyl/first carbon:
- $N$ (atomic number $7$) > $C$ (atomic number $6$) > $H$ (atomic number $1$).
$4$. Thus,the priority order is: $CONH_2 > COCH_3 > CHO > CH_2OH$.

(C) $1$. Identify the principal functional group: The ether group $(-OC_2H_5)$ is present.
$2$. Number the cyclohexane ring: To give the substituents the lowest possible locants,we start numbering at the carbon attached to the ethoxy group as $1$. Then,we proceed towards the dimethyl-substituted carbon to give it the lowest possible number,which is $1$. Thus,the carbon with two methyl groups is at position $1$.
$3$. The $IUPAC$ name is $2-$ethoxy$-1, 1-$dimethylcyclohexane. Therefore,the correct option is $C$.

(A) $1$. Identify the longest carbon chain containing the principal functional groups (hydroxyl groups). The chain has $6$ carbons,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 right to left gives the hydroxyl groups at positions $1$ and $3$,and the methyl group at position $5$.
$3$. The structure is $(CH_3)_2CH-CH_2-CH(OH)-CH(CH_2OH)-CH_3$. The longest chain is $6$ carbons long. The substituents are a methyl group at $C-5$ and two hydroxyl groups at $C-1$ and $C-3$.
$4$. The correct $IUPAC$ name is $2,5-$dimethylhexane$-1,3-$diol.

(A) Isophthalic acid is the common name for benzene-$1,3$-dicarboxylic acid.
In this structure,two carboxylic acid groups $(-COOH)$ are attached to the benzene ring at the $1$ and $3$ positions.
Therefore,the correct $IUPAC$ name is benzene-$1,3$-dicarboxylic acid.
The correct option is $A$.

(D) The given structure is $CH_3-CH=CH-CH_2Cl$.
$1$. Identify the longest carbon chain containing the double bond: The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain to give the lowest possible number to the double bond: Numbering from the right gives the double bond position at $C-2$.
$3$. Identify the substituent: There is a chloro group at $C-1$.
$4$. Combine these to get the $IUPAC$ name: $1-$chloro but$-2-$ene.

(C) The given compound is $HC \equiv C-CH=CH-CH=CH_2$.
According to $IUPAC$ nomenclature rules,when both double and triple bonds are present,the double bond is given preference in numbering if the set of locants is the same.
Numbering from the right side gives the double bonds positions $1$ and $3$,and the triple bond position $5$.
Numbering from the left side gives the triple bond position $1$,and the double bonds positions $3$ and $5$.
Comparing the sets of locants: $(1, 3, 5)$ vs $(1, 3, 5)$.
Since the sets are identical,the double bond is given the lower locant.
Therefore,the correct numbering starts from the right: $CH_2=CH-CH=CH-C \equiv CH$ is not the structure,but $CH_2=CH-CH=CH-C \equiv CH$ is the reverse of the given structure.
Given structure: $HC \equiv C-CH=CH-CH=CH_2$.
Numbering from right to left: $C_1=CH_2, C_2=CH, C_3=CH, C_4=CH, C_5=C, C_6=CH$.
This gives the name Hexa$-1,3-$dien$-5-$yne.

(A) The given structure is a hydrocarbon with a double bond between two carbon atoms.
To name it according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $4$ carbon atoms,so the parent alkane is butane,and the alkene is but$-2-$ene.
$2$. Number the chain from the end that gives the double bond the lowest locant. In this case,numbering from either side gives the double bond the position $2$.
$3$. Identify the substituents. There are two methyl groups attached at positions $2$ and $3$.
$4$. Combining these,the $IUPAC$ name is $2,3-$dimethylbut$-2-$ene.

(C) $1$. 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 carbon of the carboxylic acid group as $C-1$.
$3$. The chain is $CH_3(5)-CO(4)-CH_2(3)-CH_2(2)-COOH(1)$.
$4$. The ketone group is at position $4$,so it is named as an oxo substituent.
$5$. The parent chain has $5$ carbons,so the root is pentane. Adding the suffix for the acid gives pentanoic acid.
$6$. Combining these,the $IUPAC$ name is $4-$oxopentanoic acid.

(B) To determine the $IUPAC$ name,we identify the substituents on the benzene ring: an ethyl group $(-C_2H_5)$,a fluoro group $(-F)$,and a nitro group $(-NO_2)$.
According to $IUPAC$ rules,we number the ring to give the lowest possible locants to the substituents.
If we start numbering from the fluoro group as $1$,the nitro group is at $2$,and the ethyl group is at $4$.
The set of locants is $(1, 2, 4)$.
Alphabetical order of substituents: ethyl,fluoro,nitro.
Therefore,the name is $4-$ethyl$-1-$fluoro$-2-$nitrobenzene.

(A) The given structure is $CH_3-CH=CH-CH_2-Br$.
According to $IUPAC$ nomenclature rules,the principal functional group or the multiple bond should get the lowest possible number.
Here,the double bond is present,so we number the chain starting from the end that gives the double bond the lowest locant.
Numbering from the right side: $CH_3(4)-CH(3)=CH(2)-CH_2(1)-Br$.
The double bond starts at position $2$ and the bromine atom is at position $1$.
Therefore,the $IUPAC$ name is $1-$bromobut$-2-$ene.

(A) $1$. Identify the principal functional group: The $-COOH$ group is the principal functional group,so the parent chain is named as a pentanoic acid.
$2$. Number the chain: Start numbering from the carbon of the $-COOH$ group as $C-1$. The chain is numbered to give the lowest possible locants to the substituents.
$3$. Identify substituents: There is a methyl group at $C-2$ and a hydroxy group at $C-4$.
$4$. Alphabetical order: 'Hydroxy' comes before 'methyl'.
$5$. Final name: Combining these,the $IUPAC$ name is $4-$hydroxy$-2-$methylpentanoic acid.

(A) The given compound is $CH_3COCH_2CH(OH)CH_3$.
$1$. Identify the principal functional group: The ketone group $(-CO-)$ has higher priority than the hydroxyl group $(-OH)$.
$2$. Number the carbon chain starting from the end that gives the ketone group the lowest possible number: The ketone carbon gets position $2$.
$3$. The chain has $5$ carbons,so the parent alkane is pentane. With the ketone group,it is pentan$-2-$one.
$4$. The hydroxyl group is at position $4$,so it is named as a hydroxy substituent.
$5$. Combining these,the $IUPAC$ name is $4-$hydroxy$2-$pentanone.

(A) The given compound is $CH_2=CH-CH_2-COOH$.
$1$. Identify the longest carbon chain containing the principal functional group (carboxylic acid). The chain has $4$ carbon atoms,so the root word is $but$.
$2$. Number the chain starting from the carboxylic acid carbon as $C-1$. Thus,the double bond starts at $C-3$.
$3$. The suffix for the carboxylic acid is $-oic \ acid$ and for the double bond is $-ene$.
$4$. Combining these,the $IUPAC$ name is $but-3-enoic \ acid$.

(A) $1$. Identify the longest carbon chain containing the functional group (aldehyde). The chain has $5$ carbon atoms,so the parent alkane is pentane,and the aldehyde suffix is $-al$,making it pentanal.
$2$. Number the chain starting from the aldehyde carbon as $C-1$. The $C-2$ position has a methyl group,and the $C-3$ position has a bromine atom.
$3$. Arrange the substituents alphabetically: bromo comes before methyl.
$4$. The $IUPAC$ name is $3-$bromo$-2-$methylpentanal.

(C) The given compound is $(CH_3)_3 CCl$.
To find the $IUPAC$ name,we first identify the longest carbon chain containing the functional group.
The longest chain has $3$ carbon atoms,so the parent alkane is propane.
Numbering the chain from the end that gives the lowest number to the substituent,we get the chlorine atom and the methyl group at the $2^{nd}$ position.
Thus,the $IUPAC$ name is $2$-chloro-$2$-methylpropane.

(D) Succinic acid is a dicarboxylic acid with the formula $HOOC-(CH_2)_2-COOH$,which is $HOOC-CH_2-CH_2-COOH$.
Malonic acid is a dicarboxylic acid with the formula $HOOC-CH_2-COOH$.
Comparing these structures with the given options,option $(D)$ correctly represents succinic acid $(x)$ and malonic acid $(y)$.

(C) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond. The chain has $10$ carbons,so the parent alkane is decane.
$2$. Number the chain starting from the end that gives the $-OH$ group the lowest possible locant. Numbering from right to left gives the $-OH$ group at position $4$.
$3$. The substituents are: bromo at $7$,ethyl at $5$,and methyl at $2$.
$4$. Alphabetical order of substituents: bromo,ethyl,methyl.
$5$. Combining these,the $IUPAC$ name is $7-$bromo$-5-$ethyl$-2-$methyldec$-5-$en$-4-$ol.

(A) According to the $IUPAC$ priority rules for functional groups,the order of priority is:
$1$. Acid halides $(-COCl)$
$2$. Amides $(-CONH_2)$
$3$. Aldehydes $(-CHO)$
$4$. Alcohols $(-OH)$
Therefore,the correct decreasing order is $-COCl > -CONH_2 > -CHO > -OH$.

(A) Mesityl oxide is an $\alpha,\beta-$unsaturated ketone with the structure $(CH_3)_2C=CHCOCH_3$. The longest carbon chain containing the ketone group has $5$ carbons. Numbering from the end closer to the ketone group gives the name $4-$Methylpent$-3-$en$-2-$one.
Oxalic acid is the simplest dicarboxylic acid with the formula $(COOH)_2$. Its $IUPAC$ name is Ethanedioic acid.
Therefore,the correct $IUPAC$ names are $4-$Methylpent$-3-$en$-2-$one and Ethanedioic acid.

(B) $1$. Identify the longest carbon chain. The longest chain contains $6$ carbon atoms,so the parent alkane is hexane.
$2$. Number the chain from the end that gives the substituents the lowest possible locants. Numbering from left to right gives substituents at positions $2$ and $4$. Numbering from right to left gives substituents at positions $3$ and $5$. Thus,the correct numbering is from left to right.
$3$. The substituents are a methyl group at position $2$ and an ethyl group at position $4$.
$4$. According to $IUPAC$ rules,substituents are listed in alphabetical order. Therefore,ethyl comes before methyl.
$5$. Combining these,the $IUPAC$ name is $4-$Ethyl$-2-$methylhexane.

(C) For compound $(I)$: The longest carbon chain containing the nitro group has $4$ carbons. Numbering starts from the end closer to the nitro group. The phenyl group is at position $1$. Thus,the name is $2-$nitro$-1-$phenylbutane.
For compound $(II)$: The ring is cyclohexane. Substituents are at positions $1$ and $2$. According to alphabetical order,ethyl comes before methyl and propyl. Numbering is done to give the lowest locants to substituents. The correct numbering gives $1-$ethyl$-1-$methyl$-2-$propylcyclohexane.

(D) $1$. Identify the longest carbon chain containing the principal functional group (ketone),double bond,and triple bond. The chain has $9$ carbons,so the parent alkane is nonane.
$2$. Number the chain from the end that gives the lowest locants to the functional groups. The ketone group gets priority,so we number from right to left to give the ketone the lowest possible locant $(5)$.
$3$. The substituents are $4-$bromo,$6-$chloro,and $3,7-$diethyl.
$4$. The double bond is at position $7$ and the triple bond is at position $1$.
$5$. Combining these,the $IUPAC$ name is $4-$Bromo$-6-$chloro$-3,7-$diethylnon$-7-$en$-1-$yn$-5-$one.

(A) According to the $IUPAC$ priority rules for functional groups,the order of priority is determined by the oxidation state and chemical reactivity.
The correct decreasing order of priority is: $-COOH > -SO_3H > -CN > -CHO > -OH$.

(C) The parent compound is toluene (methylbenzene). The methyl group is assigned position $1$. The substituents $-Cl$ and $-NO_2$ are then numbered to give the lowest possible locants. Starting from the methyl group at position $1$,the chlorine atom is at position $2$ and the nitro group is at position $4$. Thus,the $IUPAC$ name is $2-$chloro$-1-$methyl$-4-$nitrobenzene.

(A) The principal functional group is $-CN$,which gets the priority $1$ in the benzene ring,making the parent compound benzonitrile.
Next,we number the ring to give the substituents the lowest possible locants.
The $-Br$ group is at position $2$ and the $-OH$ group is at position $3$.
Thus,the $IUPAC$ name is $2-$bromo$-3-$hydroxybenzonitrile.

(A) $1$. Identify the principal functional group: The $-OH$ group has higher priority than $-NH_2$ and $-CH_3$ groups,so the parent compound is phenol.
$2$. Numbering the ring: Assign position $1$ to the carbon attached to the $-OH$ group. Number the ring to give the lowest possible locants to the substituents.
$3$. If we number clockwise or counter-clockwise,the substituents $-NH_2$ and $-CH_3$ are at positions $3$ and $5$.
$4$. Alphabetical order: Amino $(A)$ comes before Methyl $(M)$.
$5$. Therefore,the $IUPAC$ name is $3-$amino$-5-$methylphenol.

(A) $1$. Identify the principal functional group: The $-CN$ group is the principal functional group,so the suffix is 'carbonitrile'.
$2$. Number the ring: The carbon atom attached to the $-CN$ group is assigned position $1$.
$3$. Numbering direction: To give the lowest locants to the substituents,we number the ring clockwise. Thus,the substituents (ethyl and methyl) are at position $2$.
$4$. Alphabetical order: 'Ethyl' comes before 'methyl'. Therefore,the name is $2-$ethyl$-2-$methylcyclobutane$-1-$carbonitrile.

(A) The given compound is $CH_3-CH(CH_2CH_3)-CHO$,which can be written as $CH_3-CH_2-CH(CH_3)-CHO$.
First,identify the longest carbon chain containing the aldehyde group. The longest chain has $4$ carbon atoms,so the parent alkane is butane.
Number the chain starting from the aldehyde carbon as $C-1$. The methyl group is attached to the $C-2$ position.
Therefore,the $IUPAC$ name is $2-$methylbutanal.

(B) The $IUPAC$ name $3-$oxo$-2-$methylhex$-4-$enal indicates the following features:
$1$. The parent chain is a hexane chain ($6$ carbons) with an aldehyde group $(-CHO)$ at position $1$.
$2$. There is a methyl group $(-CH_3)$ at position $2$.
$3$. There is an oxo group $(=O)$ at position $3$.
$4$. There is a double bond at position $4$ (between carbons $4$ and $5$).
The structure is: $CH_3-CH=CH-C(=O)-CH(CH_3)-CHO$.
Comparing this with the options,the structure is represented by option $B$.

(B) The structure of the tertiary butyl group is $(CH_3)_3C-$.
To name this according to the $IUPAC$ system,we identify the longest carbon chain attached to the main parent chain.
The carbon atom directly attached to the parent chain is assigned position $1$.
Thus,the structure becomes $CH_3-C(CH_3)_2-$,where two methyl groups are attached to the first carbon atom.
Therefore,the $IUPAC$ name is $1,1-$dimethylethyl.