Nomenclature of organic compounds Questions in English

(B) According to the $IUPAC$ priority order for functional groups,the priority sequence is: $-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > -CO > -OH > -NH_2 > -OR > -R > -X$.
Comparing the given options:
$1$. $-SO_3H$ (Sulfonic acid)
$2$. $-COCl$ (Acid chloride)
$3$. $-CHO$ (Aldehyde)
$4$. $-OH$ (Alcohol)
Among the given choices,$-SO_3H$ appears earliest in the priority list,thus it has the highest priority.

(C) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,so the parent alkane is hexane,and the alkene is hex$-3-$ene.
$2$. Number the chain from the end that gives the double bond the lowest locant. Numbering from left to right gives the double bond at position $3$.
$3$. Identify the substituents: $A$ bromine atom $(-Br)$ is at position $4$ and a methyl group $(-CH_3)$ is at position $3$.
$4$. Combine the parts alphabetically: $4-$Bromo$-3-$methylhex$-3-$ene.

(D) $1$. Identify the principal functional group: The $-COOH$ group (carboxylic acid) has higher priority than the $-CHO$ (aldehyde) and $-CH_3$ (methyl) groups. Therefore,the parent chain is a benzoic acid derivative.
$2$. Number the ring: Assign position $1$ to the carbon attached to the $-COOH$ group. Number the ring to give the lowest possible locants to the substituents. Moving towards the $-CH_3$ group gives it position $2$,and the $-CHO$ group gets position $5$.
$3$. Name the substituents: The $-CHO$ group at position $5$ is named as 'formyl',and the $-CH_3$ group at position $2$ is named as 'methyl'.
$4$. Combine the parts: The name is $5-$formyl$-2-$methylbenzoic acid.

(C) The given structure is a cyclobutane ring with a methoxy group $(-OCH_3)$ at position $1$ and two methyl groups $(-CH_3)$ at position $2$.
According to $IUPAC$ nomenclature rules for cyclic ethers:
$1$. The ring is numbered starting from the carbon attached to the methoxy group as $1$.
$2$. The next carbon is numbered $2$ to give the lowest possible locants to the substituents.
$3$. Thus,the methoxy group is at position $1$ and two methyl groups are at position $2$.
$4$. The name is $1-$Methoxy-$2,2-$dimethylcyclobutane.

(C) Pyrogallol is a common name for $1,2,3$-trihydroxybenzene.
According to $IUPAC$ nomenclature rules,when three hydroxyl $(-OH)$ groups are attached to a benzene ring,the compound is named as a benzene derivative with the suffix $-triol$.
The positions of the $-OH$ groups are $1, 2,$ and $3$.
Therefore,the correct $IUPAC$ name is Benzene-$1,2,3$-triol.

(D) In the carbinol system,methyl alcohol $(CH_3OH)$ is considered the parent compound and is called $carbinol$.
Other alcohols are named as derivatives of $carbinol$ by replacing the hydrogen atoms of the $CH_3$ group with alkyl groups.
$tert-butyl$ alcohol has the structure $(CH_3)_3C-OH$.
Here,all three hydrogen atoms of the $CH_3$ group in $carbinol$ are replaced by three methyl groups.
Therefore,it is named as $trimethyl$ $carbinol$.

(D) According to $IUPAC$ nomenclature rules,the priority order for functional groups determines the principal functional group. \\ The priority sequence is $-COOH > -CN > -OH > -C \equiv C-$. \\ Therefore,among the given options,$-COOH$ is the principal functional group.

(D) The structure of $(CH_3)_4C$ is a central carbon atom bonded to four methyl groups.
To determine the $IUPAC$ name,we first identify the longest carbon chain,which contains $3$ carbon atoms (propane).
There are two methyl groups attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2,2-$dimethylpropane.

(B) To determine the $IUPAC$ name,follow these steps:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,so it is a hexene derivative.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from left to right gives the double bond the position $3$.
$3$. Identify substituents: There is a bromo group at position $3$ and a methyl group at position $4$.
$4$. Combine these to get the name: $3-$Bromo$-4-$methylhex$-3-$ene.

(C) $1$. Identify the parent chain: The parent ring is a cyclobutane ring.
$2$. Numbering the ring: Numbering should start such that the substituents get the lowest possible locants. The methoxy group $(-OCH_3)$ and the two methyl groups $(-CH_3)$ are attached to the ring.
$3$. Assigning locants: If we assign the carbon with the two methyl groups as position $1$,the methoxy group is at position $2$. This gives the locants $1, 1, 2$.
$4$. Alphabetical order: 'Methoxy' comes before 'methyl'.
$5$. Combining the parts: The name is $2-$Methoxy$-1,1-$dimethylcyclobutane.

(D) The structural formula for propylene glycerol is $CH_2(OH)-CH(OH)-CH_2(OH)$.
This compound contains three carbon atoms in the main chain and three hydroxyl $(-OH)$ groups attached to each carbon atom.
According to $IUPAC$ nomenclature rules,the parent alkane is propane,and the presence of three $-OH$ groups is indicated by the suffix $-triol$.
The positions of the hydroxyl groups are at carbons $1$,$2$,and $3$.
Therefore,the $IUPAC$ name is Propane-$1,2,3$-triol.

(A) The principal functional group is decided based on the following $IUPAC$ priority order:
$-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > -C=C- > -C \equiv C-$.
Comparing the given options:
$1$. $-CONH_2$
$2$. $-CN$
$3$. $-OH$
$4$. $-C \equiv C-$
According to the priority order,$-CONH_2$ appears first among the given choices,making it the principal functional group.

(B) The structure of Acrylic acid is $CH_2=CH-COOH$.
It contains a three-carbon chain with a carboxylic acid functional group and a double bond.
The carbon atom of the carboxylic acid group is assigned position $1$.
The double bond starts at position $2$.
Therefore,the $IUPAC$ name is Prop$-2-$enoic acid.

(C) $1$. Identify the longest carbon chain containing the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane,and the alkene is pent$-2-$ene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from right to left gives the double bond position at $C-2$.
$3$. Identify the substituents: a chloro group at $C-4$ and a methyl group at $C-3$.
$4$. Combine these to get the $IUPAC$ name: $4-$chloro$-3-$methylpent$-2-$ene.

(B) . The given structure is a benzene ring attached to a $-SO_3^-Na^+$ group.
This compound is the sodium salt of benzene sulphonic acid,known as sodium benzene sulphonate.

(D) $1$. Identify the parent carbon chain: The compound is a cyclic structure with $4$ carbon atoms,which is a cyclobutane ring.
$2$. Numbering the ring: Start numbering from the carbon atom with the substituents to give them the lowest possible locants. The carbon with two methyl groups is assigned position $1$,and the carbon with the methoxy group is assigned position $3$.
$3$. Naming the substituents: There are two methyl groups at position $1$ and one methoxy group at position $3$.
$4$. Combining the parts: The name is $3-$methoxy$-1, 1-$dimethylcyclobutane.

(C) The $IUPAC$ name propan$-1-$ol indicates a chain of $3$ carbon atoms with an alcohol $(-OH)$ functional group attached to the first carbon atom.
In a bond line formula,each vertex and endpoint represents a carbon atom.
For propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$,we need a $3$-carbon chain where the $-OH$ group is at the end.
Option $C$ correctly represents this structure,where the terminal carbon is bonded to the $-OH$ group.

(A) According to the $IUPAC$ nomenclature rules for polyfunctional compounds,the priority order for selecting the principal functional group is as follows:
$-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > -C=C- > -C\equiv C-$
Comparing the given options with this standard order,the correct priority sequence is $-COOH > -SO_3H > -COOR$.

(C) The principal functional group is the carboxylic acid $(-COOH)$ group,which is assigned position $1$ on the benzene ring.
Following the numbering to give the lowest possible locants to the substituents,the hydroxyl group $(-OH)$ is at position $3$ and the methyl group $(-CH_3)$ is at position $4$.
Therefore,the $IUPAC$ name is $3-$hydroxy$-4-$methylbenzoic acid.

(D) $1$. Identify the principal functional group. The $-COOH$ group has higher priority than the $-OH$ group and the alkyl group. Therefore,the parent compound is benzoic acid.
$2$. Number the benzene ring starting from the carbon attached to the $-COOH$ group as $C-1$.
$3$. Assign positions to the substituents: the $-OH$ group is at position $3$ and the ethyl group $(-C_2H_5)$ is at position $5$.
$4$. Arrange the substituents in alphabetical order: 'e' for ethyl comes before 'h' for hydroxy.
$5$. Combining these,the $IUPAC$ name is $3-$hydroxy$-5-$ethylbenzoic acid. However,looking at the options provided,option $D$ ($5-$Ethyl$-3-$hydroxybenzoic acid) is the correct nomenclature based on the numbering shown in the provided solution image.

(C) $1$. The given compound is a five-membered cyclic ring (cyclopentane) with an $-OH$ group and two methyl groups attached.
$2$. According to $IUPAC$ nomenclature rules,the carbon atom attached to the principal functional group $(-OH)$ is assigned the number $1$.
$3$. Numbering the ring to give the lowest possible locants to the substituents (methyl groups),we go from the $-OH$ carbon $(C-1)$ towards the methyl groups.
$4$. The methyl groups are at positions $2$ and $5$. Thus,the name is $2,5-$dimethylcyclopentanol.

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

(B) $1$. 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 (ethyl and methyl groups),we number the ring starting from the carbon with the $-OH$ group as $1$. Moving clockwise,the ethyl group is at position $2$ and the methyl group is at position $5$.
$3$. Alphabetical order: Substituents are listed alphabetically. Since 'ethyl' comes before 'methyl',the name is $2-$ethyl$-5-$methylcyclopentanol.

(B) $1$. Identify the longest carbon chain containing the functional group $(-NH_2)$ and the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. Number the chain starting from the end closer to the $-NH_2$ group to give it the lowest possible locant. Thus,the $-NH_2$ group is at position $2$.
$3$. The double bond starts at position $3$. Therefore,the suffix for the alkene is $-3-$ene and for the amine is $-2-$amine.
$4$. Combining these,the $IUPAC$ name is Pent$-3-$en$-2-$amine.

(B) The structure of allylamine is $CH_2=CH-CH_2-NH_2$.
In this molecule,the principal functional group is the amine $(-NH_2)$,which is assigned the lowest possible number.
The carbon chain contains $3$ carbon atoms with a double bond starting at position $2$.
Therefore,the $IUPAC$ name is Prop-$2$-en-$1$-amine.

(C) The given molecule is $CH_3-CH(NH_2)-CH_2-COOH$.
$1$. The principal functional group is the carboxylic acid group $(-COOH)$,so the parent chain is a butanoic acid derivative.
$2$. Numbering starts from the $-COOH$ carbon as $C-1$. Thus,the amino group $(-NH_2)$ is at $C-3$.
$3$. The correct $IUPAC$ name is $3-$aminobutanoic acid.
$4$. Since the amino group is on the $\beta-$carbon relative to the carboxyl group,it is a $\beta-$amino acid.
Therefore,statement $C$ is true.

(C) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond. The chain has $8$ carbons,so the parent alkane is octane.
$2$. Number the chain from the end that gives the lowest locant to the principal functional group $(-OH)$. Numbering from right to left gives the $-OH$ group at position $3$.
$3$. The double bond starts at carbon $4$,so it is named as oct$-4-$en.
$4$. There is a methyl group at position $6$.
$5$. Combining these,the $IUPAC$ name is $6-$methyloct$-4-$en$-3-$ol.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $7$ carbon atoms,so the parent alkane is heptane,and with the double bond,it is hept$-3-$ene.
$2$. Number the chain from the end that gives the lowest locant to the double bond. Numbering from the right gives the double bond position $3$.
$3$. Identify the substituents: there is a bromo group at position $2$ and methyl groups at positions $4$ and $5$.
$4$. Combining these,the $IUPAC$ name is $2-$Bromo$-4,5-$dimethylhept$-3-$ene.

(A) $1$. Identify the principal functional group: The compound contains a ketone group,so the suffix is $-one$ and the parent chain is a cyclohexane ring,making it a cyclohexanone.
$2$. Number the ring: The carbonyl carbon of the ketone is assigned position $1$. We number the ring to give the substituents the lowest possible locants. Moving clockwise,the chlorine atom is at position $2$ and the methyl group is at position $4$.
$3$. Assemble the name: The substituents are $2-$chloro and $4-$methyl. Alphabetical order is followed for naming substituents ($C$ before $M$).
$4$. Thus,the $IUPAC$ name is $2-$chloro$-4-$methylcyclohexanone.

(C) The priority order for functional groups in $IUPAC$ nomenclature is as follows:
$-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > -C=C- > -C\equiv C-$
Comparing the given options:
$1$. $-COOR$ (Ester)
$2$. $-COCl$ (Acyl chloride)
$3$. $>C=O$ (Ketone)
$4$. $-CONH_2$ (Amide)
According to the priority order,the ketone group $(>C=O)$ has the lowest priority among the given choices.

(B) The chemical structure of crotonyl alcohol is $CH_3CH=CHCH_2OH$.
In the $IUPAC$ nomenclature,the longest carbon chain containing the functional group $(-OH)$ and the double bond is selected.
The numbering starts from the carbon attached to the $-OH$ group to give it the lowest possible locant.
Thus,the chain has $4$ carbons,the double bond starts at position $2$,and the alcohol group is at position $1$.
The correct $IUPAC$ name is $but-2-en-1-ol$.

(D) $1$. The structure is $CH_3-CH(OH)-C(=O)-CH_2-CH_3$. The ketone group $(>C=O)$ has higher priority than the hydroxyl group $(-OH)$ according to $IUPAC$ nomenclature rules. Thus,the ketone is the principal functional group and $-OH$ is a substituent. Option $(B)$ is false.
$2$. The $-OH$ group is attached to the $C-2$ carbon,which is bonded to four single bonds,making it $sp^3$ hybridized. Option $(A)$ is false.
$3$. Counting the atoms: $5$ carbons,$10$ hydrogens,and $2$ oxygens. The molecular formula is $C_5H_{10}O_2$. Option $(C)$ is false.
$4$. The longest carbon chain containing the ketone group has $5$ carbons (pentane). Numbering from the left gives the hydroxyl group at position $2$ and the ketone at position $3$. The $IUPAC$ name is $2-$hydroxypentan$-3-$one. Option $(D)$ is true.

(B) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $7$ carbon atoms,so the parent alkane is heptane,and with the double bond,it is hept$-3-$ene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from the right gives the double bond position $3$.
$3$. Identify the substituents: there is an iodo group at position $2$ and methyl groups at positions $4$ and $5$.
$4$. Combine these to get the $IUPAC$ name: $2-$Iodo$-4,5-$dimethylhept$-3-$ene.

(C) In the carbinol system,alcohols are considered as derivatives of methyl alcohol $(CH_3OH)$,which is called carbinol.
Isopropyl alcohol has the structure $(CH_3)_2CHOH$.
In this structure,two hydrogen atoms of the carbinol group $(-CH_2OH)$ are replaced by two methyl groups $(-CH_3)$.
Therefore,it is named as dimethyl carbinol.

(C) The given structure is $(CH_3)_3N$,which consists of a nitrogen atom bonded to three methyl groups.
According to $IUPAC$ nomenclature for tertiary amines,the longest carbon chain attached to the nitrogen is selected as the parent alkane,and the other alkyl groups are named as $N-$substituents.
Here,the parent chain is methane $(CH_4)$,and there are two methyl groups attached to the nitrogen atom.
Therefore,the $IUPAC$ name is $N, N-$dimethylmethanamine.

(C) $1$. Identify the longest carbon chain: The longest chain contains $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain: Numbering should be done such that the substituents get the lowest possible locants. Numbering from left to right gives the bromine at position $2$ and the methyl group at position $3$.
$3$. Assign names to substituents: There is a bromo group at position $2$ and a methyl group at position $3$.
$4$. Combine: The $IUPAC$ name is $2-$bromo$-3-$methylbutane.

(D) The $IUPAC$ name prop$-2-$en$-1-$amine indicates a three-carbon chain (prop) with a double bond at the second position ($-2-$en) and an amine group at the first position ($-1-$amine).
The structure is $CH_2=CH-CH_2-NH_2$.

(A) $1$. The principal functional group is the ketone,which gets the lowest possible locant. Thus,the carbonyl carbon is assigned position $1$.
$2$. Numbering the ring to give the substituents the lowest possible locants,we move towards the chlorine atom to give it position $2$.
$3$. The methyl group is then at position $4$.
$4$. Combining these,the $IUPAC$ name is $2-$chloro$-4-$methylcyclohexanone.

(C) The structure $CH_3CH(CH_3)CH_3$ is $iso-$butane (or $2-$methylpropane),not $n-$butane.
The structure of $n-$butane is $CH_3CH_2CH_2CH_3$.
Therefore,option $(C)$ is the incorrect match.

(A) $1$. Identify the principal functional group: The $-COOH$ group has the highest priority,so the parent compound is benzoic acid.
$2$. Number the ring: Start numbering from the carbon attached to the $-COOH$ group as $C-1$. Number the ring to give the lowest possible locants to the substituents.
$3$. Assign positions: The $-OH$ group is at position $3$ and the $-CH_3$ group is at position $4$.
$4$. Name the compound: Combining these,the $IUPAC$ name is $3-$hydroxy$-4-$methylbenzoic acid.

(B) $1$. Identify the principal functional group: The hydroxyl group $(-OH)$ is the principal functional group,so the parent chain is a cycloheptanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the substituents the lowest possible locants,we number the ring starting from the carbon with the $-OH$ group. Moving clockwise,the ethyl group is at position $3$ and the methyl group is at position $6$.
$3$. Alphabetical order: Ethyl comes before methyl. Thus,the name is $3-$ethyl$-6-$methylcycloheptan$-1-$ol.

(C) The carbinol system names alcohols as derivatives of methanol,which is called $carbinol$ $(CH_3OH)$.
For isobutyl alcohol,the structure is $(CH_3)_2CH-CH_2OH$.
To name this in the carbinol system,we identify the group attached to the $CH_2OH$ unit.
Here,the group attached to the $-CH_2OH$ group is an isopropyl group,$(CH_3)_2CH-$.
Therefore,the name is isopropyl carbinol.

(D) $1$. Identify the principal functional group $(PFG)$. Here,the carboxylic acid group $(-COOH)$ has higher priority than the aldehyde $(-CHO)$ and methyl $(-CH_3)$ groups. Therefore,the parent chain is benzoic acid.
$2$. Number the benzene ring starting from the carbon attached to the $-COOH$ group as $1$.
$3$. Proceed in the direction that gives the lowest locants to the other substituents. Moving towards the methyl group gives it position $2$,and the formyl group gets position $5$.
$4$. The substituents are $5-$formyl and $2-$methyl.
$5$. Combining these,the $IUPAC$ name is $5-$formyl$-2-$methylbenzoic acid.

(B) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $7$ carbon atoms,so the parent alkane is heptane,and the alkene is heptene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from the right gives the double bond position $3$.
$3$. Identify the substituents: there is a chloro group $(-Cl)$ at position $3$ and a methyl group $(-CH_3)$ at position $5$.
$4$. Combine these to get the $IUPAC$ name: $3-$chloro$-5-$methylhept$-3-$ene.

(B) The chemical formula of glyoxal is $CHO-CHO$.
The structure consists of two carbon atoms,each attached to an aldehyde group $(-CHO)$.
Since there are two carbon atoms in the parent chain,the prefix is 'eth-'.
The suffix for a dialdehyde is '-dial'.
Therefore,the $IUPAC$ name is Ethanedial.

(C) The given compound is $(CH_3)_4 C$,which can be represented as a central carbon atom bonded to four methyl groups.
To determine the $IUPAC$ name,we first identify the longest carbon chain,which contains $3$ carbon atoms (propane).
There are two methyl groups attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2, 2-$dimethylpropane.

(B) $1$. Identify the longest carbon chain containing the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane,and with the double bond,it is pentene.
$2$. Number the chain from the end that gives the double bond the lowest possible locant. Numbering from the left gives the double bond at position $2$.
$3$. Identify the substituents: there is a chloro group at position $4$ and a methyl group at position $3$.
$4$. Combining these,the $IUPAC$ name is $4-$chloro$-3-$methylpent$-2-$ene.