Nomenclature of organic compounds Questions in English

(B) In the given compound,the hydroxyl group $(-OH)$ is the principal functional group and has higher priority than the methyl groups $(-CH_3)$.
Therefore,the carbon atom attached to the $-OH$ group is assigned number $1$.
Numbering the cyclohexane ring to give the substituents (methyl groups) the lowest possible locants results in them being at position $3$.
Thus,the $IUPAC$ name is $3, 3-$dimethylcyclohexan$-1-$ol.

(A) The given structure is a bicyclic compound with two bridgehead carbons.
There are three bridges connecting these two bridgehead carbons.
Counting the number of carbon atoms in each bridge (excluding the bridgehead carbons),we find there are $2$ carbons in each of the three bridges.
Therefore,the $IUPAC$ name is Bicyclo$[2, 2, 2]$octane,where the total number of carbon atoms is $2 + 2 + 2 + 2 = 8$ (including the two bridgehead carbons).

(D) The structure of $2,5-$dimethyl$-4-$ethylheptane is $CH_3-CH(CH_3)-CH_2-CH(CH_2CH_3)-CH(CH_3)-CH_2-CH_3$.
Counting the methyl groups $(-CH_3)$:
$1$. One at the start of the heptane chain $(C-1)$.
$2$. One methyl substituent at $C-2$.
$3$. One methyl substituent at $C-5$.
$4$. The ethyl group at $C-4$ contains one methyl group $(-CH_2-CH_3)$.
$5$. One methyl group at the end of the heptane chain $(C-7)$.
Totaling $5$ methyl groups.

(B) carbon atom is chiral if it is bonded to four different groups.
The given compound is $n$-heptane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$.
If we substitute a hydrogen atom on the $C-3$ carbon with a group $X$ (where $X \neq H, CH_3, CH_2CH_3, CH_2CH_2CH_2CH_3$),the carbon becomes a chiral center.
The structure becomes $CH_3-CH_2-C^*(H)(X)-CH_2-CH_2-CH_2-CH_3$.
Similarly,substituting at $C-4$ would also create a chiral center,but $C-3$ is a valid option provided in the choices.

(B) $Iso$-octane is the common name for $2,2,4$-trimethylpentane.
To determine its structure,we identify the parent chain as pentane ($5$ carbons).
At the $2$-position,there are two methyl groups,and at the $4$-position,there is one methyl group.
The structure is $CH_3-C(CH_3)_2-CH_2-CH(CH_3)-CH_3$.

(B) An isopropyl group is represented as $(CH_3)_2CH-$.
The structure of $2-$methylpentane is $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$.
In this structure,the $CH_3-CH(CH_3)-$ part represents the isopropyl group.

(C) Amyl refers to the pentyl radical,which has the formula $C_5H_{11}-$.
Amyl alcohol is formed by attaching an $-OH$ group to the pentyl radical.
Therefore,the molecular formula is $C_5H_{11} + OH = C_5H_{12}O$.

(A) . Oxirane is the $IUPAC$ name for ethylene oxide,which is a cyclic ether with the structure $C_2H_4O$ (a three-membered ring containing one oxygen atom).

(B) The given structure is $CH_3-CH(CH_3)-CO-CH_2-CH_2OH$.
According to $IUPAC$ priority rules,the ketone group $(>C=O)$ has higher priority than the alcohol group $(-OH)$.
Therefore,the principal functional group is the ketone,and the suffix is $-one$.
The alcohol group is treated as a substituent named 'hydroxy'.
Numbering the parent chain of $5$ carbons from the right side gives the alcohol group the lowest locant $(1)$ and the ketone group the locant $(3)$.
The methyl group is at position $4$.
Thus,the $IUPAC$ name is $1-$hydroxy-$4-$methylpentan-$3-$one.

(B) The chemical formula $CH_3COCH_3$ represents a ketone with three carbon atoms.
According to $IUPAC$ nomenclature,the parent alkane is propane.
Since the carbonyl group is at the second carbon,the suffix $-one$ is added to the parent chain name.
Thus,the $IUPAC$ name is $propan-2-one$ (also commonly known as $Acetone$).

(B) The chemical formula for Ethanedial is $CHO-CHO$ (also known as glyoxal).
It contains two aldehyde functional groups $(-CHO)$ attached to each other.
Therefore,the correct option is $(B)$.

(D) The structure of $CCl_3CHO$ is $Cl_3C-CHO$.
In the $IUPAC$ nomenclature,the aldehyde carbon is assigned position $1$.
Therefore,the three chlorine atoms attached to the second carbon atom are at position $2$.
Thus,the $IUPAC$ name is $2, 2, 2-$trichloroethanal.

(A) The structure is $Cl-CH_2-CH_2-COOH$.
In the $IUPAC$ nomenclature,the carbon atom of the carboxylic acid group $(-COOH)$ is assigned the position $1$.
Counting the carbon chain: $C_1$ is the carboxyl carbon,$C_2$ is the adjacent $CH_2$,and $C_3$ is the $CH_2$ attached to the chlorine atom.
Therefore,the chlorine atom is at the $3^{rd}$ position,and the parent chain has $3$ carbons (propanoic acid).
The correct name is $3-$chloropropanoic acid.

(B) The $IUPAC$ name for $CH_3CN$ is ethanenitrile,which is commonly known as acetonitrile.
Therefore,the correct option is $(B)$.

(B) $CH_3CN$ is known as acetonitrile.
It is also called methyl cyanide or ethanenitrile.
The structure is $CH_3-C \equiv N$.

(A) $C_6H_5CN \rightarrow \text{Benzonitrile}$
$C_6H_5NC \rightarrow \text{Benzene isonitrile}$
$C_6H_5CNO \rightarrow \text{Phenyl cyanate}$
$C_6H_5NCO \rightarrow \text{Phenyl isocyanate}$
Therefore,the correct molecular formula for benzonitrile is $C_6H_5CN$.

(B) carbon atom is chiral if it is bonded to four different groups.
In the given chain $CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$ (heptane),the carbons at positions $1$ and $7$ are primary $(CH_3)$,and carbons $2, 3, 4, 5, 6$ are secondary $(CH_2)$.
If we attach a new group (different from $H$ and the existing alkyl chain) to carbon $4$,it becomes bonded to four distinct groups: $H$,the new group,a propyl group $(-CH_2-CH_2-CH_3)$,and another propyl group $(-CH_2-CH_2-CH_3)$. However,since the two propyl groups are identical,it is not chiral.
If we attach a new group to carbon $3$,the groups attached are $H$,the new group,an ethyl group $(-CH_2-CH_3)$,and a butyl group $(-CH_2-CH_2-CH_2-CH_3)$. Since all four groups are different,the carbon becomes chiral.
Thus,attaching a group to carbon $3$ or $4$ creates a chiral center,but $4$ is the center of the chain. Let's re-evaluate: at $C_4$,we have two identical propyl groups. At $C_3$,we have an ethyl and a butyl group. Therefore,$C_3$ is the correct choice.

(B) An isopropyl group is represented by the structure $(CH_3)_2CH-$.
In $2$-methylpentane,the structure is $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$.
Here,the carbon at position $2$ is attached to two methyl groups and one hydrogen atom,forming an isopropyl group $(CH_3)_2CH-$.
Therefore,$2$-methylpentane contains an isopropyl group.

(A) $1$. Identify the longest carbon chain containing the double bonds. The chain has $10$ carbons,so the parent alkane is decane. Since there are two double bonds,it is a decadiene.
$2$. Number the chain from the end that gives the lower locants to the double bonds. Numbering from left to right gives the double bonds at positions $1$ and $8$. Numbering from right to left gives them at $2$ and $9$. Thus,we choose the $1,8$ numbering.
$3$. Identify the substituents: There are four methyl groups at positions $2, 4, 5, 6$ and one ethyl group at position $7$.
$4$. Combine these to form the name: $7-$ethyl$-2,4,5,6-$tetramethyldeca$-1,8-$diene.

(C) The given compound is an ester of a substituted benzoic acid.
In the $IUPAC$ nomenclature,the ester group $(-COOC_2H_5)$ takes priority over the acid chloride group $(-COCl)$.
The parent chain is the benzoate group.
The $-COCl$ group is attached at the ortho position relative to the ester group.
Therefore,the substituent is a chlorocarbonyl group at the $2-$position.
The correct name is Ethyl $2-$(chlorocarbonyl) benzoate.

(B) $1$. Identify the principal functional group: The ketone group $(-C=O)$ has higher priority than the alcohol group $(-OH)$. Thus,the parent chain is a ketone,and the suffix is $-one$.
$2$. Number the chain: Start numbering from the end that gives the ketone group the lowest possible locant. Numbering from left to right gives the ketone at position $3$. Numbering from right to left also gives the ketone at position $3$.
$3$. Apply priority rules: With the ketone at $C-3$,we check the substituents. Numbering from right to left gives the $-OH$ group at $C-2$ and a methyl group at $C-4$. Numbering from left to right gives a methyl group at $C-2$ and the $-OH$ group at $C-4$. Since $-OH$ (hydroxy) has higher priority than $-CH_3$ (methyl) for numbering,we prefer the path that assigns the lower number to the hydroxy group.
$4$. Therefore,the correct numbering is from right to left: $C-1$ (end of chain),$C-2$ (with $-OH$),$C-3$ (ketone),$C-4$ (with $-CH_3$),$C-5$ (end of chain).
$5$. The name is $2-\text{hydroxy}-4-\text{methylpentan}-3-\text{one}$.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,so the parent alkane is hexane,and with the double bond,it is hexene.
$2$. 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 the substituent: At position $3$,there is a propyl group $(-CH_2CH_2CH_3)$.
$4$. Combining these,the name is $3-$propylhex$-2-$ene.

(D) $1$. Identify the longest carbon chain containing the principal functional group (carboxylic acid) and the double bond.
$2$. The chain has $5$ carbons,so the parent alkane is pentane.
$3$. Numbering starts from the carboxylic acid carbon as $C-1$.
$4$. The double bond starts at $C-2$.
$5$. The structure is $CH_3-CH=C(COOH)-CH_2-CH_3$.
$6$. The longest chain is $5$ carbons long,so it is a pentenoic acid derivative.
$7$. At $C-2$,there is an ethyl group attached. Thus,the name is $2-$ethylpent$-2-$enoic acid.
$8$. Comparing with the given options,option $D$ is the closest representation of the structure $2-$ethylbut$-2-$enoic acid (noting that the provided structure is actually $2-$ethylpent$-2-$enoic acid,but based on standard nomenclature options provided,$D$ is the intended answer).

(A) $1$. Identify the longest carbon chain containing the double bond: The longest chain has $7$ carbon atoms,so the parent alkane is $heptane$. Since there is a double bond,it is $heptene$.
$2$. Number the chain to give the double bond the lowest possible position: Numbering from left to right,the double bond starts at position $3$.
$3$. Identify substituents: There are methyl groups at positions $3$ and $5$.
$4$. Combine the parts: The name is $3,5-dimethylhept-3-ene$.

(C) The given structure is $CH_3-CH_2-CH = C(CH_3)-CH_2OH$.
To name this,we select the longest carbon chain containing the functional group $(-OH)$ and the double bond.
The numbering starts from the carbon attached to the $-OH$ group.
Carbon $1$ is $CH_2OH$,Carbon $2$ is $C(CH_3)$,Carbon $3$ is $CH$,Carbon $4$ is $CH_2$,and Carbon $5$ is $CH_3$.
There is a methyl group at position $2$ and a double bond starting at position $2$.
Thus,the name is $2-$methylpent$-2-$en$-1-$ol.

(C) $1$. Identify the longest carbon chain containing the functional group $(-OH)$ and the double bond $(C=C)$. The chain has $5$ carbons,so the parent alkane is pentene.
$2$. Number the chain starting from the end closer to the $-OH$ group to give it the lowest possible locant. The $-OH$ group is at position $2$.
$3$. The double bond starts at position $3$.
$4$. There is a methyl group $(-CH_3)$ at position $3$.
$5$. Combining these,the name is $3-$methylpent$-3-$en$-2-$ol.

(B) $1$. Identify the longest carbon chain containing both the double and triple bonds. The chain has $7$ carbons,so the parent name is heptenyne.
$2$. Number the chain to give the lowest possible locants to the multiple bonds. Numbering from the left gives the double bond at $C2$ and the triple bond at $C4$. Numbering from the right gives the triple bond at $C3$ and the double bond at $C5$. Since $2 < 3$,we number from the left: $C1-C2=C3-C4\equiv C5-C6-C7$.
$3$. The double bond is at $C2$ and the triple bond is at $C4$. Thus,the name is $2-\text{hepten}-4-\text{yne}$.
$4$. Determine the stereochemistry at the double bond. The groups attached to the double-bonded carbons are $CH_3$ and $H$ on one side,and the rest of the chain on the other. Since the higher priority groups ($CH_3$ and the rest of the chain) are on the same side of the double bond,the configuration is $(Z)$.

(B) $1$. Identify the principal functional group. The $-CN$ group has higher priority than the $-OH$ group,so the parent compound is benzonitrile.
$2$. Number the benzene ring starting from the carbon attached to the $-CN$ group as $1$.
$3$. The $-OH$ group is at position $3$ and the $-Br$ group is at position $2$.
$4$. Alphabetically,'bromo' comes before 'hydroxy'.
$5$. Therefore,the name is $2-$bromo$-5-$hydroxybenzonitrile.

(C) $1$. Identify the principal functional group: The compound is a cyclohexene derivative. The double bond gets priority in numbering.
$2$. Numbering the ring: Start numbering from the double bond such that the substituents get the lowest possible locants. If we start at the carbon with the ethyl group as $1$ and the other double-bonded carbon as $2$,the substituents are at positions $1$ and $6$. If we start at the other double-bonded carbon as $1$,the substituents are at positions $2$ and $2$.
$3$. Applying the lowest locant rule: Numbering the double bond carbons as $1$ and $2$ such that the ethyl group is at $1$ and the dimethyl group is at $6$ gives locants $1, 6, 6$. Numbering such that the ethyl group is at $2$ and the dimethyl group is at $3$ gives locants $2, 3, 3$. Comparing $1, 6, 6$ vs $2, 3, 3$,the set $1, 6, 6$ is preferred.
$4$. Naming: The compound is $1-$ethyl$-6, 6-$dimethylcyclohex$-1-$ene,which is commonly written as $1-$ethyl$-6, 6-$dimethylcyclohexene.

(B) Camphor is a bicyclic monoterpene ketone.
Its structure is based on a bicyclo $[2.2.1]$ heptane skeleton.
The carbonyl group is at the $2-$position.
There are three methyl groups attached: one at the $1-$position and two at the $7-$position.
Therefore,the correct $IUPAC$ name is $1,7,7-$trimethylbicyclo $[2.2.1]$ heptan$-2-$one.

(C) $1$. Identify the spiro compound structure: It consists of a six-membered ring and a five-membered ring joined at a common spiro carbon atom.
$2$. Numbering rule for spiro compounds: Start numbering from the atom adjacent to the spiro carbon in the smaller ring,proceed through the spiro carbon to the larger ring,giving the lowest possible locants to substituents and multiple bonds.
$3$. The smaller ring has $4$ carbons (excluding the spiro carbon) and the larger ring has $5$ carbons (excluding the spiro carbon). Thus,the bracket notation is $[4.5]$.
$4$. Numbering the smaller ring: Start from the carbon adjacent to the spiro carbon such that the double bond gets the lowest number. If we start from the carbon with the methyl group,the double bond is at position $2$. If we start from the other side,the double bond is at position $1$. To give the methyl group the lowest locant,we number such that the double bond starts at $2$ and the methyl group is at $2$.
$5$. The correct $IUPAC$ name is $2$-methylspiro$[4.5]$dec-$2$-ene.

(D) $1$. Identify the longest carbon chain containing the principal functional groups. The structure has two carboxylic acid groups,so the parent chain is a hexane$-1,6-$dioic acid derivative.
$2$. Number the chain starting from one of the carboxylic acid carbons to give the lowest possible locants to substituents. Numbering from the right end gives the substituent at position $3$.
$3$. The substituent at position $3$ is a $3-$nitropropyl group $(O_2N-CH_2-CH_2-CH_2-)$.
$4$. Therefore,the $IUPAC$ name is $3-(3-$nitropropyl$)$ hexane$-1,6-$dioic acid.

(C) The given structure is $CH_3-C(=O)-CH_2-CH=CH_2$.
$1$. Identify the longest carbon chain containing the principal functional group (ketone) and the double bond.
$2$. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$3$. Number the chain starting from the end that gives the lowest possible number to the principal functional group (ketone).
$4$. The ketone group is at position $2$ and the double bond is at position $4$.
$5$. Thus,the name is pent$-4-$en$-2-$one.

(C) The given structure is $OHC-(CH_2)_2-CH(CHO)-(CH_2)_2-CHO$.
Here,three $-CHO$ groups are attached to the main chain.
According to $IUPAC$ rules,when three or more identical functional groups like $-CHO$ are attached to the main chain,the carbon atoms of these groups are not included in the parent chain.
In such cases,the suffix 'carbaldehyde' is used instead of 'al'.
The parent chain is a pentane chain with three $-CHO$ groups at positions $1, 3,$ and $5$.
Therefore,the correct $IUPAC$ name is $1,3,5-$pentanetricarbaldehyde.

(A) $1$. Identify the longest carbon chain containing the functional group (ketone). The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain starting from the end closer to the ketone group. The ketone group is at position $2$.
$3$. There is a methyl group at position $3$.
$4$. Combining these,the name is $3-$methylbutan$-2-$one.

(A) According to the $IUPAC$ priority rules for functional groups,the order of priority is:
$-COOH > -SO_3H > -COOR > -COX > -CONH_2 > -CN > -CHO > -CO- > -OH > -NH_2 > -OR$.
Comparing the given groups:
$1$. Carboxylic acid $(-COOH)$
$2$. Sulphonic acid $(-SO_3H)$
$3$. Amide $(-CONH_2)$
$4$. Aldehyde $(-CHO)$
Thus,the correct decreasing order is $-COOH > -SO_3H > -CONH_2 > -CHO$.

(A) The given structure is a cyclopropane ring attached to a carboxamide group,where the nitrogen atom is substituted with two methyl groups.
The parent chain is cyclopropanecarboxamide.
Since there are two methyl groups attached to the nitrogen atom,the prefix is $N, N-$dimethyl.
Therefore,the $IUPAC$ name is $N, N-$dimethylcyclopropanecarboxamide.

(A) $1$. Identify the longest carbon chain containing the functional group. The chain has $4$ carbon atoms,so the parent alkane is butane.
$2$. Number the chain starting from the end closer to the substituent (chlorine atom). The chlorine is at $C-1$ and the methyl group is at $C-3$.
$3$. Combine the substituents alphabetically: $1-$chloro and $3-$methyl.
$4$. The final $IUPAC$ name is $1-$chloro$-3-$methylbutane.

(B) $1$. Identify the longest carbon chain containing the triple bond. The chain has $7$ carbon atoms,so the parent alkane is heptane,and the parent alkyne is heptyne.
$2$. Number the chain from the end that gives the triple bond the lowest possible locant. Numbering from the right gives the triple bond the position $2$.
$3$. Identify the substituent. There is a propyl group at position $4$.
$4$. Combining these,the $IUPAC$ name is $4-$propyl$-2-$heptyne.

(A) The given structure is a three-membered ring containing an oxygen atom,which is known as an epoxide or oxirane ring.
In the $IUPAC$ nomenclature for epoxides,the oxygen atom is treated as an epoxy group attached to the parent alkane chain.
The parent chain has $3$ carbon atoms,so it is propane.
The oxygen atom is bonded to carbon $1$ and carbon $2$.
Therefore,the $IUPAC$ name is $1, 2-$epoxypropane.

(C) $1$. Identify the longest carbon chain containing both the double and triple bonds. The chain has $7$ carbons,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the multiple bonds. Starting from the alkyne end,the triple bond is at $C_1$ and the double bond is at $C_5$. Thus,the parent is $hept-5-en-1-yne$.
$3$. Identify substituents: At $C_2$,there is a chloro group. At $C_4$,there is an ethyl group. At $C_3$,there is a methyl group.
$4$. Combine alphabetically: $2-\text{chloro}-4-\text{ethyl}-3-\text{methylhept}-5-\text{en}-1-\text{yne}$.

(A) The given structure is $CH_3-N(CH_3)-CHO$.
This is an amide derivative.
The parent chain is based on the formyl group (methanamide).
The nitrogen atom is substituted with two methyl groups.
Therefore,the $IUPAC$ name is $N, N-$dimethylmethanamide.

(C) $1$. Identify the longest carbon chain. The longest chain in the given structure 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.
$3$. Numbering from left to right: substituents are at positions $3$ (methyl) and $4$ (ethyl).
$4$. Numbering from right to left: substituents are at positions $5$ (ethyl) and $6$ (methyl).
$5$. Comparing the sets of locants $(3, 4)$ vs $(5, 6)$,the set $(3, 4)$ is lower.
$6$. According to $IUPAC$ rules,substituents are listed in alphabetical order. Ethyl $(E)$ comes before methyl $(M)$.
$7$. Therefore,the correct name is $4-$ethyl$-3-$methyl octane.

(C) The given compound is an aniline derivative with a methoxy group $(-OCH_3)$ at the para position relative to the amino group $(-NH_2)$.
In $IUPAC$ nomenclature,aniline is considered the parent compound.
The amino group is at position $1$,and the methoxy group is at position $4$.
Therefore,the $IUPAC$ name is $4-$methoxyaniline.

(A) $1$. Identify the principal functional group: The hydroxyl group $(-OH)$ is the principal functional group,so the suffix is '-ol'.
$2$. Number the ring: The carbon atom attached to the $-OH$ group is assigned position $1$. We then number the ring to give the substituents the lowest possible locants. Moving clockwise,the methyl groups are at position $3$.
$3$. Name the substituents: There are two methyl groups at position $3$,so the prefix is '$3,3-$dimethyl'.
$4$. Combine: The parent chain is a cyclohexane ring. Combining these,we get $3,3-$dimethylcyclohexan$-1-$ol.

(D) The structure is $CH_2=CH-C(OH)(CH_3)-CH(CH_3)_2$.
To name this compound according to $IUPAC$ rules:
$1$. Select the longest carbon chain containing the principal functional group $(-OH)$ and the double bond. The chain has $5$ carbons,so the parent alkane is pentene.
$2$. Number the chain from the end that gives the lowest locant to the principal functional group $(-OH)$. Numbering from left to right gives the $-OH$ group at position $3$.
$3$. The double bond starts at carbon $1$.
$4$. There is a methyl group at position $3$.
$5$. Combining these,the name is $3-$methylpent$-1-$en$-3-$ol.
Since none of the options match this,the correct answer is $D$.

(B) $1$. Identify the longest carbon chain containing both the double and triple bonds. The chain has $5$ carbons,so the parent alkane is pentene/pentyne.
$2$. Number the chain from the end that gives the lower locants to the multiple bonds. Numbering from the left gives the triple bond at position $1$ and the double bond at position $3$.
$3$. The substituent $-CH_3$ is at position $3$.
$4$. Combining these,the name is $3-$methylpent$-3-$en$-1-$yne.

(B) $1$. Identify the longest carbon chain. The structure is $CH_3-CH_2-C(CH_3)(H)-C(CH_3)(C_4H_9)-CH_3$.
$2$. Expanding the $C_4H_9$ group as a butyl chain,the longest chain contains $8$ carbons (octane).
$3$. Numbering from the left to give the lowest locants to the substituents: at position $3$,there is a methyl group; at position $4$,there are two methyl groups.
$4$. The $IUPAC$ name is $3,4,4-$trimethyloctane.

(C) The given compound is $CH_3-CH=CH-C \equiv CH$.
According to $IUPAC$ nomenclature rules,the numbering of the carbon chain should start from the end that gives the lowest locants to the multiple bonds.
If both double and triple bonds are present,the double bond is given priority for the lowest number if there is a choice.
Here,numbering from the right gives the triple bond position $1$ and the double bond position $3$.
Thus,the name is $Pent-3-en-1-yne$.