
The IUPAC name of the compound ${(C{H_3})_2}CH - CH \equiv CH - CHOH - C{H_3}$ is?
Answer
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Hint: Functional group: In the hydrogen the atoms or groups which are other than carbon and hydrogen, are known as function groups. For example: chloride if chlorine is present in the compound. And if hydroxide group i.e. $OH$ is present then the name will be alcohol.
Complete step by step solution:
First of all we will talk about the alkanes, alkenes and alkynes.
Alkanes: The compounds which are formed by carbon and hydrogen and have only a single bond between the carbon-carbon atoms, are known as alkanes. For example: The first member of the alkane family is ethane $({H_3}C - C{H_3})$. The general formula of the alkane group is ${C_n}{H_{(2n + 2)}}$.
Alkenes: The compounds which are formed by carbon and hydrogen and have at least one double bond along with a single bond between the carbon-carbon atoms, are known as alkenes. For example: The first member of the alkene family is ethene $({H_2}C = C{H_2})$. The general formula of the alkene group is ${C_n}{H_{2n}}$.
Alkynes: The compounds which are formed by carbon and hydrogen and have at least one triple bond along with a single bond between the carbon-carbon atoms, are known as alkynes. For example: The first member of the alkyne family is ethyne $(HC \equiv CH)$. The general formula of the alkyne group is ${C_n}{H_{(2n - 2)}}$.
The first step in the IUPAC naming system is to identify the longest carbon chain.
Now here in the question the number of carbon in the compound is $7$.
And the longest carbon chain contains only $6$ carbons. So word hex will be there and also there is one three bond i.e. alkyne group is there at third position from right side. So the parent chain will be hex $ - 3 - $yne.
And the numbering to the compound is done in the way so that carbon containing the functional group should have less number. Here if we do numbering from right then we get an alcohol containing group at $2$ position and alkyne group at $3$ position. So the numbering should be from the right side.
Now at position $5$ there is one methyl group attached.
So the final IUPAC name of this compound is as: $5 - {\text{methyl hex}} - 3 - en - 2 - ol$ (because the suffix to alcohol groups used is –ol).
Note: Suffix to some functional group are as: for carboxylic acid suffix used is –oic acid, for alcohols suffix used is alkyl alcohol or in place of e from alkane add ol. For example: if an alcohol group is present in methane then the IUPAC name of the compound will be methyl alcohol.
Complete step by step solution:
First of all we will talk about the alkanes, alkenes and alkynes.
Alkanes: The compounds which are formed by carbon and hydrogen and have only a single bond between the carbon-carbon atoms, are known as alkanes. For example: The first member of the alkane family is ethane $({H_3}C - C{H_3})$. The general formula of the alkane group is ${C_n}{H_{(2n + 2)}}$.
Alkenes: The compounds which are formed by carbon and hydrogen and have at least one double bond along with a single bond between the carbon-carbon atoms, are known as alkenes. For example: The first member of the alkene family is ethene $({H_2}C = C{H_2})$. The general formula of the alkene group is ${C_n}{H_{2n}}$.
Alkynes: The compounds which are formed by carbon and hydrogen and have at least one triple bond along with a single bond between the carbon-carbon atoms, are known as alkynes. For example: The first member of the alkyne family is ethyne $(HC \equiv CH)$. The general formula of the alkyne group is ${C_n}{H_{(2n - 2)}}$.
| Number of carbon atom in alkane | Name of the parent chain |
| One | Methane |
| Two | Ethane |
| Three | Propane |
| Four | Butane |
| Five | Pentane |
| Six | Hexane |
| Seven | Heptane |
The first step in the IUPAC naming system is to identify the longest carbon chain.
Now here in the question the number of carbon in the compound is $7$.
And the longest carbon chain contains only $6$ carbons. So word hex will be there and also there is one three bond i.e. alkyne group is there at third position from right side. So the parent chain will be hex $ - 3 - $yne.
And the numbering to the compound is done in the way so that carbon containing the functional group should have less number. Here if we do numbering from right then we get an alcohol containing group at $2$ position and alkyne group at $3$ position. So the numbering should be from the right side.
Now at position $5$ there is one methyl group attached.
So the final IUPAC name of this compound is as: $5 - {\text{methyl hex}} - 3 - en - 2 - ol$ (because the suffix to alcohol groups used is –ol).
Note: Suffix to some functional group are as: for carboxylic acid suffix used is –oic acid, for alcohols suffix used is alkyl alcohol or in place of e from alkane add ol. For example: if an alcohol group is present in methane then the IUPAC name of the compound will be methyl alcohol.
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