Welcome back, everyone name the al kind given below. As the problem suggests, we are given an alky. And this is because our hydrocarbon contains a triple bond. And we have to recall that the first step in naming Alkins is to identify the parent, which is the longest continuous carbon chain. And now it is really important to understand that we're going to enumerate that parent in a way such that we minimize the lo of the triple bond. We immediately notice that if we start enumerating our longest continuous carbon chain from left to right, we will be able to get the lowest possible lon for the triple bond, which is one. And now we can go along our chain and get our 2nd, 3rd, 4th, 5th and 6th carbon. If we move up, that would be the same, right? Because carbon number four is bonded to the same to substituents. So our parent is hex sign, that's because six has a prefix of hex, right? And for an L kind, the suffix is YME. However, we want to modify it slightly and specify the position of the triple bond because it starts at carbon number one, we're going to turn it into one hex sign. And our next step is to identify the substituents. Specifically, we're going to consider branching. What we notice is that we have two substituents bonded to carbon. Number 41 of them is the simplest possible methyl substituent since we have one carbon, right, with three implicit hydrogens, and the next one is called an ethyl substituent since it is a she remembered LK group, so we can essentially say that carbon four contains a methyl substitute. So we're going to use four dash methyl and four dash sl we want to arrange our subscriptions alphabetically in the name. So we're going with for fl first because we're comparing E against M followed by four muscle. And now let's finish our name with the parent which is one heck sign. So the complete name of the given Alky would be four ethyl, four methyl, one hex sign. Thank you for watching.

Ch.21 - Organic Chemistry

Tro - Chemistry: A Molecular Approach 4th Edition

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Ch.21 - Organic Chemistry

Problem 54c

Chapter 21, Problem 54c

Name each alkyne. c.

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Identify the longest carbon chain that includes the triple bond. This will determine the base name of the alkyne.

Number the carbon chain starting from the end nearest to the triple bond to give the lowest possible number to the triple bond.

Identify and name any substituents attached to the carbon chain.

Combine the names of the substituents with the base name of the alkyne, placing the substituents in alphabetical order and using numbers to indicate their positions.

Ensure the name follows IUPAC nomenclature rules, using the suffix '-yne' to indicate the presence of a triple bond.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Alkynes

Alkynes are a class of hydrocarbons characterized by at least one carbon-carbon triple bond (C≡C). They follow the general formula CnH2n-2, where 'n' is the number of carbon atoms. Alkynes are unsaturated compounds, meaning they contain fewer hydrogen atoms than alkanes and alkenes, which can affect their reactivity and properties.

Nomenclature of Alkynes

The nomenclature of alkynes follows the IUPAC naming conventions, where the longest carbon chain containing the triple bond is identified. The name is derived from the base name of the alkane with the same number of carbons, with the suffix '-yne' indicating the presence of a triple bond. Additionally, the position of the triple bond is indicated by a number, which corresponds to the first carbon of the triple bond in the chain.

Isomerism in Alkynes

Alkynes can exhibit structural isomerism, where compounds with the same molecular formula have different arrangements of atoms. This includes chain isomerism, where the carbon skeleton differs, and position isomerism, where the location of the triple bond varies. Understanding isomerism is crucial for naming and distinguishing between different alkyne compounds.