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Question: Give IUPAC names for the following compounds.

a.

b.

c.

d.

e.

f.

Short Answer

Expert verified

Answer

a) 4-phenylpent-2-yne

b) (E)-3-methylhept-2-en-4-yne

c) 2,2,5-trimethylhept-3-yne

d) 4,4-dibromopent-2-yne

e) (S)-3-methylhex-4-yn-3-ol

f) but-2-yn-1-ylcyclopentane

Step by step solution

01

About IUPAC nomenclature

The rules governed by the International Union of Pure and Applied Chemistry to name the organic compounds systematically is IUPAC nomenclature. The naming depends on the name of the parent chain, substituents attached, configurations, etc.

02

Naming of the compound (a)

The parent chain is made up of five carbons. Thus, the word ‘pent’ is used in the prefix. The numbering is given in such a way that the triple bond gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The triple bond is at second carbon and the phenyl group is at fourth carbon. Thus, the IUPAC name of this compound is 4-phenylpent-2-yne.

03

Naming of compound (b)

The parent chain is made up of seven carbons. Thus, the word ‘hept’ is used in the prefix. The numbering is given in such a way that the double bond gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The double bond is present at the second carbon and the triple bond is present at the fourth carbon. There is one methyl group at the third carbon atom. The configuration at the double bond is E because the higher priority groups are on the opposite side of the double bond. Thus, the IUPAC name of the given compound is (E)-3-methylhept-2-en-4-yne.

04

Naming of compound (c)

The parent chain is made up of seven carbons. Thus, the word ‘hept’ is used in the prefix. The numbering is given in such a way that the triple bond gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The triple bond is present at the third carbon. There are two methyl groups at the second carbon atom and one methyl group at the fifth carbon atom. Thus, the IUPAC name of the given compound is 2,2,5-trimethylhept-3-yne.

05

Naming of compound (d)

The parent chain is made up of five carbons. Thus, the word ‘pent’ is used in the prefix. The numbering is given in such a way that the triple bond gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The triple bond is present at the second carbon atom. There are two bromo groups at the fourth carbon atom. Thus, the IUPAC name of the given compound is 4,4-dibromopent-2-yne.

06

Naming of compound (e)

The parent chain is made up of six carbons. Thus, the word ‘hex’ is used in the prefix. The numbering is given in such a way that the functional group (OH) gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The functional group OH is present at the third carbon and the methyl group is present at the third carbon atom. The triple bond is present at the fourth carbon atom. Thus, the IUPAC name of the given compound is 3-methylhex-4-yn-3-ol.

07

Naming of compound (f)

The parent chain is made up of four carbons. Thus, the word ‘but’ is used in the prefix. The numbering is given in such a way that the functional group cyclopentyl ring gets the lowest possible number. The numbering is shown below.

Numbering of alkyne

The cyclopentyl group is present at the first carbon and the triple bond is present at the second carbon. Thus, the IUPAC name of the given compound is 1-cyclopentylbut-2-yne.

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Most popular questions from this chapter

Question:Using cyclooctyne as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.)

(a)cis-cyclooctene

(b)cyclooctane

(c)trans-1,2-dibromocyclooctane

(d)cyclooctanone

(e)1,1-dibromocyclooctane

(f)3-bromocyclooctene

(g)cyclooctane-1,2-dione

(h)

(i)

The hydroboration–oxidation of internal alkynes produces ketones.

(a) When hydroboration–oxidation is applied to but-2-yne, a single pure product is obtained. Determine the structure of this product, and show the intermediates in its formation.

(b) When hydroboration–oxidation is applied to pent-2-yne, two products are obtained. Show why a mixture of products should be expected with any unsymmetrical internal alkyne.

Question: Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why. (a) hex-1-yne (b) hex-2-yne (c) hex-3-yne (d) 4-methylhex-2-yne (e) 5-methylhex-2-yne (f) cyclodecyne.

Predict the major product(s) of the following reactions:

(a) phenylacetylene + 2 HBr

(b) hex-1-yne + 2 HCl

(c) cyclooctyne + 2 HBr

(d) hex-2-yne + 2 HCl

Question: Predict the products formed when CH3CH2-C≡C:Na+ reacts with the following compounds.

(a)ethyl bromide

(b)tert-butyl bromide

(c)formaldehyde

(d)cyclohexanone

(e)CH3CH2CH2CHO

(f)cyclohexanol

(g)butan-2-one,CH3CH2COCH3

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