Chapter 7, Problem 1PP

Interpretation Introduction

Interpretation:

The IUPAC names of each of the given compounds are to be given by using $E-Z$ designation.

Concept introduction:

The priority of groups in nomenclature is assigned according to Cahn-Ingold-Prelog (CIP Rule) convention rules:

The higher the atomic mass, the higher will be the priority.

If priority cannot be assigned according to atomic mass, then assign the priority according to first point of difference.

If both the priority groups are on the same side of double-bonded carbon atom, then it is known as $\left(Z\right)$ configuration.

But, if both the priority groups are diagonal to each other, then it is $\left(E\right)$ configuration.

Rule for R/S Nomenclature:

4B(+)3B(-)2B(+)1B(-)

4B(+) means if 4^th priority group is below then from 1 to 2 to 3 if leads a clockwise rotation then its R and if anticlockwise then it is S.

3B(-) means if 3^rd priority group is below then from 1 to 2 to 4 if leads a clockwise rotation then its S and if anticlockwise then it is R.

Answer to Problem 1PP

Solution:

$\left(E\right)$-1-Bromo-1-chloropent-1-ene.

$\left(E\right)$-2-Bromo-1-chloro-1-iodobut-1-ene.

$\left(Z\right)$-3,5-Dimethylhex-2-ene.

$\left(Z\right)$-1-Chloro-1-iodo-2-methylbut-1-ene.

$\left(Z,4S\right)$-3,4-Dimethylhex-2-ene.

$\left(Z,3S\right)$-1-Bromo-2-chloro-3-methylhex-1-ene.

Explanation of Solution

a)

The IUPAC name with $\left(E-Z\right)$ configuration can be done using the following steps:

Assign the priority to four groups, –Br gets first priority, pentyl group gets second, –Cl gets third, and –H gets fourth. Since both the priority groups are diagonal to each other, it is $\left(E\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is pentane and hence pent- will be used and double bond is at the second position. Therefore, pent-1-ene is used.

Chloro and bromo groups are at first position. So, 1-bromo-1-chloropent-1-ene will be used.

Hence, the IUPAC name of the compound will be $\left(E\right)$-1-bromo-1-chloropent-1-ene.

b)

The IUPAC name with $\left(E-Z\right)$ configuration can be done using the following steps:

Assign the priority to four groups, –I gets first priority, –Br gets second, –Cl gets third, and ethyl group gets fourth. Since both the priority groups are diagonal to each other, it is $\left(E\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is butyl and hence but- will be used and double bond is at the first position. Therefore, but-1-ene is used.

Chloro and iodo groups are at first position and bromo group is at second position. So, 2-bromo-1-chloro-1-iodobut-1-ene will be used.

Hence, the IUPAC name of the compound will be $\left(E\right)$-2-bromo-1-chloro-1-iodobut-1-ene.

c)

The IUPAC name with $\left(E-Z\right)$ configuration can be done using the following steps:

Assign the priority to four groups, 2-methylpropyl gets first priority, both methyl groups get second, and –H gets third. Since both the priority groups are on the same side, it is $\left(Z\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is hexyl and hence hex- will be used and double bond is at the second position. Therefore, hex-2-ene is used.

There are two methyl groups at third and fifth positions. So, 3,5-dimethylhex-2-ene will be used.

Hence, the IUPAC name of the compound will be $\left(Z\right)$-3,5-dimethylhex-2-ene.

d)

The IUPAC name with $\left(E-Z\right)$ configuration can be done using the following steps:

Assign the priority to four groups, –I gets first priority, –Cl gets second, ethyl group gets third, and methyl group gets fourth. Since both the priority groups are on the same side, it is $\left(Z\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is butyl and hence but- will be used and double bond is at the first position. Therefore, but-1-ene is used.

There is one methyl group at second position, and chloro and iodo groups at first position. So, 1-chloro-1-iodo-2 methylbut-1-ene will be used.

Hence, the IUPAC name of the compound will be $\left(Z\right)$-1-chloro-1-iodo-2-methylbut-1-ene.

e)

The IUPAC name with $\left(E-Z\right)$ configuration can be done using the following steps:

Assign the priority to four groups, 1-methylpropyl gets first priority, both methyl groups get second, and –H gets third. Since both the priority groups are on the same side, it is $\left(Z\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is hexyl and hence hex- will be used and double bond is at the second position. Therefore, hex-2-ene is used.

There are two methyl groups at third and fourth position. So, 3,4-dimethylhex-2-ene will be used.

Carbon-4 is a chiral carbon. Assigning priority to the groups, 2-butene gets first priority, ethyl gets second, methyl gets third, and hydrogen gets fourth.

Move 1-2-3, it is in anticlockwise direction. So, it is $\left(S\right)$ configuration. Hence, $\left(4S\right)$ will be used.

Hence, the IUPAC name of the compound will be $\left(Z,4S\right)$-3,4-dimethylhex-2-ene.

f)

The IUPAC name can be done using the following steps:

Assign the priority to four groups, –Br gets first priority, –Cl gets second, 1-methylbutyl gets third, and –H gets fourth. Since both the priority groups are on the same side, it is $\left(Z\right)$ configuration.

Select the longest carbon chain. The parent hydrocarbon chain is hexyl and hence hex- will be used and double bond is at the first position. Therefore, hex-1-ene is used.

There is one methyl group at third position, and chloro at second and bromo at first position. So, 1-bromo-2-chloro-3-methylhex-1-ene will be used.

Carbon-3 is a chiral carbon. Assigning priority to the groups, 1-bromo-2-chloroethene gets first priority, propyl gets second, methyl gets third, and hydrogen gets fourth.

Rotate the molecule such that –H is at a horizontal position and then move 1-2-3, it is in anticlockwise direction. So, it is $\left(S\right)$ configuration. Hence, $\left(3S\right)$ will be used.

Hence, the IUPAC name of the compound will be $\left(Z,3S\right)$-1-bromo-2-chloro-3-methylhex-1-ene.