Chapter 10, Problem 10.110QP

(a)

Interpretation Introduction

Interpretation:

Monobromination products obtained when propane treated with bromine has to given and the percentage of the monobromo products has to be expected.

Concept Introduction:

Halogenation reaction is the one where atom or atoms of halogens get substituted in a carbon chain.  Halogenation is a type of substitution reaction.

In alkanes, substitution reaction means, a group or atom being substituted for one of the hydrogen atoms.  As all the four bonds that can be formed from carbon in case of alkanes is four single bonds, halogenation reaction in case of alkane will be substitution only.  Product obtained on halogenation of alkane is alkyl halides.

IUPAC rules for naming alkanes:

There are about five rules that has to be followed for naming an alkane and they are,

• The longest continuous carbon chain in the compound has to be identified.  This is known as parent compound.  From this the parent name is obtained.  Suffix “–ane” (for alkane) is added at the end of the prefix which gives information about the number of carbon atoms.
• Numbering has to be done so that the lowest number is given to the first group that is encountered in the parent chain.
• Naming and numbering has to be given for each atom or group that is attached to the parent chain.  Numbering has to be done in a way that substituents get the least numbering.
• If the same substitution is present in the parent chain more than once, a separate prefix is added which tells about the number of times the substituent occurs.  Prefixes used are di-, tri-, tetra-, penta- etc.
• Name of the substituents has to be placed in an alphabetical order before the parent compound name.

Explanation of Solution

Propane on reaction with bromine in presence of light undergoes bromination.   Two monobrominated products are obtained in case of propane.  This can be represented as,

IUPAC names:

First monobromo derivative:

In the given compound, the longest carbon chain is found to contain three carbon atoms.  Therefore, the parent alkane name is propane.

Numbering of carbon atoms has to be done in a way that the substituents present in the longest carbon chain get the least numbering.

The substituent present in the given compound is a bromine atom.  Number has to be added before the substituent indicating the carbon in which it is attached.  Therefore, IUPAC name is obtained as,

Parent chain is propane and the substituent present is 1-bromo.  Hence, the IUPAC name is given as 1-bromopropane.

Second monobromo derivative:

In the given compound, the longest carbon chain is found to contain three carbon atoms.  Therefore, the parent alkane name is propane.

Numbering of carbon atoms has to be done in a way that the substituents present in the longest carbon chain get the least numbering.

The substituent present in the given compound is a bromine atom.  Number has to be added before the substituent indicating the carbon in which it is attached.  Therefore, IUPAC name is obtained as,

Parent chain is propane and the substituent present is 2-bromo.  Hence, the IUPAC name is given as 2-bromopropane.

Percentage of monobrominated products formed:

Total number of hydrogen atoms in propane is eight.  Six equivalents of hydrogen is required for the formation of 1-bromopropane and two equivalents of hydrogen is required for the formation of 2-bromopropane.  Therefore, the ratio of formation of 1-bromopropane to 2-bromopropane can be given as $\text{6:2}$.  Percentage of the product formed can be calculated as shown below,

  $\begin{array}{l}\text{Ratio}\text{of}\text{1-bromopropane}\text{=}\frac{6}{8}\times 100\\ =75\%\\ \text{Ratio}\text{of}\text{2-bromopropane}\text{=}\frac{2}{8}\times 100\\ =25\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

Monobromination products obtained when 2-methylpropane treated with bromine has to given and the percentage of the monobromo products has to be expected.

Concept Introduction:

Refer part (a).

Explanation of Solution

2-methylpropane on reaction with bromine in presence of light undergoes bromination.   Two monobrominated products are obtained in case of propane.  This can be represented as,

IUPAC names:

First monobromo derivative:

In the given compound, the longest carbon chain is found to contain three carbon atoms.  Therefore, the parent alkane name is propane.

Numbering of carbon atoms has to be done in a way that the substituents present in the longest carbon chain get the least numbering.

The substituent present in the given compound are a bromine atom and methyl group.  Number has to be added before the substituent indicating the carbon in which it is attached.  Therefore, IUPAC name is obtained as,

Parent chain is propane and the substituent present is 1-bromo-2-methyl.  Hence, the IUPAC name is given as 1-bromo-2-methylpropane.

Second monobromo derivative:

In the given compound, the longest carbon chain is found to contain three carbon atoms.  Therefore, the parent alkane name is propane.

Numbering of carbon atoms has to be done in a way that the substituents present in the longest carbon chain get the least numbering.

The substituent present in the given compound are a bromine atom and methyl group.  Number has to be added before the substituent indicating the carbon in which it is attached.  Therefore, IUPAC name is obtained as,

Parent chain is propane and the substituent present is 2-bromo-2-methyl.  Hence, the IUPAC name is given as 2-bromo-2-methylpropane.

Percentage of monobrominated products formed:

Total number of hydrogen atoms in propane is ten.  Nine equivalents of hydrogen is required for the formation of 1-bromo-2-methylpropane and on equivalents of hydrogen is required for the formation of 2-bromo-2-methyl propane.  Therefore, the ratio of formation of 1-bromo-2-methyl propane to 2-bromo-2-methyl propane can be given as $\text{9:1}$.  Percentage of the product formed can be calculated as shown below,

  $\begin{array}{l}\text{Ratio}\text{of}\text{1-bromo-2-methylpropane}\text{=}\frac{9}{10}\times 100\\ =90\%\\ \text{Ratio}\text{of}\text{2-bromo-2-methylpropane}\text{=}\frac{1}{10}\times 100\\ =10\%\end{array}$