Chapter 21, Problem 159IP

 (a)

Interpretation Introduction

Interpretation: The percent by mass of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is to be calculated. The hybridization of the starred carbon atom in the given figure changed from reactants to products is to be stated. Also, the product is to be named.

Concept introduction: Grignard reagent is an organometallic compound which is prepared by treating magnesium or metal atom with a strong acid. The Grignard reaction is defined as an organometallic reaction in which the vinyl, alkyl and aryl-metal halides were added to a carbonyl carbon atom and form an aldehyde or ketone.

To determine: The percent by mass of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$.

Answer to Problem 159IP

The mass percent of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is $\text{37}\text{.5\%}$.

Explanation of Solution

Given

Electronic configuration of ${\text{M}}^{+2}$ is $\left[\text{Ar}\right]{\text{3d}}^{10}$.

So, atomic number of M is $30$.

In periodic table atomic number is found to be of the element Zinc.

The molar mass of Zinc is $65.\text{39 g/mol}$.

The molar mass of compound $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is calculated by using the molar masses of $2$ carbon, $5$ hydrogen, $1$ Zinc (M) and $1$ Bromine.

The molar mass of carbon is $12\text{g/mol}$.

The molar mass of hydrogen is $1\text{g/mol}$.

The molar mass of Zinc is $65.\text{39 g/mol}$.

The molar mass of Bromine is $\text{79}\text{.96 g/mol}$.

The molar mass of compound $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is calculated as,

$\begin{array}{c}{\text{Molar mass of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MBr}=12\times 1+1\times 3+12\times 1+1\times 2+65.39\times 1+174.35\times 1\\ =12+3+12+2+65.39+174.35\\ =174.3\text{5 g/mol}\end{array}$.

Therefore the total molar mass of compound $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is $174.3\text{5 g/mol}$.

The mass percent of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is calculated by the formula,

${\text{percent by mass of M in CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MBr}=\left(\frac{\text{Molar mass of M}}{\text{Total Molar mass of compound}}\right)\times 100$.

Substitute the value of the molar mass of M and total molar mass of compound, to calculate percent by mass of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ in the above equation.

$\begin{array}{c}{\text{percent by mass of M in CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MBr}=\left(\frac{\text{Molar mass of M}}{\text{Total Molar mass of compound}}\right)\times 100\\ =\left(\frac{65.\text{39 g/mol}}{174.\text{35 g/mol}}\right)\times 100\\ =\underset{\_}{37.5\%}\end{array}$

Therefore, the mass percent of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is $\underset{\_}{37.5\%}$.

Conclusion

The mass percent of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is $\underset{\_}{37.5\%}$.

(b)

Interpretation Introduction

Interpretation: The percent by mass of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is to be calculated. The hybridization of the starred carbon atom in the given figure changed from reactants to products is to be stated. Also, the product is to be named.

Concept introduction: Grignard reagent is an organometallic compound which is prepared by treating magnesium or metal atom with a strong acid. The Grignard reaction is defined as an organometallic reaction in which the vinyl, alkyl and aryl-metal halides were added to a carbonyl carbon atom and form an aldehyde or ketone.

To determine: The hybridization of the starred carbon atom in the given figure changed from reactants to products.

Answer to Problem 159IP

The hybridization of the starred carbon atom shown in Figure 1 is changed from $s{p}^2$ carbon to $s{p}^3$ carbon.

Explanation of Solution

The given reaction is,

Figure 1

In the given reaction the reactant contains carbonyl group having double bonded oxygen atom hence, the hybridization of the starred carbon in the reactant is $s{p}^2$. After reacting the given reactant withy Grignard reagent the carbonyl group is converted into an alcoholic group in which the carbon atom is single bonded with oxygen atom. Hence, the hybridization of the starred carbon in product is $s{p}^3$. Therefore, the hybridization of starred carbon atom is changed from $s{p}^2$ carbon to $s{p}^3$ carbon.

Conclusion

The hybridization of the starred carbon atom shown in Figure 1 is changed from $s{p}^2$ carbon to $s{p}^3$ carbon.

(c)

Interpretation Introduction

Interpretation: The percent by mass of M in $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)\text{MBr}$ is to be calculated. The hybridization of the starred carbon atom in the given figure changed from reactants to products is to be stated. Also, the product is to be named.

Concept introduction: Grignard reagent is an organometallic compound which is prepared by treating magnesium or metal atom with a strong acid. The Grignard reaction is defined as an organometallic reaction in which the vinyl, alkyl and aryl-metal halides were added to a carbonyl carbon atom and form an aldehyde or ketone.

To determine: The systematic name of the product.

Answer to Problem 159IP

The systematic name of the product is $\text{3,4-dimethyl-3-hexanol}$.

Explanation of Solution

The given compound is,

Figure 2

The longest carbon chain contains six carbon atoms numbered from the nearest functional group with an alcohol group present on third carbon. Hence, the root name of the compound is hexanol. It also contains two methyl groups present on third and fourth carbon atom respectively. Hence, the name of the compound is $\text{3,4-dimethyl-3-hexanol}$.

Conclusion

The systematic name of the product shown in Figure 2 is $\text{3,4-dimethyl-3-hexanol}$.