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3 years ago

Question 4b, c please.

Chemistry

1 answer:

lora16 [44]3 years ago

4 0

Answer:

a) Structural formula for X and Y are attached below,

b) Systematic name of alcohol is Butan-2-ol.

c) X is secondary alcohol.

Explanation:

Alcohols are considered one of the most important class of organic compounds. They act as the starting materials in many important reactions like in synthesis of esters, ethers e.t.c.

Among different reactions shown by alcohols oxidation reactions are considered very important in formation of other intermediates like ketones, aldehydes and carboxylic acids.

Below are the oxidation reactions shown by different alcohols;

1) Primary Alcohol -------[O]--------> Aldehyde --------[O]--------> Carboxylic Acid

2) Secondary Alcohol --------[O]--------> Ketone -------[O]-------> No Reaction

3) Tertiary Alcohol ---------[O]---------> No reaction

Hence, oxidation takes place on alcohols which contain reactive C-H bonds. Therefore, in given scenario the alcohol is secondary alcohol and upon oxidation it produces a ketone.

Note:

In order to finalize the correct structure of final product we can assume that it is ketone because it is stated in statement that it cannot be oxidized further. Secondly the molar mass 72 g/mol was utilized to reach the final correct structure, for this purpose 28 amu (for CO carbonyl group) was subtracted from 72 and the remaining 44 amu was alloted to 2 CH₃ (Methyl part each having 15 amu mass) and CH₂ (Methylene part having 14 amu mass) respectively.

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

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Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

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Answer:

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