What aldol product is formed when two molecules of butanal react together in the presence of base?

Chemistry

1 answer:

Ainat [17]3 years ago

4 0

It is an example of Self Condensation reaction as both the nucleophile and the electrophile are the same compounds. When Butanal is reacted with base, the hydrogen at alpha position is abstracted due to its mild acidity and enolate is formed. Enolate attacks second Butanal molecule and forms a compound containing both Hydroxyl group and carbonyl group (Beta hydroxy aldehyde). Reaction is shown below and the name of product is specified below its structure,

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Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

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Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind: