Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is
expected when compound a is treated with mcpba followed by aqueous acid (h3o+).
1 answer:
Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.
Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.
The reaction is as shown in the image.
The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid is shown in the image.
m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.
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Answer:
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Explanation:
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Answer:
0.11mol/dm³
Explanation:
The reaction expression is given as:
HCl + NaOH → NaCl + H₂O
Volume of acid = 25cm³ = 0.025dm³
Volume of base = 18.4cm³ = 0.0184dm³
Concentration of base = 0.15mol/dm³
Solution:
The concentration of hydrochloric acid = ?
To solve this problem, let us first find the number of moles of the base;
Number of moles = concentration x volume
Number of moles = 0.15mol/dm³ x 0.0184dm³ = 0.00276mol
From the balanced reaction equation;
1 mole of NaOH will combine with 1 mole of HCl
Therefore, 0.00276mol of the base will combine with 0.00276mol of HCl
So;
Concentration of acid = =
= 0.11mol/dm³