Sodium nitrite (NaNO2)reacted with 2−iodooctane to give a mixture of two constitutionally isomeric compounds of molecular formul
a C8H17NO2in a combined yield of 88%. Draw reasonable structures for these two isomers. Click the "draw structure" button to launch the drawing utility. Place the two compounds in the appropriate boxes below. draw structure ... Alkyl nitrite draw structure ... Nitroalkane
2 answers:
Answer:
On the attached picture.
Explanation:
Hello,
In the case, the reaction between 2-iodooctane and sodium nitrite, leads to the formation of an alkyl nitrite and a nitro alkane as shown on the attached picture. Once the reaction began, the salt breaks and the sodium bonds with the iodine from the 2-iodooctane to form sodium iodide, in such a way, a free radical in the second carbon is formed so the NO₂ could bond both as a nitrite and as a nitro radical; therefore, the formed species are octyl 2-nitrite and 2-nitrooctane.
Best regards.
You might be interested in
Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant: = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: = 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>()<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>
<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>
<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>