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2 years ago

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The 1H NMR spectrum of compound A (C15H14O) shows only two signals: a multiplet at 7.15 ppm and a singlet at 3.55 ppm in a 5:2 r

atio. The IR spectrum has no absorption in the 3200-4000 cm^-1 region, but strong peaks can be found near 1700 cm^-1. Compound A reacts with NaBH4 followed by acidification to give compound B of the molecular formula C15H16O. The reaction of A with CH3MgBr, and then with H3O^+, gives C, with a molecular formula of C16H18O. Suggest structures for B and C.

1 answer:

Ivan2 years ago

4 0

Answer:

Both B and C are alcohols and their structures are in the attached figure.

Explanation:

Compound A: C15H14O

NMR: 2 signals → 2 types of hydrogen surroundings; symetric molecule

7.15 ppm → indicates the presence of benzene

3.55 ppm → indicates the presence of a methyl (-CH3) substituted benzene

IR: no absorption in the 3200-4000 cm⁻¹ region → no hydroxyl groups (-OH)

strong peaks can be found near 1700 cm⁻¹ region → possible carbonyl group (-C=O) and alkenes (C=C).

C₁₅H₁₄O + NaBH₄ → C₁₅H₁₆O

NaBH₄: reduces aldehydes and ketones to alcohols.

C₁₅H₁₄O + CH₃MgBr → C₁₆H₁₈O

CH₃MgBr: reacts with ketones and aldehydes to produce terciary or secundary alcohols.

Considering all the information above, the structures attached were discovered.

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<h3>Answer:</h3>

0.699 mole CaCl₂

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To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

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In this case;

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2 years ago

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$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)

$Equation (1)+Equation (2)$ gives-

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

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