Draw Lewis structures for the following compounds and | StudySoup

Textbook Solutions for Organic Chemistry

Chapter 1 Problem 6P

Question

Draw Lewis structures for the following compounds and ions, showing appropriate formal

charges.

(a) \({\left[\mathrm{CH}_{3} \mathrm{OH}_{2}\right]^{+}}\)

(b) \(\mathrm{NH}_{4} \mathrm{Cl}\)

(c) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{NCl}\)

(d) \(\mathrm{NaOCH}_{3}\)

(e) \({ }^{+} \mathrm{CH}_{3}\)

(f) \({ }^{-} \mathrm{CH}_{3}\)

(g) \(\mathrm{NaBH}_{4}\)

(h) \(\mathrm{NaBH}_{3} \mathrm{CN}\)

(i) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}-\mathrm{BF}_{3}\)

(j) \({\left[\mathrm{HONH}_{3}\right]^{+}}\)

(k) \(\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}\)

(l) \({\left[\mathrm{H}_{2} \mathrm{C}=\mathrm{OH}\right]^{+}}\)

Solution

Step 1 of 12

a)

No of valence electrons = 4 from Carbon, 5 from hydrogen, 6 from Oxygen = 15.

Due to positive ion no of electrons become 15-1 = 14.

Formal charge on Carbon =  = 0

Formal charge on Hydrogen =  = 0

Formal charge on Oxygen = Oxygen has 6 valence electrons. 3 electrons are shared with Carbon, and 2 hydrogen atoms. Out of left 3 electrons one electron is relieved to give a positive charge.

 

Hence FC on Oxygen =

 

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full solution

Title Organic Chemistry 8 
Author L.G. Wade Jr
ISBN 9780321768414

Draw Lewis structures for the following compounds and

Chapter 1 textbook questions

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