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Chapter 1: Q57P (page 2)

Rank the bonds from the most polar to least polar.

a. C - O,C - F,C - N

b. C - Cl > C - Br > C - I

c. H - O,H - N,H - C

d. C - H,C - C,C - N

Short Answer

Expert verified

a. C-F>C-O>C-N

b. C-Cl>C-Br>C-I

c. H-O>H-N>H-C

d. C-N>C-H>C-C

Step by step solution

01

Defining bond polarity

The distribution of positive and negative electrical charge over a bond is known as bond polarity. The bond polarity can be determined by comparing the difference in electronegativity between the atoms. More the difference in electronegativity more polar is the bond.

02

Finding the order of polarity from most polar to least polar.

a. C-F>C-O>C-N

In the above series as the electronegativity order is F>O>N the bond polarity also increases with increase in electronegativity difference.

b. C-Cl>C-Br>C-I

Chlorine has the maximum electronegativity and iodine has the minimum electronegativity .Thus the polarity order is C-Cl>C-Br>C-I.


c. H-C

The electronegativity order varies as, O>N>C thus the bond polarity also increases with increase in electronegativity difference and H-O becomes the most polar while H-C becomes the least polar.

d. C-H

C-C does not have any electronegativity difference thus it is least polar .

In case C-H,C-N of nitrogen being more electronegative compared to hydrogen thus is C-N more polar compared to C-H.

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Most popular questions from this chapter

The mass spectrum for a compound with molecular weight of 102 is shown below. Its IR spectrum has a broad, strong absorption at 3600 cm–1 and a

medium absorption at 1360 cm–1.

a.Identify the compound.

b.Show the mechanism for formation of the peak at m/z= 84.

A ketone can be prepared from the reaction of a nitrile with a Grignard reagent. Describe the intermediate formed in this reaction, and show how it can be converted to a ketone.

Rank the carbanions shown in the margin from most basic to least basic.

HC=C-

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Put a number in each of the blanks.

  1. _ s orbital and _p orbitals form _ SP3orbitals.
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A tertiary amine reacts with hydrogen peroxide to form a tertiary amine oxide.

Tertiary amine oxides undergo a reaction similar to the Hofmann elimination reaction (Section 10.10), called a Cope elimination. In this reaction, a tertiary amine oxide, rather than a quaternary ammonium ion, undergoes elimination. A strong base is not needed for a Cope elimination because the amine oxide acts as its own base.

Does the Cope elimination have an alkene-like transition state or a carbanion-like transition state?
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