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Draw a Lewis structure, and classify each of the following compounds. The possible classification are as follows:

alcohol, ketone, carboxylic acid, ether, aldehyde and alkene.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

Short Answer

Expert verified

(a) aldehyde

(b) alcohol

(c) ketone

(d) ether, alkene

(e) carboxylic acid

(f) ether, alkene

(g) ketone, alkene

(h) aldehyde

(i) alcohol

Step by step solution

01

Lewis structures

All the paired and unpaired electrons must be shown while representing a Lewis structure. The paired electrons are represented as solid lines (bonds) and the unpaired electrons are represented as dots.

02

Functional group

A functional group may be defined as an atom or a group of atoms that gives some characteristic properties to a compound.

The compounds containing hydroxy (-OH) functional group are classified as alcohols. Alcohols have the general formula R-OH , where R= alkyl group.

The functional group is which two alkyl groups are bonded to an oxygen atom are known as ethers. Ethers have the general formula R-O-R', where R and R' represent alkyl groups.

The functional group for both aldehydes and ketones are the carbonyl group (C=O). In a ketone, there are two alkyl groups bonded to the carbonyl group whereas in an aldehyde, there is one alkyl group and one hydrogen atom bonded to the carbonyl group.

The functional group in which carboxyl group, -COOH group is present is referred to as carboxylic acid. Carboxylic acids have the general formula

R-COOH or R-CO2H, where R= alkyl group.

03

Alkenes

Alkenes are the hydrocarbons containing carbon-carbon double bond (C=C) .

04

Identification of compounds

(a)

From the Lewis structure, it can be seen that there is one alkyl group and one hydrogen atom bonded to the carbonyl group (C=O). Therefore, this compound is an aldehyde.

(b)

From the Lewis structure, it can be seen that the compound contains hydroxy (-OH) functional group. Therefore, this compound is an alcohol.

(c)

From the Lewis structure, it can be seen that there are two alkyl groups bonded to the carbonyl group (C=O). Therefore, this compound is a ketone.

(d)

From the Lewis structure, it can be seen that there are two alkyl groups bonded to an oxygen atom. Therefore, this compound is an ether. Also, it contains carbon-carbon double bond (C=C). Hence, it can also be classified as an alkene.

(e)

From the Lewis structure, it can be seen that the compound contains carboxyl

(-COOH ) functional group. Therefore, this compound is a carboxylic acid.

(f)

From the Lewis structure, it can be seen that there are two alkyl groups bonded to an oxygen atom. Therefore, this compound is an ether. Also, it contains carbon-carbon double bond (C=C). Hence, it can also be classified as an alkene.

(g)

From the Lewis structure, it can be seen that there are two alkyl groups bonded to the carbonyl group (C=O). Therefore, this compound is a ketone. Also, it contains carbon-carbon double bond (C=C). Hence, it can also be classified as an alkene.

(h)

From the Lewis structure, it can be seen that there is one alkyl group and one hydrogen atom bonded to the carbonyl group (C=O). Therefore, this compound is an aldehyde.

(i)

From the Lewis structure, it can be seen that the compound contains hydroxy (-OH) functional group. Therefore, this compound is an alcohol.

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Most popular questions from this chapter

The C โ‰ก N triple bond in acetonitrile has a dipole moment of about 3.6Dand a bond length of about 1.16โ„ซ. Calculate the amount of charge separation in this bond. How important is the charge separated resonance form in the structure of acetonitrile?

Consider the following compounds that vary from nearly nonacidic to strongly acidic. Draw the conjugate bases of these compounds and explain why the acidity increases so dramatically with substitution by nitro groups.

N-Methylpyrrolidine has a boiling point of81ฮฟC, and piperidine has a boiling point of106ฮฟC.

  1. Explain the large difference (25ฮฟC) in boiling point for these two isomers.
  2. Tetrahydropyran has a boiling point of 88ฮฟC, and cyclopentanol has a boiling point of 141ฮฟC. These two isomers have a boiling point difference of 53ฮฟC.Explain why the two oxygen-containing isomers have a much larger boiling point difference than two amine isomers.
  3. N,N-Dimethylformamide has a boiling point of 150ฮฟC, and N-methylacetamide has a boiling point of 206ฮฟC, for a difference of 56ฮฟC.Explain why these two nitrogen-containing isomers have a much larger boiling point difference than the two amine isomers. Also explain why these two amides have higher boiling points than any of the other four compounds shown (two amines, an ether, and an alcohol).

Consider each pair of bases, and explain which one is more basic. Draw their conjugate acids, and show which one is a stronger acid.

(a)

(b)

(c)

(d)

Classify the following hydrocarbons, and draw a lewis structure for each one. A compound may fit into more than one of the following classifications:

Alkane, alkene, alkyne, cycloalkane, cycloalkene, cycloalkyne, aromatic hydrocarbon.

  1. (CH3CH2)2CHCH(CH3)2
  2. CH3CHCHCH2CH3
  3. CH3CCCH2CH2CH3

e.

f.

g.

h.

i.

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