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Organic Chemistry 261(270, 271) Midterm-I Examination September 29, 2008 Name (print) -----------------------ID No. --------------------------

Time: 50 minutes Total number of pages: 6 Answer all questions in the space provided.

Question I. (15.0) II. (14.0) III. (14.0) IV. (15.0) V. (10.0) Total (68.0) Percentage

Mark

GOOD LUCK

2 (9.0) 1. a) Write a Lewis structure for each of the following compounds showing any unshared electron pairs. b) Calculate the formal charge on each atom other than hydrogen. Be sure to show your calculations. a) CH3OH2 EC(c) = 1/2(8) = 4 FC(c) = 4- 4 = 0

H H H C O H H

EC (O) = 1/2 (6) + 2 = 5 FC (O) = 6-5 = +1

b) (CH3)3CO

H C H H H C C O HH C H
c) (CH3)2O-BF3 H HH C F H H C O B F H F (6.0) 2.

H

H

Formal charge calculations for (b, c) as of part (a)

Which resonance form in the following pairs would contribute more to the Hybrid (more stable)? Explain the reason for your choice.

a) CH3CH CH CH OH CH3CH CH CH OH Carbon of the other structure do not meet the octet rule O CH3 More stable because of more covalent bonds
O c) CH2 C CH3 CH2 O C CH3 Negative charge resides on more electronegative atom.

O C NH2

b)

CH3

C NH2

3 (8.0 ) II a) Write a dash formula for each of the following compounds showing any unshared electron pairs. b) predict the hybridization of the indicated atom in each molecule?

a) CH3

CH N CH3

Answers:

sp3 a) CH3
sp3
O b) CH3 C C C H

sp2

sp2

sp3

CH N CH3
sp sp O C C C H sp2 sp2

b) CH3

sp3 sp
c) CH3BeCH3

c) CH3BeCH3 sp3 d) BH4

d) BH4

(6.0) 2. Which compound in each of the following pairs would have the higher boiling point? Explain a reason for your answer.

a) CH3CH2CH2OH or CH3CH2OCH3 Alcohol, because of hydrogen bonding

b)

or

O Ketone, because of dipole-dipole intractions

c)

N H

or

N CH3

Primary amine, because of hydrogen bonding

4 (8.0) III 1. a) Draw structures of three alkyl bromide with the formula C4H9Br b) Classify each as to whether it is primary, secondary, or tertiary alkyl bromide. CH3CH2CH2 CH2Br primary

CH3 CH3CH2 CHBr CH3 CH3 C Br tertiary CH3 (6.0) 2. Write a condensed structural formula for each of the following compound. O seconday

a)
O (CH3)2CHCOCH(CH3)2 or (CH3)2CH C CH(CH3)3

b)

NH
CH3 CH3CH2CH(CH3)CH2NHCH2CH3 or CH3CH2 H

CHCH2NCH2CH3

OH

OH
CH2 CH2

CH CH CH CH CH3

c)

5 (15.0) IV 1. Draw a structure for compounds that meet the following descriptions. a) Two amines with the formula C3H9N

CH3CH2CH2NH2

and

CH3NHCH2CH3

and

Many other possibilities.

b) Two ketones with the formula C5H10O

O CH3CH2
and

O CH2CH3 CH3 C CH2CH2CH3

C

Many other possibilities.

c) Two ethers with the formulas C4H8O O O

and

Many other possibilities.

d) Draw bond-line structures of two cyclic compounds with molecular formula C4H8.

e) Draw an isomer of CH3CH2CH2CH2C N
CH3 CH3CHCH2C N

with the same functional group.

and one more possibility

6 (10.0) V. What is the relationship between the members of the following pairs? That is, are they Stereoisomers, constitutional isomers, the same, or resonance structure. Explain the reason for your choice. CH3 CH a) CH2 CHCH2CH3 and H2C CH2

Answer: constitutional isomers same molecular formula, but different connectivity of atoms).

b)

NH2

and

NH2

Answer: Resonance structures- (same connectivity of atoms, but different distribution of electrons. CH3 c) CH3 C CH3 CH3 Answer: different drawing of the same molecule , same Molecular formula or (CH3)3C CH3

d)

H H3C C C

CH3 H

and

H3C H C C

CH3 H

Answer: stereoisomers (cis-trans isomers) different location of atoms in space, but same molecular formula.

7

Periodic Table of the Elements
1 1 18 2

H
1.00794 3 2 4 13 5 14 6 15 7 16 8 17 9

He
4.002602 10

Li
6.941 11

Be
9.012182 12

B
10.811 13

C
12.0107 14

N
14.0067 15

O S...
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