Review for Exam I answers




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Chemistry 1152

Review for Exam I — Answers




1
Hexane


. CH3CH2CH2CH2CH2CH3

There are other possible answers for this question.

2. a. CH4 b. CH3CH2CH2CH3 c.


d.


3. No number is needed as there is only one possible position

for it. (The number 2 position.)

4. Distillation—Convert substance to vapor and recondense it to a liquid. This is a type of purification technique.

Fractional Distillation—Distillation where the vapor is condensed at different points (or fractions) giving different types of solutions.
5. All combustion reactions give carbon dioxide and water (CO2 + H2O). If there is a deficiency of oxygen, CO or C could be formed instead.
6. Alkanes are non-polar organic compounds. Water is a polar compound. Since “Like Dissolves Like,” we would never have expected that the two should mix.
7. A sigma bond (σ) is produced by an “end-to-end” overlap of orbitals. The circular intersection of orbitals in this type of bonding allows bond rotation with no change to the bond. All single bonds are of this type of bond.
8. a. 2-bromo-3-ethyl-3-methylpentane

b. 3-methylhexane

c. 5-chloro-4-ethyl-2,3-dimethylheptane

d. 3,5-dimethyl-6-propyldecane

e. 3-bromo-1-chloro-2-iodopropane

f. 1-chloro-2,3-dimethylpentane

g. 3-ethyl-2,6-dimethylheptane

h. 2,3-dimethylhexane

i. 4-ethyl-5-methylnonane

j. 1,1-dichloropropane


9. A pi (π) bond is created by a “side-to-side” overlap of orbitals. This prevents bond rotation in unsaturated hydrocarbons.


10. (a) (b)




(c)

(d)






(e) (f) (g)




(h) (i) (j)


11. CH2=CHCH3



12. (a) (b) (c)

(d) (e) (f)



(g) (h) (i)

(j)
13. a. 2,4-dimethyl-2-hexene

b. cyclohexene

c. chloroethene

d. 2,2,5,5-tetramethyl-3-octyne

e. 3-methyl-1-butene

f. 4-ethyl-3-methyl-3-heptene

g. 3-bromocyclopentene

h. 7-iodo-4-methyl-2-heptene

i. heptane

j. 2-methylbutane

k. cyclohexane

l. methylbenzene (toluene)

m. 2-ethyl-1,3-dimethylbenzene

n. bromomethane

o. 1,4-dichlorobenzene or p-dichlorobenzene

p. 3,5-dinitro-1-propylbenzene

q. methylcycloheptane

r. 3-methyl-1-butyne

s. 3-phenylhexane
14. Add Br2. If the red-brown color of the Br2 disappears immediately, the compound was the alkene, 2-butene. If the color remains, or lingers only to fade in the light, it was the alkane, butane. Alternatively, KMnO4 could be used. If it lost its pretty purple color, it was he alkene.
15. No, both would react in the same fashion. The tests are for unsaturation so you could not differentiate between alkenes and alkynes.

1




or h
6. CH3CH2CH3 + Cl2 CH3CH2CH2—Cl + HCl
Alkane substitution reactions require heat or ultraviolet light and produce a by-product.
CH2=CHCH3 + Cl2
Alkene addition reactions are immediate without a catalyst and there is no by-product.
17. a. 3-methyl-4-octene

b. cyclopentane

c. 2-pentene

d. 1-pentene


18. (a) (b) (c)

(d) (e)

19. (a) B. It has the longest carbon chain

(b) A & D and C, E & F

(c) E. Markovnikov’s rule does not apply.

(d) E & F

(e)

20. Fractional distillation should work just fine.




21.


(a)

(b)


(c)

(d)




(e) + HCl No Reaction



(f)


(g)



Ni

(h) CH3CH=CH2 + H2 CH3CH2CH3

(i)


Pt


(j) + H2

(k)


(l)




(m) CH3CH2CH2CH2CH2CH2CH3 + O2 CO2 + H2O

(n)

(o)


(p)




(q) + H2SO4



(r) + H2SO4



(s) + KMnO4




H2SO4


(t) + HNO3




h

(u) + Br2

(v) CH2=CHCH3 + HCl


Br2

Fe


Br2

h

(w)

(x)



(y)




(z) + O2 CO2 + H2O



(aa)


(bb)


(cc)






  1. + Excess HBr

(
2 products possible


ee) CH3CH=CHCH2CH3 + HCl or

(
h


ff)


+ HCl






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