Chapter 7 7.1 One problem with elimination reactions is that mixtures of products are often formed. For example,

treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?

Solution:

Br

major product

7.2 How many alkene products, including E,Z isomers, might be obtained by dehydration of 3-methyl-3-hexanol (3-methyl-3-hydroxyhexane) with aqueous sulfuric acid?

Solution:

OH

dehydration

7.3 What product would you expect to obtain from addition of Cl2 to 1,2-dimethylcyclohexene? Show the stereochemistry of the product.

Solution:

H3C CH3

Cl Cl

H3C CH3

Cl

Cl

H3C

CH3Cl

Cl

H3C CH3

Cl

Cl

H3C

CH3Cl

Cl

7.4 Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the

stereochemistry of each, and explain why a mixture is formed. Solution:

H3C CH3

H Cl

H3C

CH3

Cl

H3C CH3

Cl

H3C CH3

ClH3C Cl

CH3

7.5 What product would you expect from the reaction of cyclopentene with NBS and water? Show the

stereochemistry. Solution:

Br Br

Br

HO

H

OH

Br

H

H

H

H

OH

Br

Br

OH

H

H

H

H

Br

OH

7.6 When an unsymmetrical alkene such as propene is treated with NBS in aqueous DMSO, the major

product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? Explain.

H3C CH

CH2NBS,H2ODMSO H3C C

HCH2Br

OH

Solution:

CH2

BrH

H3CCH2

BrH

H3C

H2OH2O

vs

The product is Markovnikoff product. 7.7 What products would you expect from oxymercuration of the following alkenes? Solution: (a)

H3CH2C

H2C C

HCH2

Hg(OAc)2

H2OH3C

H2C

H2C

HC CH3

OH2-pentanol

(b)

H3C C CH

H2C CH3

CH3

Hg(OAc)2

H2OH3C C

H2C

H2C CH3

CH3

OH

2-methyl-2-pentanol

7.8 What alkanes might the following alcohols have been prepared from? Solution: (a)

H3C C CH

H2C

H2C CH3

CH3

H2C CH2C

H2C

H2C CH3

CH3

2-methyl-1-hexene

2-methyl-2-hexene

Hg(OAc)2

H2O

Hg(OAc)2

H2O

H3C CH2C

H2C

H2C CH3

OH

OH

(b)

Hg(OAc)2

H2O

1.BH3

OH-2.H2O2

OH

cyclohexylethylene

ethylidenecyclohexane

7.9 Show the structures of the products you would obtain by hydroboration/oxidation of the following alkenes:

(a)

CH3C CHCH2CH3

CH3 (b) CH3

Solution: (a)

CH3CH CHCH2CH3

CH3 OH

(b)

CH3

OH

7.10 What alkenes might be used to prepare the following alcohols by hydroboration/oxidation? (a) (CH3)2CHCH2CH2OH

(b)

(CH3)2CHCHCH3

OH

(c) CH2OH

Solution: (a) (CH3)2CHCH CH2

(b) (CH3)2C CHCH3

(c) CH2

7.11 The following cycloalkene gives a mixture of two alcohols on hydroborotion followed by

oxidation, Draw the structures of both, and explain the result.

H3C

HCH3

H

Solution: The structure of the two cycloalcohols:

H3C

HCH3

H

1. BH3 / THF

2. OH-, H2O

H3C

H

CH3

H

H3C

HCH3

H

H

OH

OH

H

7.12 What products would you expect from the following reactions?

(a)

CH2

C Cl

Cl

Cl

H

KOH?

(b) (H3C)2HCH2CHC CHCH3 CH2I2

Zn(Cu)?

Solution:

(a)

CH2

C ClCl

Cl

H

KOH CCl Cl

HCl

(b) (H3C)2HCH2CHC CHCH3 CH2I2

Zn(Cu)ZnI2

7.13 What product would you obtain from calalytic hydrogenation of the following alkenes?

(a) H3C C

CH3

CH

H2C CH3

(b)CH3

CH3 Solution：

(a) H3C CH

CH3

H2C

H2C CH3

(b)CH3

CH3 7.14 What alkene would you start with to prepare each of the following compounds?

(a)OH

OH

H

CH3

(b)H3CH2C C

H

OH

C

OH

CH3

CH3 (c) HOH2C C

H

OH

CH

OH

H2C OH

S

olution:

(a)

H

CH3

(b)H3CH2C C

HC

CH3

CH3 (c) H2C CH

CH

CH2

7.15 What products would you expect? Solution:

(a) Aqueous acidic KMnO4 Product CH3COCH2CH2CH2CH2COOH

(b) O3

Product CH3COCH2CH2CH2CH2CHO 7.16 Propose structure for alkenes. Solution:

(a): (CH3) 2C=O and H2C=O Structure (CH3) 2C =CH2 (b): 2 equiv CH3CH2CH=O Structure CH3CH2CH=CHCH2CH3

7.17 Show the monomer units you would use to prepare the following polymers:

(a)

H2C C

H

OCH3

H2C C

H

OCH3

H2C C

H

OCH3

Solution: The monomer unit is H2C CH

OCH3

.

(b)

CH

CH

CH

CH

CH

CH

Cl Cl Cl Cl Cl Cl

Solution: The monomer unit is ClHC CHCl. 7.18 One of the chain-termination steps that sometimes occurs to interrupt polymerization is the

following reaction between two radicals:

CH2CH22 CH2CH3 + CH

CH2

Propose a mechanism for this reaction, using fishhook arrows to indicate electron flow. Solution:

HC

CH2CH3+CH

CH2

H2C CH2

H

CH2

7.19 tert-Butyl vinyl ether is polymerized commercially for use in adhesives by a cationic process.

Draw a segment of poly(tert-butyl vinyl ether), and show the mechanism of the chain-carrying step.

O

tert-butyl vinyl ether

Solution:

OH

Acid catalyst

O

O

O O

Repeat many times O

CH

H2C* *

n 7.20 Name the following alkenes, and predict the products of their reaction with (i) KMnO4 in aqueous

acid and (ii) O3, following by Zn in acetic acid:

(a) (b) Solution: (a)2,5-dimethyl-2-heptene

(i)

O

OH

O

and

(ii)

O

H

O

and

(b)3,3-dimethylcyclopentene

(i)

OH

O

HO O

(ii)

H

O

H O

7.21 Draw the structures of alkenes that would yield the following alcohols on hydration (red=O). Tell

in each case whether you would use hydroboration/oxidation.

Solution: (a) CH2

I would use oxymercuration in this case.

(b)

Both oxymercuration / demercuration and hyboration / oxidation reactions could be used. 7.22 The following alkene undergoes hydroboration/oxidation to yiele a single product rather than a

mixture. Explain the result, and draw the product showing its stereochemistry. Solution: The reactant is symmetrical, so OH adding to either of the carbon that is double bonded

will yield the same product. The single product is: OH

7.23 Predict the products of the following reactions (the aromatic ring is unreactive in all cases).

Indicate regiochemistry when relevant.

CH CH2

(a) H2/Pd

(b) Br2

(c) HBr

(d)1.OsO42.NaHSO4

(e) D2/Pd

Solution: (a)

CH CH2H2/Pd

CH

H CH

HH

(b)

CH CH2Br2

C CH

BrH

HBr

(c)

CH CH2HBr

C CH

BrHH

H

(d)

CH CH21.OsO42.NaHSO4

C CH

HO OHH H

(e)

CH CH2D2/Pd

CD

HC

D

H

H

7.24 Suggest structures for alkenes that give the following reactions products. There may be more than one answer for some cases.

(a) ? CH3CHCH2CH2CH2CH3

CH3

(b)

? CH3

CH3

(c)

?Br2 CH3CHCHCH2CHCH3

CH3

Br

Br

(d)

?HCl CH3CHCHCH2CH2CH2CH3

CH3

Cl

(e) ?

1.Hg(OAc)2,H2O2.NaBH4

CH3CH2CH2CHCH3

OH

Solution: (a)

CH3CHCH2CH2CH CH2

CH3 H3CHCH2CHC CHCH3

H3C

H3CHCHC CHCH2CH3

H3C

H3CC CHCH2CH2CH3

CH3

H2C CCH2CH2CH2CH3

CH3

(b)

CH3

CH3

CH3

CH3

(c) H3CHC CHCH2CHCH3

CH3

(d)

H2C CHCHCH2CH2CH2CH3

CH3 (e) H3CH2CH2CHC CH2 7.25 Predict the products of the following reactions, indicating both regiochemistry and stereiochemistry when appropriate:

(a)

CH3

H

1. O3

2.Zn, H3O+ ?

Solution:

CH3

H

1. O3

2.Zn, H3O+

O

O

CH3

H

(b)

KMnO4H3O+ ?

Solution:

KMnO4

H3O+

O

OH

O

OH

(c)

CH3

1. BH32. H2O2,- OH

?

Solution:

CH3

1. BH32. H2O2,- OH

H CH3

OH H

+

CH3 H

OHH

50% 50%

(d)

CH3

1. Hg(OAc)2,H2O2. NaBH4

?

Solution:

CH3

1. Hg(OAc)2,H2O2. NaBH4

CH3

OHH

H

H

H OH

CH3

50% 50% 7.26 How would you carry out the following transformations? Indicate the reagents you would use in each case.

(a)

?

H

H

OH

OH

Solution:

H

H

OH

OH

1.OsO4

2.NaHSO3,H2O

(b)

?

OH

Solution:

OH

1. Hg(Ac)22. NaBH4

(c)

?

H

H

Cl

Cl

Solution:

H

H

Cl

Cl

1. CHCl3

2. KOH

(d)

CH3

OH ?

CH3

Solution:

CH3

OH

CH3

1. H2SO4, H2O2. THF, 50 ℃

(e)

H3CHC CHCHCH3

CH3

? CH3CH

O

+ CH3CHCH

CH3

O

Solution:

H3CHC CHCHCH3

CH3

CH3CH

O

+ CH3CHCH

CH3

O

1. O32. Zn/H3O+

(f) H3CC CH2

CH3

?CH3CHCH2OH

CH3

Solution: H3CC CH2

CH3

CH3CHCH2OH

CH3

1. BH3, THF2. H2O2,OH-

7.27 What product will result from hydroboration/oxidation of 1-methylcyclopentene with deuterated

borane, BD3? Show both the stereochemistry (spatial arrangement) and the regiochemistry (orientation) of the product.

H CH3

BD3

H CH3

D2B D

H CH3

BD

D

D

δ

δ

-OHH2O2

H CH3

HO D

H CH3

D2B D

H CH3

BD

D

D

δδ

-OHH2O2

H CH3

HO D

7.28 Draw the structure of an alkene that yield only acetone, (CH3)2C=O, on ozonolysis followed by

treatment with Zn.

OO3

Zn/H3O+ O+

7.29 Draw the structure of a hydrocarbon that react with 1 molar equivalent of H2 on catalytic hydrogenation and gives only pentanal , CH3CH2CH2CH2CHO, on the ozonolysis followed by treatment with Zn. Write the reaction involved.

Solution: The structure of the hydrocarbon is showed in the following picture:

The reactions are showed as follow:

Catalyst

H2

Zn/H3O

O3

O

H2

7.30 Draw the structure of alkenes that give the following products on oxidative cleavage with KMnO4

in the acidic solution:

(a).

O

CO2+ (b).

O

+O

(c).

O +

O

Solution:

(a). (b).

(c).

7.31 Compound A has the formula C10H16. On catalytic hydrogenation over palladium, it reacts with

only 1 molar equivalent of H2. Compound A also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, B (C10H16O2).

(a) How many rings does A have? A have two rings.

(b) What are the structures of A and B?

O

O

A

B

(c) Write the reactions.

+H2palladium

O

O

+O3Zn

7.32 An unknown hydrocarbon A, with formula C6H12, reacts with 1molar equivalent of H2 over a

palladium catalyst. Hydrocarbon A also reacts with OsO4 to give a diol, B. When oxidized with KMnO4 in acidic solution, A gives two fragments. One fragment is propanoic acid, CH3CH2CO2H, and the other fragment is a ketone. C. What are the structures of A, B, and C? Write all reactions, and show your reasoning.

A

CH3

CH3HHO OH

H2C

B H3C

C

O

O

+ H2palladium

CH3

CH3HHO OH

H2C

H3COsO4

KMnO4 + H3CCH2

CO

H

O

7.33 Using an oxidative cleavage reaction explain how you would distinguish between the following

two isomeric dienes:

and Solution: When they react with KMnO4 in acid, the first one will yield two products and the second one

will only yield one product. 7.34 Compound A, C10H18O, undergoes reaction with dilute H2SO4 at 2500C to yield a mixture of two

alkenes, C10H16. The major alkene product, B, gives only cyclopentanone after ozone treatment followed by reduction with zine in acetic acid. Identify A and B, and write the reaction.

Solution: A: B:

HO

Reaction:

HOH2SO4

2500C

major

O3

Zn/H3OO

7.35 Which reaction would you expect to be faster, addition of HBr to cyclohexene or to

1-methylcyclohexene? Explain. Solution: The intermediates of the two reactions are not equal in stability. Judge by Hammond postulate,

the more stable intermediate forms faster, so addition of HBr to 1-methylcyclohexene may be faster.

7.36 Predict the products of the following reactions, and indicate regiochemistry if relevant:

(a) HBrBr

(c)CH I , Zn-Cu2 2

HH

(d)CH I , Zn-Cu2 2

H

H

7.37 Iodine azide, IN3, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a

monosubstituted alkene such as 1-butene is used, only one product results:

H3CH2CHC CH2 + I N N N CH3CH2CHCH2I

N N N

(a) Add lone-pair electrons to the structure shown for IN3, and draw a second resonance form for the

molecule.

I N N I NN N N

(b) Calculate formal charges for the atoms in both resonance structures you drew for IN3 in part

For the iodine: Iodine valence electrons = 7 Iodine bonding electrons =2 Iodine nonbonding electrons =6 Formal charge = 7 - 2/2 – 6 = 0

For the left nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =6 Nitrogen nonbonding electrons =2 Formal charge = 5 – 6/2 – 2 = 0 For the middle nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =8

Nitrogen nonbonding electrons =0 Formal charge = 5 – 8/2 – 0= 1 For the right nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =4 Nitrogen nonbonding electrons =4 Formal charge = 5 – 4/2 – 4= - 1 (c) In light of the result observed when IN3 adds to 1-butebe, what is the polarity of the I-N3bomd?

Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.

H3CH2CHC CH2 + I N N N

H3CH2CHC CH2I N N NCH3CH2CHCH2I

N N N

7.38 Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of H2 on catalytic hydrogenation and gives only butanedial on ozonolysis. O

HCH2C

H2C CH

O

Butanedial

Solution:

7.39 Simmons–Smith reaction of cyclohexene with diiodomethane gives a single cycloproprane

product, but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?

Solution: The reaction occurs as below:

+ I2HC CH3Zn(Cu)Ether

H CH3H3C H

+

7.40 In planning the synthesis of one compound from another, it’s just as important to know what not to

do as to know what to do. The reactions all have serious drawbacks to them. Explain the potential problems of each.

Solution: (a)

H3C C

CH3CHCH3

HI H3C CH

CH3CHCH3I

According to Markovnikov’s rule the reaction should be:

H3C C

CH3CHCH3

HI H3C CCH3

CH2CH3I

(b)

1.OsO42.NaHSO3

OH

H

H

OH

Actually it should be syn addition as below:

1.OsO42.NaHSO3

OH

OH

H

H

(c)

1.O32.Zn CHO

CHO

Actually the reaction should be:

1.O32.Zn

H2C CHOOHC2

(d)

CH31.BH32.H2O2, OH

CH3

OHH

H

Actually the reaction should be:

CH31.BH32.H2O2, OH

CH3

HOH

H

7.41 Which of the following alcohols could be made selectively by hydroboration/oxidation of an alkene? Explain.

(a) CH3CH2CH2CHCH3

OH

(b) (CH3)2CHC(CH3)2

OH

(c)

CH3

OH

H

H (d)

CH3

H

OH

H

Solution: (a) No selectivity (b) Selectively synthesized (c)Can’t be formed

(d) Can’t be formed 7.42 What alkenes might be used to prepare the following cyclopropane?

(a) CH(CH3)2

(b)

Cl

Cl

Solution:

(a) CH(CH3)2+ (CH3)2CHCHI2

Zn(Cu)

Ether

(b)

Cl

Cl

+ CHCl3OH

7.43 Predict the products of the following reactions. Don’t worry about the size of the molecule;

concentrate on the functional groups.

HO

CH3

CH3

Br2

HBr

1. OsO4

2. NaHSO3

1. BH3, THF2. H2O2, OH-

CH2I2, Zn (Cu)

Solution:

a) HO

CH3

CH3

Br2

HO

CH3

CH3

BrBr

b) HO

CH3

CH3

HBr

HO

CH3

CH3

Br

c) HO

CH3

CH3

OsO4

NaHSO3

HO

CH3

CH3

OHOH

d) HO

CH3

CH3

BH3,THFH2O2,OH

HO

CH3

CH3

OH

e) HO

CH3

CH3

CH2I2Zn(Cu)

HO

CH3

CH3

7.44 The sex attractant of the common housefly is a hydrocarbon with the formula C23H46. On

treatment with aqueous acidic KMnO4, two products are obtained, CH3(CH2)12CO2H and CH3(CH2)7CO2H.Propose a structure.

Solution:

H3CH2C

HC

12CH

H2C CH3

7 should be the right structure.

7.45 Compound A has the formula C8H8. It reacts rapidly with KMnO4 to give CO2 and carboxylic

acid, B (C7H6O2), but reacts with only 1molar equivalent of H2 on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of H2 are taken up and hydrocarbon C (C8H16) is produced. What are the structure of A, B, and C? Write the reactions.

SOLUTION:

A CC

H

H

H

B COH

O

CH2C

CH3

The reactions:

CC

H

H

H

H2C

CH3

KMnO4C

OH

O

CO2

CC

H

H

H

H2

palladium catalyst

H2C

CH3

CC

H

H

H

4mol H2

1mol

7.46. Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of

methyl methacrylate. Draw a representative segment of Plexiglas.

H2C C

CH3

C OCH3

O

Methyl methacrylate

SOLUTION:

The segment of Plexiglas:

H2C C

C

CH3

O

OCH3

7.47 Poly (vinyl pyrrolidone), prepared from N-vinylpyrrolidone, is used both in cosmetics and as a synthetic blood substitute. Draw a representative segment of the polymer.

N

O

N-vinylpyrrolidone Solution:

HC

H2C **

N

O

7.48 Reaction of 2-methylpropene with CH3OH in the presence of H2SO4 catalyst yields methyl

tert-butyl ether, CH3OC(CH3)3, by a mechanism analogous to that of acid-catalyst alkene hydration. Write the mechanism, using curved arrows for each step.

Solution:

CH2CH3C

CH3

+ CH3OHH HSO4

CH3CH3C

CH3

+ CH3OH

CH3CH3C

CH3

OH3C H

HSO4

CH3CH3C

CH3

OH3C

+ H2SO4

7.49 When 4-penten-1-ol is treated with aqueous Br2, a cyclic bromo ether is formed, rather than the

expected bromohydrin. Propose a mechanism, using curved arrows to show electron movement. Solution:

O

O CH2Br

H HO

O

Br

H

Br

7.50 How would you distinguish between the following pairs of compounds using simple chemical

tests? Tell what you would do and what you would see. (a) cyclopentene and cyclopentane (b) 2-hexene and benzene Solution: Bromine in carbon tetrachloride solution is reddish-brown color. Using this one to identify the double bond by electrophilic addition reaction. 7.51

Cl

CCl

Cl

C O-Na+

O Cl

CCl

Cl

C O

O- Cl

C-Cl

Cl

+O C O

Cl

C

Cl

Cl-+ +CO2+Na+

They have the same intermediate which is CCl3-.

7.52 (a) 3 (b) double bonds:2

(c) 7.53 Evidence that cleavage of 1,2-diols by HIO4 occurs through a five membered cyclic periodate

intermediate is based on kinetic data – the measurement of reaction rates. When diols A and B were prepared and the rates of their reaction with HIO4 were measured, it was found that diol A cleaved approximately 1 million times faster than diol B. Make molecular models of A and B and of potential cyclic periodate intermediates, and then explain the kinetic results.

OH

H

H

HOH

H

OH

OH

A Bcis diol trans diol

Solution:

A:

OH

H

H

HOI OH

OO

HIO4

O

OO

O

In cis diol, two –OH are in one plane, so it is easier to form a cyclic periodate inetermediate, while in trans diol, the two –OH aren’t in one plane, there is larger steric strain in the five membered cyclic periodate intermediate, it is less stable, so the rates of A is more faster.

7.54 Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and

trans-1-3-methylbromocyclohexane and cis- and trans-1-2-methylbromocyclohexane. The analogous reaction of HBr with 3-bromocyclohexane trans-1, 2-dibromocyclohe- xane as the sole product. Draw structures of the possible intermediates, and then Explain why only a single product is formed in the reaction of HBr with 3-bromocyclohexane.

Solution:

Br

H Br

BrBr

Br

Br

H

H

Br

7.55 The following reaction takes place in high yield:

CO2CH3

Hg(OAc)2

CO2CH3

HgAcO

Use your general knowledge of alkene chemistry to propose a mechanism, even though you’ve never seen this reaction before.

Solution:

CO2CH3

+AcO Hg

OAc

CO2CH3

Hg

AcO

CO2CH3

HgAcO

CO2CH3

HgAcO

OAc

OAcH

CO2CH3

HgAcO

7.56 Reaction of cyclohexene with mercury(II) acetate in CH3OH rather than H2O, followed by

treatment with NaBH4, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.

1.Hg(OAc)2, CH3OH

2.NaBH4OCH3

Solution:

OCH3

Hg OAc

OAcHg

OAc

HO CH3

HgOAc

O

H

CH3

OAc

HgOAc

O CH3

NaBH4

7.57 Hydroboration of 2-methyl-2-pentene at 25℃ followed by oxidation with alkaline H2O2 yields 2-methyl-3-pentano, but hydroboration at 160℃ followed by oxidation yields 4-methyl-1-pentanol. Explain.

1.BH3,THF,25℃

2.H2O2,OH-CH3CHCHCH2CH3

CH3

OH

2-Methyl-3-pentanol

1.BH3,THF,160℃

2.H2O2,OH-CH3CHCH2CH2CH2OH

CH3

4-Methyl-1-pentanol

CH3C CHCH2CH3

CH3

Solution: Case 1:

CH3C CHCH2CH3

CH3

3 + BH3 THF,25℃CH3CH CHCH2CH3

CH3

BH2

3 H2O2,OH-

CH3CHCHCH2CH3

CH3

OH

3

Case 2:

BH2

BH2

BH2

H2O2 / OH-

OH

7.58 Alkynes undergo many of the same reactions that alkenes do. What products would you expect

from each of the following reactions?

H3C CH

H2C

H2C C CH

CH3

(1) with 1 equiv Br2; (2) with 2 equiv H2 ,Pd/C (3)with 1 equiv HBr Solutions: (1)

H3C CH

H2C

H2C C CH

CH3

H3C CH

H2C

H2C C CH

CH31 equiv Br2Br

Br (2)

H3C CH

H2C

H2C C CH

CH3

H3C CH

H2C

H2C

H2C CH3

CH32 equiv H2 ,Pd/C

(3)

H3C CH

H2C

H2C C CH

CH3

H3C CH

H2C

H2C C CH2

CH31 equiv HBr Br

7.59 Explain the observation that hydroxylation of cis-2-butene with OsO4 yields a different product

than hydroxylation of trans -2-butene. First draw the structure and show the stereochemistry of each product, and then make molecular models.

Solution:

H3C CH

CH

CH3

CH3

CHHC

H3C

cis-But-2-ene

H3CCH

HC

CH3

trans-But-2-ene

CH3

CC

H3C

OH OHH H

Butane-2,3-diolmeso

compound

H3CC

CCH3

OH

OH

H H

Butane-2,3-diolenantiomer