ochem 1 - extra problems

8/4/2019 Ochem 1 - Extra Problems

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- -----

~ Example 1-8: For each pair of molecules below, state the relationship between them.

(a)

(5'~~

and

(b)

H - t H ' H { :H \ HsC,\H / ~

and 1- Hs

(c)

and

H H

---------- -- --- -- - -

(dl ~

~

OH and H H

H20H HPi

(e) HO CH20H

H2 and H H

H H H2

H20H HO


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Solution.

(a ) The molecules are superimosable and therefore identical:- C 100 c..\ QU~ ~(b) The molecules are identical. (The left carbon is not a chiral center.)

(C ) Diastereomers:

. (~ ) En4nt iomers:

(e )'Diast e reomers: .

H

H

CHO

H

H

------------------~ Example 1-10: What is the specific rotation of the R enantiomer of2-bromobutanoic acid? What do you think the specific rotation of a

50/50 mixture of the R enantiomer and the S enantiomer will be?

Solution.

I:'R S I

The magnitude of rotation cannot be predicted; it E;,e ~m~llY ~ed.

It just so happens that in this case the R enantiomer has the (+)rotation {while the S

enantiomer has the (-) rotation]. Jhtt be c~: this is only coincidental. (+) and (_)

say nothing about whether the absolute configuration is R orS. There is nocorrelation between the sign of rotation and the absolute configuration. \ ~~

The specific rotation of an equal (50/50) mixture of enantiomers'::'a racemic

IIUX. ture-i~ce one enantiomer cancels out the other-one. l!Il(gqpc'ilJSiURt:ln

~ ~

'4 1"\c.41

~'n~~


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~ Example 1-11: Which of the following molecules are meso compounds?

w~

(b) (c) HOCH2GHCH2OH

ICI

H3C

H3

(d)

H~

(e) (6

HO H

2HCI

(g)H H

Solution.

(a) Not meso. The chiral centers have opposite absolute configurations, but one

side of the molecule is not the mirror image of the other:

(b) Meso. The chiral centers have opposite absolute configurations, and one side of

the molecule is the mirror image of the other side:

H Sri H Sr

H,~H,.

~ ~ ~bcJ\

fX (c) Achiral, not meso. This molecule has no chiral centers. - CQva.s\t~(d) Meso. The chiral centers have opposite absolute configurations, and one side of


H

H


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(e) Not meso. There are three chiral centers but no internal plane of symme!I'Y.

(Youmay find it helpful to notice that in order for one side of the molecule to be the

mirror image of the other side, any compound that is meso must have an ~ennumber of chiral centers. In this case, then, the fact that there are an odd number of

C h ii 'a t centers automatically means the molecule cannot be meso.)

(f) Not meso. The chiral centers are of the same absolute configuration, and one

side of the molecule is not the mirror image of the other:

(g) Meso. The chiral centers have opposite absolute configurations, and one side of


------------- ----

~- Example 1-16: fu;-each of the foll~;-ing molecules, indicate the hybridization and

idealized bond angles for the indicated atoms.

/

:: \C ::lC \ 0

1..:. 11.00

~ ':.'~bC)

(b)

(c) (d)

( 9, _ ...(iv)

HN I(ii) ... r~

~_ ... (iii)

(e)


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Solution. Remember to always draw the electrons on nitrogen if they're not drawnin the structure.

(a) i) sr , 1200 ii) sr , 1200 iii) Sp2, 1200(Thelone pair is delocalized,8Oit's not counted.)iv) sp3 , 1090 v) Sp2, 1200 vi) sp2 , 1200

(b) i) spJ, 1090 ii) spJ, 1090 iii) sp3 , 1090

(c) i) spJ, 1090 ii) Sp2, 1200 iii) s r , 1090

(d) i) sp2 , 1200 ii) sr , 1200 iii)sr, 1200 Iv) s r , 1200

(e) i) Sp3, 1090 ii) sp3 , 1090 iii) sp2 , 1200 iv) sp , 1800 v) sp , 1800

(f) i) sp3 , 1090 Ii) sp , 1800 iii) sp3 , 1090 Iv) spJ, 1090

--- -------_._- ----------------~ Example 2-5: For each of the following pairs of compounds, predict which one-

the first or the second-yvill have the higher boiling point.

(a)

< )OH or

< )Br

-------- --------

(b)

0 c YH

or

(c) 0

~HNH3

or

(d)

(e)

or

or

Solution.

(a) The first. Phenol can form intermolecular hydrogen bonds, but phenyl bromide cannot.

(b) The second. The alcohol can form hydrogen bonds, but the ether cannot. (The ether

could accept a hydrogen bond with the lone pair of electrons on the oxygen, but there are no

hydrogen bond donors in this molecule.)

(c) The second. The carboxylic acid can form hydrogen bonds, but the ester cannot. The

ester has hydrogen bond acceptors, but no hydrogen bond donors.

(d) The first. Both compounds can form hydrogen bonds, but the first compound (the onew~th two -QH grOUps)can form more of them, increasing its boiling point.

(e) The first. Neither compound can form hydrogen bonds, but the bromide is a muchheavier molecule, increasing its boiling point relative to the hydrocarbon. -


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Example 2-7: For each of the following alcohol substitution reactions, predict

whether it will proceed via an SN1or an SN2mechanism.

(a)

H-Br

(b)~H-CI H./1.~

(c)H-I

(d)~ H-Br

Solution.

(a) 3 alcohol, SN1


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-- - --- ---------

~ Example 2-8: From each of the following dehydration reactions, choose the major

organic product (A, B, C, or D).

(i) (A)~ (C)~

~

H~04

heat

ffll

l WI + -

(ii)

---.heat

(BI

1.

(O)~

(iii) (A)

cr(0

0-Cf(,

H2SO4

heat

(B) (0)

0- 0--

------ -------

(iv) (A)

0-(C)

cr(YuH H~04 heat

ffllo--

(0)

~

- s o f i J . i l o n .

(i) C (LA)"~ no\- ~ ?'( . J(ii) A (Note that carbocation rearrangement would occur.)

. - r - - . . 4(iii) D

(iv) B (carbocation rearrangement)

ochem 1 - extra problems

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