each -20 po Problems 1-3 are 2 points res below for the following que Section :

Question

each -20 po Problems 1-3 are 2 points res below for the following que Section : Written 1) Like Kiain's edition problem 5.55: Refer to the structu of points. 's An answer can be F- none Any incorrect answers will result in a Cl BrNONE a. If the observed rotation of C is ()26, what is the observed rotation of B b. Which molecule(s) is/are an enantiomer(s) of A? c. Which molecule(s) is/are optically active? 2) How many 'HNMR and 1"'CNMR signals would be observed for the molecules shown below? Number of 13CNMR signalsNumber of HNMR signals observed (1 point each)observed (1 point each) Structure CH3 ) Are the following pairs of molecules identical structures, enantiomers, diastereomers, constitutional isomers or different compounds? and Br CH3 OH OH and Br enontismers Br

Explanation / Answer

Answer 1 Part a1

Since the compound is a meso isomer the optical rotation observed will be zero.

Answer 1 Part a 2

B: Br is in a different orientation. E is same are B (Meso isomer)

C is mirror image of A. D is same as C.

Answer 1 Part b

C is mirroe image of A. D is same as aC

Answer 2 Part c

A, C & D

B & E are Meso isomers for the structure

Answer 2. Part a)

C13NMR signals: 4

Because of the symmetry of the structure there are 4 types of Carbon:

i) methyl groups (-CH3 x 4)

ii) aldehyde groups (C-CHO x 2)

iii) benzene C attached to methyl group (H3C-C x 4)

iv)benzene C attached to aldehyde group.

H1NMR signals: 2

Because of the symmetry of the structure there are 2 types of H

i) H attached to methyl groups (-CH3 x 12) and

ii) H attached to aldehyde groups (-CHO x 2)

Answer 2. Part b)

C13NMR signals: 6

consider the Common name for the product is o-bromo tolunene. toluene has 5 signals for C13NMR.

i) Mehtyl group (-CH3 x 1)

ii) C where the methyl group is attached to the benzene ring (C x 1)

iii) bromo substituted C at the ortho position (C x 1)

iv) Benzene (CH) at the other ortho position (CH x 1)

v) Benzene (CH) at the 2 meta positions (CH x 2)

vi) Benzene (CH) at the para position (CH x 1)

H1NMR signals: 2

i) H attached to methyl group (-CH3 x 3) and

ii) H attached to benzene ring (-CH x 4)

Answer 3 Part a)

Both structures are identical:

If you rotate the plane of reference (along the vertical axis) and bring the Br and methyl group into the plane of the paper/screen then you would be rotating the structure counter clockwise. That would send the ethyl group to the background and bring the carbonyl group out of the plane of the paper/screen.

Now If you rotate the structure in the plane of the paper/screen counter clockwise by 180 deg you will get the second structure

Answer 3 Part b

Both are constitutional (structural) isomers

i) 1-Bromo -2- butanol

ii) 3-Bromo -2- butanol

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