Hi i'm having trouble truly understanding this and I am studying DAT Destroyer question to figure this all out. Sorry if this seems really elementary... my online digging has failed me and my textbook has not been able to fully help me understand this (or rather, i cannot seem to understand my textbook fully)
so here it is: [edited to make sense this time....also, i realized this question doesn't really have anything to do with optical activity...BUT I do have another question on OPTICAL ACTIVITY...i'm so sorry, i guess my head is just a mess with these topics and all the studying]
X Br2> Product A + Product B
CH2Br2
Which produces two diastereomers A and B?
a) methylenecyclohexane b) pentene c) 3-ethylcyclopentene d) pentene e) More than one of these
Ans. C
Reasons: book explanation (product)
a) not optically active (1-bromo-1-bromomethylcyclohexane)
b) Will have an enantiomer. Only 1 chiral carbon ((trans)-1,2-dibromopentane)
c) More than one chiral center exists, 2 diastereomers possible ((trans)-1,2-dibromo-3-ethylcyclopentane)
d) Will have an enantiomer only ((trans)-1,2-dibromocyclopentane)
My question: Why does answer d) not count? Are (cis)-1,2-dibromocyclopentane and (trans)-1,2-dibromocyclopentane equivalent in this case?
Opitcal Activity question
Which are optically inactive?
a) 4-methylhex-2,3-ene (is that right? CH3-CH=C=C(CH3)-C2H5
(Note: the red CH3 points into the page while the purple
H points out of the page)
b) 4(S)-iodocyclohexene (I points into the page, while the H on C-4 points out of the page)
c) (cis)-1,3-cyclopentanediol (both OH groups pointing into the page)
d) (2R, 3R)-3-bromo-3-hydroxy-2-methylpropanoic acid
e) Two of these
Ans. C meso, achiral --> optically inactive
Why is B considered optically active with just 1 chiral center?
Also:
the situation with answer c) seems very similar, structurally, to answer d) from the previous question above... I'm thinking perhaps the reason is that with both chiral centers present (in answer c) here and answer d) from the above question), you get the same result when going right and left around the cyclic molecule from the chiral center. But that thinking only makes sense to me when explaining how those 2 chiral centers are the same for cis. For trans, wouldn't it be different because you run into a substituent that is pointing a different direction than the other substituent?
so here it is: [edited to make sense this time....also, i realized this question doesn't really have anything to do with optical activity...BUT I do have another question on OPTICAL ACTIVITY...i'm so sorry, i guess my head is just a mess with these topics and all the studying]
X Br2> Product A + Product B
CH2Br2
Which produces two diastereomers A and B?
a) methylenecyclohexane b) pentene c) 3-ethylcyclopentene d) pentene e) More than one of these
Ans. C
Reasons: book explanation (product)
a) not optically active (1-bromo-1-bromomethylcyclohexane)
b) Will have an enantiomer. Only 1 chiral carbon ((trans)-1,2-dibromopentane)
c) More than one chiral center exists, 2 diastereomers possible ((trans)-1,2-dibromo-3-ethylcyclopentane)
d) Will have an enantiomer only ((trans)-1,2-dibromocyclopentane)
My question: Why does answer d) not count? Are (cis)-1,2-dibromocyclopentane and (trans)-1,2-dibromocyclopentane equivalent in this case?
Opitcal Activity question
Which are optically inactive?
a) 4-methylhex-2,3-ene (is that right? CH3-CH=C=C(CH3)-C2H5
(Note: the red CH3 points into the page while the purple
H points out of the page)
b) 4(S)-iodocyclohexene (I points into the page, while the H on C-4 points out of the page)
c) (cis)-1,3-cyclopentanediol (both OH groups pointing into the page)
d) (2R, 3R)-3-bromo-3-hydroxy-2-methylpropanoic acid
e) Two of these
Ans. C meso, achiral --> optically inactive
Why is B considered optically active with just 1 chiral center?
Also:
the situation with answer c) seems very similar, structurally, to answer d) from the previous question above... I'm thinking perhaps the reason is that with both chiral centers present (in answer c) here and answer d) from the above question), you get the same result when going right and left around the cyclic molecule from the chiral center. But that thinking only makes sense to me when explaining how those 2 chiral centers are the same for cis. For trans, wouldn't it be different because you run into a substituent that is pointing a different direction than the other substituent?
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